bookshelf/Bookshelf/Enderton/Set/Chapter_3.lean

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import Bookshelf.Enderton.Set.Chapter_2
import Bookshelf.Enderton.Set.OrderedPair
import Bookshelf.Enderton.Set.Relation
import Common.Logic.Basic
import Mathlib.Data.Real.Basic
import Mathlib.Data.Rel
import Mathlib.Order.RelClasses
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import Mathlib.Tactic.CasesM
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/-! # Enderton.Set.Chapter_3
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Relations and Functions
-/
namespace Enderton.Set.Chapter_3
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/-- #### Theorem 3B
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If `x ∈ C` and `y ∈ C`, then `⟨x, y⟩ ∈ 𝒫 𝒫 C`.
-/
theorem theorem_3b {C : Set α} (hx : x ∈ C) (hy : y ∈ C)
: OrderedPair x y ∈ 𝒫 𝒫 C := by
have hxs : {x} ⊆ C := Set.singleton_subset_iff.mpr hx
have hxys : {x, y} ⊆ C := Set.mem_mem_imp_pair_subset hx hy
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exact Set.mem_mem_imp_pair_subset hxs hxys
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/-- #### Exercise 3.1
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Suppose that we attempted to generalize the Kuratowski definitions of ordered
pairs to ordered triples by defining
```
⟨x, y, z⟩* = {{x}, {x, y}, {x, y, z}}.open Set
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```
Show that this definition is unsuccessful by giving examples of objects `u`,
`v`, `w`, `x`, `y`, `z` with `⟨x, y, z⟩* = ⟨u, v, w⟩*` but with either `y ≠ v`
or `z ≠ w` (or both).
-/
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theorem exercise_3_1 {x y z u v w : }
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(hx : x = 1) (hy : y = 1) (hz : z = 2)
(hu : u = 1) (hv : v = 2) (hw : w = 2)
: ({{x}, {x, y}, {x, y, z}} : Set (Set )) = {{u}, {u, v}, {u, v, w}}
∧ y ≠ v := by
apply And.intro
· rw [hx, hy, hz, hu, hv, hw]
simp
· rw [hy, hv]
simp only
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/-- #### Exercise 3.2a
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Show that `A × (B C) = (A × B) (A × C)`.
-/
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theorem exercise_3_2a {A : Set α} {B C : Set β}
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: Set.prod A (B C) = (Set.prod A B) (Set.prod A C) := by
calc Set.prod A (B C)
_ = { p | p.1 ∈ A ∧ p.2 ∈ B C } := rfl
_ = { p | p.1 ∈ A ∧ (p.2 ∈ B p.2 ∈ C) } := rfl
_ = { p | (p.1 ∈ A ∧ p.2 ∈ B) (p.1 ∈ A ∧ p.2 ∈ C) } := by
ext x
rw [Set.mem_setOf_eq]
conv => lhs; rw [and_or_left]
_ = { p | p ∈ Set.prod A B (p ∈ Set.prod A C) } := rfl
_ = (Set.prod A B) (Set.prod A C) := rfl
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/-- #### Exercise 3.2b
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Show that if `A × B = A × C` and `A ≠ ∅`, then `B = C`.
-/
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theorem exercise_3_2b {A : Set α} {B C : Set β}
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(h : Set.prod A B = Set.prod A C) (hA : Set.Nonempty A)
: B = C := by
by_cases hB : Set.Nonempty B
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· rw [Set.Subset.antisymm_iff]
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have ⟨a, ha⟩ := hA
apply And.intro
· show ∀ t, t ∈ B → t ∈ C
intro t ht
have : (a, t) ∈ Set.prod A B := ⟨ha, ht⟩
rw [h] at this
exact this.right
· show ∀ t, t ∈ C → t ∈ B
intro t ht
have : (a, t) ∈ Set.prod A C := ⟨ha, ht⟩
rw [← h] at this
exact this.right
· have nB : B = ∅ := Set.not_nonempty_iff_eq_empty.mp hB
rw [nB, Set.prod_right_emptyset_eq_emptyset, Set.ext_iff] at h
rw [nB]
by_contra nC
have ⟨a, ha⟩ := hA
have ⟨c, hc⟩ := Set.nonempty_iff_ne_empty.mpr (Ne.symm nC)
exact (h (a, c)).mpr ⟨ha, hc⟩
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/-- #### Exercise 3.3
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Show that `A × 𝓑 = {A × X | X ∈ 𝓑}`.
