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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
/-- #### Exercise 4a
/-- ### Exercise 4a
`⌊x + n⌋ = ⌊x⌋ + n` for every integer `n`.
-/
theorem exercise_4a (x : ℝ) (n : ℤ) : ⌊x + n⌋ = ⌊x⌋ + n :=
Int.floor_add_int x n
/-- #### Exercise 4b.1
/-- ### Exercise 4b.1
`⌊-x⌋ = -⌊x⌋` if `x` is an integer.
-/
theorem exercise_4b_1 (x : ℤ) : ⌊-x⌋ = -⌊x⌋ := by
simp only [Int.floor_int, id_eq]
/-- #### Exercise 4b.2
/-- ### Exercise 4b.2
`⌊-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 le_of_lt (Set.mem_Ico.mp hn').right
/-- #### Exercise 4c
/-- ### Exercise 4c
`⌊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]
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⌋`.
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⌋) :=
Real.Floor.floor_mul_eq_sum_range_floor_add_index_div n x
/-- #### Exercise 4d
/-- ### Exercise 4d
`⌊2x⌋ = ⌊x⌋ + ⌊x + 1/2⌋`
-/
@ -94,7 +94,7 @@ theorem exercise_4d (x : ℝ)
simp
rw [add_comm]
/-- #### Exercise 4e
/-- ### Exercise 4e
`⌊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]
rw [← add_assoc]
/-- #### Exercise 7b
/-- ### Exercise 7b
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
@ -125,7 +125,7 @@ theorem exercise_7b (ha : a > 0) (hb : b > 0) (hp : Nat.Coprime a b)
section
/-- #### Exercise 8
/-- ### Exercise 8
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

48
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]
_ = IsGLB S (-x) := rfl
/-- #### Theorem I.27
/-- ### Theorem I.27
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`.
@ -143,11 +143,11 @@ lemma leq_nat_abs_ceil_self (x : ℝ) : x ≤ Int.natAbs ⌈x⌉ := by
· have h' : ((Int.natAbs ⌈x⌉) : ℝ) ≥ 0 := by simp
calc x
_ ≤ 0 := le_of_lt (lt_of_not_le h)
_ ≤ ↑(Int.natAbs ⌈x⌉) := GE.ge.le h'
_ ≤ ↑(Int.natAbs ⌈x⌉) := GE.ge.le h'
/-! ## 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`.
-/
@ -159,7 +159,7 @@ theorem exists_pnat_geq_self (x : ℝ) : ∃ n : ℕ+, ↑n > x := by
_ = n := rfl
exact ⟨n, this⟩
/-- #### Theorem I.30
/-- ### Theorem I.30
If `x > 0` and if `y` is an arbitrary real number, there exists a positive
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'
exact ⟨n, p'⟩
/-- #### Theorem I.31
/-- ### Theorem I.31
If three real numbers `a`, `x`, and `y` satisfy the inequalities
`a ≤ x ≤ a + y / n` for every integer `n ≥ 1`, then `x = a`.
@ -243,7 +243,7 @@ theorem forall_pnat_frac_leq_self_leq_imp_eq {x y a : ℝ}
/--
Every member of a set `S` is less than or equal to some value `ub` if and only
if `ub` is an upper bound of `S`.
if `ub` is an upper bound of `S`.
-/
lemma mem_upper_bounds_iff_forall_le {S : Set ℝ}
: ub ∈ upperBounds S ↔ ∀ x ∈ S, x ≤ ub := by
@ -270,7 +270,7 @@ lemma mem_imp_ge_lub {x : ℝ} (h : IsLUB S s) : x ∈ upperBounds S → x ≥ s
intro 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`
has a supremum, then for some `x` in `S` we have `x > sup S - h`.
@ -294,7 +294,7 @@ theorem sup_imp_exists_gt_sup_sub_delta {S : Set ℝ} {s h : ℝ} (hp : h > 0)
/--
Every member of a set `S` is greater than or equal to some value `lb` if and
only if `lb` is a lower bound of `S`.
only if `lb` is a lower bound of `S`.
-/
lemma mem_lower_bounds_iff_forall_ge {S : Set ℝ}
: lb ∈ lowerBounds S ↔ ∀ x ∈ S, x ≥ lb := by
@ -321,7 +321,7 @@ lemma mem_imp_le_glb {x : ℝ} (h : IsGLB S s) : x ∈ lowerBounds S → x ≤ s
intro 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`
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)
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
`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'
_ ≤ 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
`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'
· 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
`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
simp at this
/-- #### Exercise 1
/-- ### Exercise 1
If `x` and `y` are arbitrary real numbers with `x < y`, prove that there is at
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'
_ = y := hz.right
/-- #### Exercise 2
/-- ### Exercise 2
If `x` is an arbitrary real number, prove that there are integers `m` and `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
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`.
-/
@ -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
rwa [one_mul] at hr
/-- #### Exercise 4
/-- ### Exercise 4
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
@ -540,7 +540,7 @@ theorem exercise_4 (x : ℝ) : ∃! n : ℤ, n ≤ x ∧ x < n + 1 := by
rw [← Int.floor_eq_iff] at hy
exact Eq.symm hy
/-- #### Exercise 5
/-- ### Exercise 5
If `x` is an arbitrary real number, prove that there is exactly one integer `n`
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
· exact hy.left
/-! #### Exercise 6
/-! ### Exercise 6
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.
@ -569,7 +569,7 @@ in the real-number system.
###### TODO
-/
/-! #### Exercise 7
/-! ### Exercise 7
If `x` is rational, `x ≠ 0`, and `y` irrational, prove that `x + y`, `x - y`,
`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
-/
/-! #### Exercise 8
/-! ### Exercise 8
Is the sum or product of two irrational numbers always irrational?
###### TODO
-/
/-! #### Exercise 9
/-! ### Exercise 9
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
@ -593,7 +593,7 @@ many.
###### TODO
-/
/-! #### Exercise 10
/-! ### Exercise 10
An integer `n` is called *even* if `n = 2m` for some integer `m`, and *odd* if
`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)
/-! #### Exercise 11
/-! ### Exercise 11
Prove that there is no rational number whose square is `2`.
@ -629,7 +629,7 @@ contradiction.]
###### TODO
-/
/-! #### Exercise 12
/-! ### Exercise 12
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

10
Bookshelf/Avigad/Chapter_2.lean
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@ -3,7 +3,7 @@
Dependent Type Theory
-/
/-! #### Exercise 1
/-! ### Exercise 1
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
/-! #### Exercise 2
/-! ### Exercise 2
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
/-! #### Exercise 3
/-! ### Exercise 3
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
@ -70,7 +70,7 @@ variable (c d : vec Prop 2)
end ex3
/-! #### Exercise 4
/-! ### Exercise 4
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
@ -106,4 +106,4 @@ variable (d : ex3.vec Prop 3)
end ex4
end Avigad.Chapter2
end Avigad.Chapter2

8
Bookshelf/Avigad/Chapter_3.lean
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@ -3,7 +3,7 @@
Propositions and Proofs
-/
/-! #### Exercise 1
/-! ### Exercise 1
Prove the following identities.
-/
@ -104,7 +104,7 @@ theorem imp_imp_not_imp_not : (p → q) → (¬q → ¬p) :=
end ex1
/-! #### Exercise 2
/-! ### Exercise 2
Prove the following identities. These require classical reasoning.
-/
@ -150,7 +150,7 @@ theorem imp_imp_imp : (((p → q) → p) → p) :=
end ex2
/-! #### Exercise 3
/-! ### Exercise 3
Prove `¬(p ↔ ¬p)` without using classical logic.
