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Enderton (set). Chapter 6 formatting. 

More consistent header levels throughout.
fin-dom-ran
Joshua Potter 2023-11-11 15:15:03 -07:00
parent b97b8fbbca
commit f215a3180a
19 changed files with 605 additions and 478 deletions
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18
Bookshelf/Apostol/Chapter_1_11.lean
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@ -8,21 +8,21 @@ namespace Apostol.Chapter_1_11
open BigOperators open BigOperators
/-- #### Exercise 4a /-- ### Exercise 4a
`⌊x + n⌋ = ⌊x⌋ + n` for every integer `n`. `⌊x + n⌋ = ⌊x⌋ + n` for every integer `n`.
-/ -/
theorem exercise_4a (x : ℝ) (n : ℤ) : ⌊x + n⌋ = ⌊x⌋ + n := theorem exercise_4a (x : ℝ) (n : ℤ) : ⌊x + n⌋ = ⌊x⌋ + n :=
Int.floor_add_int x n Int.floor_add_int x n
/-- #### Exercise 4b.1 /-- ### Exercise 4b.1
`⌊-x⌋ = -⌊x⌋` if `x` is an integer. `⌊-x⌋ = -⌊x⌋` if `x` is an integer.
-/ -/
theorem exercise_4b_1 (x : ℤ) : ⌊-x⌋ = -⌊x⌋ := by theorem exercise_4b_1 (x : ℤ) : ⌊-x⌋ = -⌊x⌋ := by
simp only [Int.floor_int, id_eq] simp only [Int.floor_int, id_eq]
/-- #### Exercise 4b.2 /-- ### Exercise 4b.2
`⌊-x⌋ = -⌊x⌋ - 1` otherwise. `⌊-x⌋ = -⌊x⌋ - 1` otherwise.
-/ -/
@ -42,7 +42,7 @@ theorem exercise_4b_2 (x : ℝ) (h : ∃ n : ℤ, x ∈ Set.Ioo ↑n (↑n + (1
· exact (Set.mem_Ioo.mp hn).left · exact (Set.mem_Ioo.mp hn).left
· exact le_of_lt (Set.mem_Ico.mp hn').right · exact le_of_lt (Set.mem_Ico.mp hn').right
/-- #### Exercise 4c /-- ### Exercise 4c
`⌊x + y⌋ = ⌊x⌋ + ⌊y⌋` or `⌊x⌋ + ⌊y⌋ + 1`. `⌊x + y⌋ = ⌊x⌋ + ⌊y⌋` or `⌊x⌋ + ⌊y⌋ + 1`.
-/ -/
@ -72,7 +72,7 @@ theorem exercise_4c (x y : ℝ)
rw [← sub_lt_iff_lt_add', ← sub_sub, add_sub_cancel, add_sub_cancel] rw [← sub_lt_iff_lt_add', ← sub_sub, add_sub_cancel, add_sub_cancel]
exact add_lt_add (Int.fract_lt_one x) (Int.fract_lt_one y) exact add_lt_add (Int.fract_lt_one x) (Int.fract_lt_one y)
/-- #### Exercise 5 /-- ### Exercise 5
The formulas in Exercises 4(d) and 4(e) suggest a generalization for `⌊nx⌋`. The formulas in Exercises 4(d) and 4(e) suggest a generalization for `⌊nx⌋`.
State and prove such a generalization. State and prove such a generalization.
@ -81,7 +81,7 @@ theorem exercise_5 (n : ℕ) (x : ℝ)
: ⌊n * x⌋ = Finset.sum (Finset.range n) (fun i => ⌊x + i/n⌋) := : ⌊n * x⌋ = Finset.sum (Finset.range n) (fun i => ⌊x + i/n⌋) :=
Real.Floor.floor_mul_eq_sum_range_floor_add_index_div n x Real.Floor.floor_mul_eq_sum_range_floor_add_index_div n x
/-- #### Exercise 4d /-- ### Exercise 4d
`⌊2x⌋ = ⌊x⌋ + ⌊x + 1/2⌋` `⌊2x⌋ = ⌊x⌋ + ⌊x + 1/2⌋`
-/ -/
@ -94,7 +94,7 @@ theorem exercise_4d (x : ℝ)
simp simp
rw [add_comm] rw [add_comm]
/-- #### Exercise 4e /-- ### Exercise 4e
`⌊3x⌋ = ⌊x⌋ + ⌊x + 1/3⌋ + ⌊x + 2/3⌋` `⌊3x⌋ = ⌊x⌋ + ⌊x + 1/3⌋ + ⌊x + 2/3⌋`
-/ -/
@ -108,7 +108,7 @@ theorem exercise_4e (x : ℝ)
conv => rhs; rw [← add_rotate']; arg 2; rw [add_comm] conv => rhs; rw [← add_rotate']; arg 2; rw [add_comm]
rw [← add_assoc] rw [← add_assoc]
/-- #### Exercise 7b /-- ### Exercise 7b
If `a` and `b` are positive integers with no common factor, we have the formula If `a` and `b` are positive integers with no common factor, we have the formula
`∑_{n=1}^{b-1} ⌊na / b⌋ = ((a - 1)(b - 1)) / 2`. When `b = 1`, the sum on the `∑_{n=1}^{b-1} ⌊na / b⌋ = ((a - 1)(b - 1)) / 2`. When `b = 1`, the sum on the
@ -125,7 +125,7 @@ theorem exercise_7b (ha : a > 0) (hb : b > 0) (hp : Nat.Coprime a b)
section section
/-- #### Exercise 8 /-- ### Exercise 8
Let `S` be a set of points on the real line. The *characteristic function* of Let `S` be a set of points on the real line. The *characteristic function* of
`S` is, by definition, the function `Χ` such that `Χₛ(x) = 1` for every `x` in `S` is, by definition, the function `Χ` such that `Χₛ(x) = 1` for every `x` in

42
Bookshelf/Apostol/Chapter_I_03.lean
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@ -100,7 +100,7 @@ theorem is_lub_neg_set_iff_is_glb_set_neg (S : Set ℝ)
_ = IsGreatest (lowerBounds S) (-x) := by rw [is_least_neg_set_eq_is_greatest_set_neq] _ = IsGreatest (lowerBounds S) (-x) := by rw [is_least_neg_set_eq_is_greatest_set_neq]
_ = IsGLB S (-x) := rfl _ = IsGLB S (-x) := rfl
/-- #### Theorem I.27 /-- ### Theorem I.27
Every nonempty set `S` that is bounded below has a greatest lower bound; that Every nonempty set `S` that is bounded below has a greatest lower bound; that
is, there is a real number `L` such that `L = inf S`. is, there is a real number `L` such that `L = inf S`.
@ -147,7 +147,7 @@ lemma leq_nat_abs_ceil_self (x : ℝ) : x ≤ Int.natAbs ⌈x⌉ := by
/-! ## The Archimedean property of the real-number system -/ /-! ## The Archimedean property of the real-number system -/
/-- #### Theorem I.29 /-- ### Theorem I.29
For every real `x` there exists a positive integer `n` such that `n > x`. For every real `x` there exists a positive integer `n` such that `n > x`.
-/ -/
@ -159,7 +159,7 @@ theorem exists_pnat_geq_self (x : ℝ) : ∃ n : ℕ+, ↑n > x := by
_ = n := rfl _ = n := rfl
exact ⟨n, this⟩ exact ⟨n, this⟩
/-- #### Theorem I.30 /-- ### Theorem I.30
If `x > 0` and if `y` is an arbitrary real number, there exists a positive If `x > 0` and if `y` is an arbitrary real number, there exists a positive
integer `n` such that `nx > y`. integer `n` such that `nx > y`.
@ -174,7 +174,7 @@ theorem exists_pnat_mul_self_geq_of_pos {x y : ℝ}
rw [div_mul, div_self (show x ≠ 0 from LT.lt.ne' hx), div_one] at p' rw [div_mul, div_self (show x ≠ 0 from LT.lt.ne' hx), div_one] at p'
exact ⟨n, p'⟩ exact ⟨n, p'⟩
/-- #### Theorem I.31 /-- ### Theorem I.31
If three real numbers `a`, `x`, and `y` satisfy the inequalities If three real numbers `a`, `x`, and `y` satisfy the inequalities
`a ≤ x ≤ a + y / n` for every integer `n ≥ 1`, then `x = a`. `a ≤ x ≤ a + y / n` for every integer `n ≥ 1`, then `x = a`.
@ -270,7 +270,7 @@ lemma mem_imp_ge_lub {x : ℝ} (h : IsLUB S s) : x ∈ upperBounds S → x ≥ s
intro hx intro hx
exact h.right hx exact h.right hx
/-- #### Theorem I.32a /-- ### Theorem I.32a
Let `h` be a given positive number and let `S` be a set of real numbers. If `S` Let `h` be a given positive number and let `S` be a set of real numbers. If `S`
has a supremum, then for some `x` in `S` we have `x > sup S - h`. has a supremum, then for some `x` in `S` we have `x > sup S - h`.
@ -321,7 +321,7 @@ lemma mem_imp_le_glb {x : ℝ} (h : IsGLB S s) : x ∈ lowerBounds S → x ≤ s
intro hx intro hx
exact h.right hx exact h.right hx
/-- #### Theorem I.32b /-- ### Theorem I.32b
Let `h` be a given positive number and let `S` be a set of real numbers. If `S` Let `h` be a given positive number and let `S` be a set of real numbers. If `S`
has an infimum, then for some `x` in `S` we have `x < inf S + h`. has an infimum, then for some `x` in `S` we have `x < inf S + h`.
@ -343,7 +343,7 @@ theorem inf_imp_exists_lt_inf_add_delta {S : Set ℝ} {s h : ℝ} (hp : h > 0)
exact le_of_not_gt (not_and.mp (nb x) hx) exact le_of_not_gt (not_and.mp (nb x) hx)
rwa [← mem_lower_bounds_iff_forall_ge] at nb' rwa [← mem_lower_bounds_iff_forall_ge] at nb'
/-- #### Theorem I.33a (Additive Property) /-- ### Theorem I.33a (Additive Property)
Given nonempty subsets `A` and `B` of `ℝ`, let `C` denote the set Given nonempty subsets `A` and `B` of `ℝ`, let `C` denote the set
`C = {a + b : a ∈ A, b ∈ B}`. If each of `A` and `B` has a supremum, then `C` `C = {a + b : a ∈ A, b ∈ B}`. If each of `A` and `B` has a supremum, then `C`
@ -393,7 +393,7 @@ theorem sup_minkowski_sum_eq_sup_add_sup (A B : Set ℝ) (a b : ℝ)
_ ≤ a' + b' + 1 / n := le_of_lt hab' _ ≤ a' + b' + 1 / n := le_of_lt hab'
_ ≤ c + 1 / n := add_le_add_right hc' (1 / n) _ ≤ c + 1 / n := add_le_add_right hc' (1 / n)
/-- #### Theorem I.33b (Additive Property) /-- ### Theorem I.33b (Additive Property)
Given nonempty subsets `A` and `B` of `ℝ`, let `C` denote the set Given nonempty subsets `A` and `B` of `ℝ`, let `C` denote the set
`C = {a + b : a ∈ A, b ∈ B}`. If each of `A` and `B` has an infimum, then `C` `C = {a + b : a ∈ A, b ∈ B}`. If each of `A` and `B` has an infimum, then `C`
@ -443,7 +443,7 @@ theorem inf_minkowski_sum_eq_inf_add_inf (A B : Set ℝ)
_ ≤ a + b := le_of_lt hab' _ ≤ a + b := le_of_lt hab'
· exact hc.right hlb · exact hc.right hlb
/-- #### Theorem I.34 /-- ### Theorem I.34
Given two nonempty subsets `S` and `T` of `ℝ` such that `s ≤ t` for every `s` in Given two nonempty subsets `S` and `T` of `ℝ` such that `s ≤ t` for every `s` in
`S` and every `t` in `T`. Then `S` has a supremum, and `T` has an infimum, and `S` and every `t` in `T`. Then `S` has a supremum, and `T` has an infimum, and
@ -489,7 +489,7 @@ theorem forall_mem_le_forall_mem_imp_sup_le_inf (S T : Set ℝ)
_ < x := hx.right _ < x := hx.right
simp at this simp at this
/-- #### Exercise 1 /-- ### Exercise 1
If `x` and `y` are arbitrary real numbers with `x < y`, prove that there is at If `x` and `y` are arbitrary real numbers with `x < y`, prove that there is at
least one real `z` satisfying `x < z < y`. least one real `z` satisfying `x < z < y`.
@ -506,7 +506,7 @@ theorem exercise_1 (x y : ℝ) (h : x < y) : ∃ z, x < z ∧ z < y := by
_ < x + z := (add_lt_add_iff_left x).mpr hz' _ < x + z := (add_lt_add_iff_left x).mpr hz'
_ = y := hz.right _ = y := hz.right
/-- #### Exercise 2 /-- ### Exercise 2
If `x` is an arbitrary real number, prove that there are integers `m` and `n` If `x` is an arbitrary real number, prove that there are integers `m` and `n`
such that `m < x < n`. such that `m < x < n`.
@ -514,7 +514,7 @@ such that `m < x < n`.
theorem exercise_2 (x : ℝ) : ∃ m n : ℝ, m < x ∧ x < n := by theorem exercise_2 (x : ℝ) : ∃ m n : ℝ, m < x ∧ x < n := by
refine ⟨x - 1, ⟨x + 1, ⟨?_, ?_⟩⟩⟩ <;> norm_num refine ⟨x - 1, ⟨x + 1, ⟨?_, ?_⟩⟩⟩ <;> norm_num
/-- #### Exercise 3 /-- ### Exercise 3
If `x > 0`, prove that there is a positive integer `n` such that `1 / n < x`. If `x > 0`, prove that there is a positive integer `n` such that `1 / n < x`.
-/ -/
@ -525,7 +525,7 @@ theorem exercise_3 (x : ℝ) (h : x > 0) : ∃ n : ℕ+, 1 / n < x := by
conv at hr => arg 2; rw [mul_comm, ← mul_assoc]; simp conv at hr => arg 2; rw [mul_comm, ← mul_assoc]; simp
rwa [one_mul] at hr rwa [one_mul] at hr
/-- #### Exercise 4 /-- ### Exercise 4
If `x` is an arbitrary real number, prove that there is exactly one integer `n` If `x` is an arbitrary real number, prove that there is exactly one integer `n`
which satisfies the inequalities `n ≤ x < n + 1`. This `n` is called the which satisfies the inequalities `n ≤ x < n + 1`. This `n` is called the
@ -540,7 +540,7 @@ theorem exercise_4 (x : ℝ) : ∃! n : ℤ, n ≤ x ∧ x < n + 1 := by
rw [← Int.floor_eq_iff] at hy rw [← Int.floor_eq_iff] at hy
exact Eq.symm hy exact Eq.symm hy
/-- #### Exercise 5 /-- ### Exercise 5
If `x` is an arbitrary real number, prove that there is exactly one integer `n` If `x` is an arbitrary real number, prove that there is exactly one integer `n`
which satisfies `x ≤ n < x + 1`. which satisfies `x ≤ n < x + 1`.
@ -559,7 +559,7 @@ theorem exercise_5 (x : ℝ) : ∃! n : ℤ, x ≤ n ∧ n < x + 1 := by
rwa [add_sub_cancel] at this rwa [add_sub_cancel] at this
· exact hy.left · exact hy.left
/-! #### Exercise 6 /-! ### Exercise 6
If `x` and `y` are arbitrary real numbers, `x < y`, prove that there exists at If `x` and `y` are arbitrary real numbers, `x < y`, prove that there exists at
least one rational number `r` satisfying `x < r < y`, and hence infinitely many. least one rational number `r` satisfying `x < r < y`, and hence infinitely many.
@ -569,7 +569,7 @@ in the real-number system.
###### TODO ###### TODO
-/ -/
/-! #### Exercise 7 /-! ### Exercise 7
If `x` is rational, `x ≠ 0`, and `y` irrational, prove that `x + y`, `x - y`, If `x` is rational, `x ≠ 0`, and `y` irrational, prove that `x + y`, `x - y`,
`xy`, `x / y`, and `y / x` are all irrational. `xy`, `x / y`, and `y / x` are all irrational.
@ -577,14 +577,14 @@ If `x` is rational, `x ≠ 0`, and `y` irrational, prove that `x + y`, `x - y`,
###### TODO ###### TODO
-/ -/
/-! #### Exercise 8 /-! ### Exercise 8
Is the sum or product of two irrational numbers always irrational? Is the sum or product of two irrational numbers always irrational?
###### TODO ###### TODO
-/ -/
/-! #### Exercise 9 /-! ### Exercise 9
If `x` and `y` are arbitrary real numbers, `x < y`, prove that there exists at If `x` and `y` are arbitrary real numbers, `x < y`, prove that there exists at
least one irrational number `z` satisfying `x < z < y`, and hence infinitely least one irrational number `z` satisfying `x < z < y`, and hence infinitely
@ -593,7 +593,7 @@ many.
###### TODO ###### TODO
-/ -/
/-! #### Exercise 10 /-! ### Exercise 10
An integer `n` is called *even* if `n = 2m` for some integer `m`, and *odd* if An integer `n` is called *even* if `n = 2m` for some integer `m`, and *odd* if
`n + 1` is even. Prove the following statements: `n + 1` is even. Prove the following statements:
@ -618,7 +618,7 @@ def isEven (n : ℤ) := ∃ m : ℤ, n = 2 * m
def isOdd (n : ℤ) := isEven (n + 1) def isOdd (n : ℤ) := isEven (n + 1)
/-! #### Exercise 11 /-! ### Exercise 11
Prove that there is no rational number whose square is `2`. Prove that there is no rational number whose square is `2`.
@ -629,7 +629,7 @@ contradiction.]
###### TODO ###### TODO
-/ -/
/-! #### Exercise 12 /-! ### Exercise 12
The Archimedean property of the real-number system was deduced as a consequence The Archimedean property of the real-number system was deduced as a consequence
of the least-upper-bound axiom. Prove that the set of rational numbers satisfies of the least-upper-bound axiom. Prove that the set of rational numbers satisfies

8
Bookshelf/Avigad/Chapter_2.lean
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@ -3,7 +3,7 @@
Dependent Type Theory Dependent Type Theory
-/ -/
/-! #### Exercise 1 /-! ### Exercise 1
Define the function `Do_Twice`, as described in Section 2.4. Define the function `Do_Twice`, as described in Section 2.4.
-/ -/
@ -20,7 +20,7 @@ def doTwiceTwice (f : (Nat → Nat) → (Nat → Nat)) (x : Nat → Nat) := f (f
end ex1 end ex1
/-! #### Exercise 2 /-! ### Exercise 2
Define the functions `curry` and `uncurry`, as described in Section 2.4. Define the functions `curry` and `uncurry`, as described in Section 2.4.
-/ -/
@ -35,7 +35,7 @@ def uncurry (f : α → β → γ) : (α × β → γ) :=
end ex2 end ex2
/-! #### Exercise 3 /-! ### Exercise 3
Above, we used the example `vec α n` for vectors of elements of type `α` of Above, we used the example `vec α n` for vectors of elements of type `α` of
length `n`. Declare a constant `vec_add` that could represent a function that length `n`. Declare a constant `vec_add` that could represent a function that
@ -70,7 +70,7 @@ variable (c d : vec Prop 2)
end ex3 end ex3
/-! #### Exercise 4 /-! ### Exercise 4
Similarly, declare a constant `matrix` so that `matrix α m n` could represent Similarly, declare a constant `matrix` so that `matrix α m n` could represent
the type of `m` by `n` matrices. Declare some constants to represent functions the type of `m` by `n` matrices. Declare some constants to represent functions