-/
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theorem exercise_3_3 {A : Set (Set α)} {𝓑 : Set (Set β)}
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: Set.prod A (⋃₀ 𝓑) = ⋃₀ {Set.prod A X | X ∈ 𝓑} := by
calc Set.prod A (⋃₀ 𝓑)
_ = { p | p.1 ∈ A ∧ p.2 ∈ ⋃₀ 𝓑} := rfl
_ = { p | p.1 ∈ A ∧ ∃ b ∈ 𝓑, p.2 ∈ b } := rfl
_ = { p | ∃ b ∈ 𝓑, p.1 ∈ A ∧ p.2 ∈ b } := by
ext x
rw [Set.mem_setOf_eq]
apply Iff.intro
· intro ⟨h₁, b, h₂⟩
exact ⟨b, h₂.left, h₁, h₂.right⟩
· intro ⟨b, h₁, h₂, h₃⟩
exact ⟨h₂, b, h₁, h₃⟩
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_ = ⋃₀ { Set.prod A p | p ∈ 𝓑 } := by
ext x
rw [Set.mem_setOf_eq]
unfold Set.sUnion sSup Set.instSupSetSet
simp only [Set.mem_setOf_eq, exists_exists_and_eq_and]
apply Iff.intro
· intro ⟨b, h₁, h₂, h₃⟩
exact ⟨b, h₁, h₂, h₃⟩
· intro ⟨b, h₁, h₂, h₃⟩
exact ⟨b, h₁, h₂, h₃⟩
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/-- #### Exercise 3.5a
Assume that `A` and `B` are given sets, and show that there exists a set `C`
such that for any `y`,
```
y ∈ C ↔ y = {x} × B for some x in A.
```
In other words, show that `{{x} × B | x ∈ A}` is a set.
-/
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theorem exercise_3_5a {A : Set α} {B : Set β}
: ∃ C : Set (Set (α × β)),
y ∈ C ↔ ∃ x ∈ A, y = Set.prod {x} B := by
let C := {y ∈ 𝒫 (Set.prod A B) | ∃ a ∈ A, ∀ x, (x ∈ y ↔ ∃ b ∈ B, x = (a, b))}
refine ⟨C, ?_⟩
apply Iff.intro
· intro hC
simp only [Set.mem_setOf_eq] at hC
have ⟨_, ⟨a, ⟨ha, h⟩⟩⟩ := hC
refine ⟨a, ⟨ha, ?_⟩⟩
ext x
apply Iff.intro
· intro hxy
unfold Set.prod
simp only [Set.mem_singleton_iff, Set.mem_setOf_eq]
have ⟨b, ⟨hb, hx⟩⟩ := (h x).mp hxy
rw [Prod.ext_iff] at hx
simp only at hx
rw [← hx.right] at hb
exact ⟨hx.left, hb⟩
· intro hx
simp only [Set.mem_singleton_iff, Set.mem_setOf_eq] at hx
have := (h (a, x.snd)).mpr ⟨x.snd, ⟨hx.right, rfl⟩⟩
have hxab : x = (a, x.snd) := by
ext <;> simp
exact hx.left
rwa [← hxab] at this
· intro ⟨x, ⟨hx, hy⟩⟩
show y ∈ 𝒫 Set.prod A B ∧ ∃ a, a ∈ A ∧
∀ (x : α × β), x ∈ y ↔ ∃ b, b ∈ B ∧ x = (a, b)
apply And.intro
· simp only [Set.mem_powerset_iff]
rw [hy]
unfold Set.prod
simp only [
Set.mem_singleton_iff,
Set.setOf_subset_setOf,
and_imp,
Prod.forall
]
intro a b ha hb
exact ⟨by rw [ha]; exact hx, hb⟩
· refine ⟨x, ⟨hx, ?_⟩⟩
intro p
apply Iff.intro
· intro hab
rw [hy] at hab
unfold Set.prod at hab
simp only [Set.mem_singleton_iff, Set.mem_setOf_eq] at hab
exact ⟨p.2, ⟨hab.right, by ext; exact hab.left; simp⟩⟩
· intro ⟨b, ⟨hb, hab⟩⟩
rw [hy]
unfold Set.prod
simp only [Set.mem_singleton_iff, Set.mem_setOf_eq]
rw [Prod.ext_iff] at hab
simp only at hab
rw [hab.right]
exact ⟨hab.left, hb⟩
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/-- #### Exercise 3.5b
With `A`, `B`, and `C` as above, show that `A × B = C`.