-/
@ -164,4 +164,4 @@ theorem iff_not_self (hp : p) : ¬(p ↔ ¬p) :=
end ex3
end Avigad.Chapter3
end Avigad.Chapter3

12
Bookshelf/Avigad/Chapter_4.lean
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@ -40,7 +40,7 @@ theorem forall_or_distrib
end ex1
/-! #### Exercise 2
/-! ### Exercise 2
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
@ -78,7 +78,7 @@ theorem forall_swap : (∀ x, r → p x) ↔ (r → ∀ x, p x) :=
end ex2
/-! #### Exercise 3
/-! ### Exercise 3
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.
@ -101,7 +101,7 @@ theorem barber_paradox (h : ∀ x : men, shaves barber x ↔ ¬shaves x x) : Fal
end ex3
/-! #### Exercise 4
/-! ### Exercise 4
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
@ -143,7 +143,7 @@ def Fermat'sLastTheorem : Prop :=
end ex4
/-! #### Exercise 5
/-! ### Exercise 5
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
/-! #### Exercise 6
/-! ### Exercise 6
Give a calculational proof of the theorem `log_mul` below.
-/
@ -258,4 +258,4 @@ theorem log_mul {x y : Float} (hx : x > 0) (hy : y > 0) :
end ex6
end Avigad.Chapter4
end Avigad.Chapter4

18
Bookshelf/Avigad/Chapter_5.lean
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@ -13,7 +13,7 @@ namespace Avigad.Chapter5
namespace ex1
/-! ##### Exercises 3.1 -/
/-! #### Exercises 3.1 -/
section ex3_1
@ -154,7 +154,7 @@ theorem imp_imp_not_imp_not : (p → q) → (¬q → ¬p) := by
end ex3_1
/-! ##### Exercises 3.2 -/
/-! #### Exercises 3.2 -/
section ex3_2
@ -223,7 +223,7 @@ theorem imp_imp_imp : (((p → q) → p) → p) := by
end ex3_2
/-! ##### Exercises 3.3 -/
/-! #### Exercises 3.3 -/
section ex3_3
@ -235,7 +235,7 @@ theorem iff_not_self (hp : p) : ¬(p ↔ ¬p) := by
end ex3_3
/-! ##### Exercises 4.1 -/
/-! #### Exercises 4.1 -/
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
/-! ##### Exercises 4.2 -/
/-! #### Exercises 4.2 -/
section ex4_2
@ -316,7 +316,7 @@ theorem forall_swap : (∀ x, r → p x) ↔ (r → ∀ x, p x) := by
end ex4_2
/-! ##### Exercises 4.3 -/
/-! #### Exercises 4.3 -/
section ex4_3
@ -336,7 +336,7 @@ theorem barber_paradox (h : ∀ x : men, shaves barber x ↔ ¬ shaves x x)
end ex4_3
/-! ##### Exercises 4.5 -/
/-! #### Exercises 4.5 -/
section ex4_5
@ -448,7 +448,7 @@ end ex4_5
end ex1
/-! #### Exercise 2
/-! ### Exercise 2
Use tactic combinators to obtain a one line proof of the following:
-/
@ -461,4 +461,4 @@ theorem or_and_or_and_or (p q r : Prop) (hp : p)
end ex2
end Avigad.Chapter5
end Avigad.Chapter5

8
Bookshelf/Avigad/Chapter_7.lean
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@ -5,7 +5,7 @@ Inductive Types
namespace Avigad.Chapter7
/-! #### Exercise 1
/-! ### Exercise 1
Try defining other operations on the natural numbers, such as multiplication,
the predecessor function (with `pred 0 = 0`), truncated subtraction (with
@ -77,7 +77,7 @@ end Nat
end ex1
/-! #### Exercise 2
/-! ### Exercise 2
Define some operations on lists, like a `length` function or the `reverse`
function. Prove some properties, such as the following:
@ -178,7 +178,7 @@ theorem reverse_reverse (t : List α)
end ex2
/-! #### Exercise 3
/-! ### Exercise 3
Define an inductive data type consisting of terms built up from the following
constructors:
@ -217,4 +217,4 @@ def eval_foo : Foo → List Nat → Nat
end ex3
end Avigad.Chapter7
end Avigad.Chapter7

16
Bookshelf/Avigad/Chapter_8.lean
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@ -5,7 +5,7 @@ Induction and Recursion
namespace Avigad.Chapter8
/-! #### Exercise 1
/-! ### Exercise 1
Open a namespace `Hidden` to avoid naming conflicts, and use the equation
compiler to define addition, multiplication, and exponentiation on the natural
@ -29,7 +29,7 @@ def exp : Nat → Nat → Nat
end ex1
/-! #### Exercise 2
/-! ### Exercise 2
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
@ -48,7 +48,7 @@ def reverse : List α → List α
end ex2
/-! #### Exercise 3
/-! ### Exercise 3
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
@ -86,7 +86,7 @@ noncomputable def brecOn {motive : Nat → Sort u}
end ex3
/-! #### Exercise 4
/-! ### Exercise 4
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
@ -113,7 +113,7 @@ end Vector
end ex4
/-! #### Exercise 5
/-! ### Exercise 5
Consider the following type of arithmetic expressions.
-/
@ -149,7 +149,7 @@ def sampleVal : Nat → Nat
-- Try it out. You should get 47 here.
#eval eval sampleVal sampleExpr
/-! ##### Constant Fusion
/-! ### Constant Fusion
Implement "constant fusion," a procedure that simplifies subterms like `5 + 7
to `12`. Using the auxiliary function `simpConst`, define a function "fuse": to
@ -189,7 +189,7 @@ theorem simpConst_eq (v : Nat → Nat)
| times _ (var _)
| times _ (plus _ _)
| times _ (times _ _) => simp only
theorem fuse_eq (v : Nat → Nat)
: ∀ e : Expr, eval v (fuse e) = eval v e := by
@ -206,4 +206,4 @@ theorem fuse_eq (v : Nat → Nat)
end ex5
end Avigad.Chapter8
end Avigad.Chapter8

22
Bookshelf/Enderton/Logic/Chapter_1.lean
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@ -131,7 +131,7 @@ lemma no_neg_sentential_count_eq_binary_count {φ : Wff} (h : ¬φ.hasNotSymbol)
unfold sententialSymbolCount binarySymbolCount
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
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
/-! #### Exercise 1.1.2
/-! ### Exercise 1.1.2
Show that there are no wffs of length `2`, `3`, or `6`, but that any other
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
/-- #### Exercise 1.1.3
/-- ### Exercise 1.1.3
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
@ -320,7 +320,7 @@ theorem exercise_1_1_3 (φ : Wff)
rw [ih₁, ih₂]
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
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]
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
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)
/-! #### Exercise 1.2.1
/-! ### Exercise 1.2.1
Show that neither of the following two formulas tautologically implies the
other:
@ -430,7 +430,7 @@ theorem exercise_1_2_2a (P Q : Prop)
: (((P → Q) → P) → P) := by
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
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
/-- #### Exercise 1.2.3 (a)
/-- ### Exercise 1.2.3 (a)
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
tauto
/-- #### Exercise 1.2.3 (b)
/-- ### Exercise 1.2.3 (b)
Determine whether or not `((P ∧ Q) → R))` tautologically implies
`((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
tauto
/-! #### Exercise 1.2.5
/-! ### Exercise 1.2.5
Prove or refute each of the following assertions:
@ -519,7 +519,7 @@ theorem exercise_1_2_6b
: (False ∨ True) ∧ ¬ False := by
simp
/-! #### Exercise 1.2.15
/-! ### Exercise 1.2.15
Of the following three formulas, which tautologically implies which?