6
Bookshelf/Avigad/Chapter_3.lean
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@ -3,7 +3,7 @@
Propositions and Proofs Propositions and Proofs
-/ -/
/-! #### Exercise 1 /-! ### Exercise 1
Prove the following identities. Prove the following identities.
-/ -/
@ -104,7 +104,7 @@ theorem imp_imp_not_imp_not : (p → q) → (¬q → ¬p) :=
end ex1 end ex1
/-! #### Exercise 2 /-! ### Exercise 2
Prove the following identities. These require classical reasoning. Prove the following identities. These require classical reasoning.
-/ -/
@ -150,7 +150,7 @@ theorem imp_imp_imp : (((p → q) → p) → p) :=
end ex2 end ex2
/-! #### Exercise 3 /-! ### Exercise 3
Prove `¬(p ↔ ¬p)` without using classical logic. Prove `¬(p ↔ ¬p)` without using classical logic.
-/ -/

10
Bookshelf/Avigad/Chapter_4.lean
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@ -40,7 +40,7 @@ theorem forall_or_distrib
end ex1 end ex1
/-! #### Exercise 2 /-! ### Exercise 2
It is often possible to bring a component of a formula outside a universal It is often possible to bring a component of a formula outside a universal
quantifier, when it does not depend on the quantified variable. Try proving quantifier, when it does not depend on the quantified variable. Try proving
@ -78,7 +78,7 @@ theorem forall_swap : (∀ x, r → p x) ↔ (r → ∀ x, p x) :=
end ex2 end ex2
/-! #### Exercise 3 /-! ### Exercise 3
Consider the "barber paradox," that is, the claim that in a certain town there Consider the "barber paradox," that is, the claim that in a certain town there
is a (male) barber that shaves all and only the men who do not shave themselves. is a (male) barber that shaves all and only the men who do not shave themselves.
@ -101,7 +101,7 @@ theorem barber_paradox (h : ∀ x : men, shaves barber x ↔ ¬shaves x x) : Fal
end ex3 end ex3
/-! #### Exercise 4 /-! ### Exercise 4
Remember that, without any parameters, an expression of type `Prop` is just an Remember that, without any parameters, an expression of type `Prop` is just an
assertion. Fill in the definitions of `prime` and `Fermat_prime` below, and assertion. Fill in the definitions of `prime` and `Fermat_prime` below, and
@ -143,7 +143,7 @@ def Fermat'sLastTheorem : Prop :=
end ex4 end ex4
/-! #### Exercise 5 /-! ### Exercise 5
Prove as many of the identities listed in Section 4.4 as you can. Prove as many of the identities listed in Section 4.4 as you can.
-/ -/
@ -228,7 +228,7 @@ theorem exists_self_iff_self_exists (a : α) : (∃ x, r → p x) ↔ (r → ∃
end ex5 end ex5
/-! #### Exercise 6 /-! ### Exercise 6
Give a calculational proof of the theorem `log_mul` below. Give a calculational proof of the theorem `log_mul` below.
-/ -/

16
Bookshelf/Avigad/Chapter_5.lean
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@ -13,7 +13,7 @@ namespace Avigad.Chapter5
namespace ex1 namespace ex1
/-! ##### Exercises 3.1 -/ /-! #### Exercises 3.1 -/
section ex3_1 section ex3_1
@ -154,7 +154,7 @@ theorem imp_imp_not_imp_not : (p → q) → (¬q → ¬p) := by
end ex3_1 end ex3_1
/-! ##### Exercises 3.2 -/ /-! #### Exercises 3.2 -/
section ex3_2 section ex3_2
@ -223,7 +223,7 @@ theorem imp_imp_imp : (((p → q) → p) → p) := by
end ex3_2 end ex3_2
/-! ##### Exercises 3.3 -/ /-! #### Exercises 3.3 -/
section ex3_3 section ex3_3
@ -235,7 +235,7 @@ theorem iff_not_self (hp : p) : ¬(p ↔ ¬p) := by
end ex3_3 end ex3_3
/-! ##### Exercises 4.1 -/ /-! #### Exercises 4.1 -/
section ex4_1 section ex4_1
@ -264,7 +264,7 @@ theorem forall_or_distrib : (∀ x, p x) ∨ (∀ x, q x) → ∀ x, p x ∨ q x
end ex4_1 end ex4_1
/-! ##### Exercises 4.2 -/ /-! #### Exercises 4.2 -/
section ex4_2 section ex4_2
@ -316,7 +316,7 @@ theorem forall_swap : (∀ x, r → p x) ↔ (r → ∀ x, p x) := by
end ex4_2 end ex4_2
/-! ##### Exercises 4.3 -/ /-! #### Exercises 4.3 -/
section ex4_3 section ex4_3
@ -336,7 +336,7 @@ theorem barber_paradox (h : ∀ x : men, shaves barber x ↔ ¬ shaves x x)
end ex4_3 end ex4_3
/-! ##### Exercises 4.5 -/ /-! #### Exercises 4.5 -/
section ex4_5 section ex4_5
@ -448,7 +448,7 @@ end ex4_5
end ex1 end ex1
/-! #### Exercise 2 /-! ### Exercise 2
Use tactic combinators to obtain a one line proof of the following: Use tactic combinators to obtain a one line proof of the following:
-/ -/

6
Bookshelf/Avigad/Chapter_7.lean
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@ -5,7 +5,7 @@ Inductive Types
namespace Avigad.Chapter7 namespace Avigad.Chapter7
/-! #### Exercise 1 /-! ### Exercise 1
Try defining other operations on the natural numbers, such as multiplication, Try defining other operations on the natural numbers, such as multiplication,
the predecessor function (with `pred 0 = 0`), truncated subtraction (with the predecessor function (with `pred 0 = 0`), truncated subtraction (with
@ -77,7 +77,7 @@ end Nat
end ex1 end ex1
/-! #### Exercise 2 /-! ### Exercise 2
Define some operations on lists, like a `length` function or the `reverse` Define some operations on lists, like a `length` function or the `reverse`
function. Prove some properties, such as the following: function. Prove some properties, such as the following:
@ -178,7 +178,7 @@ theorem reverse_reverse (t : List α)
end ex2 end ex2
/-! #### Exercise 3 /-! ### Exercise 3
Define an inductive data type consisting of terms built up from the following Define an inductive data type consisting of terms built up from the following
constructors: constructors:

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@ -5,7 +5,7 @@ Induction and Recursion
namespace Avigad.Chapter8 namespace Avigad.Chapter8
/-! #### Exercise 1 /-! ### Exercise 1
Open a namespace `Hidden` to avoid naming conflicts, and use the equation Open a namespace `Hidden` to avoid naming conflicts, and use the equation
compiler to define addition, multiplication, and exponentiation on the natural compiler to define addition, multiplication, and exponentiation on the natural
@ -29,7 +29,7 @@ def exp : Nat → Nat → Nat
end ex1 end ex1
/-! #### Exercise 2 /-! ### Exercise 2
Similarly, use the equation compiler to define some basic operations on lists Similarly, use the equation compiler to define some basic operations on lists
(like the reverse function) and prove theorems about lists by induction (such as (like the reverse function) and prove theorems about lists by induction (such as
@ -48,7 +48,7 @@ def reverse : List α → List α
end ex2 end ex2
/-! #### Exercise 3 /-! ### Exercise 3
Define your own function to carry out course-of-value recursion on the natural Define your own function to carry out course-of-value recursion on the natural
numbers. Similarly, see if you can figure out how to define `WellFounded.fix` on numbers. Similarly, see if you can figure out how to define `WellFounded.fix` on
@ -86,7 +86,7 @@ noncomputable def brecOn {motive : Nat → Sort u}
end ex3 end ex3
/-! #### Exercise 4 /-! ### Exercise 4
Following the examples in Section Dependent Pattern Matching, define a function Following the examples in Section Dependent Pattern Matching, define a function
that will append two vectors. This is tricky; you will have to define an that will append two vectors. This is tricky; you will have to define an
@ -113,7 +113,7 @@ end Vector
end ex4 end ex4
/-! #### Exercise 5 /-! ### Exercise 5
Consider the following type of arithmetic expressions. Consider the following type of arithmetic expressions.
-/ -/
@ -149,7 +149,7 @@ def sampleVal : Nat → Nat
-- Try it out. You should get 47 here. -- Try it out. You should get 47 here.
#eval eval sampleVal sampleExpr #eval eval sampleVal sampleExpr
/-! ##### Constant Fusion /-! ### Constant Fusion
Implement "constant fusion," a procedure that simplifies subterms like `5 + 7 Implement "constant fusion," a procedure that simplifies subterms like `5 + 7
to `12`. Using the auxiliary function `simpConst`, define a function "fuse": to to `12`. Using the auxiliary function `simpConst`, define a function "fuse": to

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@ -131,7 +131,7 @@ lemma no_neg_sentential_count_eq_binary_count {φ : Wff} (h : ¬φ.hasNotSymbol)
unfold sententialSymbolCount binarySymbolCount unfold sententialSymbolCount binarySymbolCount
rw [ih₁ h.left, ih₂ h.right] rw [ih₁ h.left, ih₂ h.right]
/-- #### Parentheses Count /-- ### Parentheses Count
Let `φ` be a well-formed formula and `c` be the number of places at which a Let `φ` be a well-formed formula and `c` be the number of places at which a
sentential connective symbol exists. Then there is `2c` parentheses in `φ`. sentential connective symbol exists. Then there is `2c` parentheses in `φ`.
@ -180,7 +180,7 @@ theorem length_eq_sum_symbol_count (φ : Wff)
end Wff end Wff
/-! #### Exercise 1.1.2 /-! ### Exercise 1.1.2
Show that there are no wffs of length `2`, `3`, or `6`, but that any other Show that there are no wffs of length `2`, `3`, or `6`, but that any other
positive length is possible. positive length is possible.
@ -296,7 +296,7 @@ theorem exercise_1_1_2_ii (n : ℕ) (hn : n ≠ 2 ∧ n ≠ 3 ∧ n ≠ 6)
end Exercise_1_1_2 end Exercise_1_1_2
/-- #### Exercise 1.1.3 /-- ### Exercise 1.1.3
Let `α` be a wff; let `c` be the number of places at which binary connective Let `α` be a wff; let `c` be the number of places at which binary connective
symbols (`∧`, `∨`, `→`, `↔`) occur in `α`; let `s` be the number of places at symbols (`∧`, `∨`, `→`, `↔`) occur in `α`; let `s` be the number of places at
@ -320,7 +320,7 @@ theorem exercise_1_1_3 (φ : Wff)
rw [ih₁, ih₂] rw [ih₁, ih₂]
ring ring
/-- #### Exercise 1.1.5 (a) /-- ### Exercise 1.1.5 (a)
Suppose that `α` is a wff not containing the negation symbol `¬`. Show that the Suppose that `α` is a wff not containing the negation symbol `¬`. Show that the
length of `α` (i.e., the number of symbols in the string) is odd. length of `α` (i.e., the number of symbols in the string) is odd.
@ -355,7 +355,7 @@ theorem exercise_1_1_5a (α : Wff) (hα : ¬α.hasNotSymbol)
rw [hk₁, hk₂] rw [hk₁, hk₂]
ring ring
/-- #### Exercise 1.1.5 (b) /-- ### Exercise 1.1.5 (b)
Suppose that `α` is a wff not containing the negation symbol `¬`. Show that more Suppose that `α` is a wff not containing the negation symbol `¬`. Show that more
than a quarter of the symbols are sentence symbols. than a quarter of the symbols are sentence symbols.
@ -390,7 +390,7 @@ theorem exercise_1_1_5b (α : Wff) (hα : ¬α.hasNotSymbol)
] ]
exact inv_lt_one (by norm_num) exact inv_lt_one (by norm_num)
/-! #### Exercise 1.2.1 /-! ### Exercise 1.2.1
Show that neither of the following two formulas tautologically implies the Show that neither of the following two formulas tautologically implies the
other: other:
@ -430,7 +430,7 @@ theorem exercise_1_2_2a (P Q : Prop)
: (((P → Q) → P) → P) := by : (((P → Q) → P) → P) := by
tauto tauto
/-! #### Exercise 1.2.2 (b) /-! ### Exercise 1.2.2 (b)
Define `σₖ` recursively as follows: `σ₀ = (P → Q)` and `σₖ₊₁ = (σₖ → P)`. For Define `σₖ` recursively as follows: `σ₀ = (P → Q)` and `σₖ₊₁ = (σₖ → P)`. For
which values of `k` is `σₖ` a tautology? (Part (a) corresponds to `k = 2`.) which values of `k` is `σₖ` a tautology? (Part (a) corresponds to `k = 2`.)
@ -486,7 +486,7 @@ theorem exercise_1_2_2b_iii {k : ℕ} (h : Odd k)
end Exercise_1_2_2 end Exercise_1_2_2
/-- #### Exercise 1.2.3 (a) /-- ### Exercise 1.2.3 (a)
Determine whether or not `((P → Q)) ∨ (Q → P)` is a tautology. Determine whether or not `((P → Q)) ∨ (Q → P)` is a tautology.
-/ -/
@ -494,7 +494,7 @@ theorem exercise_1_2_3a (P Q : Prop)
: ((P → Q) ∨ (Q → P)) := by : ((P → Q) ∨ (Q → P)) := by
tauto tauto
/-- #### Exercise 1.2.3 (b) /-- ### Exercise 1.2.3 (b)
Determine whether or not `((P ∧ Q) → R))` tautologically implies Determine whether or not `((P ∧ Q) → R))` tautologically implies
`((P → R) ∨ (Q → R))`. `((P → R) ∨ (Q → R))`.
@ -503,7 +503,7 @@ theorem exercise_1_2_3b (P Q R : Prop)
: ((P ∧ Q) → R) ↔ ((P → R) ∨ (Q → R)) := by : ((P ∧ Q) → R) ↔ ((P → R) ∨ (Q → R)) := by
tauto tauto
/-! #### Exercise 1.2.5 /-! ### Exercise 1.2.5
Prove or refute each of the following assertions: Prove or refute each of the following assertions:
@ -519,7 +519,7 @@ theorem exercise_1_2_6b
: (False ∨ True) ∧ ¬ False := by : (False ∨ True) ∧ ¬ False := by
simp simp
/-! #### Exercise 1.2.15 /-! ### Exercise 1.2.15
Of the following three formulas, which tautologically implies which? Of the following three formulas, which tautologically implies which?
(a) `(A ↔ B)` (a) `(A ↔ B)`

2
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@ -8999,7 +8999,7 @@
\begin{proof} \begin{proof}
Let $S$ be a \nameref{ref:finite-set} and $S'$ be a Let $S$ be a \nameref{ref:finite-set} and $S'$ be a
\nameref{ref:proper-subset} $S'$ of $S$. \nameref{ref:proper-subset} of $S$.
Then there exists some set $T$, disjoint from $S'$, such that Then there exists some set $T$, disjoint from $S'$, such that
$S' \cup T = S$. $S' \cup T = S$.
By definition of a \nameref{ref:finite-set}, $S$ is By definition of a \nameref{ref:finite-set}, $S$ is

10
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@ -19,7 +19,7 @@ The `∅` does not equal the singleton set containing `∅`.
lemma empty_ne_singleton_empty (h : ∅ = ({∅} : Set (Set α))) : False := lemma empty_ne_singleton_empty (h : ∅ = ({∅} : Set (Set α))) : False :=
absurd h (Ne.symm $ Set.singleton_ne_empty (∅ : Set α)) absurd h (Ne.symm $ Set.singleton_ne_empty (∅ : Set α))
/-- #### Exercise 1.1a /-- ### Exercise 1.1a
`{∅} ___ {∅, {∅}}` `{∅} ___ {∅, {∅}}`
-/ -/
@ -27,7 +27,7 @@ theorem exercise_1_1a
: {∅} ∈ ({∅, {∅}} : Set (Set (Set α))) : {∅} ∈ ({∅, {∅}} : Set (Set (Set α)))
∧ {∅} ⊆ ({∅, {∅}} : Set (Set (Set α))) := ⟨by simp, by simp⟩ ∧ {∅} ⊆ ({∅, {∅}} : Set (Set (Set α))) := ⟨by simp, by simp⟩
/-- #### Exercise 1.1b /-- ### Exercise 1.1b
`{∅} ___ {∅, {{∅}}}` `{∅} ___ {∅, {{∅}}}`
-/ -/
@ -39,7 +39,7 @@ theorem exercise_1_1b
simp at h simp at h
exact empty_ne_singleton_empty h exact empty_ne_singleton_empty h
/-- #### Exercise 1.1c /-- ### Exercise 1.1c
`{{∅}} ___ {∅, {∅}}` `{{∅}} ___ {∅, {∅}}`
-/ -/
@ -47,7 +47,7 @@ theorem exercise_1_1c
: {{∅}} ∉ ({∅, {∅}} : Set (Set (Set (Set α)))) : {{∅}} ∉ ({∅, {∅}} : Set (Set (Set (Set α))))
∧ {{∅}} ⊆ ({∅, {∅}} : Set (Set (Set (Set α)))) := ⟨by simp, by simp⟩ ∧ {{∅}} ⊆ ({∅, {∅}} : Set (Set (Set (Set α)))) := ⟨by simp, by simp⟩
/-- #### Exercise 1.1d /-- ### Exercise 1.1d
`{{∅}} ___ {∅, {{∅}}}` `{{∅}} ___ {∅, {{∅}}}`
-/ -/
@ -59,7 +59,7 @@ theorem exercise_1_1d
simp at h simp at h
exact empty_ne_singleton_empty h exact empty_ne_singleton_empty h
/-- #### Exercise 1.1e /-- ### Exercise 1.1e
`{{∅}} ___ {∅, {∅, {∅}}}` `{{∅}} ___ {∅, {∅, {∅}}}`
-/ -/