-/
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theorem exercise_3_5b {A : Set α} (B : Set β)
: Set.prod A B = ⋃₀ {Set.prod ({x} : Set α) B | x ∈ A} := by
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rw [Set.Subset.antisymm_iff]
apply And.intro
· show ∀ t, t ∈ Set.prod A B → t ∈ ⋃₀ {Set.prod {x} B | x ∈ A}
intro t h
simp only [Set.mem_setOf_eq] at h
unfold Set.sUnion sSup Set.instSupSetSet
simp only [Set.mem_setOf_eq, exists_exists_and_eq_and]
unfold Set.prod at h
simp only [Set.mem_setOf_eq] at h
refine ⟨t.fst, ⟨h.left, ?_⟩⟩
unfold Set.prod
simp only [Set.mem_singleton_iff, Set.mem_setOf_eq, true_and]
exact h.right
· show ∀ t, t ∈ ⋃₀ {Set.prod {x} B | x ∈ A} → t ∈ Set.prod A B
unfold Set.prod
intro t ht
simp only [
Set.mem_singleton_iff,
Set.mem_sUnion,
Set.mem_setOf_eq,
exists_exists_and_eq_and
] at ht
have ⟨a, ⟨h, ⟨ha, hb⟩⟩⟩ := ht
simp only [Set.mem_setOf_eq]
rw [← ha] at h
exact ⟨h, hb⟩
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/-- #### Theorem 3D
If `⟨x, y⟩ ∈ A`, then `x` and `y` belong to ` A`.
-/
theorem theorem_3d {A : Set (Set (Set α))} (h : OrderedPair x y ∈ A)
: x ∈ ⋃₀ (⋃₀ A) ∧ y ∈ ⋃₀ (⋃₀ A) := by
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have hp := Chapter_2.exercise_2_3 (OrderedPair x y) h
unfold OrderedPair at hp
have hq : {x, y} ∈ ⋃₀ A := hp (by simp)
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have : {x, y} ⊆ ⋃₀ ⋃₀ A := Chapter_2.exercise_2_3 {x, y} hq
exact ⟨this (by simp), this (by simp)⟩
section Relation
open Set.Relation
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/-- #### Exercise 3.6
Show that a set `A` is a relation **iff** `A ⊆ dom A × ran A`.
-/
theorem exercise_3_6 {A : Set.HRelation α β}
: A ⊆ Set.prod (dom A) (ran A) := by
show ∀ t, t ∈ A → t ∈ Set.prod (Prod.fst '' A) (Prod.snd '' A)
intro (a, b) ht
unfold Set.prod
simp only [
Set.mem_image,
Prod.exists,
exists_and_right,
exists_eq_right,
Set.mem_setOf_eq
]
exact ⟨⟨b, ht⟩, ⟨a, ht⟩⟩
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/-- #### Exercise 3.7
Show that if `R` is a relation, then `fld R = R`.
-/
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theorem exercise_3_7 {R : Set.Relation α}
: R.fld = ⋃₀ ⋃₀ R.toOrderedPairs := by
let img := R.toOrderedPairs
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rw [Set.Subset.antisymm_iff]
apply And.intro
· show ∀ x, x ∈ R.fld → x ∈ ⋃₀ ⋃₀ img
intro x hx
apply Or.elim hx
· intro hd
unfold Set.Relation.dom Prod.fst at hd
simp only [
Set.mem_image, Prod.exists, exists_and_right, exists_eq_right
] at hd
have ⟨y, hp⟩ := hd
have hm : OrderedPair x y ∈ Set.image (fun p => OrderedPair p.1 p.2) R := by
unfold Set.image
simp only [Prod.exists, Set.mem_setOf_eq]
exact ⟨x, ⟨y, ⟨hp, rfl⟩⟩⟩
unfold OrderedPair at hm
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have : {x} ∈ ⋃₀ img := Chapter_2.exercise_2_3 {{x}, {x, y}} hm (by simp)
exact (Chapter_2.exercise_2_3 {x} this) (show x ∈ {x} by rfl)
· intro hr
unfold Set.Relation.ran Prod.snd at hr
simp only [Set.mem_image, Prod.exists, exists_eq_right] at hr
have ⟨t, ht⟩ := hr
have hm : OrderedPair t x ∈ Set.image (fun p => OrderedPair p.1 p.2) R := by
simp only [Set.mem_image, Prod.exists]
exact ⟨t, ⟨x, ⟨ht, rfl⟩⟩⟩
unfold OrderedPair at hm
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have : {t, x} ∈ ⋃₀ img := Chapter_2.exercise_2_3 {{t}, {t, x}} hm
(show {t, x} ∈ {{t}, {t, x}} by simp)
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exact Chapter_2.exercise_2_3 {t, x} this (show x ∈ {t, x} by simp)
· show ∀ t, t ∈ ⋃₀ ⋃₀ img → t ∈ Set.Relation.fld R
intro t ht
have ⟨T, hT⟩ : ∃ T ∈ ⋃₀ img, t ∈ T := ht
have ⟨T', hT'⟩ : ∃ T' ∈ img, T ∈ T' := hT.left
dsimp only at hT'
unfold Set.Relation.toOrderedPairs at hT'
simp only [Set.mem_image, Prod.exists] at hT'
have ⟨x, ⟨y, ⟨p, hp⟩⟩⟩ := hT'.left
have hr := hT'.right
rw [← hp] at hr
unfold OrderedPair at hr
simp only [Set.mem_singleton_iff, Set.mem_insert_iff] at hr
-- Use `exercise_6_6` to prove that if `t = x` then `t ∈ dom R` and if
-- `t = y` then `t ∈ ran R`.