(a) `(A ↔ B)`

2
Bookshelf/Enderton/Set.tex
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@ -8999,7 +8999,7 @@
\begin{proof}
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
$S' \cup T = S$.
By definition of a \nameref{ref:finite-set}, $S$ is

10
Bookshelf/Enderton/Set/Chapter_1.lean
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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 :=
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))) := ⟨by simp, by simp⟩
/-- #### Exercise 1.1b
/-- ### Exercise 1.1b
`{∅} ___ {∅, {{∅}}}`
-/
@ -39,7 +39,7 @@ theorem exercise_1_1b
simp at 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)))) := ⟨by simp, by simp⟩
/-- #### Exercise 1.1d
/-- ### Exercise 1.1d
`{{∅}} ___ {∅, {{∅}}}`
-/
@ -59,7 +59,7 @@ theorem exercise_1_1d
simp at h
exact empty_ne_singleton_empty h
/-- #### Exercise 1.1e
/-- ### Exercise 1.1e
`{{∅}} ___ {∅, {∅, {∅}}}`
-/

68
Bookshelf/Enderton/Set/Chapter_2.lean
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@ -10,7 +10,7 @@ Axioms and Operations
namespace Enderton.Set.Chapter_2
/-! #### Commutative Laws
/-! ### Commutative Laws
For any sets `A` and `B`,
```
@ -40,7 +40,7 @@ theorem commutative_law_ii (A B : Set α)
#check Set.inter_comm
/-! #### Associative Laws
/-! ### Associative Laws
For any sets `A`, `B`, and `C`,
```
@ -75,7 +75,7 @@ theorem associative_law_ii (A B C : Set α)
#check Set.inter_assoc
/-! #### Distributive Laws
/-! ### Distributive Laws
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
/-! #### De Morgan's Laws
/-! ### De Morgan's Laws
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
/-! #### Identities Involving ∅
/-! ### Identities Involving ∅
For any set `A`,
```
@ -185,7 +185,7 @@ theorem emptyset_identity_iii (A C : Set α)
#check Set.inter_diff_self
/-! #### Monotonicity
/-! ### Monotonicity
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
/-! #### Anti-monotonicity
/-! ### Anti-monotonicity
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
/-- #### Intersection/Difference Associativity
/-- ### Intersection/Difference Associativity
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
/-- #### Exercise 2.1
/-- ### Exercise 2.1
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.
@ -300,7 +300,7 @@ theorem exercise_2_1 {A B C : Set ℤ}
· rw [hC] at 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`.
-/
@ -339,7 +339,7 @@ theorem exercise_2_2 {A B : Set (Set ℕ)}
have h₂ := h₁ 2
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
example in this section.)
@ -352,7 +352,7 @@ theorem exercise_2_3 {A : Set (Set α)}
rw [Set.mem_setOf_eq]
exact ⟨x, ⟨hx, hy⟩⟩
/-- #### Exercise 2.4
/-- ### Exercise 2.4
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]
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`.
-/
@ -376,7 +376,7 @@ theorem exercise_2_5 {𝓐 : Set (Set α)} (h : ∀ x ∈ 𝓐, x ⊆ B)
have ⟨t, ⟨ht𝓐, hyt⟩⟩ := hy
exact (h t ht𝓐) hyt
/-- #### Exercise 2.6a
/-- ### Exercise 2.6a
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]
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?
-/
@ -412,7 +412,7 @@ theorem exercise_2_6b
conv => rhs; rw [hB, exercise_2_6a]
exact hB
/-- #### Exercise 2.7a
/-- ### Exercise 2.7a
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 _
exact hA
/-- #### Exercise 2.7b (i)
/-- ### Exercise 2.7b (i)
Show that `𝓟 A ∪ 𝓟 B ⊆ 𝓟 (A ∪ B)`.
-/
@ -447,7 +447,7 @@ theorem exercise_2_7b_i
rw [Set.mem_setOf_eq]
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)`.?
-/
@ -499,7 +499,7 @@ theorem exercise_2_7b_ii
refine Or.inl (Set.Subset.trans hx ?_)
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`.
-/
@ -527,7 +527,7 @@ theorem exercise_2_9 (ha : a = {1}) (hB : B = {{1}})
have := h 1
simp at this
/-- #### Exercise 2.10
/-- ### Exercise 2.10
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]
exact h₂
/-- #### Exercise 2.11 (i)
/-- ### Exercise 2.11 (i)
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
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`.
-/
@ -651,7 +651,7 @@ lemma left_diff_eq_singleton_one : (A \ B) \ C = {1} := by
| inl 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
different from `(A - B) - C`.
@ -668,7 +668,7 @@ theorem exercise_2_14 : A \ (B \ C) ≠ (A \ B) \ C := by
end
/-- #### Exercise 2.15 (a)
/-- ### Exercise 2.15 (a)
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
/-- #### Exercise 2.15 (b)
/-- ### Exercise 2.15 (b)
Show that `A + (B + C) = (A + B) + C`.
-/
@ -749,7 +749,7 @@ theorem exercise_2_15b (A B C : Set α)
#check symmDiff_assoc
/-- #### Exercise 2.16
/-- ### Exercise 2.16
Simplify:
`[(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]
_ = B \ A := by rw [Set.union_diff_left]
/-! #### Exercise 2.17
/-! ### Exercise 2.17
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)
: 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`?
-/
@ -810,7 +810,7 @@ theorem exercise_2_19 {A B : Set α}
have := h ∅
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`.
Show that `B = C`.
@ -836,7 +836,7 @@ theorem exercise_2_20 {A B C : Set α}
rw [← hu] at this
exact Or.elim this (absurd · hA) (by simp)
/-- #### Exercise 2.21
/-- ### Exercise 2.21
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
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`.
-/
@ -889,7 +889,7 @@ theorem exercise_2_22 {A B : Set (Set α)}
· intro 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 }`.
-/
@ -908,7 +908,7 @@ theorem exercise_2_24a {𝓐 : Set (Set α)}
_ = { x | ∀ t ∈ { 𝒫 X | X ∈ 𝓐 }, x ∈ t} := by simp
_ = ⋂₀ { 𝒫 X | X ∈ 𝓐 } := rfl
/-- #### Exercise 2.24b
/-- ### Exercise 2.24b
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]
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
what conditions does equality hold?

124
Bookshelf/Enderton/Set/Chapter_3.lean
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@ -17,7 +17,7 @@ namespace Enderton.Set.Chapter_3
open Set.Relation
/-- #### Lemma 3B
/-- ### Lemma 3B
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
/-- #### Theorem 3D
/-- ### Theorem 3D
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)⟩
/-- #### 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`.
-/
@ -75,7 +75,7 @@ theorem theorem_3g_i {F : Set.HRelation α β}
unfold isOneToOne at hF
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`.
-/
@ -89,7 +89,7 @@ theorem theorem_3g_ii {F : Set.HRelation α β}
unfold isOneToOne at hF
exact (single_rooted_eq_unique hF.right hx hx₁).symm
/-- #### Theorem 3H
/-- ### Theorem 3H
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]
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
`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
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
`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]
exact hy
/-- #### Theorem 3K (a)
/-- ### Theorem 3K (a)
The following hold for any sets. (`F` need not be a function.)
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]
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 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]
exact ⟨u, hu⟩
/-! #### Theorem 3K (c)
/-! ### Theorem 3K (c)
The following hold for any sets. (`F` need not be a function.)
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 ⟨hv₁, hv₂⟩
/-! #### Corollary 3L
/-! ### Corollary 3L
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
exact (theorem_3k_c_ii hG').symm
/-- #### Theorem 3M
/-- ### Theorem 3M
If `R` is a symmetric and transitive relation, then `R` is an equivalence
relation on `fld R`.