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@ -10,7 +10,7 @@ Axioms and Operations
namespace Enderton.Set.Chapter_2 namespace Enderton.Set.Chapter_2
/-! #### Commutative Laws /-! ### Commutative Laws
For any sets `A` and `B`, For any sets `A` and `B`,
``` ```
@ -40,7 +40,7 @@ theorem commutative_law_ii (A B : Set α)
#check Set.inter_comm #check Set.inter_comm
/-! #### Associative Laws /-! ### Associative Laws
For any sets `A`, `B`, and `C`, For any sets `A`, `B`, and `C`,
``` ```
@ -75,7 +75,7 @@ theorem associative_law_ii (A B C : Set α)
#check Set.inter_assoc #check Set.inter_assoc
/-! #### Distributive Laws /-! ### Distributive Laws
For any sets `A`, `B`, and `C`, For any sets `A`, `B`, and `C`,
``` ```
@ -108,7 +108,7 @@ theorem distributive_law_ii (A B C : Set α)
#check Set.union_distrib_left #check Set.union_distrib_left
/-! #### De Morgan's Laws /-! ### De Morgan's Laws
For any sets `A`, `B`, and `C`, For any sets `A`, `B`, and `C`,
``` ```
@ -149,7 +149,7 @@ theorem de_morgans_law_ii (A B C : Set α)
#check Set.diff_inter #check Set.diff_inter
/-! #### Identities Involving ∅ /-! ### Identities Involving ∅
For any set `A`, For any set `A`,
``` ```
@ -185,7 +185,7 @@ theorem emptyset_identity_iii (A C : Set α)
#check Set.inter_diff_self #check Set.inter_diff_self
/-! #### Monotonicity /-! ### Monotonicity
For any sets `A`, `B`, and `C`, For any sets `A`, `B`, and `C`,
``` ```
@ -229,7 +229,7 @@ theorem monotonicity_iii (A B : Set (Set α)) (h : A ⊆ B)
#check Set.sUnion_mono #check Set.sUnion_mono
/-! #### Anti-monotonicity /-! ### Anti-monotonicity
For any sets `A`, `B`, and `C`, For any sets `A`, `B`, and `C`,
``` ```
@ -261,7 +261,7 @@ theorem anti_monotonicity_ii (A B : Set (Set α)) (h : A ⊆ B)
#check Set.sInter_subset_sInter #check Set.sInter_subset_sInter
/-- #### Intersection/Difference Associativity /-- ### Intersection/Difference Associativity
Let `A`, `B`, and `C` be sets. Then `A ∩ (B - C) = (A ∩ B) - C`. Let `A`, `B`, and `C` be sets. Then `A ∩ (B - C) = (A ∩ B) - C`.
-/ -/
@ -278,7 +278,7 @@ theorem inter_diff_assoc (A B C : Set α)
#check Set.inter_diff_assoc #check Set.inter_diff_assoc
/-- #### Exercise 2.1 /-- ### Exercise 2.1
Assume that `A` is the set of integers divisible by `4`. Similarly assume that Assume that `A` is the set of integers divisible by `4`. Similarly assume that
`B` and `C` are the sets of integers divisible by `9` and `10`, respectively. `B` and `C` are the sets of integers divisible by `9` and `10`, respectively.
@ -300,7 +300,7 @@ theorem exercise_2_1 {A B C : Set ℤ}
· rw [hC] at hc · rw [hC] at hc
exact Set.mem_setOf.mp hc exact Set.mem_setOf.mp hc
/-- #### Exercise 2.2 /-- ### Exercise 2.2
Give an example of sets `A` and `B` for which `⋃ A = ⋃ B` but `A ≠ B`. Give an example of sets `A` and `B` for which `⋃ A = ⋃ B` but `A ≠ B`.
-/ -/
@ -339,7 +339,7 @@ theorem exercise_2_2 {A B : Set (Set ℕ)}
have h₂ := h₁ 2 have h₂ := h₁ 2
simp at h₂ simp at h₂
/-- #### Exercise 2.3 /-- ### Exercise 2.3
Show that every member of a set `A` is a subset of `U A`. (This was stated as an Show that every member of a set `A` is a subset of `U A`. (This was stated as an
example in this section.) example in this section.)
@ -352,7 +352,7 @@ theorem exercise_2_3 {A : Set (Set α)}
rw [Set.mem_setOf_eq] rw [Set.mem_setOf_eq]
exact ⟨x, ⟨hx, hy⟩⟩ exact ⟨x, ⟨hx, hy⟩⟩
/-- #### Exercise 2.4 /-- ### Exercise 2.4
Show that if `A ⊆ B`, then `⋃ A ⊆ ⋃ B`. Show that if `A ⊆ B`, then `⋃ A ⊆ ⋃ B`.
-/ -/
@ -364,7 +364,7 @@ theorem exercise_2_4 {A B : Set (Set α)} (h : A ⊆ B) : ⋃₀ A ⊆ ⋃₀ B
rw [Set.mem_setOf_eq] rw [Set.mem_setOf_eq]
exact ⟨t, ⟨h ht, hxt⟩⟩ exact ⟨t, ⟨h ht, hxt⟩⟩
/-- #### Exercise 2.5 /-- ### Exercise 2.5
Assume that every member of `𝓐` is a subset of `B`. Show that `⋃ 𝓐 ⊆ B`. Assume that every member of `𝓐` is a subset of `B`. Show that `⋃ 𝓐 ⊆ B`.
-/ -/
@ -376,7 +376,7 @@ theorem exercise_2_5 {𝓐 : Set (Set α)} (h : ∀ x ∈ 𝓐, x ⊆ B)
have ⟨t, ⟨ht𝓐, hyt⟩⟩ := hy have ⟨t, ⟨ht𝓐, hyt⟩⟩ := hy
exact (h t ht𝓐) hyt exact (h t ht𝓐) hyt
/-- #### Exercise 2.6a /-- ### Exercise 2.6a
Show that for any set `A`, `⋃ 𝓟 A = A`. Show that for any set `A`, `⋃ 𝓟 A = A`.
-/ -/
@ -393,7 +393,7 @@ theorem exercise_2_6a : ⋃₀ (𝒫 A) = A := by
rw [Set.mem_setOf_eq] rw [Set.mem_setOf_eq]
exact ⟨A, ⟨by rw [Set.mem_setOf_eq], hx⟩⟩ exact ⟨A, ⟨by rw [Set.mem_setOf_eq], hx⟩⟩
/-- #### Exercise 2.6b /-- ### Exercise 2.6b
Show that `A ⊆ 𝓟 ⋃ A`. Under what conditions does equality hold? Show that `A ⊆ 𝓟 ⋃ A`. Under what conditions does equality hold?
-/ -/
@ -412,7 +412,7 @@ theorem exercise_2_6b
conv => rhs; rw [hB, exercise_2_6a] conv => rhs; rw [hB, exercise_2_6a]
exact hB exact hB
/-- #### Exercise 2.7a /-- ### Exercise 2.7a
Show that for any sets `A` and `B`, `𝓟 A ∩ 𝓟 B = 𝓟 (A ∩ B)`. Show that for any sets `A` and `B`, `𝓟 A ∩ 𝓟 B = 𝓟 (A ∩ B)`.
-/ -/
@ -430,7 +430,7 @@ theorem exercise_2_7A
intro x hA _ intro x hA _
exact hA exact hA
/-- #### Exercise 2.7b (i) /-- ### Exercise 2.7b (i)
Show that `𝓟 A ∪ 𝓟 B ⊆ 𝓟 (A ∪ B)`. Show that `𝓟 A ∪ 𝓟 B ⊆ 𝓟 (A ∪ B)`.
-/ -/
@ -447,7 +447,7 @@ theorem exercise_2_7b_i
rw [Set.mem_setOf_eq] rw [Set.mem_setOf_eq]
exact Set.subset_union_of_subset_right hB A exact Set.subset_union_of_subset_right hB A
/-- #### Exercise 2.7b (ii) /-- ### Exercise 2.7b (ii)
Under what conditions does `𝓟 A ∪ 𝓟 B = 𝓟 (A ∪ B)`.? Under what conditions does `𝓟 A ∪ 𝓟 B = 𝓟 (A ∪ B)`.?
-/ -/
@ -499,7 +499,7 @@ theorem exercise_2_7b_ii
refine Or.inl (Set.Subset.trans hx ?_) refine Or.inl (Set.Subset.trans hx ?_)
exact subset_of_eq (Set.right_subset_union_eq_self hB) exact subset_of_eq (Set.right_subset_union_eq_self hB)
/-- #### Exercise 2.9 /-- ### Exercise 2.9
Give an example of sets `a` and `B` for which `a ∈ B` but `𝓟 a ∉ 𝓟 B`. Give an example of sets `a` and `B` for which `a ∈ B` but `𝓟 a ∉ 𝓟 B`.
-/ -/
@ -527,7 +527,7 @@ theorem exercise_2_9 (ha : a = {1}) (hB : B = {{1}})
have := h 1 have := h 1
simp at this simp at this
/-- #### Exercise 2.10 /-- ### Exercise 2.10
Show that if `a ∈ B`, then `𝓟 a ∈ 𝓟 𝓟 ⋃ B`. Show that if `a ∈ B`, then `𝓟 a ∈ 𝓟 𝓟 ⋃ B`.
-/ -/
@ -540,7 +540,7 @@ theorem exercise_2_10 {B : Set (Set α)} (ha : a ∈ B)
rw [← hb, Set.mem_setOf_eq] rw [← hb, Set.mem_setOf_eq]
exact h₂ exact h₂
/-- #### Exercise 2.11 (i) /-- ### Exercise 2.11 (i)
Show that for any sets `A` and `B`, `A = (A ∩ B) ∪ (A - B)`. Show that for any sets `A` and `B`, `A = (A ∩ B) ∪ (A - B)`.
-/ -/
@ -557,7 +557,7 @@ theorem exercise_2_11_i {A B : Set α}
· intro hx · intro hx
exact ⟨hx, em (B x)⟩ exact ⟨hx, em (B x)⟩
/-- #### Exercise 2.11 (ii) /-- ### Exercise 2.11 (ii)
Show that for any sets `A` and `B`, `A ∪ (B - A) = A ∪ B`. Show that for any sets `A` and `B`, `A ∪ (B - A) = A ∪ B`.
-/ -/
@ -651,7 +651,7 @@ lemma left_diff_eq_singleton_one : (A \ B) \ C = {1} := by
| inl y => rw [hx] at y; simp at y | inl y => rw [hx] at y; simp at y
| inr y => rw [hx] at y; simp at y | inr y => rw [hx] at y; simp at y
/-- #### Exercise 2.14 /-- ### Exercise 2.14
Show by example that for some sets `A`, `B`, and `C`, the set `A - (B - C)` is Show by example that for some sets `A`, `B`, and `C`, the set `A - (B - C)` is
different from `(A - B) - C`. different from `(A - B) - C`.
@ -668,7 +668,7 @@ theorem exercise_2_14 : A \ (B \ C) ≠ (A \ B) \ C := by
end end
/-- #### Exercise 2.15 (a) /-- ### Exercise 2.15 (a)
Show that `A ∩ (B + C) = (A ∩ B) + (A ∩ C)`. Show that `A ∩ (B + C) = (A ∩ B) + (A ∩ C)`.
-/ -/
@ -696,7 +696,7 @@ theorem exercise_2_15a (A B C : Set α)
#check Set.inter_symmDiff_distrib_left #check Set.inter_symmDiff_distrib_left
/-- #### Exercise 2.15 (b) /-- ### Exercise 2.15 (b)
Show that `A + (B + C) = (A + B) + C`. Show that `A + (B + C) = (A + B) + C`.
-/ -/
@ -749,7 +749,7 @@ theorem exercise_2_15b (A B C : Set α)
#check symmDiff_assoc #check symmDiff_assoc
/-- #### Exercise 2.16 /-- ### Exercise 2.16
Simplify: Simplify:
`[(A ∪ B ∪ C) ∩ (A ∪ B)] - [(A ∪ (B - C)) ∩ A]` `[(A ∪ B ∪ C) ∩ (A ∪ B)] - [(A ∪ (B - C)) ∩ A]`
@ -761,7 +761,7 @@ theorem exercise_2_16 {A B C : Set α}
_ = (A ∪ B) \ A := by rw [Set.union_inter_cancel_left] _ = (A ∪ B) \ A := by rw [Set.union_inter_cancel_left]
_ = B \ A := by rw [Set.union_diff_left] _ = B \ A := by rw [Set.union_diff_left]
/-! #### Exercise 2.17 /-! ### Exercise 2.17
Show that the following four conditions are equivalent. Show that the following four conditions are equivalent.
@ -797,7 +797,7 @@ theorem exercise_2_17_iii {A B : Set α} (h : A ∪ B = B)
theorem exercise_2_17_iv {A B : Set α} (h : A ∩ B = A) theorem exercise_2_17_iv {A B : Set α} (h : A ∩ B = A)
: A ⊆ B := Set.inter_eq_left.mp h : A ⊆ B := Set.inter_eq_left.mp h
/-- #### Exercise 2.19 /-- ### Exercise 2.19
Is `𝒫 (A - B)` always equal to `𝒫 A - 𝒫 B`? Is it ever equal to `𝒫 A - 𝒫 B`? Is `𝒫 (A - B)` always equal to `𝒫 A - 𝒫 B`? Is it ever equal to `𝒫 A - 𝒫 B`?
-/ -/
@ -810,7 +810,7 @@ theorem exercise_2_19 {A B : Set α}
have := h ∅ have := h ∅
exact absurd (this.mp he) ne exact absurd (this.mp he) ne
/-- #### Exercise 2.20 /-- ### Exercise 2.20
Let `A`, `B`, and `C` be sets such that `A ∪ B = A ∪ C` and `A ∩ B = A ∩ C`. Let `A`, `B`, and `C` be sets such that `A ∪ B = A ∪ C` and `A ∩ B = A ∩ C`.
Show that `B = C`. Show that `B = C`.
@ -836,7 +836,7 @@ theorem exercise_2_20 {A B C : Set α}
rw [← hu] at this rw [← hu] at this
exact Or.elim this (absurd · hA) (by simp) exact Or.elim this (absurd · hA) (by simp)
/-- #### Exercise 2.21 /-- ### Exercise 2.21
Show that `⋃ (A ∪ B) = (⋃ A) ∪ (⋃ B)`. Show that `⋃ (A ∪ B) = (⋃ A) ∪ (⋃ B)`.
-/ -/
@ -860,7 +860,7 @@ theorem exercise_2_21 {A B : Set (Set α)}
have ⟨t, ht⟩ : ∃ t, t ∈ B ∧ x ∈ t := hB have ⟨t, ht⟩ : ∃ t, t ∈ B ∧ x ∈ t := hB
exact ⟨t, ⟨Set.mem_union_right A ht.left, ht.right⟩⟩ exact ⟨t, ⟨Set.mem_union_right A ht.left, ht.right⟩⟩
/-- #### Exercise 2.22 /-- ### Exercise 2.22
Show that if `A` and `B` are nonempty sets, then `⋂ (A ∪ B) = ⋂ A ∩ ⋂ B`. Show that if `A` and `B` are nonempty sets, then `⋂ (A ∪ B) = ⋂ A ∩ ⋂ B`.
-/ -/
@ -889,7 +889,7 @@ theorem exercise_2_22 {A B : Set (Set α)}
· intro hB · intro hB
exact (this t).right hB exact (this t).right hB
/-- #### Exercise 2.24a /-- ### Exercise 2.24a
Show that is `𝓐` is nonempty, then `𝒫 (⋂ 𝓐) = ⋂ { 𝒫 X | X ∈ 𝓐 }`. Show that is `𝓐` is nonempty, then `𝒫 (⋂ 𝓐) = ⋂ { 𝒫 X | X ∈ 𝓐 }`.
-/ -/
@ -908,7 +908,7 @@ theorem exercise_2_24a {𝓐 : Set (Set α)}
_ = { x | ∀ t ∈ { 𝒫 X | X ∈ 𝓐 }, x ∈ t} := by simp _ = { x | ∀ t ∈ { 𝒫 X | X ∈ 𝓐 }, x ∈ t} := by simp
_ = ⋂₀ { 𝒫 X | X ∈ 𝓐 } := rfl _ = ⋂₀ { 𝒫 X | X ∈ 𝓐 } := rfl
/-- #### Exercise 2.24b /-- ### Exercise 2.24b
Show that Show that
``` ```
@ -950,7 +950,7 @@ theorem exercise_2_24b {𝓐 : Set (Set α)}
simp only [Set.mem_setOf_eq, exists_exists_and_eq_and, Set.mem_powerset_iff] simp only [Set.mem_setOf_eq, exists_exists_and_eq_and, Set.mem_powerset_iff]
exact ⟨⋃₀ 𝓐, ⟨hA, hx⟩⟩ exact ⟨⋃₀ 𝓐, ⟨hA, hx⟩⟩
/-- #### Exercise 2.25 /-- ### Exercise 2.25
Is `A ∪ (⋃ 𝓑)` always the same as `⋃ { A ∪ X | X ∈ 𝓑 }`? If not, then under Is `A ∪ (⋃ 𝓑)` always the same as `⋃ { A ∪ X | X ∈ 𝓑 }`? If not, then under
what conditions does equality hold? what conditions does equality hold?