have hxy_mem : t = x t = y → t ∈ Set.Relation.fld R := by
intro ht
have hz : R ⊆ Set.prod (dom R) (ran R) := exercise_3_6
have : (x, y) ∈ Set.prod (dom R) (ran R) := hz p
unfold Set.prod at this
simp at this
apply Or.elim ht
· intro ht'
rw [← ht'] at this
exact Or.inl this.left
· intro ht'
rw [← ht'] at this
exact Or.inr this.right
-- Eliminate `T = {x} T = {x, y}`.
apply Or.elim hr
· intro hx
have := hT.right
rw [hx] at this
simp only [Set.mem_singleton_iff] at this
exact hxy_mem (Or.inl this)
· intro hxy
have := hT.right
rw [hxy] at this
simp only [Set.mem_singleton_iff, Set.mem_insert_iff] at this
exact hxy_mem this
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/-- #### Exercise 3.8 (i)
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Show that for any set `𝓐`:
```
dom A = { dom R | R ∈ 𝓐 }
```
-/
theorem exercise_3_8_i {A : Set (Set.HRelation α β)}
: dom (⋃₀ A) = ⋃₀ { dom R | R ∈ A } := by
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ext x
unfold dom Prod.fst
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simp only [
Set.mem_image,
Set.mem_sUnion,
Prod.exists,
exists_and_right,
exists_eq_right,
Set.mem_setOf_eq,
exists_exists_and_eq_and
]
apply Iff.intro
· intro ⟨y, t, ht, hx⟩
exact ⟨t, ht, y, hx⟩
· intro ⟨t, ht, y, hx⟩
exact ⟨y, t, ht, hx⟩
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/-- #### Exercise 3.8 (ii)
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Show that for any set `𝓐`:
```
ran A = { ran R | R ∈ 𝓐 }
```
-/
theorem exercise_3_8_ii {A : Set (Set.HRelation α β)}
: ran (⋃₀ A) = ⋃₀ { ran R | R ∈ A } := by
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ext x
unfold ran Prod.snd
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simp only [
Set.mem_image,
Set.mem_sUnion,
Prod.exists,
exists_eq_right,
Set.mem_setOf_eq,
exists_exists_and_eq_and
]
apply Iff.intro
· intro ⟨t, ⟨y, ⟨hy, ht⟩⟩⟩
exact ⟨y, ⟨hy, ⟨t, ht⟩⟩⟩
· intro ⟨y, ⟨hy, ⟨t, ht⟩⟩⟩
exact ⟨t, ⟨y, ⟨hy, ht⟩⟩⟩
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/-- #### Exercise 3.9 (i)
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Discuss the result of replacing the union operation by the intersection
operation in the preceding problem.
```
dom A = { dom R | R ∈ 𝓐 }
```
-/
theorem exercise_3_9_i {A : Set (Set.HRelation α β)}
: dom (⋂₀ A) ⊆ ⋂₀ { dom R | R ∈ A } := by
show ∀ x, x ∈ dom (⋂₀ A) → x ∈ ⋂₀ { dom R | R ∈ A }
unfold dom Prod.fst
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simp only [
Set.mem_image,
Set.mem_sInter,
Prod.exists,
exists_and_right,
exists_eq_right,
Set.mem_setOf_eq,
forall_exists_index,
and_imp,
forall_apply_eq_imp_iff₂
]
intro _ y hy R hR
exact ⟨y, hy R hR⟩
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/-- #### Exercise 3.9 (ii)
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Discuss the result of replacing the union operation by the intersection
operation in the preceding problem.