@ -497,7 +497,7 @@ theorem theorem_3m {R : Set.Relation α}
have := hS ht
exact hT this ht
/-- #### Theorem 3R
/-- ### Theorem 3R
Let `R` be a linear ordering on `A`.
@ -521,7 +521,7 @@ theorem theorem_3r {R : Rel α α} (hR : IsStrictTotalOrder α R)
right
exact h₂
/-- #### Exercise 3.1
/-- ### Exercise 3.1
Suppose that we attempted to generalize the Kuratowski definitions of ordered
pairs to ordered triples by defining
@ -544,7 +544,7 @@ theorem exercise_3_1 {x y z u v w : ℕ}
· rw [hy, hv]
simp only
/-- #### Exercise 3.2a
/-- ### Exercise 3.2a
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
_ = (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`.
-/
@ -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)
exact (h (a, c)).mpr ⟨ha, hc⟩
/-- #### Exercise 3.3
/-- ### Exercise 3.3
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⟩
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`
such that for any `y`,
@ -685,7 +685,7 @@ theorem exercise_3_5a {A : Set α} {B : Set β}
rw [hab.right]
exact ⟨hab.left, hb⟩
/-- #### Exercise 3.5b
/-- ### Exercise 3.5b
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⟩
/-- #### Exercise 3.6
/-- ### Exercise 3.6
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⟩⟩
/-- #### Exercise 3.7
/-- ### Exercise 3.7
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
exact hxy_mem this
/-- #### Exercise 3.8 (i)
/-- ### Exercise 3.8 (i)
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⟩
exact ⟨y, t, ht, hx⟩
/-- #### Exercise 3.8 (ii)
/-- ### Exercise 3.8 (ii)
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⟩⟩⟩
exact ⟨t, ⟨y, ⟨hy, ht⟩⟩⟩
/-- #### Exercise 3.9 (i)
/-- ### Exercise 3.9 (i)
Discuss the result of replacing the union operation by the intersection
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
exact ⟨y, hy R hR⟩
/-- #### Exercise 3.9 (ii)
/-- ### Exercise 3.9 (ii)
Discuss the result of replacing the union operation by the intersection
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
exact ⟨y, hy R hR⟩
/-- #### Exercise 3.12
/-- ### Exercise 3.12
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]
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
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]
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.
-/
@ -981,7 +981,7 @@ theorem exercise_3_14_a {f g : Set.HRelation α β}
: isSingleValued (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**
`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
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
`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
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.
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)
(exercise_3_17_i hF.right hG.right)
/-! #### Exercise 3.18
/-! ### Exercise 3.18
Let `R` be the set
```
@ -1252,7 +1252,7 @@ theorem exercise_3_18_v
end Exercise_3_18
/-! #### Exercise 3.19
/-! ### Exercise 3.19
Let
```
@ -1502,7 +1502,7 @@ theorem exercise_3_19_x
end Exercise_3_19
/-- #### Exercise 3.20
/-- ### Exercise 3.20
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 ∩ (Set.prod A (ran F)) := rfl
/-- #### Exercise 3.22 (a)
/-- ### Exercise 3.22 (a)
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
exact ⟨u, this, hu.right⟩
/-- #### Exercise 3.22 (b)
/-- ### Exercise 3.22 (b)
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
_ = image F (image G A) := rfl
/-- #### Exercise 3.22 (c)
/-- ### Exercise 3.22 (c)
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
_ = (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
`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
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
`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
_ = 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 }`.
-/
@ -1671,7 +1671,7 @@ theorem exercise_3_24 {F : Set.HRelation α β} {A : Set β}
· intro ⟨y, hy⟩
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
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
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
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 } :=
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
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⟩
/-- #### Exercise 3.28
/-- ### Exercise 3.28
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`
@ -1832,7 +1832,7 @@ theorem exercise_3_28 {A : Set α} {B : Set β}
have hz := mem_pair_imp_snd_mem_ran hb.right
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
```
@ -1875,7 +1875,7 @@ theorem exercise_3_29 {f : Set.HRelation α β} {G : Set.HRelation β (Set α)}
rw [heq] at this
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:
```
@ -1892,7 +1892,7 @@ variable {F : Set α → Set α} {A B C : Set α}
(hB : B = ⋂₀ { X | X ⊆ A ∧ F X ⊆ 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`.
-/
@ -1988,7 +1988,7 @@ theorem exercise_3_30_a : F B = B ∧ F C = C := by
· rw [Set.Subset.antisymm_iff]
exact ⟨hC_subset, hC_supset⟩
/-- ##### Exercise 3.30 (b)
/-- #### Exercise 3.30 (b)
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
/-- #### Exercise 3.32 (a)
/-- ### Exercise 3.32 (a)
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
exact hR hp
/-- #### Exercise 3.32 (b)
/-- ### Exercise 3.32 (b)
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⟩
exact hR this
/-- #### Exercise 3.33
/-- ### Exercise 3.33
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]
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
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
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
relation. Is `⋃ 𝓐` a transitive relation?
@ -2157,7 +2157,7 @@ theorem exercise_3_34_b {𝓐 : Set (Set.Relation ℕ)}
simp at this
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}⟧`.
-/
@ -2168,7 +2168,7 @@ theorem exercise_3_35 {R : Set.Relation α} {x : α}
_ = { t | ∃ u ∈ ({x} : Set α), (u, t) ∈ R } := by simp
_ = 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
`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]
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
follows:
@ -2311,7 +2311,7 @@ theorem exercise_3_37 {P : Set (Set α)} {A : Set α}
simp only [Set.mem_setOf_eq]
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
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]
_ = 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
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]
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
`⟨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]
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
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 *
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
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)
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.
Define the binary relation `<_L` on the Cartesian product `A × B` by:

66
Bookshelf/Enderton/Set/Chapter_4.lean
View File

@ -11,7 +11,7 @@ Natural Numbers
namespace Enderton.Set.Chapter_4
/-- #### Theorem 4C
/-- ### Theorem 4C
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
/-- #### Theorem 4I
/-- ### Theorem 4I
For natural numbers `m` and `n`,
```
@ -34,7 +34,7 @@ m + n⁺ = (m + n)⁺
theorem theorem_4i (m n : ℕ)
: m + 0 = m ∧ m + n.succ = (m + n).succ := ⟨rfl, rfl⟩
/-- #### Theorem 4J
/-- ### Theorem 4J
For natural numbers `m` and `n`,
```
@ -45,7 +45,7 @@ m ⬝ n⁺ = m ⬝ n + m .
theorem theorem_4j (m n : ℕ)
: 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`.
-/
@ -60,7 +60,7 @@ lemma left_additive_identity (n : ℕ)
#check Nat.zero_add
/-- #### Lemma 2
/-- ### Lemma 2
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
/-- #### Theorem 4K-1
/-- ### Theorem 4K-1
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
/-- #### Theorem 4K-2
/-- ### Theorem 4K-2
Commutative law for addition. For `m, n ∈ ω`,
```
@ -119,7 +119,7 @@ theorem theorem_4k_2 {m n : ℕ}
#check Nat.add_comm
/-- #### Zero Multiplicand
/-- ### Zero Multiplicand
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
/-- #### Successor Distribution
/-- ### Successor Distribution
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
/-- #### Theorem 4K-3
/-- ### Theorem 4K-3
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.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⁺`.
-/
@ -197,7 +197,7 @@ theorem succ_identity (m : ℕ)
#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`.
-/
@ -213,9 +213,9 @@ theorem right_mul_id (m : ℕ)
#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`.