124
Bookshelf/Enderton/Set/Chapter_3.lean
View File

@ -17,7 +17,7 @@ namespace Enderton.Set.Chapter_3
open Set.Relation open Set.Relation
/-- #### Lemma 3B /-- ### Lemma 3B
If `x ∈ C` and `y ∈ C`, then `⟨x, y⟩ ∈ 𝒫 𝒫 C`. If `x ∈ C` and `y ∈ C`, then `⟨x, y⟩ ∈ 𝒫 𝒫 C`.
-/ -/
@ -35,7 +35,7 @@ lemma lemma_3b {C : Set α} (hx : x ∈ C) (hy : y ∈ C)
-/ -/
exact Set.mem_mem_imp_pair_subset hxs hxys exact Set.mem_mem_imp_pair_subset hxs hxys
/-- #### Theorem 3D /-- ### Theorem 3D
If `⟨x, y⟩ ∈ A`, then `x` and `y` belong to `⋃ ⋃ A`. If `⟨x, y⟩ ∈ A`, then `x` and `y` belong to `⋃ ⋃ A`.
-/ -/
@ -61,7 +61,7 @@ theorem theorem_3d {A : Set (Set (Set α))} (h : OrderedPair x y ∈ A)
exact ⟨this (by simp), this (by simp)⟩ exact ⟨this (by simp), this (by simp)⟩
/-- #### Theorem 3G (i) /-- ### Theorem 3G (i)
Assume that `F` is a one-to-one function. If `x ∈ dom F`, then `F⁻¹(F(x)) = x`. Assume that `F` is a one-to-one function. If `x ∈ dom F`, then `F⁻¹(F(x)) = x`.
-/ -/
@ -75,7 +75,7 @@ theorem theorem_3g_i {F : Set.HRelation α β}
unfold isOneToOne at hF unfold isOneToOne at hF
exact (single_valued_eq_unique hF.left hy hy₁).symm exact (single_valued_eq_unique hF.left hy hy₁).symm
/-- #### Theorem 3G (ii) /-- ### Theorem 3G (ii)
Assume that `F` is a one-to-one function. If `y ∈ ran F`, then `F(F⁻¹(y)) = y`. Assume that `F` is a one-to-one function. If `y ∈ ran F`, then `F(F⁻¹(y)) = y`.
-/ -/
@ -89,7 +89,7 @@ theorem theorem_3g_ii {F : Set.HRelation α β}
unfold isOneToOne at hF unfold isOneToOne at hF
exact (single_rooted_eq_unique hF.right hx hx₁).symm exact (single_rooted_eq_unique hF.right hx hx₁).symm
/-- #### Theorem 3H /-- ### Theorem 3H
Assume that `F` and `G` are functions. Then Assume that `F` and `G` are functions. Then
``` ```
@ -128,7 +128,7 @@ theorem theorem_3h_dom {F : Set.HRelation β γ} {G : Set.HRelation α β}
simp only [Set.mem_setOf_eq] simp only [Set.mem_setOf_eq]
exact ⟨a, ha.left.left, hb⟩ exact ⟨a, ha.left.left, hb⟩
/-- #### Theorem 3J (a) /-- ### Theorem 3J (a)
Assume that `F : A → B`, and that `A` is nonempty. There exists a function Assume that `F : A → B`, and that `A` is nonempty. There exists a function
`G : B → A` (a "left inverse") such that `G ∘ F` is the identity function on `A` `G : B → A` (a "left inverse") such that `G ∘ F` is the identity function on `A`
@ -281,7 +281,7 @@ theorem theorem_3j_a {F : Set.HRelation α β}
rw [← single_valued_eq_unique hF.is_func hx₂.right ht₂.left] at ht₂ rw [← single_valued_eq_unique hF.is_func hx₂.right ht₂.left] at ht₂
exact single_valued_eq_unique hG₁.is_func ht₂.right ht₁.right exact single_valued_eq_unique hG₁.is_func ht₂.right ht₁.right
/-- #### Theorem 3J (b) /-- ### Theorem 3J (b)
Assume that `F : A → B`, and that `A` is nonempty. There exists a function Assume that `F : A → B`, and that `A` is nonempty. There exists a function
`H : B → A` (a "right inverse") such that `F ∘ H` is the identity function on `H : B → A` (a "right inverse") such that `F ∘ H` is the identity function on
@ -300,7 +300,7 @@ theorem theorem_3j_b {F : Set.HRelation α β} (hF : mapsInto F A B)
simp only [Set.mem_setOf_eq, Prod.exists, exists_eq_right, Set.setOf_mem_eq] simp only [Set.mem_setOf_eq, Prod.exists, exists_eq_right, Set.setOf_mem_eq]
exact hy exact hy
/-- #### Theorem 3K (a) /-- ### Theorem 3K (a)
The following hold for any sets. (`F` need not be a function.) The following hold for any sets. (`F` need not be a function.)
The image of a union is the union of the images: The image of a union is the union of the images:
@ -335,7 +335,7 @@ theorem theorem_3k_a {F : Set.HRelation α β} {𝓐 : Set (Set α)}
simp only [Set.mem_sUnion, Set.mem_setOf_eq] simp only [Set.mem_sUnion, Set.mem_setOf_eq]
exact ⟨u, ⟨A, hA.left, hu.left⟩, hu.right⟩ exact ⟨u, ⟨A, hA.left, hu.left⟩, hu.right⟩
/-! #### Theorem 3K (b) /-! ### Theorem 3K (b)
The following hold for any sets. (`F` need not be a function.) The following hold for any sets. (`F` need not be a function.)
The image of an intersection is included in the intersection of the images: The image of an intersection is included in the intersection of the images:
@ -395,7 +395,7 @@ theorem theorem_3k_b_ii {F : Set.HRelation α β} {𝓐 : Set (Set α)}
simp only [Set.mem_sInter, Set.mem_setOf_eq] simp only [Set.mem_sInter, Set.mem_setOf_eq]
exact ⟨u, hu⟩ exact ⟨u, hu⟩
/-! #### Theorem 3K (c) /-! ### Theorem 3K (c)
The following hold for any sets. (`F` need not be a function.) The following hold for any sets. (`F` need not be a function.)
The image of a difference includes the difference of the images: The image of a difference includes the difference of the images:
@ -449,7 +449,7 @@ theorem theorem_3k_c_ii {F : Set.HRelation α β} {A B : Set α}
exact absurd hu₁.left hu.left.right exact absurd hu₁.left hu.left.right
exact ⟨hv₁, hv₂⟩ exact ⟨hv₁, hv₂⟩
/-! #### Corollary 3L /-! ### Corollary 3L
For any function `G` and sets `A`, `B`, and `𝓐`: For any function `G` and sets `A`, `B`, and `𝓐`:
@ -477,7 +477,7 @@ theorem corollary_3l_iii {G : Set.HRelation β α} {A B : Set α}
single_valued_self_iff_single_rooted_inv.mp hG single_valued_self_iff_single_rooted_inv.mp hG
exact (theorem_3k_c_ii hG').symm exact (theorem_3k_c_ii hG').symm
/-- #### Theorem 3M /-- ### Theorem 3M
If `R` is a symmetric and transitive relation, then `R` is an equivalence If `R` is a symmetric and transitive relation, then `R` is an equivalence
relation on `fld R`. relation on `fld R`.
@ -497,7 +497,7 @@ theorem theorem_3m {R : Set.Relation α}
have := hS ht have := hS ht
exact hT this ht exact hT this ht
/-- #### Theorem 3R /-- ### Theorem 3R
Let `R` be a linear ordering on `A`. Let `R` be a linear ordering on `A`.
@ -521,7 +521,7 @@ theorem theorem_3r {R : Rel α α} (hR : IsStrictTotalOrder α R)
right right
exact h₂ exact h₂
/-- #### Exercise 3.1 /-- ### Exercise 3.1
Suppose that we attempted to generalize the Kuratowski definitions of ordered Suppose that we attempted to generalize the Kuratowski definitions of ordered
pairs to ordered triples by defining pairs to ordered triples by defining
@ -544,7 +544,7 @@ theorem exercise_3_1 {x y z u v w : ℕ}
· rw [hy, hv] · rw [hy, hv]
simp only simp only
/-- #### Exercise 3.2a /-- ### Exercise 3.2a
Show that `A × (B ∪ C) = (A × B) ∪ (A × C)`. Show that `A × (B ∪ C) = (A × B) ∪ (A × C)`.
-/ -/
@ -560,7 +560,7 @@ theorem exercise_3_2a {A : Set α} {B C : Set β}
_ = { p | p ∈ Set.prod A B ∨ (p ∈ Set.prod A C) } := rfl _ = { p | p ∈ Set.prod A B ∨ (p ∈ Set.prod A C) } := rfl
_ = (Set.prod A B) ∪ (Set.prod A C) := rfl _ = (Set.prod A B) ∪ (Set.prod A C) := rfl
/-- #### Exercise 3.2b /-- ### Exercise 3.2b
Show that if `A × B = A × C` and `A ≠ ∅`, then `B = C`. Show that if `A × B = A × C` and `A ≠ ∅`, then `B = C`.
-/ -/
@ -589,7 +589,7 @@ theorem exercise_3_2b {A : Set α} {B C : Set β}
have ⟨c, hc⟩ := Set.nonempty_iff_ne_empty.mpr (Ne.symm nC) have ⟨c, hc⟩ := Set.nonempty_iff_ne_empty.mpr (Ne.symm nC)
exact (h (a, c)).mpr ⟨ha, hc⟩ exact (h (a, c)).mpr ⟨ha, hc⟩
/-- #### Exercise 3.3 /-- ### Exercise 3.3
Show that `A × ⋃ 𝓑 = ⋃ {A × X | X ∈ 𝓑}`. Show that `A × ⋃ 𝓑 = ⋃ {A × X | X ∈ 𝓑}`.
-/ -/
@ -617,7 +617,7 @@ theorem exercise_3_3 {A : Set (Set α)} {𝓑 : Set (Set β)}
· intro ⟨b, h₁, h₂, h₃⟩ · intro ⟨b, h₁, h₂, h₃⟩
exact ⟨b, h₁, h₂, h₃⟩ exact ⟨b, h₁, h₂, h₃⟩
/-- #### Exercise 3.5a /-- ### Exercise 3.5a
Assume that `A` and `B` are given sets, and show that there exists a set `C` Assume that `A` and `B` are given sets, and show that there exists a set `C`
such that for any `y`, such that for any `y`,
@ -685,7 +685,7 @@ theorem exercise_3_5a {A : Set α} {B : Set β}
rw [hab.right] rw [hab.right]
exact ⟨hab.left, hb⟩ exact ⟨hab.left, hb⟩
/-- #### Exercise 3.5b /-- ### Exercise 3.5b
With `A`, `B`, and `C` as above, show that `A × B = ∪ C`. With `A`, `B`, and `C` as above, show that `A × B = ∪ C`.
-/ -/
@ -718,7 +718,7 @@ theorem exercise_3_5b {A : Set α} (B : Set β)
exact ⟨h, hb⟩ exact ⟨h, hb⟩
/-- #### Exercise 3.6 /-- ### Exercise 3.6
Show that a set `A` is a relation **iff** `A ⊆ dom A × ran A`. Show that a set `A` is a relation **iff** `A ⊆ dom A × ran A`.
-/ -/
@ -736,7 +736,7 @@ theorem exercise_3_6 {A : Set.HRelation α β}
] ]
exact ⟨⟨b, ht⟩, ⟨a, ht⟩⟩ exact ⟨⟨b, ht⟩, ⟨a, ht⟩⟩
/-- #### Exercise 3.7 /-- ### Exercise 3.7
Show that if `R` is a relation, then `fld R = ⋃ ⋃ R`. Show that if `R` is a relation, then `fld R = ⋃ ⋃ R`.
-/ -/
@ -815,7 +815,7 @@ theorem exercise_3_7 {R : Set.Relation α}
simp only [Set.mem_singleton_iff, Set.mem_insert_iff] at this simp only [Set.mem_singleton_iff, Set.mem_insert_iff] at this
exact hxy_mem this exact hxy_mem this
/-- #### Exercise 3.8 (i) /-- ### Exercise 3.8 (i)
Show that for any set `𝓐`: Show that for any set `𝓐`:
``` ```
@ -841,7 +841,7 @@ theorem exercise_3_8_i {A : Set (Set.HRelation α β)}
· intro ⟨t, ht, y, hx⟩ · intro ⟨t, ht, y, hx⟩
exact ⟨y, t, ht, hx⟩ exact ⟨y, t, ht, hx⟩
/-- #### Exercise 3.8 (ii) /-- ### Exercise 3.8 (ii)
Show that for any set `𝓐`: Show that for any set `𝓐`:
``` ```
@ -866,7 +866,7 @@ theorem exercise_3_8_ii {A : Set (Set.HRelation α β)}
· intro ⟨y, ⟨hy, ⟨t, ht⟩⟩⟩ · intro ⟨y, ⟨hy, ⟨t, ht⟩⟩⟩
exact ⟨t, ⟨y, ⟨hy, ht⟩⟩⟩ exact ⟨t, ⟨y, ⟨hy, ht⟩⟩⟩
/-- #### Exercise 3.9 (i) /-- ### Exercise 3.9 (i)
Discuss the result of replacing the union operation by the intersection Discuss the result of replacing the union operation by the intersection
operation in the preceding problem. operation in the preceding problem.
@ -892,7 +892,7 @@ theorem exercise_3_9_i {A : Set (Set.HRelation α β)}
intro _ y hy R hR intro _ y hy R hR
exact ⟨y, hy R hR⟩ exact ⟨y, hy R hR⟩
/-- #### Exercise 3.9 (ii) /-- ### Exercise 3.9 (ii)
Discuss the result of replacing the union operation by the intersection Discuss the result of replacing the union operation by the intersection
operation in the preceding problem. operation in the preceding problem.
@ -918,7 +918,7 @@ theorem exercise_3_9_ii {A : Set (Set.HRelation α β)}
intro _ y hy R hR intro _ y hy R hR
exact ⟨y, hy R hR⟩ exact ⟨y, hy R hR⟩
/-- #### Exercise 3.12 /-- ### Exercise 3.12
Assume that `f` and `g` are functions and show that Assume that `f` and `g` are functions and show that
``` ```
@ -948,7 +948,7 @@ theorem exercise_3_12 {f g : Set.HRelation α β}
rw [single_valued_eq_unique hf hp hy₁.left.left] rw [single_valued_eq_unique hf hp hy₁.left.left]
exact hy₁.left.right exact hy₁.left.right
/-- #### Exercise 3.13 /-- ### Exercise 3.13
Assume that `f` and `g` are functions with `f ⊆ g` and `dom g ⊆ dom f`. Show Assume that `f` and `g` are functions with `f ⊆ g` and `dom g ⊆ dom f`. Show
that `f = g`. that `f = g`.
@ -972,7 +972,7 @@ theorem exercise_3_13 {f g : Set.HRelation α β}
rw [single_valued_eq_unique hg hp hx.left.right] rw [single_valued_eq_unique hg hp hx.left.right]
exact hx.left.left exact hx.left.left
/-- #### Exercise 3.14 (a) /-- ### Exercise 3.14 (a)
Assume that `f` and `g` are functions. Show that `f ∩ g` is a function. Assume that `f` and `g` are functions. Show that `f ∩ g` is a function.
-/ -/
@ -981,7 +981,7 @@ theorem exercise_3_14_a {f g : Set.HRelation α β}
: isSingleValued (f ∩ g) := : isSingleValued (f ∩ g) :=
single_valued_subset hf (Set.inter_subset_left f g) single_valued_subset hf (Set.inter_subset_left f g)
/-- #### Exercise 3.14 (b) /-- ### Exercise 3.14 (b)
Assume that `f` and `g` are functions. Show that `f ∪ g` is a function **iff** Assume that `f` and `g` are functions. Show that `f ∪ g` is a function **iff**
`f(x) = g(x)` for every `x` in `(dom f) ∩ (dom g)`. `f(x) = g(x)` for every `x` in `(dom f) ∩ (dom g)`.
@ -1061,7 +1061,7 @@ theorem exercise_3_14_b {f g : Set.HRelation α β}
· intro hz · intro hz
exact absurd (mem_pair_imp_fst_mem_dom hz) hgx exact absurd (mem_pair_imp_fst_mem_dom hz) hgx
/-- #### Exercise 3.15 /-- ### Exercise 3.15
Let `𝓐` be a set of functions such that for any `f` and `g` in `𝓐`, either Let `𝓐` be a set of functions such that for any `f` and `g` in `𝓐`, either
`f ⊆ g` or `g ⊆ f`. Show that `⋃ 𝓐` is a function. `f ⊆ g` or `g ⊆ f`. Show that `⋃ 𝓐` is a function.
@ -1084,7 +1084,7 @@ theorem exercise_3_15 {𝓐 : Set (Set.HRelation α β)}
have := hg' hg.right have := hg' hg.right
exact single_valued_eq_unique (h𝓐 f hf.left) this hf.right exact single_valued_eq_unique (h𝓐 f hf.left) this hf.right
/-! #### Exercise 3.17 /-! ### Exercise 3.17
Show that the composition of two single-rooted sets is again single-rooted. Show that the composition of two single-rooted sets is again single-rooted.
Conclude that the composition of two one-to-one functions is again one-to-one. Conclude that the composition of two one-to-one functions is again one-to-one.
@ -1119,7 +1119,7 @@ theorem exercise_3_17_ii {F : Set.HRelation β γ} {G : Set.HRelation α β}
(single_valued_comp_is_single_valued hF.left hG.left) (single_valued_comp_is_single_valued hF.left hG.left)
(exercise_3_17_i hF.right hG.right) (exercise_3_17_i hF.right hG.right)
/-! #### Exercise 3.18 /-! ### Exercise 3.18
Let `R` be the set Let `R` be the set
``` ```
@ -1252,7 +1252,7 @@ theorem exercise_3_18_v
end Exercise_3_18 end Exercise_3_18
/-! #### Exercise 3.19 /-! ### Exercise 3.19
Let Let
``` ```
@ -1502,7 +1502,7 @@ theorem exercise_3_19_x
end Exercise_3_19 end Exercise_3_19
/-- #### Exercise 3.20 /-- ### Exercise 3.20
Show that `F ↾ A = F ∩ (A × ran F)`. Show that `F ↾ A = F ∩ (A × ran F)`.
-/ -/
@ -1522,7 +1522,7 @@ theorem exercise_3_20 {F : Set.HRelation α β} {A : Set α}
_ = F ∩ {p | p.fst ∈ A ∧ p.snd ∈ ran F} := rfl _ = F ∩ {p | p.fst ∈ A ∧ p.snd ∈ ran F} := rfl
_ = F ∩ (Set.prod A (ran F)) := rfl _ = F ∩ (Set.prod A (ran F)) := rfl
/-- #### Exercise 3.22 (a) /-- ### Exercise 3.22 (a)
Show that the following is correct for any sets. Show that the following is correct for any sets.
``` ```
@ -1537,7 +1537,7 @@ theorem exercise_3_22_a {A B : Set α} {F : Set.HRelation α β} (h : A ⊆ B)
have := h hu.left have := h hu.left
exact ⟨u, this, hu.right⟩ exact ⟨u, this, hu.right⟩
/-- #### Exercise 3.22 (b) /-- ### Exercise 3.22 (b)
Show that the following is correct for any sets. Show that the following is correct for any sets.
``` ```
@ -1567,7 +1567,7 @@ theorem exercise_3_22_b {A B : Set α} {F : Set.HRelation α β}
_ = { v | ∃ a ∈ image G A, (a, v) ∈ F } := rfl _ = { v | ∃ a ∈ image G A, (a, v) ∈ F } := rfl
_ = image F (image G A) := rfl _ = image F (image G A) := rfl
/-- #### Exercise 3.22 (c) /-- ### Exercise 3.22 (c)
Show that the following is correct for any sets. Show that the following is correct for any sets.
``` ```
@ -1585,7 +1585,7 @@ theorem exercise_3_22_c {A B : Set α} {Q : Set.Relation α}
_ = { p | p ∈ Q ∧ p.1 ∈ A} ∪ { p | p ∈ Q ∧ p.1 ∈ B } := rfl _ = { p | p ∈ Q ∧ p.1 ∈ A} ∪ { p | p ∈ Q ∧ p.1 ∈ B } := rfl
_ = (restriction Q A) ∪ (restriction Q B) := rfl _ = (restriction Q A) ∪ (restriction Q B) := rfl
/-- #### Exercise 3.23 (i) /-- ### Exercise 3.23 (i)
Let `I` be the identity function on the set `A`. Show that for any sets `B` and Let `I` be the identity function on the set `A`. Show that for any sets `B` and
`C`, `B ∘ I = B ↾ A`. `C`, `B ∘ I = B ↾ A`.
@ -1609,7 +1609,7 @@ theorem exercise_3_23_i {A : Set α} {B : Set.HRelation α β} {I : Set.Relation
intro (x, y) hp intro (x, y) hp
refine ⟨x, ⟨hp.right, rfl⟩, hp.left⟩ refine ⟨x, ⟨hp.right, rfl⟩, hp.left⟩
/-- #### Exercise 3.23 (ii) /-- ### Exercise 3.23 (ii)
Let `I` be the identity function on the set `A`. Show that for any sets `B` and Let `I` be the identity function on the set `A`. Show that for any sets `B` and
`C`, `I⟦C⟧ = A ∩ C`. `C`, `I⟦C⟧ = A ∩ C`.
@ -1641,7 +1641,7 @@ theorem exercise_3_23_ii {A C : Set α} {I : Set.Relation α}
_ = C ∩ A := rfl _ = C ∩ A := rfl
_ = A ∩ C := Set.inter_comm C A _ = A ∩ C := Set.inter_comm C A
/-- #### Exercise 3.24 /-- ### Exercise 3.24
Show that for a function `F`, `F⁻¹⟦A⟧ = { x ∈ dom F | F(x) ∈ A }`. Show that for a function `F`, `F⁻¹⟦A⟧ = { x ∈ dom F | F(x) ∈ A }`.
-/ -/
@ -1671,7 +1671,7 @@ theorem exercise_3_24 {F : Set.HRelation α β} {A : Set β}
· intro ⟨y, hy⟩ · intro ⟨y, hy⟩
exact ⟨y, hy.left⟩ exact ⟨y, hy.left⟩
/-- #### Exercise 3.25 (b) /-- ### Exercise 3.25 (b)
Show that the result of part (a) holds for any function `G`, not necessarily Show that the result of part (a) holds for any function `G`, not necessarily
one-to-one. one-to-one.
@ -1694,7 +1694,7 @@ theorem exercise_3_25_b {G : Set.HRelation α β} (hG : isSingleValued G)
have ⟨t, ht⟩ := ran_exists h.left have ⟨t, ht⟩ := ran_exists h.left
exact ⟨t, ⟨t, x, ht, rfl, rfl⟩, by rwa [← h.right]⟩ exact ⟨t, ⟨t, x, ht, rfl, rfl⟩, by rwa [← h.right]⟩
/-- #### Exercise 3.25 (a) /-- ### Exercise 3.25 (a)
Assume that `G` is a one-to-one function. Show that `G ∘ G⁻¹` is the identity Assume that `G` is a one-to-one function. Show that `G ∘ G⁻¹` is the identity
function on `ran G`. function on `ran G`.
@ -1703,7 +1703,7 @@ theorem exercise_3_25_a {G : Set.HRelation α β} (hG : isOneToOne G)
: comp G (inv G) = { p | p.1 ∈ ran G ∧ p.1 = p.2 } := : comp G (inv G) = { p | p.1 ∈ ran G ∧ p.1 = p.2 } :=
exercise_3_25_b hG.left exercise_3_25_b hG.left
/-- #### Exercise 3.27 /-- ### Exercise 3.27
Show that `dom (F ∘ G) = G⁻¹⟦dom F⟧` for any sets `F` and `G`. (`F` and `G` need Show that `dom (F ∘ G) = G⁻¹⟦dom F⟧` for any sets `F` and `G`. (`F` and `G` need
not be functions.) not be functions.)
@ -1737,7 +1737,7 @@ theorem exercise_3_27 {F : Set.HRelation β γ} {G : Set.HRelation α β}
] ]
exact ⟨t, u, hu.right, ht⟩ exact ⟨t, u, hu.right, ht⟩
/-- #### Exercise 3.28 /-- ### Exercise 3.28