```
ran A = { ran R | R ∈ 𝓐 }
```
-/
theorem exercise_3_9_ii {A : Set (Set.HRelation α β)}
: ran (⋂₀ A) ⊆ ⋂₀ { ran R | R ∈ A } := by
show ∀ x, x ∈ ran (⋂₀ A) → x ∈ ⋂₀ { ran R | R ∈ A }
unfold ran Prod.snd
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simp only [
Set.mem_image,
Set.mem_sInter,
Prod.exists,
exists_and_right,
exists_eq_right,
Set.mem_setOf_eq,
forall_exists_index,
and_imp,
forall_apply_eq_imp_iff₂
]
intro _ y hy R hR
exact ⟨y, hy R hR⟩
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/-- #### Theorem 3G (i)
Assume that `F` is a one-to-one function. If `x ∈ dom F`, then `F⁻¹(F(x)) = x`.
-/
theorem theorem_3g_i {F : Set.HRelation α β}
(hF : isOneToOne F) (hx : x ∈ dom F)
: ∃! y, (x, y) ∈ F ∧ (y, x) ∈ inv F := by
simp only [mem_self_comm_mem_inv, and_self]
have ⟨y, hy⟩ := dom_exists hx
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refine ⟨y, hy, ?_⟩
intro y₁ hy₁
unfold isOneToOne at hF
exact (single_valued_eq_unique hF.left hy hy₁).symm
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/-- #### Theorem 3G (ii)
Assume that `F` is a one-to-one function. If `y ∈ ran F`, then `F(F⁻¹(y)) = y`.
-/
theorem theorem_3g_ii {F : Set.HRelation α β}
(hF : isOneToOne F) (hy : y ∈ ran F)
: ∃! x, (x, y) ∈ F ∧ (y, x) ∈ inv F := by
simp only [mem_self_comm_mem_inv, and_self]
have ⟨x, hx⟩ := ran_exists hy
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refine ⟨x, hx, ?_⟩
intro x₁ hx₁
unfold isOneToOne at hF
exact (single_rooted_eq_unique hF.right hx hx₁).symm
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/-- #### Theorem 3H
Assume that `F` and `G` are functions. Then
```
dom (F ∘ G) = {x ∈ dom G | G(x) ∈ dom F}.
```
-/
theorem theorem_3h_dom {F : Set.HRelation β γ} {G : Set.HRelation α β}
(_ : isSingleValued F) (hG : isSingleValued G)
: dom (comp F G) = {x ∈ dom G | ∃! y, (x, y) ∈ G ∧ y ∈ dom F} := by
let rhs := {x ∈ dom G | ∃! y, (x, y) ∈ G ∧ y ∈ dom F }
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rw [Set.Subset.antisymm_iff]
apply And.intro
· show ∀ t, t ∈ dom (comp F G) → t ∈ rhs
intro t ht
simp only [Set.mem_setOf_eq]
have ⟨z, hz⟩ := dom_exists ht
refine ⟨dom_comp_imp_dom_self ht, ?_⟩
simp only [Set.mem_setOf_eq] at hz
have ⟨a, ha⟩ := hz
unfold dom
simp only [Set.mem_image, Prod.exists, exists_and_right, exists_eq_right]
unfold ExistsUnique
simp only [and_imp, forall_exists_index]
refine ⟨a, ⟨ha.left, z, ha.right⟩, ?_⟩
intro y₁ hy₁
exact fun _ _ => single_valued_eq_unique hG hy₁ ha.left
· show ∀ t, t ∈ rhs → t ∈ dom (comp F G)
intro t ht
simp only [Set.mem_setOf_eq] at ht
unfold dom
simp only [Set.mem_image, Prod.exists, exists_and_right, exists_eq_right]
have ⟨a, ha⟩ := ht.right
simp at ha
have ⟨b, hb⟩ := dom_exists ha.left.right
refine ⟨b, ?_⟩
unfold comp
simp only [Set.mem_setOf_eq]
exact ⟨a, ha.left.left, hb⟩
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/-- #### Theorem 3J (a)
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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`