-/
theorem theorem_4k_5 (m n : ℕ)
: m * n = n * m := by
@ -232,7 +232,7 @@ theorem theorem_4k_5 (m n : ℕ)
#check Nat.mul_comm
/-- #### Theorem 4K-4
/-- ### Theorem 4K-4
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
/-- #### Lemma 4L(b)
/-- ### Lemma 4L(b)
No natural number is a member of itself.
-/
@ -269,7 +269,7 @@ lemma lemma_4l_b (n : ℕ)
#check Nat.lt_irrefl
/-- #### Lemma 10
/-- ### Lemma 10
For every natural number `n ≠ 0`, `0 ∈ n`.
-/
@ -285,7 +285,7 @@ theorem zero_least_nat (n : ℕ)
#check Nat.pos_of_ne_zero
/-! #### Theorem 4N
/-! ### Theorem 4N
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
/-! #### Corollary 4P
/-! ### Corollary 4P
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
/-- #### Well Ordering of ω
/-- ### Well Ordering of ω
Let `A` be a nonempty subset of `ω`. Then there is some `m ∈ A` such that
`m ≤ n` for all `n ∈ A`.
@ -401,7 +401,7 @@ theorem well_ordering_nat {A : Set ℕ} (hA : Set.Nonempty A)
rw [this] at h
simp only [sdiff_self, Set.bot_eq_empty] at h
exact absurd h.symm (Set.Nonempty.ne_empty hA)
-- Use strong induction to prove every element of `ω` is in the complement.
have : ∀ n : ℕ, (∀ m, m < n → m ∈ A.compl) := by
intro n
@ -438,7 +438,7 @@ theorem well_ordering_nat {A : Set ℕ} (hA : Set.Nonempty A)
#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
less than `n` is in `A`, then `n ∈ A`. Then `A = ω`.
@ -455,21 +455,21 @@ theorem strong_induction_principle_nat (A : Set ℕ)
by_contra nh
have ⟨m, hm⟩ := well_ordering_nat (Set.nmem_singleton_empty.mp nh)
refine absurd (h m ?_) hm.left
-- Show that every number less than `m` is in `A`.
intro x hx
by_contra nx
have : x < x := Nat.lt_of_lt_of_le hx (hm.right x nx)
simp at this
/-- #### Exercise 4.1
/-- ### Exercise 4.1
Show that `1 ≠ 3` i.e., that `∅⁺ ≠ ∅⁺⁺⁺`.
-/
theorem exercise_4_1 : 1 ≠ 3 := by
simp
/-- #### Exercise 4.13
/-- ### Exercise 4.13
Let `m` and `n` be natural numbers such that `m ⬝ n = 0`. Show that either
`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
/-- #### Exercise 4.14
/-- ### Exercise 4.14
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'⟩
exact absurd this h
/-- #### Exercise 4.17
/-- ### Exercise 4.17
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.succ := rfl
/-- #### Exercise 4.19
/-- ### Exercise 4.19
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`.
@ -600,7 +600,7 @@ theorem exercise_4_19 (m d : ℕ) (hd : d ≠ 0)
_ < d := hr
simp at this
/-- #### Exercise 4.22
/-- ### Exercise 4.22
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.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
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, ?_⟩
rw [hm₁]
/-- #### Exercise 4.24
/-- ### Exercise 4.24
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
exact (theorem_4n_i m p n).mpr hr
/-- #### Exercise 4.25
/-- ### Exercise 4.25
Assume that `n ∈ m` and `q ∈ p`. Show that
```
@ -685,4 +685,4 @@ theorem exercise_4_25 (m n p q : ℕ) (h₁ : n < m) (h₂ : q < p)
conv at h₁ => right; rw [add_comm]
exact h₁
end Enderton.Set.Chapter_4
end Enderton.Set.Chapter_4

541
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
/-- #### Theorem 6B
/-- ### Theorem 6B
No set is equinumerous to its powerset.
-/
theorem theorem_6b (A : Set α)
: A ≉ 𝒫 A := by
/-
> Let `A` be an arbitrary set and `f: A → 𝒫 A`.
-/
rw [Set.not_equinumerous_def]
intro f hf
unfold Set.BijOn at hf
/-
> Define `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
have hφ := hf.right.right (show φ ∈ 𝒫 A by simp)
have ⟨a, ha⟩ := hφ
@ -82,23 +95,18 @@ lemma pigeonhole_principle_aux (n : ℕ)
-/
induction n with
/-
## (i)
> By definition, `0 = ∅`.
> #### (i)
> By definition, `0 = ∅`. Then `0` has no proper subsets. Hence `0 ∈ S`
> vacuously.
-/
| zero =>
intro _ hM
unfold Set.Iio 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
/-
> Hence `0 ∈ S` vacuously.
-/
exact False.elim hM
/-
## (ii)
> #### (ii)
> Suppose `n ∈ S` and `M ⊂ n⁺`. Furthermore, let `f: M → n⁺` be a one-to-one
> function.
-/
@ -111,19 +119,19 @@ lemma pigeonhole_principle_aux (n : ℕ)
· rw [hM', Set.SurjOn_emptyset_Iio_iff_eq_zero] 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:
-/
by_cases h : ¬ ∃ t, t ∈ M ∧ f t = n
/-
### Case 1
> ##### Case 1
> `n ∉ ran f`.
> Then `f` is not onto `n⁺`.
-/
· have ⟨t, ht⟩ := hf_surj (show n ∈ _ by simp)
exact absurd ⟨t, ht⟩ h
/-
### Case 2
> ##### Case 2
> `n ∈ ran 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_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}
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]
exact ng_surj
/-
> 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
> `a ∈ n⁺`. Hence `f` is not onto `n⁺`.
> 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 `a ∈ n⁺`. Hence `f` is not onto `n⁺`.
-/
refine absurd (hf_surj $ calc a
_ < n := ha₁
@ -294,14 +302,14 @@ lemma pigeonhole_principle_aux (n : ℕ)
· refine ⟨y, hy₁, ?_⟩
rwa [if_neg hc₁, if_neg hc₂]
/-
### Subconclusion
> ##### Subconclusion
> The foregoing cases are exhaustive. Hence `n⁺ ∈ S`.
## (iii)
> By (i) and (ii), `S` is an inductive set. By Theorem 4B, `S = ω`. Thus for all
> natural numbers `n`, there is no one-to-one correspondence between `n` and a
> proper subset of `n`. In other words, no natural number is equinumerous to a
> proper subset of itself.
>
> #### (iii)
> By *(i)* and *(ii)*, `S` is an inductive set. By *Theorem 4B*, `S = ω`. Thus
> for all natural numbers `n`, there is no one-to-one correspondence between `n`
> and a proper subset of `n`. In other words, no natural number is equinumerous
> 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⟩
exact absurd hf.right.right this
/-- #### Corollary 6C
/-- ### Corollary 6C
No finite set is equinumerous to a proper subset of itself.
-/
theorem corollary_6c [DecidableEq α] [Nonempty α]
{S S' : Set α} (hS : Set.Finite S) (h : S' ⊂ S)
: 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'
have hT : S = S' ∪ (S \ S') := by
simp only [Set.union_diff_self]
exact (Set.left_subset_union_eq_self (subset_of_ssubset h)).symm
-- `hF : S' ∪ T ≈ S`.
-- `hG : S ≈ n`.
-- `hH : S' ∪ T ≈ n`.
/-
> By *Theorem 6A*, `S' ∪ T ≈ S` which, by the same theorem, implies
> `S' ∪ T ≈ n`.