Assume that `f` is a one-to-one function from `A` into `B`, and that `G` is the Assume that `f` is a one-to-one function from `A` into `B`, and that `G` is the
function with `dom G = 𝒫 A` defined by the equation `G(X) = f⟦X⟧`. Show that `G` function with `dom G = 𝒫 A` defined by the equation `G(X) = f⟦X⟧`. Show that `G`
@ -1832,7 +1832,7 @@ theorem exercise_3_28 {A : Set α} {B : Set β}
have hz := mem_pair_imp_snd_mem_ran hb.right have hz := mem_pair_imp_snd_mem_ran hb.right
exact hf.right.ran_ss hz exact hf.right.ran_ss hz
/-- #### Exercise 3.29 /-- ### Exercise 3.29
Assume that `f : A → B` and define a function `G : B → 𝒫 A` by Assume that `f : A → B` and define a function `G : B → 𝒫 A` by
``` ```
@ -1875,7 +1875,7 @@ theorem exercise_3_29 {f : Set.HRelation α β} {G : Set.HRelation β (Set α)}
rw [heq] at this rw [heq] at this
exact single_valued_eq_unique hf.is_func this.right ht exact single_valued_eq_unique hf.is_func this.right ht
/-! #### Exercise 3.30 /-! ### Exercise 3.30
Assume that `F : 𝒫 A → 𝒫 A` and that `F` has the monotonicity property: Assume that `F : 𝒫 A → 𝒫 A` and that `F` has the monotonicity property:
``` ```
@ -1892,7 +1892,7 @@ variable {F : Set α → Set α} {A B C : Set α}
(hB : B = ⋂₀ { X | X ⊆ A ∧ F X ⊆ X }) (hB : B = ⋂₀ { X | X ⊆ A ∧ F X ⊆ X })
(hC : C = ⋃₀ { X | X ⊆ A ∧ X ⊆ F X }) (hC : C = ⋃₀ { X | X ⊆ A ∧ X ⊆ F X })
/-- ##### Exercise 3.30 (a) /-- #### Exercise 3.30 (a)
Show that `F(B) = B` and `F(C) = C`. Show that `F(B) = B` and `F(C) = C`.
-/ -/
@ -1988,7 +1988,7 @@ theorem exercise_3_30_a : F B = B ∧ F C = C := by
· rw [Set.Subset.antisymm_iff] · rw [Set.Subset.antisymm_iff]
exact ⟨hC_subset, hC_supset⟩ exact ⟨hC_subset, hC_supset⟩
/-- ##### Exercise 3.30 (b) /-- #### Exercise 3.30 (b)
Show that if `F(X) = X`, then `B ⊆ X ⊆ C`. Show that if `F(X) = X`, then `B ⊆ X ⊆ C`.
-/ -/
@ -2010,7 +2010,7 @@ theorem exercise_3_30_b : ∀ X, X ⊆ A ∧ F X = X → B ⊆ X ∧ X ⊆ C :=
end Exercise_3_30 end Exercise_3_30
/-- #### Exercise 3.32 (a) /-- ### Exercise 3.32 (a)
Show that `R` is symmetric **iff** `R⁻¹ ⊆ R`. Show that `R` is symmetric **iff** `R⁻¹ ⊆ R`.
-/ -/
@ -2028,7 +2028,7 @@ theorem exercise_3_32_a {R : Set.Relation α}
rw [← mem_self_comm_mem_inv] at hp rw [← mem_self_comm_mem_inv] at hp
exact hR hp exact hR hp
/-- #### Exercise 3.32 (b) /-- ### Exercise 3.32 (b)
Show that `R` is transitive **iff** `R ∘ R ⊆ R`. Show that `R` is transitive **iff** `R ∘ R ⊆ R`.
-/ -/
@ -2045,7 +2045,7 @@ theorem exercise_3_32_b {R : Set.Relation α}
have : (x, z) ∈ comp R R := ⟨y, hx, hz⟩ have : (x, z) ∈ comp R R := ⟨y, hx, hz⟩
exact hR this exact hR this
/-- #### Exercise 3.33 /-- ### Exercise 3.33
Show that `R` is a symmetric and transitive relation **iff** `R = R⁻¹ ∘ R`. Show that `R` is a symmetric and transitive relation **iff** `R = R⁻¹ ∘ R`.
-/ -/
@ -2095,7 +2095,7 @@ theorem exercise_3_33 {R : Set.Relation α}
rw [h, hR] rw [h, hR]
exact ⟨y, hx, this⟩ exact ⟨y, hx, this⟩
/-- #### Exercise 3.34 (a) /-- ### Exercise 3.34 (a)
Assume that `𝓐` is a nonempty set, every member of which is a transitive Assume that `𝓐` is a nonempty set, every member of which is a transitive
relation. Is the set `⋂ 𝓐` a transitive relation? relation. Is the set `⋂ 𝓐` a transitive relation?
@ -2110,7 +2110,7 @@ theorem exercise_3_34_a {𝓐 : Set (Set.Relation α)}
have hy' := hy A hA have hy' := hy A hA
exact h𝓐 A hA hx' hy' exact h𝓐 A hA hx' hy'
/-- #### Exercise 3.34 (b) /-- ### Exercise 3.34 (b)
Assume that `𝓐` is a nonempty set, every member of which is a transitive Assume that `𝓐` is a nonempty set, every member of which is a transitive
relation. Is `⋃ 𝓐` a transitive relation? relation. Is `⋃ 𝓐` a transitive relation?
@ -2157,7 +2157,7 @@ theorem exercise_3_34_b {𝓐 : Set (Set.Relation ℕ)}
simp at this simp at this
exact absurd (h h₁ h₂) h₃ exact absurd (h h₁ h₂) h₃
/-- #### Exercise 3.35 /-- ### Exercise 3.35
Show that for any `R` and `x`, we have `[x]_R = R⟦{x}⟧`. Show that for any `R` and `x`, we have `[x]_R = R⟦{x}⟧`.
-/ -/
@ -2168,7 +2168,7 @@ theorem exercise_3_35 {R : Set.Relation α} {x : α}
_ = { t | ∃ u ∈ ({x} : Set α), (u, t) ∈ R } := by simp _ = { t | ∃ u ∈ ({x} : Set α), (u, t) ∈ R } := by simp
_ = image R {x} := rfl _ = image R {x} := rfl
/-- #### Exercise 3.36 /-- ### Exercise 3.36
Assume that `f : A → B` and that `R` is an equivalence relation on `B`. Define Assume that `f : A → B` and that `R` is an equivalence relation on `B`. Define
`Q` to be the set `{⟨x, y⟩ ∈ A × A | ⟨f(x), f(y)⟩ ∈ R}`. Show that `Q` is an `Q` to be the set `{⟨x, y⟩ ∈ A × A | ⟨f(x), f(y)⟩ ∈ R}`. Show that `Q` is an
@ -2234,7 +2234,7 @@ theorem exercise_3_36 {f : Set.HRelation α β}
simp only [exists_and_left, Set.mem_setOf_eq] simp only [exists_and_left, Set.mem_setOf_eq]
exact ⟨fx, hfx, fz, hfz, hR.trans h₁ h₂⟩ exact ⟨fx, hfx, fz, hfz, hR.trans h₁ h₂⟩
/-- #### Exercise 3.37 /-- ### Exercise 3.37
Assume that `P` is a partition of a set `A`. Define the relation `Rₚ` as Assume that `P` is a partition of a set `A`. Define the relation `Rₚ` as
follows: follows:
@ -2311,7 +2311,7 @@ theorem exercise_3_37 {P : Set (Set α)} {A : Set α}
simp only [Set.mem_setOf_eq] simp only [Set.mem_setOf_eq]
exact ⟨B₁, hB₁.left, hB₁.right.left, by rw [hB]; exact hB₂.right.right⟩ exact ⟨B₁, hB₁.left, hB₁.right.left, by rw [hB]; exact hB₂.right.right⟩
/-- #### Exercise 3.38 /-- ### Exercise 3.38
Theorem 3P shows that `A / R` is a partition of `A` whenever `R` is an Theorem 3P shows that `A / R` is a partition of `A` whenever `R` is an
equivalence relation on `A`. Show that if we start with the equivalence relation equivalence relation on `A`. Show that if we start with the equivalence relation
@ -2379,7 +2379,7 @@ theorem exercise_3_38 {P : Set (Set α)} {A : Set α}
simp only [Set.mem_setOf_eq] simp only [Set.mem_setOf_eq]
_ = neighborhood Rₚ x := rfl _ = neighborhood Rₚ x := rfl
/-- #### Exercise 3.39 /-- ### Exercise 3.39
Assume that we start with an equivalence relation `R` on `A` and define `P` to Assume that we start with an equivalence relation `R` on `A` and define `P` to
be the partition `A / R`. Show that `Rₚ`, as defined in Exercise 37, is just be the partition `A / R`. Show that `Rₚ`, as defined in Exercise 37, is just
@ -2417,7 +2417,7 @@ theorem exercise_3_39 {P : Set (Set α)} {R Rₚ : Set.Relation α} {A : Set α}
rw [hP] rw [hP]
exact ⟨x, hxA, rfl⟩ exact ⟨x, hxA, rfl⟩
/-- #### Exercise 3.41 (a) /-- ### Exercise 3.41 (a)
Let `ℝ` be the set of real numbers and define the realtion `Q` on `ℝ × ℝ` by Let `ℝ` be the set of real numbers and define the realtion `Q` on `ℝ × ℝ` by
`⟨u, v⟩ Q ⟨x, y⟩` **iff** `u + y = x + v`. Show that `Q` is an equivalence `⟨u, v⟩ Q ⟨x, y⟩` **iff** `u + y = x + v`. Show that `Q` is an equivalence
@ -2470,7 +2470,7 @@ theorem exercise_3_41_a {Q : Set.Relation (ℝ × ℝ)}
conv => right; rw [add_comm] conv => right; rw [add_comm]
exact this exact this
/-- #### Exercise 3.43 /-- ### Exercise 3.43
Assume that `R` is a linear ordering on a set `A`. Show that `R⁻¹` is also a Assume that `R` is a linear ordering on a set `A`. Show that `R⁻¹` is also a
linear ordering on `A`. linear ordering on `A`.
@ -2493,7 +2493,7 @@ theorem exercise_3_43 {R : Rel α α} (hR : IsStrictTotalOrder α R)
unfold Rel.inv flip at * unfold Rel.inv flip at *
exact hR.trans c b a hac hab exact hR.trans c b a hac hab
/-! #### Exercise 3.44 /-! ### Exercise 3.44
Assume that `<` is a linear ordering on a set `A`. Assume that `f : A → A` and Assume that `<` is a linear ordering on a set `A`. Assume that `f : A → A` and
that `f` has the property that whenever `x < y`, then `f(x) < f(y)`. Show that that `f` has the property that whenever `x < y`, then `f(x) < f(y)`. Show that
@ -2537,7 +2537,7 @@ theorem exercise_3_44_ii {R : Rel α α} (hR : IsStrictTotalOrder α R)
have := hR.trans (f x) (f y) (f x) h (hf y x h₂) have := hR.trans (f x) (f y) (f x) h (hf y x h₂)
exact absurd this (hR.irrefl (f x)) exact absurd this (hR.irrefl (f x))
/-- #### Exercise 3.45 /-- ### Exercise 3.45
Assume that `<_A` and `<_B` are linear orderings on `A` and `B`, respectively. Assume that `<_A` and `<_B` are linear orderings on `A` and `B`, respectively.
Define the binary relation `<_L` on the Cartesian product `A × B` by: Define the binary relation `<_L` on the Cartesian product `A × B` by:

58
Bookshelf/Enderton/Set/Chapter_4.lean
View File

@ -11,7 +11,7 @@ Natural Numbers
namespace Enderton.Set.Chapter_4 namespace Enderton.Set.Chapter_4
/-- #### Theorem 4C /-- ### Theorem 4C
Every natural number except `0` is the successor of some natural number. Every natural number except `0` is the successor of some natural number.
-/ -/
@ -23,7 +23,7 @@ theorem theorem_4c (n : ℕ)
#check Nat.exists_eq_succ_of_ne_zero #check Nat.exists_eq_succ_of_ne_zero
/-- #### Theorem 4I /-- ### Theorem 4I
For natural numbers `m` and `n`, For natural numbers `m` and `n`,
``` ```
@ -34,7 +34,7 @@ m + n⁺ = (m + n)⁺
theorem theorem_4i (m n : ℕ) theorem theorem_4i (m n : ℕ)
: m + 0 = m ∧ m + n.succ = (m + n).succ := ⟨rfl, rfl⟩ : m + 0 = m ∧ m + n.succ = (m + n).succ := ⟨rfl, rfl⟩
/-- #### Theorem 4J /-- ### Theorem 4J
For natural numbers `m` and `n`, For natural numbers `m` and `n`,
``` ```
@ -45,7 +45,7 @@ m ⬝ n⁺ = m ⬝ n + m .
theorem theorem_4j (m n : ℕ) theorem theorem_4j (m n : ℕ)
: m * 0 = 0 ∧ m * n.succ = m * n + m := ⟨rfl, rfl⟩ : m * 0 = 0 ∧ m * n.succ = m * n + m := ⟨rfl, rfl⟩
/-- #### Left Additive Identity /-- ### Left Additive Identity
For all `n ∈ ω`, `A₀(n) = n`. In other words, `0 + n = n`. For all `n ∈ ω`, `A₀(n) = n`. In other words, `0 + n = n`.
-/ -/
@ -60,7 +60,7 @@ lemma left_additive_identity (n : ℕ)
#check Nat.zero_add #check Nat.zero_add
/-- #### Lemma 2 /-- ### Lemma 2
For all `m, n ∈ ω`, `Aₘ₊(n) = Aₘ(n⁺)`. In other words, `m⁺ + n = m + n⁺`. For all `m, n ∈ ω`, `Aₘ₊(n) = Aₘ(n⁺)`. In other words, `m⁺ + n = m + n⁺`.
-/ -/
@ -76,7 +76,7 @@ lemma lemma_2 (m n : ℕ)
#check Nat.succ_add_eq_succ_add #check Nat.succ_add_eq_succ_add
/-- #### Theorem 4K-1 /-- ### Theorem 4K-1
Associatve law for addition. For `m, n, p ∈ ω`, Associatve law for addition. For `m, n, p ∈ ω`,
``` ```
@ -99,7 +99,7 @@ theorem theorem_4k_1 {m n p : ℕ}
#check Nat.add_assoc #check Nat.add_assoc
/-- #### Theorem 4K-2 /-- ### Theorem 4K-2
Commutative law for addition. For `m, n ∈ ω`, Commutative law for addition. For `m, n ∈ ω`,
``` ```
@ -119,7 +119,7 @@ theorem theorem_4k_2 {m n : ℕ}
#check Nat.add_comm #check Nat.add_comm
/-- #### Zero Multiplicand /-- ### Zero Multiplicand
For all `n ∈ ω`, `M₀(n) = 0`. In other words, `0 ⬝ n = 0`. For all `n ∈ ω`, `M₀(n) = 0`. In other words, `0 ⬝ n = 0`.
-/ -/
@ -135,7 +135,7 @@ theorem zero_multiplicand (n : ℕ)
#check Nat.zero_mul #check Nat.zero_mul
/-- #### Successor Distribution /-- ### Successor Distribution
For all `m, n ∈ ω`, `Mₘ₊(n) = Mₘ(n) + n`. In other words, For all `m, n ∈ ω`, `Mₘ₊(n) = Mₘ(n) + n`. In other words,
``` ```
@ -159,7 +159,7 @@ theorem succ_distrib (m n : ℕ)
#check Nat.succ_mul #check Nat.succ_mul
/-- #### Theorem 4K-3 /-- ### Theorem 4K-3
Distributive law. For `m, n, p ∈ ω`, Distributive law. For `m, n, p ∈ ω`,
``` ```
@ -181,7 +181,7 @@ theorem theorem_4k_3 (m n p : ℕ)
_ = (m * n + n) + (m * p + p) := by rw [theorem_4k_1, theorem_4k_1] _ = (m * n + n) + (m * p + p) := by rw [theorem_4k_1, theorem_4k_1]
_ = m.succ * n + m.succ * p := by rw [succ_distrib, succ_distrib] _ = m.succ * n + m.succ * p := by rw [succ_distrib, succ_distrib]
/-- #### Successor Identity /-- ### Successor Identity
For all `m ∈ ω`, `Aₘ(1) = m⁺`. In other words, `m + 1 = m⁺`. For all `m ∈ ω`, `Aₘ(1) = m⁺`. In other words, `m + 1 = m⁺`.
-/ -/
@ -197,7 +197,7 @@ theorem succ_identity (m : ℕ)
#check Nat.succ_eq_one_add #check Nat.succ_eq_one_add
/-- #### Right Multiplicative Identity /-- ### Right Multiplicative Identity
For all `m ∈ ω`, `Mₘ(1) = m`. In other words, `m ⬝ 1 = m`. For all `m ∈ ω`, `Mₘ(1) = m`. In other words, `m ⬝ 1 = m`.
-/ -/
@ -213,7 +213,7 @@ theorem right_mul_id (m : ℕ)
#check Nat.mul_one #check Nat.mul_one
/-- #### Theorem 4K-5 /-- ### Theorem 4K-5
Commutative law for multiplication. For `m, n ∈ ω`, `m ⬝ n = n ⬝ m`. Commutative law for multiplication. For `m, n ∈ ω`, `m ⬝ n = n ⬝ m`.
-/ -/
@ -232,7 +232,7 @@ theorem theorem_4k_5 (m n : ℕ)
#check Nat.mul_comm #check Nat.mul_comm
/-- #### Theorem 4K-4 /-- ### Theorem 4K-4
Associative law for multiplication. For `m, n, p ∈ ω`, Associative law for multiplication. For `m, n, p ∈ ω`,
``` ```
@ -254,7 +254,7 @@ theorem theorem_4k_4 (m n p : ℕ)
#check Nat.mul_assoc #check Nat.mul_assoc
/-- #### Lemma 4L(b) /-- ### Lemma 4L(b)
No natural number is a member of itself. No natural number is a member of itself.
-/ -/
@ -269,7 +269,7 @@ lemma lemma_4l_b (n : ℕ)
#check Nat.lt_irrefl #check Nat.lt_irrefl
/-- #### Lemma 10 /-- ### Lemma 10
For every natural number `n ≠ 0`, `0 ∈ n`. For every natural number `n ≠ 0`, `0 ∈ n`.
-/ -/
@ -285,7 +285,7 @@ theorem zero_least_nat (n : ℕ)
#check Nat.pos_of_ne_zero #check Nat.pos_of_ne_zero
/-! #### Theorem 4N /-! ### Theorem 4N
For any natural numbers `n`, `m`, and `p`, For any natural numbers `n`, `m`, and `p`,
``` ```
@ -361,7 +361,7 @@ theorem theorem_4n_ii (m n p : ℕ)
#check Nat.mul_lt_mul_of_pos_right #check Nat.mul_lt_mul_of_pos_right
/-! #### Corollary 4P /-! ### Corollary 4P
The following cancellation laws hold for `m`, `n`, and `p` in `ω`: The following cancellation laws hold for `m`, `n`, and `p` in `ω`:
``` ```
@ -384,7 +384,7 @@ theorem corollary_4p_i (m n p : ℕ) (h : m + p = n + p)
#check Nat.add_right_cancel #check Nat.add_right_cancel
/-- #### Well Ordering of ω /-- ### Well Ordering of ω
Let `A` be a nonempty subset of `ω`. Then there is some `m ∈ A` such that Let `A` be a nonempty subset of `ω`. Then there is some `m ∈ A` such that
`m ≤ n` for all `n ∈ A`. `m ≤ n` for all `n ∈ A`.
@ -438,7 +438,7 @@ theorem well_ordering_nat {A : Set ℕ} (hA : Set.Nonempty A)
#check WellOrder #check WellOrder
/-- #### Strong Induction Principle for ω /-- ### Strong Induction Principle for ω
Let `A` be a subset of `ω`, and assume that for every `n ∈ ω`, if every number Let `A` be a subset of `ω`, and assume that for every `n ∈ ω`, if every number
less than `n` is in `A`, then `n ∈ A`. Then `A = ω`. less than `n` is in `A`, then `n ∈ A`. Then `A = ω`.
@ -462,14 +462,14 @@ theorem strong_induction_principle_nat (A : Set ℕ)
have : x < x := Nat.lt_of_lt_of_le hx (hm.right x nx) have : x < x := Nat.lt_of_lt_of_le hx (hm.right x nx)
simp at this simp at this
/-- #### Exercise 4.1 /-- ### Exercise 4.1
Show that `1 ≠ 3` i.e., that `∅⁺ ≠ ∅⁺⁺⁺`. Show that `1 ≠ 3` i.e., that `∅⁺ ≠ ∅⁺⁺⁺`.
-/ -/
theorem exercise_4_1 : 1 ≠ 3 := by theorem exercise_4_1 : 1 ≠ 3 := by
simp simp
/-- #### Exercise 4.13 /-- ### Exercise 4.13
Let `m` and `n` be natural numbers such that `m ⬝ n = 0`. Show that either Let `m` and `n` be natural numbers such that `m ⬝ n = 0`. Show that either
`m = 0` or `n = 0`. `m = 0` or `n = 0`.
@ -498,7 +498,7 @@ Call a natural number *odd* if it has the form `(2 ⬝ p) + 1` for some `p`.
-/ -/
def odd (n : ℕ) : Prop := ∃ p, (2 * p) + 1 = n def odd (n : ℕ) : Prop := ∃ p, (2 * p) + 1 = n
/-- #### Exercise 4.14 /-- ### Exercise 4.14
Show that each natural number is either even or odd, but never both. Show that each natural number is either even or odd, but never both.
-/ -/
@ -549,7 +549,7 @@ theorem exercise_4_14 (n : ℕ)
have : even n := ⟨q, hq'⟩ have : even n := ⟨q, hq'⟩
exact absurd this h exact absurd this h
/-- #### Exercise 4.17 /-- ### Exercise 4.17
Prove that `mⁿ⁺ᵖ = mⁿ ⬝ mᵖ.` Prove that `mⁿ⁺ᵖ = mⁿ ⬝ mᵖ.`
-/ -/
@ -567,7 +567,7 @@ theorem exercise_4_17 (m n p : ℕ)
_ = m ^ n * (m ^ p * m) := by rw [theorem_4k_4] _ = m ^ n * (m ^ p * m) := by rw [theorem_4k_4]
_ = m ^ n * m ^ p.succ := rfl _ = m ^ n * m ^ p.succ := rfl
/-- #### Exercise 4.19 /-- ### Exercise 4.19
Prove that if `m` is a natural number and `d` is a nonzero number, then there Prove that if `m` is a natural number and `d` is a nonzero number, then there
exist numbers `q` and `r` such that `m = (d ⬝ q) + r` and `r` is less than `d`. exist numbers `q` and `r` such that `m = (d ⬝ q) + r` and `r` is less than `d`.
@ -600,7 +600,7 @@ theorem exercise_4_19 (m d : ℕ) (hd : d ≠ 0)
_ < d := hr _ < d := hr
simp at this simp at this
/-- #### Exercise 4.22 /-- ### Exercise 4.22
Show that for any natural numbers `m` and `p` we have `m ∈ m + p⁺`. Show that for any natural numbers `m` and `p` we have `m ∈ m + p⁺`.
-/ -/
@ -612,7 +612,7 @@ theorem exercise_4_22 (m p : ℕ)
_ < m + p.succ := ih _ < m + p.succ := ih
_ < m + p.succ.succ := Nat.lt.base (m + p.succ) _ < m + p.succ.succ := Nat.lt.base (m + p.succ)
/-- #### Exercise 4.23 /-- ### Exercise 4.23
Assume that `m` and `n` are natural numbers with `m` less than `n`. Show that Assume that `m` and `n` are natural numbers with `m` less than `n`. Show that
there is some `p` in `ω` for which `m + p⁺ = n`. (It follows from this and the there is some `p` in `ω` for which `m + p⁺ = n`. (It follows from this and the
@ -637,7 +637,7 @@ theorem exercise_4_23 {m n : ℕ} (hm : m < n)
refine ⟨0, ?_⟩ refine ⟨0, ?_⟩
rw [hm₁] rw [hm₁]
/-- #### Exercise 4.24 /-- ### Exercise 4.24
Assume that `m + n = p + q`. Show that Assume that `m + n = p + q`. Show that
``` ```
@ -660,7 +660,7 @@ theorem exercise_4_24 (m n p q : ℕ) (h : m + n = p + q)
rw [← h] at hr rw [← h] at hr
exact (theorem_4n_i m p n).mpr hr exact (theorem_4n_i m p n).mpr hr
/-- #### Exercise 4.25 /-- ### Exercise 4.25
Assume that `n ∈ m` and `q ∈ p`. Show that Assume that `n ∈ m` and `q ∈ p`. Show that
``` ```

337
Bookshelf/Enderton/Set/Chapter_6.lean
View File

@ -20,16 +20,29 @@ former provides noncomputable utilities around obtaining inverse functions
namespace Enderton.Set.Chapter_6 namespace Enderton.Set.Chapter_6
/-- #### Theorem 6B /-- ### Theorem 6B
No set is equinumerous to its powerset. No set is equinumerous to its powerset.
-/ -/
theorem theorem_6b (A : Set α) theorem theorem_6b (A : Set α)
: A ≉ 𝒫 A := by : A ≉ 𝒫 A := by
/-
> Let `A` be an arbitrary set and `f: A → 𝒫 A`.
-/
rw [Set.not_equinumerous_def] rw [Set.not_equinumerous_def]
intro f hf intro f hf
unfold Set.BijOn at hf unfold Set.BijOn at hf
/-
> Define `φ = {a ∈ A | a ∉ f(a)}`.
-/
let φ := { a ∈ A | a ∉ f a } let φ := { a ∈ A | a ∉ f a }
/-
> Clearly `φ ∈ 𝒫 A`. Furthermore, for all `a ∈ A`, `φ ≠ f(a)` since `a ∈ φ` if
> and only if `a ∉ f(a)`. Thus `f` cannot be onto `𝒫 A`. Since `f` was
> arbitrarily chosen, there exists no one-to-one correspondence between `A` and
> `𝒫 A`. Since `A` was arbitrarily chosen, there is no set equinumerous to its
> powerset.
-/
suffices ∀ a ∈ A, f a ≠ φ by suffices ∀ a ∈ A, f a ≠ φ by
have hφ := hf.right.right (show φ ∈ 𝒫 A by simp) have hφ := hf.right.right (show φ ∈ 𝒫 A by simp)
have ⟨a, ha⟩ := hφ have ⟨a, ha⟩ := hφ
@ -82,23 +95,18 @@ lemma pigeonhole_principle_aux (n : ℕ)
-/ -/
induction n with induction n with
/- /-
## (i) > #### (i)
> By definition, `0 = ∅`. > By definition, `0 = ∅`. Then `0` has no proper subsets. Hence `0 ∈ S`
> vacuously.
-/ -/
| zero => | zero =>
intro _ hM intro _ hM
unfold Set.Iio at hM unfold Set.Iio at hM
simp only [Nat.zero_eq, not_lt_zero', Set.setOf_false] at hM simp only [Nat.zero_eq, not_lt_zero', Set.setOf_false] at hM
/-
> Then `0` has no proper subsets.
-/
rw [Set.ssubset_empty_iff_false] at hM rw [Set.ssubset_empty_iff_false] at hM
/-
> Hence `0 ∈ S` vacuously.
-/
exact False.elim hM exact False.elim hM
/- /-
## (ii) > #### (ii)
> Suppose `n ∈ S` and `M ⊂ n⁺`. Furthermore, let `f: M → n⁺` be a one-to-one > Suppose `n ∈ S` and `M ⊂ n⁺`. Furthermore, let `f: M → n⁺` be a one-to-one
> function. > function.
-/ -/
@ -111,19 +119,19 @@ lemma pigeonhole_principle_aux (n : ℕ)
· rw [hM', Set.SurjOn_emptyset_Iio_iff_eq_zero] at hf_surj · rw [hM', Set.SurjOn_emptyset_Iio_iff_eq_zero] at hf_surj
simp at hf_surj simp at hf_surj
/- /-
> Otherwise `M ≠ 0`. Because `M` is finite, the trichotomy law for `ω` implies > Otherwise `M ≠ 0`. Because `M` is finite, the *Trichotomy Law for `ω`* implies
> the existence of a largest member `p ∈ M`. There are two cases to consider: > the existence of a largest member `p ∈ M`. There are two cases to consider:
-/ -/
by_cases h : ¬ ∃ t, t ∈ M ∧ f t = n by_cases h : ¬ ∃ t, t ∈ M ∧ f t = n
/- /-
### Case 1 > ##### Case 1
> `n ∉ ran f`. > `n ∉ ran f`.
> Then `f` is not onto `n⁺`. > Then `f` is not onto `n⁺`.
-/ -/
· have ⟨t, ht⟩ := hf_surj (show n ∈ _ by simp) · have ⟨t, ht⟩ := hf_surj (show n ∈ _ by simp)
exact absurd ⟨t, ht⟩ h exact absurd ⟨t, ht⟩ h
/- /-
### Case 2 > ##### Case 2
> `n ∈ ran f`. > `n ∈ ran f`.
> Then there exists some `t ∈ M` such that `⟨t, n⟩ ∈ f`. > Then there exists some `t ∈ M` such that `⟨t, n⟩ ∈ f`.
-/ -/
@ -159,7 +167,7 @@ lemma pigeonhole_principle_aux (n : ℕ)
have hg_maps := Set.Function.swap_MapsTo_self hp₁ ht₁ hf_maps have hg_maps := Set.Function.swap_MapsTo_self hp₁ ht₁ hf_maps
have hg_inj := Set.Function.swap_InjOn_self hp₁ ht₁ hf_inj have hg_inj := Set.Function.swap_InjOn_self hp₁ ht₁ hf_inj
/- /-
> Then (1) indicates `g` must not be onto `n`. > Then *(1)* indicates `g` must not be onto `n`.
-/ -/
let M' := M \ {p} let M' := M \ {p}
have hM' : M' ⊂ Set.Iio n := by have hM' : M' ⊂ Set.Iio n := by
@ -240,9 +248,9 @@ lemma pigeonhole_principle_aux (n : ℕ)
simp only [Set.mem_Iio, Set.mem_setOf_eq, not_exists, not_and] simp only [Set.mem_Iio, Set.mem_setOf_eq, not_exists, not_and]
exact ng_surj exact ng_surj
/- /-
> By the trichotomy law for `ω`, `a ≠ n`. Therefore `a ∉ ran f'`. > By the *Trichotomy Law for `ω`*, `a ≠ n`. Therefore `a ∉ ran f'`.
> `ran f' = ran f` meaning `a ∉ ran f`. Because `a ∈ n ∈n⁺`, Theorem 4F implies > `ran f' = ran f` meaning `a ∉ ran f`. Because `a ∈ n ∈ n⁺`, *Theorem 4F*
> `a ∈ n⁺`. Hence `f` is not onto `n⁺`. > implies `a ∈ n⁺`. Hence `f` is not onto `n⁺`.
-/ -/
refine absurd (hf_surj $ calc a refine absurd (hf_surj $ calc a
_ < n := ha₁ _ < n := ha₁
@ -294,14 +302,14 @@ lemma pigeonhole_principle_aux (n : ℕ)
· refine ⟨y, hy₁, ?_⟩ · refine ⟨y, hy₁, ?_⟩
rwa [if_neg hc₁, if_neg hc₂] rwa [if_neg hc₁, if_neg hc₂]
/- /-
### Subconclusion > ##### Subconclusion
> The foregoing cases are exhaustive. Hence `n⁺ ∈ S`. > The foregoing cases are exhaustive. Hence `n⁺ ∈ S`.
>
## (iii) > #### (iii)
> By (i) and (ii), `S` is an inductive set. By Theorem 4B, `S = ω`. Thus for all > By *(i)* and *(ii)*, `S` is an inductive set. By *Theorem 4B*, `S = ω`. Thus
> natural numbers `n`, there is no one-to-one correspondence between `n` and a > for all natural numbers `n`, there is no one-to-one correspondence between `n`
> proper subset of `n`. In other words, no natural number is equinumerous to a > and a proper subset of `n`. In other words, no natural number is equinumerous
> proper subset of itself. > to a proper subset of itself.
-/ -/
/-- /--
@ -314,34 +322,43 @@ theorem pigeonhole_principle {n : ℕ}
have := pigeonhole_principle_aux n M hM f ⟨hf.left, hf.right.left⟩ have := pigeonhole_principle_aux n M hM f ⟨hf.left, hf.right.left⟩
exact absurd hf.right.right this exact absurd hf.right.right this
/-- #### Corollary 6C /-- ### Corollary 6C
No finite set is equinumerous to a proper subset of itself. No finite set is equinumerous to a proper subset of itself.
-/ -/
theorem corollary_6c [DecidableEq α] [Nonempty α] theorem corollary_6c [DecidableEq α] [Nonempty α]
{S S' : Set α} (hS : Set.Finite S) (h : S' ⊂ S) {S S' : Set α} (hS : Set.Finite S) (h : S' ⊂ S)
: S ≉ S' := by : S ≉ S' := by
/-
> Let `S` be a finite set and `S'` be a proper subset of `S`. Then there exists
> some set `T`, disjoint from `S'`, such that `S' ∪ T = S`. By definition of a
> finite set, `S` is equinumerous to a natural number `n`.
-/
let T := S \ S' let T := S \ S'
have hT : S = S' ∪ (S \ S') := by have hT : S = S' ∪ (S \ S') := by
simp only [Set.union_diff_self] simp only [Set.union_diff_self]
exact (Set.left_subset_union_eq_self (subset_of_ssubset h)).symm exact (Set.left_subset_union_eq_self (subset_of_ssubset h)).symm
/-
-- `hF : S' ∪ T ≈ S`. > By *Theorem 6A*, `S' ∪ T ≈ S` which, by the same theorem, implies
-- `hG : S ≈ n`. > `S' ∪ T ≈ n`.
-- `hH : S' ∪ T ≈ n`. -/
have hF := Set.equinumerous_refl S have hF := Set.equinumerous_refl S
conv at hF => arg 1; rw [hT] conv at hF => arg 1; rw [hT]
have ⟨n, hG⟩ := Set.finite_iff_equinumerous_nat.mp hS have ⟨n, hG⟩ := Set.finite_iff_equinumerous_nat.mp hS
have ⟨H, hH⟩ := Set.equinumerous_trans hF hG /-
> Let `f` be a one-to-one correspondence between `S' ∪ T` and `n`.
-- Restrict `H` to `S'` to yield a bijection between `S'` and a proper subset -/
-- of `n`. have ⟨f, hf⟩ := Set.equinumerous_trans hF hG
let R := (Set.Iio n) \ (H '' T) /-
have hR : Set.BijOn H S' R := by > Then `f ↾ S'` is a one-to-one correspondence between `S'` and a proper subset
> of `n`.
-/
let R := (Set.Iio n) \ (f '' T)
have hR : Set.BijOn f S' R := by
refine ⟨?_, ?_, ?_⟩ refine ⟨?_, ?_, ?_⟩
· -- `Set.MapsTo H S' R` · -- `Set.MapsTo H S' R`
intro x hx intro x hx
refine ⟨hH.left $ Set.mem_union_left T hx, ?_⟩ refine ⟨hf.left $ Set.mem_union_left T hx, ?_⟩
unfold Set.image unfold Set.image
by_contra nx by_contra nx
simp only [Finset.mem_coe, Set.mem_setOf_eq] at nx simp only [Finset.mem_coe, Set.mem_setOf_eq] at nx
@ -349,7 +366,7 @@ theorem corollary_6c [DecidableEq α] [Nonempty α]
have ⟨a, ha₁, ha₂⟩ := nx have ⟨a, ha₁, ha₂⟩ := nx
have hc₁ : a ∈ S' ∪ T := Set.mem_union_right S' ha₁ have hc₁ : a ∈ S' ∪ T := Set.mem_union_right S' ha₁
have hc₂ : x ∈ S' ∪ T := Set.mem_union_left T hx have hc₂ : x ∈ S' ∪ T := Set.mem_union_left T hx
rw [hH.right.left hc₁ hc₂ ha₂] at ha₁ rw [hf.right.left hc₁ hc₂ ha₂] at ha₁
have hx₁ : {x} ⊆ S' := Set.singleton_subset_iff.mpr hx have hx₁ : {x} ⊆ S' := Set.singleton_subset_iff.mpr hx
have hx₂ : {x} ⊆ T := Set.singleton_subset_iff.mpr ha₁ have hx₂ : {x} ⊆ T := Set.singleton_subset_iff.mpr ha₁
@ -364,14 +381,14 @@ theorem corollary_6c [DecidableEq α] [Nonempty α]
intro x₁ hx₁ x₂ hx₂ h intro x₁ hx₁ x₂ hx₂ h
have hc₁ : x₁ ∈ S' ∪ T := Set.mem_union_left T hx₁ have hc₁ : x₁ ∈ S' ∪ T := Set.mem_union_left T hx₁
have hc₂ : x₂ ∈ S' ∪ T := Set.mem_union_left T hx₂ have hc₂ : x₂ ∈ S' ∪ T := Set.mem_union_left T hx₂
exact hH.right.left hc₁ hc₂ h exact hf.right.left hc₁ hc₂ h
· -- `Set.SurjOn H S' R` · -- `Set.SurjOn H S' R`
show ∀ r, r ∈ R → r ∈ H '' S' show ∀ r, r ∈ R → r ∈ f '' S'
intro r hr intro r hr
unfold Set.image unfold Set.image
simp only [Set.mem_setOf_eq] simp only [Set.mem_setOf_eq]
dsimp only at hr dsimp only at hr
have := hH.right.right hr.left have := hf.right.right hr.left
simp only [Set.mem_image, Set.mem_union] at this simp only [Set.mem_image, Set.mem_union] at this
have ⟨x, hx⟩ := this have ⟨x, hx⟩ := this
apply Or.elim hx.left apply Or.elim hx.left
@ -382,9 +399,14 @@ theorem corollary_6c [DecidableEq α] [Nonempty α]
rw [← hx.right] rw [← hx.right]
simp only [Set.mem_image, Finset.mem_coe] simp only [Set.mem_image, Finset.mem_coe]
exact ⟨x, hx', rfl⟩ exact ⟨x, hx', rfl⟩
/-
intro hf > By the *Pigeonhole Principle*, `n` is not equinumerous to any proper subset of
have hf₁ : S ≈ R := Set.equinumerous_trans hf ⟨H, hR⟩ > `n`. Therefore *Theorem 6A* implies `S'` cannot be equinumerous to `n`, which,
> by the same theorem, implies `S'` cannot be equinumerous to `S`. Hence no
> finite set is equinumerous to a proper subset of itself.
-/
intro hf'
have hf₁ : S ≈ R := Set.equinumerous_trans hf' ⟨f, hR⟩
have hf₂ : R ≈ Set.Iio n := by have hf₂ : R ≈ Set.Iio n := by
have ⟨k, hk⟩ := Set.equinumerous_symm hf₁ have ⟨k, hk⟩ := Set.equinumerous_symm hf₁
exact Set.equinumerous_trans ⟨k, hk⟩ hG exact Set.equinumerous_trans ⟨k, hk⟩ hG
@ -398,30 +420,83 @@ theorem corollary_6c [DecidableEq α] [Nonempty α]
· show ¬ ∀ r, r ∈ Set.Iio n → r ∈ R · show ¬ ∀ r, r ∈ Set.Iio n → r ∈ R
intro nr intro nr
have ⟨t, ht₁⟩ : Set.Nonempty T := Set.diff_ssubset_nonempty h have ⟨t, ht₁⟩ : Set.Nonempty T := Set.diff_ssubset_nonempty h
have ht₂ : H t ∈ Set.Iio n := hH.left (Set.mem_union_right S' ht₁) have ht₂ : f t ∈ Set.Iio n := hf.left (Set.mem_union_right S' ht₁)
have ht₃ : H t ∈ R := nr (H t) ht₂ have ht₃ : f t ∈ R := nr (f t) ht₂
exact absurd ⟨t, ht₁, rfl⟩ ht₃.right exact absurd ⟨t, ht₁, rfl⟩ ht₃.right
/-- #### Corollary 6D (a) /-- ### Corollary 6D (a)
Any set equinumerous to a proper subset of itself is infinite. Any set equinumerous to a proper subset of itself is infinite.
-/ -/
theorem corollary_6d_a [DecidableEq α] [Nonempty α] theorem corollary_6d_a [DecidableEq α] [Nonempty α]
{S S' : Set α} (hS : S' ⊂ S) (hf : S ≈ S') {S S' : Set α} (hS : S' ⊂ S) (hf : S ≈ S')
: Set.Infinite S := by : Set.Infinite S := by
/-
> Let `S` be a set equinumerous to proper subset `S'` of itself. Then `S` cannot
> be a finite set by *Corollary 6C*. By definition, `S` is an infinite set.
-/
by_contra nS by_contra nS
simp only [Set.not_infinite] at nS simp only [Set.not_infinite] at nS
exact absurd hf (corollary_6c nS hS) exact absurd hf (corollary_6c nS hS)
/-- #### Corollary 6D (b) /-- ### Corollary 6D (b)
The set `ω` is infinite. The set `ω` is infinite.
-/ -/
theorem corollary_6d_b theorem corollary_6d_b
: Set.Infinite (@Set.univ ℕ) := by : Set.Infinite (@Set.univ ℕ) := by
/-
> Consider set `S = {n ∈ ω | n is even}`. We prove that (i) `S` is equinumerous
> to `ω` and (ii) that `ω` is infinite.
-/
let S : Set ℕ := { 2 * n | n ∈ @Set.univ ℕ } let S : Set ℕ := { 2 * n | n ∈ @Set.univ ℕ }
let f x := 2 * x let f x := 2 * x
suffices Set.BijOn f (@Set.univ ℕ) S by /-
> #### (i)
> Define `f : ω → S` given by `f(n) = 2 ⬝ n`. Notice `f` is well-defined by the
> definition of an even natural number, introduced in *Exercise 4.14*. We first
> show `f` is one-to-one and then that `f` is onto.
-/
have : Set.BijOn f (@Set.univ ℕ) S := by
refine ⟨by simp, ?_, ?_⟩