-/
have hF := Set.equinumerous_refl S
conv at hF => arg 1; rw [hT]
have ⟨n, hG⟩ := Set.finite_iff_equinumerous_nat.mp hS
have ⟨H, hH⟩ := Set.equinumerous_trans hF hG
-- Restrict `H` to `S'` to yield a bijection between `S'` and a proper subset
-- of `n`.
let R := (Set.Iio n) \ (H '' T)
have hR : Set.BijOn H S' R := by
/-
> Let `f` be a one-to-one correspondence between `S' ∪ T` and `n`.
-/
have ⟨f, hf⟩ := Set.equinumerous_trans hF hG
/-
> 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 ⟨?_, ?_, ?_⟩
· -- `Set.MapsTo H S' R`
intro x hx
refine ⟨hH.left $ Set.mem_union_left T hx, ?_⟩
refine ⟨hf.left $ Set.mem_union_left T hx, ?_⟩
unfold Set.image
by_contra 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 hc₁ : a ∈ S' ∪ T := Set.mem_union_right S' ha₁
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 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
have hc₁ : x₁ ∈ S' ∪ T := Set.mem_union_left T hx₁
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`
show ∀ r, r ∈ R → r ∈ H '' S'
show ∀ r, r ∈ R → r ∈ f '' S'
intro r hr
unfold Set.image
simp only [Set.mem_setOf_eq]
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
have ⟨x, hx⟩ := this
apply Or.elim hx.left
@ -382,9 +399,14 @@ theorem corollary_6c [DecidableEq α] [Nonempty α]
rw [← hx.right]
simp only [Set.mem_image, Finset.mem_coe]
exact ⟨x, hx', rfl⟩
intro hf
have hf₁ : S ≈ R := Set.equinumerous_trans hf ⟨H, hR⟩
/-
> By the *Pigeonhole Principle*, `n` is not equinumerous to any proper subset of
> `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 ⟨k, hk⟩ := Set.equinumerous_symm hf₁
exact Set.equinumerous_trans ⟨k, hk⟩ hG
@ -398,106 +420,185 @@ theorem corollary_6c [DecidableEq α] [Nonempty α]
· show ¬ ∀ r, r ∈ Set.Iio n → r ∈ R
intro nr
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 ht₃ : H t ∈ R := nr (H t) ht₂
have ht₂ : f t ∈ Set.Iio n := hf.left (Set.mem_union_right S' ht₁)
have ht₃ : f t ∈ R := nr (f t) ht₂
exact absurd ⟨t, ht₁, rfl⟩ ht₃.right
/-- #### Corollary 6D (a)
/-- ### Corollary 6D (a)
Any set equinumerous to a proper subset of itself is infinite.
-/
theorem corollary_6d_a [DecidableEq α] [Nonempty α]
{S S' : Set α} (hS : S' ⊂ S) (hf : S ≈ S')
: 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
simp only [Set.not_infinite] at nS
exact absurd hf (corollary_6c nS hS)
/-- #### Corollary 6D (b)
/-- ### Corollary 6D (b)
The set `ω` is infinite.
-/
theorem corollary_6d_b
: 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 f x := 2 * x
suffices Set.BijOn f (@Set.univ ℕ) S by
refine corollary_6d_a ?_ ⟨f, this⟩
rw [Set.ssubset_def]
apply And.intro
· simp
· show ¬ ∀ x, x ∈ Set.univ → x ∈ S
simp only [
Set.mem_univ,
true_and,
Set.mem_setOf_eq,
forall_true_left,
not_forall,
not_exists
]
refine ⟨1, ?_⟩
intro x nx
simp only [mul_eq_one, false_and] at nx
/-
> #### (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⟩
rw [Set.ssubset_def]
apply And.intro
· simp
· show ¬ ∀ x, x ∈ Set.univ → x ∈ S
simp only [
Set.mem_univ,
true_and,
Set.mem_setOf_eq,
forall_true_left,
not_forall,
not_exists
]
refine ⟨1, ?_⟩
intro x nx
simp only [mul_eq_one, false_and] at nx
refine ⟨by simp, ?_, ?_⟩
· -- `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
/-- ### Corollary 6E
Any finite set is equinumerous to a unique natural number.
-/
theorem corollary_6e [Nonempty α] (S : Set α) (hS : Set.Finite S)
: ∃! 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
refine ⟨n, hf, ?_⟩
/-
> Suppose `S` is equinumerous to another natural number `m`.
-/
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
| Or.inr (Or.inl r) => exact r
| 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)
/-
> If `n < m`, then `m ≈ S` and `S ≈ n`. By *Theorem 6A*, it follows `m ≈ n`. But
> the *Pigeonhole Principle* indicates no natural number is equinumerous to a
> proper subset of itself, a contradiction.
-/
| Or.inr (Or.inr r) =>
have hh := Set.equinumerous_symm hf
have hk := Set.equinumerous_trans hh hg
have hnm : Set.Iio n ⊂ Set.Iio m := Set.Iio_nat_lt_ssubset r
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`
less than `n`.
-/
lemma lemma_6f {n : ℕ}
: ∀ {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
/-
> #### (i)
> By definition, `0 = ∅`. Thus `0` has no proper subsets. Hence `0 ∈ S`
> vacuously.
-/
| zero =>
intro C hC
unfold Set.Iio at hC
simp only [Nat.zero_eq, not_lt_zero', Set.setOf_false] at hC
rw [Set.ssubset_empty_iff_false] at 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 =>
/-
> 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
: ∀ S, S ⊆ Set.Iio n →
∃ m, m < n + 1 ∧ S ≈ Set.Iio m := by
@ -513,120 +614,150 @@ lemma lemma_6f {n : ℕ}
· -- `S = Set.Iio n`
intro h
exact ⟨n, by simp, Set.eq_imp_equinumerous h⟩
intro C hC
by_cases hn : n ∈ C
· -- Since `C` is a proper subset of `n⁺`, the set `n⁺ - C` is nonempty.
have hC₁ : Set.Nonempty (Set.Iio (n + 1) \ C) := by
rw [Set.ssubset_def] at hC
have : ¬ ∀ x, x ∈ Set.Iio (n + 1) → x ∈ C := hC.right
simp only [Set.mem_Iio, not_forall, exists_prop] at this
exact this
-- `p` is the least element of `n⁺ - C`.
have ⟨p, hp⟩ := Chapter_4.well_ordering_nat hC₁
let C' := (C \ {n}) ∪ {p}
have hC'₁ : C' ⊆ Set.Iio n := by
show ∀ x, x ∈ C' → x ∈ Set.Iio n
intro x hx
match @trichotomous ℕ LT.lt _ x n with
| Or.inl r => exact r
| Or.inr (Or.inl r) =>
rw [r] at hx
apply Or.elim hx
· intro nx
simp at nx
· intro nx
simp only [Set.mem_singleton_iff] at nx
rw [nx] at hn
exact absurd hn hp.left.right
| Or.inr (Or.inr r) =>
apply Or.elim hx
· intro ⟨h₁, h₂⟩
have h₃ := subset_of_ssubset hC h₁
simp only [Set.mem_singleton_iff, Set.mem_Iio] at h₂ h₃
exact Or.elim (Nat.lt_or_eq_of_lt h₃) id (absurd · h₂)
· intro h
simp only [Set.mem_singleton_iff] at h
have := hp.left.left
rw [← h] at this
exact Or.elim (Nat.lt_or_eq_of_lt this)
id (absurd · (Nat.ne_of_lt r).symm)
have ⟨m, hm₁, hm₂⟩ := h_subset_equinumerous C' hC'₁
suffices C' ≈ C from
⟨m, hm₁, Set.equinumerous_trans (Set.equinumerous_symm this) hm₂⟩
-- Proves `f` is a one-to-one correspondence between `C'` and `C`.
let f x := if x = p then n else x
refine ⟨f, ?_, ?_, ?_⟩
· -- `Set.MapsTo f C' C`
intro x hx
dsimp only
by_cases hxp : x = p
· rw [if_pos hxp]
exact hn
· rw [if_neg hxp]
apply Or.elim hx
· exact fun x => x.left
· intro hx₁
simp only [Set.mem_singleton_iff] at hx₁
exact absurd hx₁ hxp
· -- `Set.InjOn f C'`
intro x₁ hx₁ x₂ hx₂ hf
dsimp only at hf
by_cases hx₁p : x₁ = p
· by_cases hx₂p : x₂ = p
· rw [hx₁p, hx₂p]
· rw [if_pos hx₁p, if_neg hx₂p] at hf
apply Or.elim hx₂
· intro nx
exact absurd hf.symm nx.right
· intro nx
simp only [Set.mem_singleton_iff] at nx
exact absurd nx hx₂p
· by_cases hx₂p : x₂ = p
· rw [if_neg hx₁p, if_pos hx₂p] at hf
apply Or.elim hx₁
· intro nx
exact absurd hf nx.right
· intro nx
simp only [Set.mem_singleton_iff] at nx
exact absurd nx hx₁p
· rwa [if_neg hx₁p, if_neg hx₂p] at hf
· -- `Set.SurjOn f C' C`
show ∀ x, x ∈ C → x ∈ f '' C'
intro x hx
simp only [
Set.union_singleton,
Set.mem_diff,
Set.mem_singleton_iff,
Set.mem_image,
Set.mem_insert_iff,
exists_eq_or_imp,
ite_true
]
by_cases nx₁ : x = n
· left
exact nx₁.symm
· right
by_cases nx₂ : x = p
· have := hp.left.right
rw [← nx₂] at this
exact absurd hx this
· exact ⟨x, ⟨hx, nx₁⟩, by rwa [if_neg]⟩
/-
> There are two cases to consider:
-/
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
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
/-
> ##### 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
rw [Set.ssubset_def] at hC
have : ¬ ∀ x, x ∈ Set.Iio (n + 1) → x ∈ C := hC.right
simp only [Set.mem_Iio, not_forall, exists_prop] at this
exact this
-- `p` is the least element of `n⁺ - C`.
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}
have hC'₁ : C' ⊆ Set.Iio n := by
show ∀ x, x ∈ C' → x ∈ Set.Iio n
intro x hx
match @trichotomous ℕ LT.lt _ x n with
| Or.inl r => exact r
| Or.inr (Or.inl r) =>
rw [r] at hx
apply Or.elim hx
· intro nx
simp at nx
· intro nx
simp only [Set.mem_singleton_iff] at nx
rw [nx] at hn
exact absurd hn hp.left.right
| Or.inr (Or.inr r) =>
apply Or.elim hx
· intro ⟨h₁, h₂⟩
have h₃ := subset_of_ssubset hC h₁
simp only [Set.mem_singleton_iff, Set.mem_Iio] at h₂ h₃
exact Or.elim (Nat.lt_or_eq_of_lt h₃) id (absurd · h₂)
· intro h
simp only [Set.mem_singleton_iff] at h
have := hp.left.left
rw [← h] at this
exact Or.elim (Nat.lt_or_eq_of_lt this)
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
/-
> 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
⟨m, hm₁, Set.equinumerous_trans (Set.equinumerous_symm this) hm₂⟩
/-
> 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
refine ⟨f, ?_, ?_, ?_⟩
· -- `Set.MapsTo f C' C`
intro x hx
dsimp only
by_cases hxp : x = p
· rw [if_pos hxp]
exact hn
· rw [if_neg hxp]
apply Or.elim hx
· exact fun x => x.left
· intro hx₁
simp only [Set.mem_singleton_iff] at hx₁
exact absurd hx₁ hxp
· -- `Set.InjOn f C'`
intro x₁ hx₁ x₂ hx₂ hf
dsimp only at hf
by_cases hx₁p : x₁ = p
· by_cases hx₂p : x₂ = p
· rw [hx₁p, hx₂p]
· rw [if_pos hx₁p, if_neg hx₂p] at hf
apply Or.elim hx₂
· intro nx
exact absurd hf.symm nx.right
· intro nx
simp only [Set.mem_singleton_iff] at nx
exact absurd nx hx₂p
· by_cases hx₂p : x₂ = p
· rw [if_neg hx₁p, if_pos hx₂p] at hf
apply Or.elim hx₁
· intro nx
exact absurd hf nx.right
· intro nx
simp only [Set.mem_singleton_iff] at nx
exact absurd nx hx₁p
· rwa [if_neg hx₁p, if_neg hx₂p] at hf
· -- `Set.SurjOn f C' C`
show ∀ x, x ∈ C → x ∈ f '' C'
intro x hx
simp only [
Set.union_singleton,
Set.mem_diff,
Set.mem_singleton_iff,
Set.mem_image,
Set.mem_insert_iff,
exists_eq_or_imp,
ite_true
]
by_cases nx₁ : x = n
· left
exact nx₁.symm
· right
by_cases nx₂ : x = p
· have := hp.left.right
rw [← nx₂] at this
exact absurd hx this
· 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`.
-/
/-- #### Corollary 6G
/-- ### Corollary 6G
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
rwa [h]
/-- #### Subset Size
/-- ### Subset Size
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`.
@ -751,7 +882,7 @@ lemma subset_size [DecidableEq α] [Nonempty α] {A B : Set α}
simp at h₁
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
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
/-- #### Set Difference Size
/-- ### Set Difference Size
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`.
@ -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) :=
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**
`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
sorry
/-- #### Exercise 6.8
/-- ### Exercise 6.8
Prove that the union of two finites sets is finite, without any use of
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
sorry
/-- #### Exercise 6.9
/-- ### Exercise 6.9
Prove that the Cartesian product of two finites sets is finite, without any use
of arithmetic.

20
Bookshelf/Enderton/Set/OrderedPair.lean
View File

@ -28,8 +28,7 @@ theorem ext_iff {x y u v : α}
] at hu 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
rw [Set.singleton_eq_singleton_iff] at hu_x
rw [hu_x] at huv_x
@ -41,8 +40,7 @@ theorem ext_iff {x y u v : α}
rw [← hx_v] at 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
rw [Set.singleton_eq_singleton_iff] at hu_x
rw [hu_x] at huv_xy
@ -58,8 +56,7 @@ theorem ext_iff {x y u v : α}
] at this
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
rw [Set.ext_iff] at huv_x hu_xy
have hu_x := huv_x u
@ -71,7 +68,7 @@ theorem ext_iff {x y u v : α}
true_iff,
or_true,
iff_true
] at hu_x hy_u
] at hu_x hy_u
apply Or.elim huv
· intro huv_x
rw [← hu_x, Set.ext_iff] at huv_x
@ -93,19 +90,18 @@ theorem ext_iff {x y u v : α}
· intro hv_y
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
rw [Set.ext_iff] at huv_xy hu_xy
have hx_u := hu_xy x
have hy_u := hu_xy y
simp only [
Set.mem_singleton_iff, Set.mem_insert_iff, true_or, iff_true, or_true
] at hx_u hy_u
] at hx_u hy_u
have := huv_xy v
simp only [
Set.mem_singleton_iff, Set.mem_insert_iff, or_true, true_iff
] at this
] at this
apply Or.elim this
· intro hv_x
rw [hv_x, hx_u]
@ -116,4 +112,4 @@ theorem ext_iff {x y u v : α}
· intro h
rw [h.left, h.right]
end OrderedPair
end OrderedPair

8
Bookshelf/Enderton/Set/Relation.lean
View File

@ -168,7 +168,7 @@ theorem ran_inv_eq_dom_self {F : HRelation α β}
The restriction of a `Relation` to a `Set`.