/-
> Suppose `f(n₁) = f(n₁) = 2 ⬝ n₁`. We must prove that `n₁ = n₂`.
-/
· -- `Set.InjOn f Set.univ`
intro n₁ _ n₂ _ hf
/-
> By the *Trichotomy Law for `ω`*, exactly one of the following may occur:
> `n₁ = n₂`, `n₁ < n₂`, or `n₂ < n₁`. If `n₁ < n₂`, then *Theorem 4N* implies
> `n₁ ⬝ 2 < n₂ ⬝ 2`. *Theorem 4K-5* then indicates `2 ⬝ n₁ < 2 ⬝ n₂`, a
> contradiction to `2 ⬝ n₁ = 2 ⬝ n₂`. A parallel argument holds for when
> `n₂ < n₁`. Thus `n₁ = n₂`.
-/
match @trichotomous ℕ LT.lt _ n₁ n₂ with
| Or.inr (Or.inl r) => exact r
| Or.inl r =>
have := (Chapter_4.theorem_4n_ii n₁ n₂ 1).mp r
conv at this => left; rw [mul_comm]
conv at this => right; rw [mul_comm]
exact absurd hf (Nat.ne_of_lt this)
| Or.inr (Or.inr r) =>
have := (Chapter_4.theorem_4n_ii n₂ n₁ 1).mp r
conv at this => left; rw [mul_comm]
conv at this => right; rw [mul_comm]
exact absurd hf.symm (Nat.ne_of_lt this)
/-
> Next, let `m ∈ S`. That is, `m` is an even number. By definition, there exists
> some `n ∈ ω` such that `m = 2 ⬝ n`. Thus `f(n) = m`.
-/
· -- `Set.SurjOn f Set.univ S`
show ∀ x, x ∈ S → x ∈ f '' Set.univ
intro x hx
unfold Set.image
simp only [Set.mem_univ, true_and, Set.mem_setOf_eq] at hx ⊢
exact hx
/-
> By *(i)*, `ω` is equinumerous to a subset of itself. By *Corollary 6D (a)*,
> `ω` is infinite.
-/
refine corollary_6d_a ?_ ⟨f, this⟩ refine corollary_6d_a ?_ ⟨f, this⟩
rw [Set.ssubset_def] rw [Set.ssubset_def]
apply And.intro apply And.intro
@ -439,65 +514,91 @@ theorem corollary_6d_b
intro x nx intro x nx
simp only [mul_eq_one, false_and] at nx simp only [mul_eq_one, false_and] at nx
refine ⟨by simp, ?_, ?_⟩ /-- ### Corollary 6E
· -- `Set.InjOn f Set.univ`
intro n₁ _ n₂ _ hf
match @trichotomous ℕ LT.lt _ n₁ n₂ with
| Or.inr (Or.inl r) => exact r
| Or.inl r =>
have := (Chapter_4.theorem_4n_ii n₁ n₂ 1).mp r
conv at this => left; rw [mul_comm]
conv at this => right; rw [mul_comm]
exact absurd hf (Nat.ne_of_lt this)
| Or.inr (Or.inr r) =>
have := (Chapter_4.theorem_4n_ii n₂ n₁ 1).mp r
conv at this => left; rw [mul_comm]
conv at this => right; rw [mul_comm]
exact absurd hf.symm (Nat.ne_of_lt this)
· -- `Set.SurjOn f Set.univ S`
show ∀ x, x ∈ S → x ∈ f '' Set.univ
intro x hx
unfold Set.image
simp only [Set.mem_univ, true_and, Set.mem_setOf_eq] at hx ⊢
exact hx
/-- #### Corollary 6E
Any finite set is equinumerous to a unique natural number. Any finite set is equinumerous to a unique natural number.
-/ -/
theorem corollary_6e [Nonempty α] (S : Set α) (hS : Set.Finite S) theorem corollary_6e [Nonempty α] (S : Set α) (hS : Set.Finite S)
: ∃! n : ℕ, S ≈ Set.Iio n := by : ∃! n : ℕ, S ≈ Set.Iio n := by
/-
> Let `S` be a finite set. By definition `S` is equinumerous to a natural number
> `n`.
-/
have ⟨n, hf⟩ := Set.finite_iff_equinumerous_nat.mp hS have ⟨n, hf⟩ := Set.finite_iff_equinumerous_nat.mp hS
refine ⟨n, hf, ?_⟩ refine ⟨n, hf, ?_⟩
/-
> Suppose `S` is equinumerous to another natural number `m`.
-/
intro m hg intro m hg
/-
> By the *Trichotomy Law for `ω`*, exactly one of three situations is possible:
> `n = m`, `n < m`, or `m < n`.
-/
match @trichotomous ℕ LT.lt _ m n with match @trichotomous ℕ LT.lt _ m n with
| Or.inr (Or.inl r) => exact r /-
| Or.inl r => > If `n < m`, then `m ≈ S` and `S ≈ n`. By *Theorem 6A*, it follows `m ≈ n`. But
have hh := Set.equinumerous_symm hg > the *Pigeonhole Principle* indicates no natural number is equinumerous to a
have hk := Set.equinumerous_trans hh hf > proper subset of itself, a contradiction.
have hmn : Set.Iio m ⊂ Set.Iio n := Set.Iio_nat_lt_ssubset r -/
exact absurd hk (pigeonhole_principle hmn)
| Or.inr (Or.inr r) => | Or.inr (Or.inr r) =>
have hh := Set.equinumerous_symm hf have hh := Set.equinumerous_symm hf
have hk := Set.equinumerous_trans hh hg have hk := Set.equinumerous_trans hh hg
have hnm : Set.Iio n ⊂ Set.Iio m := Set.Iio_nat_lt_ssubset r have hnm : Set.Iio n ⊂ Set.Iio m := Set.Iio_nat_lt_ssubset r
exact absurd hk (pigeonhole_principle hnm) exact absurd hk (pigeonhole_principle hnm)
/-
> If `m < n`, a parallel argument applies.
-/
| Or.inl r =>
have hh := Set.equinumerous_symm hg
have hk := Set.equinumerous_trans hh hf
have hmn : Set.Iio m ⊂ Set.Iio n := Set.Iio_nat_lt_ssubset r
exact absurd hk (pigeonhole_principle hmn)
/-
> Hence `n = m`, proving every finite set is equinumerous to a unique natural
> number.
-/
| Or.inr (Or.inl r) => exact r
/-- #### Lemma 6F
/-- ### Lemma 6F
If `C` is a proper subset of a natural number `n`, then `C ≈ m` for some `m` If `C` is a proper subset of a natural number `n`, then `C ≈ m` for some `m`
less than `n`. less than `n`.
-/ -/
lemma lemma_6f {n : ℕ} lemma lemma_6f {n : ℕ}
: ∀ {C}, C ⊂ Set.Iio n → ∃ m, m < n ∧ C ≈ Set.Iio m := by : ∀ {C}, C ⊂ Set.Iio n → ∃ m, m < n ∧ C ≈ Set.Iio m := by
/-
> Let
>
> `S = {n ∈ ω | ∀C ⊂ n, ∃m < n such that C ≈ m}`. (2)
>
> We prove that (i) `0 ∈ S` and (ii) if `n ∈ S` then `n⁺ ∈ S`. Afterward we
> prove (iii) the lemma statement.
-/
induction n with induction n with
/-
> #### (i)
> By definition, `0 = ∅`. Thus `0` has no proper subsets. Hence `0 ∈ S`
> vacuously.
-/
| zero => | zero =>
intro C hC intro C hC
unfold Set.Iio at hC unfold Set.Iio at hC
simp only [Nat.zero_eq, not_lt_zero', Set.setOf_false] at hC simp only [Nat.zero_eq, not_lt_zero', Set.setOf_false] at hC
rw [Set.ssubset_empty_iff_false] at hC rw [Set.ssubset_empty_iff_false] at hC
exact False.elim hC exact False.elim hC
/-
> #### (ii)
> Suppose `n ∈ S` and consider `n⁺`. By definition of the successor,
> `n⁺ = n ∪ {n}`. There are two cases to consider:
-/
| succ n ih => | succ n ih =>
/-
> Let `C` be an arbitrary, proper subset of `n⁺`.
-/
intro C hC
-- A useful theorem we use in a couple of places.
have h_subset_equinumerous have h_subset_equinumerous
: ∀ S, S ⊆ Set.Iio n → : ∀ S, S ⊆ Set.Iio n →
∃ m, m < n + 1 ∧ S ≈ Set.Iio m := by ∃ m, m < n + 1 ∧ S ≈ Set.Iio m := by
@ -513,10 +614,31 @@ lemma lemma_6f {n : ℕ}
· -- `S = Set.Iio n` · -- `S = Set.Iio n`
intro h intro h
exact ⟨n, by simp, Set.eq_imp_equinumerous h⟩ exact ⟨n, by simp, Set.eq_imp_equinumerous h⟩
/-
intro C hC > There are two cases to consider:
by_cases hn : n ∈ C -/
· -- Since `C` is a proper subset of `n⁺`, the set `n⁺ - C` is nonempty. by_cases hn : n ∉ C
/-
> ##### Case 1
> Suppose `n ∉ C`. Then `C ⊆ n`. If `C` is a proper subset of `n`, *(2)* implies
> `C` is equinumerous to some `m < n < n⁺`. If `C = n`, then *Theorem 6A*
> implies `C` is equinumerous to `n < n⁺`.
-/
· refine h_subset_equinumerous C ?_
show ∀ x, x ∈ C → x ∈ Set.Iio n
intro x hx
apply Or.elim (Nat.lt_or_eq_of_lt (subset_of_ssubset hC hx))
· exact id
· intro hx₁
rw [hx₁] at hx
exact absurd hx hn
/-
> ##### Case 2
> Suppose `n ∈ C`. Since `C` is a proper subset of `n⁺`, the set `n⁺ - C` is
> nonempty. By the *Well Ordering of `ω`*, `n⁺ - C` has a least element, say
> `p` (which does not equal `n`).
-/
simp only [not_not] at hn
have hC₁ : Set.Nonempty (Set.Iio (n + 1) \ C) := by have hC₁ : Set.Nonempty (Set.Iio (n + 1) \ C) := by
rw [Set.ssubset_def] at hC rw [Set.ssubset_def] at hC
have : ¬ ∀ x, x ∈ Set.Iio (n + 1) → x ∈ C := hC.right have : ¬ ∀ x, x ∈ Set.Iio (n + 1) → x ∈ C := hC.right
@ -524,7 +646,9 @@ lemma lemma_6f {n : ℕ}
exact this exact this
-- `p` is the least element of `n⁺ - C`. -- `p` is the least element of `n⁺ - C`.
have ⟨p, hp⟩ := Chapter_4.well_ordering_nat hC₁ have ⟨p, hp⟩ := Chapter_4.well_ordering_nat hC₁
/-
> Consider now set `C' = (C - {n}) ∪ {p}`. By construction, `C' ⊆ n`.
-/
let C' := (C \ {n}) ∪ {p} let C' := (C \ {n}) ∪ {p}
have hC'₁ : C' ⊆ Set.Iio n := by have hC'₁ : C' ⊆ Set.Iio n := by
show ∀ x, x ∈ C' → x ∈ Set.Iio n show ∀ x, x ∈ C' → x ∈ Set.Iio n
@ -552,12 +676,23 @@ lemma lemma_6f {n : ℕ}
rw [← h] at this rw [← h] at this
exact Or.elim (Nat.lt_or_eq_of_lt this) exact Or.elim (Nat.lt_or_eq_of_lt this)
id (absurd · (Nat.ne_of_lt r).symm) id (absurd · (Nat.ne_of_lt r).symm)
/-
> As seen in *Case 1*, `C'` is equinumerous to some `m < n⁺`.
-/
have ⟨m, hm₁, hm₂⟩ := h_subset_equinumerous C' hC'₁ have ⟨m, hm₁, hm₂⟩ := h_subset_equinumerous C' hC'₁
/-
> It suffices to show there exists a one-to-one correspondence between `C'` and
> `C`, since then *Theorem 6A* implies `C` is equinumerous to `m` as well.
-/
suffices C' ≈ C from suffices C' ≈ C from
⟨m, hm₁, Set.equinumerous_trans (Set.equinumerous_symm this) hm₂⟩ ⟨m, hm₁, Set.equinumerous_trans (Set.equinumerous_symm this) hm₂⟩
/-
-- Proves `f` is a one-to-one correspondence between `C'` and `C`. > Function `f : C' → C` given by
>
> `f(x) = if x = p then n else x`
>
> is trivially one-to-one and onto as expected.
-/
let f x := if x = p then n else x let f x := if x = p then n else x
refine ⟨f, ?_, ?_, ?_⟩ refine ⟨f, ?_, ?_, ?_⟩
· -- `Set.MapsTo f C' C` · -- `Set.MapsTo f C' C`
@ -615,18 +750,14 @@ lemma lemma_6f {n : ℕ}
rw [← nx₂] at this rw [← nx₂] at this
exact absurd hx this exact absurd hx this
· exact ⟨x, ⟨hx, nx₁⟩, by rwa [if_neg]⟩ · exact ⟨x, ⟨hx, nx₁⟩, by rwa [if_neg]⟩
/-
> #### (iii)
> By *(i)* and *(ii)*, `S` is an inductive set. By *Theorem 4B*, `S = ω`.
> Therefore, for every proper subset `C` of a natural number `n`, there exists
> some `m < n` such that `C ≈ n`.
-/
· refine h_subset_equinumerous C ?_ /-- ### Corollary 6G
show ∀ x, x ∈ C → x ∈ Set.Iio n
intro x hx
unfold Set.Iio
apply Or.elim (Nat.lt_or_eq_of_lt (subset_of_ssubset hC hx))
· exact id
· intro hx₁
rw [hx₁] at hx
exact absurd hx hn
/-- #### Corollary 6G
Any subset of a finite set is finite. Any subset of a finite set is finite.
-/ -/
@ -693,7 +824,7 @@ theorem corollary_6g {S S' : Set α} (hS : Set.Finite S) (hS' : S' ⊆ S)
· intro h · intro h
rwa [h] rwa [h]
/-- #### Subset Size /-- ### Subset Size
Let `A` be a finite set and `B ⊂ A`. Then there exist natural numbers `m, n ∈ ω` Let `A` be a finite set and `B ⊂ A`. Then there exist natural numbers `m, n ∈ ω`
such that `B ≈ m`, `A ≈ n`, and `m ≤ n`. such that `B ≈ m`, `A ≈ n`, and `m ≤ n`.
@ -751,7 +882,7 @@ lemma subset_size [DecidableEq α] [Nonempty α] {A B : Set α}
simp at h₁ simp at h₁
exact absurd ⟨h, hh⟩ (corollary_6c hA this) exact absurd ⟨h, hh⟩ (corollary_6c hA this)
/-- #### Finite Domain and Range Size /-- ### Finite Domain and Range Size
Let `A` and `B` be finite sets and `f : A → B` be a function. Then there exist Let `A` and `B` be finite sets and `f : A → B` be a function. Then there exist
natural numbers `m, n ∈ ω` such that `dom f ≈ m`, `ran f ≈ n`, and `m ≥ n`. natural numbers `m, n ∈ ω` such that `dom f ≈ m`, `ran f ≈ n`, and `m ≥ n`.
@ -778,7 +909,7 @@ theorem finite_dom_ran_size [Nonempty α] {A B : Set α}
sorry sorry
/-- #### Set Difference Size /-- ### Set Difference Size
Let `A ≈ m` for some natural number `m` and `B ⊆ A`. Then there exists some Let `A ≈ m` for some natural number `m` and `B ⊆ A`. Then there exists some
`n ∈ ω` such that `B ≈ n` and `A - B ≈ m - n`. `n ∈ ω` such that `B ≈ n` and `A - B ≈ m - n`.
@ -1036,7 +1167,7 @@ lemma sdiff_size [DecidableEq α] [Nonempty α] {A B : Set α}
: ∃ n : ℕ, n ≤ m ∧ B ≈ Set.Iio n ∧ A \ B ≈ (Set.Iio m) \ (Set.Iio n) := : ∃ n : ℕ, n ≤ m ∧ B ≈ Set.Iio n ∧ A \ B ≈ (Set.Iio m) \ (Set.Iio n) :=
sdiff_size_aux A hA B hB sdiff_size_aux A hA B hB
/-- #### Exercise 6.7 /-- ### Exercise 6.7
Assume that `A` is finite and `f : A → A`. Show that `f` is one-to-one **iff** Assume that `A` is finite and `f : A → A`. Show that `f` is one-to-one **iff**
`ran f = A`. `ran f = A`.
@ -1085,7 +1216,7 @@ theorem exercise_6_7 [DecidableEq α] [Nonempty α] {A : Set α} {f : α → α}
have h₂ : A \ {y} ≈ Set.Iio (n₁ - 1) := sorry have h₂ : A \ {y} ≈ Set.Iio (n₁ - 1) := sorry
sorry sorry
/-- #### Exercise 6.8 /-- ### Exercise 6.8
Prove that the union of two finites sets is finite, without any use of Prove that the union of two finites sets is finite, without any use of
arithmetic. arithmetic.
@ -1094,7 +1225,7 @@ theorem exercise_6_8 {A B : Set α} (hA : Set.Finite A) (hB : Set.Finite B)
: Set.Finite (A ∪ B) := by : Set.Finite (A ∪ B) := by
sorry sorry
/-- #### Exercise 6.9 /-- ### Exercise 6.9
Prove that the Cartesian product of two finites sets is finite, without any use Prove that the Cartesian product of two finites sets is finite, without any use
of arithmetic. of arithmetic.

12
Bookshelf/Enderton/Set/OrderedPair.lean
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@ -28,8 +28,7 @@ theorem ext_iff {x y u v : α}
] at hu huv ] at hu huv
apply Or.elim hu <;> apply Or.elim huv apply Or.elim hu <;> apply Or.elim huv
· -- #### Case 1 · -- `{u} = {x}` and `{u, v} = {x}`.
-- `{u} = {x}` and `{u, v} = {x}`.
intro huv_x hu_x intro huv_x hu_x
rw [Set.singleton_eq_singleton_iff] at hu_x rw [Set.singleton_eq_singleton_iff] at hu_x
rw [hu_x] at huv_x rw [hu_x] at huv_x
@ -41,8 +40,7 @@ theorem ext_iff {x y u v : α}
rw [← hx_v] at this rw [← hx_v] at this
exact ⟨hu_x.symm, this⟩ exact ⟨hu_x.symm, this⟩
· -- #### Case 2 · -- `{u} = {x}` and `{u, v} = {x, y}`.
-- `{u} = {x}` and `{u, v} = {x, y}`.
intro huv_xy hu_x intro huv_xy hu_x
rw [Set.singleton_eq_singleton_iff] at hu_x rw [Set.singleton_eq_singleton_iff] at hu_x
rw [hu_x] at huv_xy rw [hu_x] at huv_xy
@ -58,8 +56,7 @@ theorem ext_iff {x y u v : α}
] at this ] at this
exact ⟨hu_x.symm, Or.elim this (absurd ·.symm hx_v) (·.symm)⟩ exact ⟨hu_x.symm, Or.elim this (absurd ·.symm hx_v) (·.symm)⟩
· -- #### Case 3 · -- `{u} = {x, y}` and `{u, v} = {x}`.
-- `{u} = {x, y}` and `{u, v} = {x}`.
intro huv_x hu_xy intro huv_x hu_xy
rw [Set.ext_iff] at huv_x hu_xy rw [Set.ext_iff] at huv_x hu_xy
have hu_x := huv_x u have hu_x := huv_x u
@ -93,8 +90,7 @@ theorem ext_iff {x y u v : α}
· intro hv_y · intro hv_y
exact ⟨hu_x.symm, hv_y.symm⟩ exact ⟨hu_x.symm, hv_y.symm⟩
· -- #### Case 4 · -- `{u} = {x, y}` and `{u, v} = {x, y}`.
-- `{u} = {x, y}` and `{u, v} = {x, y}`.
intro huv_xy hu_xy intro huv_xy hu_xy
rw [Set.ext_iff] at huv_xy hu_xy rw [Set.ext_iff] at huv_xy hu_xy
have hx_u := hu_xy x have hx_u := hu_xy x

76
Bookshelf/Smullyan/Aviary.lean
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@ -9,7 +9,7 @@ considering they still have the same limitation described above during actual
use. Their inclusion here serves more as pseudo-documentation than anything. use. Their inclusion here serves more as pseudo-documentation than anything.
-/ -/
/-- #### Bald Eagle /-- ### Bald Eagle
`E'xy₁y₂y₃z₁z₂z₃ = x(y₁y₂y₃)(z₁z₂z₃)` `E'xy₁y₂y₃z₁z₂z₃ = x(y₁y₂y₃)(z₁z₂z₃)`
-/ -/
@ -17,31 +17,31 @@ def E' (x : α → β → γ)
(y₁ : δ → ε → α) (y₂ : δ) (y₃ : ε) (y₁ : δ → ε → α) (y₂ : δ) (y₃ : ε)
(z₁ : ζ → η → β) (z₂ : ζ) (z₃ : η) := x (y₁ y₂ y₃) (z₁ z₂ z₃) (z₁ : ζ → η → β) (z₂ : ζ) (z₃ : η) := x (y₁ y₂ y₃) (z₁ z₂ z₃)
/-- #### Becard /-- ### Becard
`B₃xyzw = x(y(zw))` `B₃xyzw = x(y(zw))`
-/ -/
def B₃ (x : α → ε) (y : β → α) (z : γ → β) (w : γ) := x (y (z w)) def B₃ (x : α → ε) (y : β → α) (z : γ → β) (w : γ) := x (y (z w))
/-- #### Blackbird /-- ### Blackbird
`B₁xyzw = x(yzw)` `B₁xyzw = x(yzw)`
-/ -/
def B₁ (x : α → ε) (y : β → γ → α) (z : β) (w : γ) := x (y z w) def B₁ (x : α → ε) (y : β → γ → α) (z : β) (w : γ) := x (y z w)
/-- #### Bluebird /-- ### Bluebird
`Bxyz = x(yz)` `Bxyz = x(yz)`
-/ -/
def B (x : α → γ) (y : β → α) (z : β) := x (y z) def B (x : α → γ) (y : β → α) (z : β) := x (y z)
/-- #### Bunting /-- ### Bunting
`B₂xyzwv = x(yzwv)` `B₂xyzwv = x(yzwv)`
-/ -/
def B₂ (x : α → ζ) (y : β → γ → ε → α) (z : β) (w : γ) (v : ε) := x (y z w v) def B₂ (x : α → ζ) (y : β → γ → ε → α) (z : β) (w : γ) (v : ε) := x (y z w v)
/-- #### Cardinal Once Removed /-- ### Cardinal Once Removed
`C*xyzw = xywz` `C*xyzw = xywz`
-/ -/
@ -49,48 +49,48 @@ def C_star (x : α → β → γ → δ) (y : α) (z : γ) (w : β) := x y w z
notation "C*" => C_star notation "C*" => C_star
/-- #### Cardinal /-- ### Cardinal
`Cxyz = xzy` `Cxyz = xzy`
-/ -/
def C (x : α → β → δ) (y : β) (z : α) := x z y def C (x : α → β → δ) (y : β) (z : α) := x z y
/-- #### Converse Warbler /-- ### Converse Warbler
`W'xy = yxx` `W'xy = yxx`
-/ -/
def W' (x : α) (y : α → α → β) := y x x def W' (x : α) (y : α → α → β) := y x x
/-- #### Dickcissel /-- ### Dickcissel
`D₁xyzwv = xyz(wv)` `D₁xyzwv = xyz(wv)`
-/ -/
def D₁ (x : α → β → δ → ε) (y : α) (z : β) (w : γ → δ) (v : γ) := x y z (w v) def D₁ (x : α → β → δ → ε) (y : α) (z : β) (w : γ → δ) (v : γ) := x y z (w v)
/-! #### Double Mockingbird /-! ### Double Mockingbird
`M₂xy = xy(xy)` `M₂xy = xy(xy)`
-/ -/
/-- #### Dove /-- ### Dove
`Dxyzw = xy(zw)` `Dxyzw = xy(zw)`
-/ -/
def D (x : α → γ → δ) (y : α) (z : β → γ) (w : β) := x y (z w) def D (x : α → γ → δ) (y : α) (z : β → γ) (w : β) := x y (z w)
/-- #### Dovekie /-- ### Dovekie
`D₂xyzwv = x(yz)(wv)` `D₂xyzwv = x(yz)(wv)`
-/ -/
def D₂ (x : α → δ → ε) (y : β → α) (z : β) (w : γ → δ) (v : γ) := x (y z) (w v) def D₂ (x : α → δ → ε) (y : β → α) (z : β) (w : γ → δ) (v : γ) := x (y z) (w v)
/-- #### Eagle /-- ### Eagle
`Exyzwv = xy(zwv)` `Exyzwv = xy(zwv)`
-/ -/
def E (x : α → δ → ε) (y : α) (z : β → γ → δ) (w : β) (v : γ) := x y (z w v) def E (x : α → δ → ε) (y : α) (z : β → γ → δ) (w : β) (v : γ) := x y (z w v)
/-- #### Finch Once Removed /-- ### Finch Once Removed
`F*xyzw = xwzy` `F*xyzw = xwzy`
-/ -/
@ -98,95 +98,95 @@ def F_star (x : α → β → γ → δ) (y : γ) (z : β) (w : α) := x w z y
notation "F*" => F_star notation "F*" => F_star
/-- #### Finch /-- ### Finch
`Fxyz = zyx` `Fxyz = zyx`
-/ -/
def F (x : α) (y : β) (z : β → α → γ) := z y x def F (x : α) (y : β) (z : β → α → γ) := z y x
/-- #### Goldfinch /-- ### Goldfinch
`Gxyzw = xw(yz)` `Gxyzw = xw(yz)`
-/ -/
def G (x : α → γ → δ) (y : β → γ) (z : β) (w : α) := x w (y z) def G (x : α → γ → δ) (y : β → γ) (z : β) (w : α) := x w (y z)
/-- #### Hummingbird /-- ### Hummingbird
`Hxyz = xyzy` `Hxyz = xyzy`
-/ -/
def H (x : α → β → α → γ) (y : α) (z : β) := x y z y def H (x : α → β → α → γ) (y : α) (z : β) := x y z y
/-- #### Identity Bird /-- ### Identity Bird
`Ix = x` `Ix = x`
-/ -/
def I (x : α) : α := x def I (x : α) : α := x
/-- #### Kestrel /-- ### Kestrel
`Kxy = x` `Kxy = x`
-/ -/
def K (x : α) (_ : β) := x def K (x : α) (_ : β) := x
/-! #### Lark /-! ### Lark
`Lxy = x(yy)` `Lxy = x(yy)`
-/ -/
/-! #### Mockingbird /-! ### Mockingbird
`Mx = xx` `Mx = xx`
-/ -/
/-- #### Owl /-- ### Owl
`Oxy = y(xy)` `Oxy = y(xy)`
-/ -/
def O (x : (α → β) → α) (y : α → β) := y (x y) def O (x : (α → β) → α) (y : α → β) := y (x y)
/-- #### Phoenix /-- ### Phoenix
`Φxyzw = x(yw)(zw)` `Φxyzw = x(yw)(zw)`
-/ -/
def Φ (x : β → γ → δ) (y : α → β) (z : α → γ) (w : α) := x (y w) (z w) def Φ (x : β → γ → δ) (y : α → β) (z : α → γ) (w : α) := x (y w) (z w)
/-- #### Psi Bird /-- ### Psi Bird
`Ψxyzw = x(yz)(yw)` `Ψxyzw = x(yz)(yw)`
-/ -/
def Ψ (x : α → α → γ) (y : β → α) (z : β) (w : β) := x (y z) (y w) def Ψ (x : α → α → γ) (y : β → α) (z : β) (w : β) := x (y z) (y w)
/-- #### Quacky Bird /-- ### Quacky Bird
`Q₄xyz = z(yx)` `Q₄xyz = z(yx)`
-/ -/
def Q₄ (x : α) (y : α → β) (z : β → γ) := z (y x) def Q₄ (x : α) (y : α → β) (z : β → γ) := z (y x)
/-- #### Queer Bird /-- ### Queer Bird
`Qxyz = y(xz)` `Qxyz = y(xz)`
-/ -/
def Q (x : α → β) (y : β → γ) (z : α) := y (x z) def Q (x : α → β) (y : β → γ) (z : α) := y (x z)
/-- #### Quirky Bird /-- ### Quirky Bird
`Q₃xyz = z(xy)` `Q₃xyz = z(xy)`
-/ -/
def Q₃ (x : α → β) (y : α) (z : β → γ) := z (x y) def Q₃ (x : α → β) (y : α) (z : β → γ) := z (x y)
/-- #### Quixotic Bird /-- ### Quixotic Bird
`Q₁xyz = x(zy)` `Q₁xyz = x(zy)`
-/ -/
def Q₁ (x : α → γ) (y : β) (z : β → α) := x (z y) def Q₁ (x : α → γ) (y : β) (z : β → α) := x (z y)
/-- #### Quizzical Bird /-- ### Quizzical Bird
`Q₂xyz = y(zx)` `Q₂xyz = y(zx)`
-/ -/
def Q₂ (x : α) (y : β → γ) (z : α → β) := y (z x) def Q₂ (x : α) (y : β → γ) (z : α → β) := y (z x)
/-- #### Robin Once Removed /-- ### Robin Once Removed
`R*xyzw = xzwy` `R*xyzw = xzwy`
-/ -/
@ -194,36 +194,36 @@ def R_star (x : α → β → γ → δ) (y : γ) (z : α) (w : β) := x z w y
notation "R*" => R_star notation "R*" => R_star
/-- #### Robin /-- ### Robin
`Rxyz = yzx` `Rxyz = yzx`
-/ -/
def R (x : α) (y : β → α → γ) (z : β) := y z x def R (x : α) (y : β → α → γ) (z : β) := y z x
/-- #### Sage Bird /-- ### Sage Bird
`Θx = x(Θx)` `Θx = x(Θx)`
-/ -/
partial def Θ [Inhabited α] (x : α → α) := x (Θ x) partial def Θ [Inhabited α] (x : α → α) := x (Θ x)
/-- #### Starling /-- ### Starling
`Sxyz = xz(yz)` `Sxyz = xz(yz)`
-/ -/
def S (x : α → β → γ) (y : α → β) (z : α) := x z (y z) def S (x : α → β → γ) (y : α → β) (z : α) := x z (y z)
/-- #### Thrush /-- ### Thrush
`Txy = yx` `Txy = yx`
-/ -/
def T (x : α) (y : α → β) := y x def T (x : α) (y : α → β) := y x
/-! #### Turing Bird /-! ### Turing Bird
`Uxy = y(xxy)` `Uxy = y(xxy)`
-/ -/
/-- #### Vireo Once Removed /-- ### Vireo Once Removed
`V*xyzw = xwyz` `V*xyzw = xwyz`
-/ -/
@ -231,13 +231,13 @@ def V_star (x : α → β → γ → δ) (y : β) (z : γ) (w : α) := x w y z
notation "V*" => V_star notation "V*" => V_star
/-- #### Vireo /-- ### Vireo
`Vxyz = zxy` `Vxyz = zxy`
-/ -/
def V (x : α) (y : β) (z : α → β → γ) := z x y def V (x : α) (y : β) (z : α → β → γ) := z x y
/-- #### Warbler /-- ### Warbler
`Wxy = xyy` `Wxy = xyy`
-/ -/

6
Common/Set/Equinumerous.lean
View File

@ -8,7 +8,7 @@ Additional theorems around finite sets.
namespace Set namespace Set
/-! ### Definitions -/ /-! ## Definitions -/
/-- /--
A set `A` is equinumerous to a set `B` (written `A ≈ B`) if and only if there is A set `A` is equinumerous to a set `B` (written `A ≈ B`) if and only if there is
@ -79,7 +79,7 @@ theorem eq_imp_equinumerous {A B : Set α} (h : A = B)
conv at this => right; rw [h] conv at this => right; rw [h]
exact this exact this
/-! ### Finite Sets -/ /-! ## Finite Sets -/
/-- /--
A set is finite if and only if it is equinumerous to a natural number. A set is finite if and only if it is equinumerous to a natural number.
@ -87,7 +87,7 @@ A set is finite if and only if it is equinumerous to a natural number.
axiom finite_iff_equinumerous_nat {α : Type _} {S : Set α} axiom finite_iff_equinumerous_nat {α : Type _} {S : Set α}
: Set.Finite S ↔ ∃ n : ℕ, S ≈ Set.Iio n : Set.Finite S ↔ ∃ n : ℕ, S ≈ Set.Iio n
/-! ### Emptyset -/ /-! ## Emptyset -/
/-- /--
Any set equinumerous to the emptyset is the emptyset. Any set equinumerous to the emptyset is the emptyset.