-/
def restriction (R : HRelation α β) (A : Set α) : HRelation α β :=
{ p ∈ R | p.1 ∈ A }
{ p ∈ R | p.1 ∈ A }
/-! ## Image -/
@ -580,7 +580,7 @@ theorem neighborhood_eq_iff_mem_relate {R : Set.Relation α} {A : Set α}
/--
Assume that `R` is an equivalence relation on `A`. If two sets `x` and `y`
belong to the same neighborhood, then `xRy`.
belong to the same neighborhood, then `xRy`.
-/
theorem neighborhood_mem_imp_relate {R : Set.Relation α} {A : Set α}
(hR : isEquivalence R A)
@ -596,7 +596,7 @@ A **partition** `Π` of a set `A` is a set of nonempty subsets of `A` that is
disjoint and exhaustive.
-/
structure Partition (P : Set (Set α)) (A : Set α) : Prop where
p_subset : ∀ p ∈ P, p ⊆ A
p_subset : ∀ p ∈ P, p ⊆ A
nonempty : ∀ p ∈ P, Set.Nonempty p
disjoint : ∀ a ∈ P, ∀ b, b ∈ P → a ≠ b → a ∩ b = ∅
exhaustive : ∀ a ∈ A, ∃ p, p ∈ P ∧ a ∈ p
@ -666,4 +666,4 @@ theorem modEquiv_partition {A : Set α} {R : Relation α} (hR : isEquivalence R
end Relation
end Set
end Set

76
Bookshelf/Smullyan/Aviary.lean
View File

@ -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.
-/
/-- #### Bald Eagle
/-- ### Bald Eagle
`E'xy₁y₂y₃z₁z₂z₃ = x(y₁y₂y₃)(z₁z₂z₃)`
-/
@ -17,31 +17,31 @@ def E' (x : α → β → γ)
(y₁ : δ → ε → α) (y₂ : δ) (y₃ : ε)
(z₁ : ζ → η → β) (z₂ : ζ) (z₃ : η) := x (y₁ y₂ y₃) (z₁ z₂ z₃)
/-- #### Becard
/-- ### Becard
`B₃xyzw = x(y(zw))`
-/
def B₃ (x : α → ε) (y : β → α) (z : γ → β) (w : γ) := x (y (z w))
/-- #### Blackbird
/-- ### Blackbird
`B₁xyzw = x(yzw)`
-/
def B₁ (x : α → ε) (y : β → γ → α) (z : β) (w : γ) := x (y z w)
/-- #### Bluebird
/-- ### Bluebird
`Bxyz = x(yz)`
-/
def B (x : α → γ) (y : β → α) (z : β) := x (y z)
/-- #### Bunting
/-- ### Bunting
`B₂xyzwv = x(yzwv)`
-/
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`
-/
@ -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
/-- #### Cardinal
/-- ### Cardinal
`Cxyz = xzy`
-/
def C (x : α → β → δ) (y : β) (z : α) := x z y
/-- #### Converse Warbler
/-- ### Converse Warbler
`W'xy = yxx`
-/
def W' (x : α) (y : α → α → β) := y x x
/-- #### Dickcissel
/-- ### Dickcissel
`D₁xyzwv = xyz(wv)`
-/
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)`
-/
/-- #### Dove
/-- ### Dove
`Dxyzw = xy(zw)`
-/
def D (x : α → γ → δ) (y : α) (z : β → γ) (w : β) := x y (z w)
/-- #### Dovekie
/-- ### Dovekie
`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)
/-- #### Eagle
/-- ### Eagle
`Exyzwv = xy(zwv)`
-/
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`
-/
@ -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
/-- #### Finch
/-- ### Finch
`Fxyz = zyx`
-/
def F (x : α) (y : β) (z : β → α → γ) := z y x
/-- #### Goldfinch
/-- ### Goldfinch
`Gxyzw = xw(yz)`
-/
def G (x : α → γ → δ) (y : β → γ) (z : β) (w : α) := x w (y z)
/-- #### Hummingbird
/-- ### Hummingbird
`Hxyz = xyzy`
-/
def H (x : α → β → α → γ) (y : α) (z : β) := x y z y
/-- #### Identity Bird
/-- ### Identity Bird
`Ix = x`
-/
def I (x : α) : α := x
/-- #### Kestrel
/-- ### Kestrel
`Kxy = x`
-/
def K (x : α) (_ : β) := x
/-! #### Lark
/-! ### Lark
`Lxy = x(yy)`
-/
/-! #### Mockingbird
/-! ### Mockingbird
`Mx = xx`
-/
/-- #### Owl
/-- ### Owl
`Oxy = y(xy)`
-/
def O (x : (α → β) → α) (y : α → β) := y (x y)
/-- #### Phoenix
/-- ### Phoenix
`Φxyzw = x(yw)(zw)`
-/
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)`
-/
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)`
-/
def Q₄ (x : α) (y : α → β) (z : β → γ) := z (y x)
/-- #### Queer Bird
/-- ### Queer Bird
`Qxyz = y(xz)`
-/
def Q (x : α → β) (y : β → γ) (z : α) := y (x z)
/-- #### Quirky Bird
/-- ### Quirky Bird
`Q₃xyz = z(xy)`
-/
def Q₃ (x : α → β) (y : α) (z : β → γ) := z (x y)
/-- #### Quixotic Bird
/-- ### Quixotic Bird
`Q₁xyz = x(zy)`
-/
def Q₁ (x : α → γ) (y : β) (z : β → α) := x (z y)
/-- #### Quizzical Bird
/-- ### Quizzical Bird
`Q₂xyz = y(zx)`
-/
def Q₂ (x : α) (y : β → γ) (z : α → β) := y (z x)
/-- #### Robin Once Removed
/-- ### Robin Once Removed
`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
/-- #### Robin
/-- ### Robin
`Rxyz = yzx`
-/
def R (x : α) (y : β → α → γ) (z : β) := y z x
/-- #### Sage Bird
/-- ### Sage Bird
`Θx = x(Θx)`
-/
partial def Θ [Inhabited α] (x : α → α) := x (Θ x)
/-- #### Starling
/-- ### Starling
`Sxyz = xz(yz)`
-/
def S (x : α → β → γ) (y : α → β) (z : α) := x z (y z)
/-- #### Thrush
/-- ### Thrush
`Txy = yx`
-/
def T (x : α) (y : α → β) := y x
/-! #### Turing Bird
/-! ### Turing Bird
`Uxy = y(xxy)`
-/
/-- #### Vireo Once Removed
/-- ### Vireo Once Removed
`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
/-- #### Vireo
/-- ### Vireo
`Vxyz = zxy`
-/
def V (x : α) (y : β) (z : α → β → γ) := z x y
/-- #### Warbler
/-- ### Warbler
`Wxy = xyy`
-/

8
Common/Set/Equinumerous.lean
View File

@ -8,7 +8,7 @@ Additional theorems around finite sets.
namespace Set
/-! ### Definitions -/
/-! ## Definitions -/
/--
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]
exact this
/-! ### Finite Sets -/
/-! ## Finite Sets -/
/--
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}
: Set.Finite S ↔ ∃ n : ℕ, S ≈ Set.Iio n
/-! ### Emptyset -/
/-! ## Emptyset -/
/--
Any set equinumerous to the emptyset is the emptyset.
@ -128,4 +128,4 @@ theorem equinumerous_emptyset_emptyset [Bot β]
· unfold SurjOn
simp
end Set
end Set