**iff** `F` is one-to-one.
-/
theorem theorem_3j_a {F : Set.HRelation α β} {A : Set α} {B : Set β}
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(hF : mapsInto F A B) (hA : Set.Nonempty A)
: (∃ G : Set.HRelation β α,
mapsInto G B A ∧
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(comp G F = { p | p.1 ∈ A ∧ p.1 = p.2 })) ↔ isOneToOne F := by
apply Iff.intro
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· intro ⟨G, hG⟩
refine ⟨hF.is_func, ?_⟩
intro y hy
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have ⟨x₁, hx₁⟩ := ran_exists hy
refine ⟨x₁, ⟨mem_pair_imp_fst_mem_dom hx₁, hx₁⟩, ?_⟩
intro x₂ hx₂
have hG' : y ∈ dom G := by
rw [hG.left.dom_eq]
exact hF.ran_ss hy
have ⟨z, hz⟩ := dom_exists hG'
have := hG.right
unfold comp at this
rw [Set.ext_iff] at this
have h₁ := (this (x₁, z)).mp ⟨y, hx₁, hz⟩
have h₂ := (this (x₂, z)).mp ⟨y, hx₂.right, hz⟩
simp only [Set.mem_setOf_eq] at h₁ h₂
rw [h₁.right, h₂.right]
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· sorry
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/-- #### Theorem 3J (b)
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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
`B` **iff** `F` maps `A` onto `B`.
-/
theorem theorem_3j_b {F : Set.HRelation α β} {A : Set α} {B : Set β}
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(hF : mapsInto F A B) (hA : Set.Nonempty A)
: (∃ H : Set.HRelation β α,
mapsInto H B A ∧
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(comp F H = { p | p.1 ∈ B ∧ p.1 = p.2 })) ↔ mapsOnto F A B := by
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sorry
/-- #### 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:
```
F⟦ 𝓐⟧ = {F⟦A⟧ | A ∈ 𝓐}
```
-/
theorem theorem_3k_a {F : Set.HRelation α β} {𝓐 : Set (Set α)}
: image F (⋃₀ 𝓐) = ⋃₀ { image F A | A ∈ 𝓐 } := by
rw [Set.Subset.antisymm_iff]
apply And.intro
· show ∀ v, v ∈ image F (⋃₀ 𝓐) → v ∈ ⋃₀ { image F A | A ∈ 𝓐 }
intro v hv
unfold image at hv
simp only [Set.mem_sUnion, Set.mem_setOf_eq] at hv
have ⟨u, hu⟩ := hv
have ⟨A, hA⟩ := hu.left
simp only [Set.mem_sUnion, Set.mem_setOf_eq, exists_exists_and_eq_and]
refine ⟨A, hA.left, ?_⟩
show v ∈ image F A
unfold image
simp only [Set.mem_setOf_eq]
exact ⟨u, hA.right, hu.right⟩
· show ∀ v, v ∈ ⋃₀ {x | ∃ A, A ∈ 𝓐 ∧ image F A = x} → v ∈ image F (⋃₀ 𝓐)
intro v hv
simp only [Set.mem_sUnion, Set.mem_setOf_eq, exists_exists_and_eq_and] at hv
have ⟨A, hA⟩ := hv
unfold image at hA
simp only [Set.mem_setOf_eq] at hA
have ⟨u, hu⟩ := hA.right
unfold image
simp only [Set.mem_sUnion, Set.mem_setOf_eq]
exact ⟨u, ⟨A, hA.left, hu.left⟩, hu.right⟩
/-! #### 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:
```
F⟦⋂ 𝓐⟧ ⊆ ⋂ {F⟦A⟧ | A ∈ 𝓐}
```
Equality holds if `F` is single-rooted.
-/
theorem theorem_3k_b_i {F : Set.HRelation α β} {𝓐 : Set (Set α)}
: image F (⋂₀ 𝓐) ⊆ ⋂₀ { image F A | A ∈ 𝓐} := by
show ∀ v, v ∈ image F (⋂₀ 𝓐) → v ∈ ⋂₀ { image F A | A ∈ 𝓐}
intro v hv
unfold image at hv
simp only [Set.mem_sInter, Set.mem_setOf_eq] at hv
have ⟨u, hu⟩ := hv
simp only [
Set.mem_sInter,
Set.mem_setOf_eq,
forall_exists_index,
and_imp,
forall_apply_eq_imp_iff₂
]
intro A hA
unfold image
simp only [Set.mem_setOf_eq]
exact ⟨u, hu.left A hA, hu.right⟩
theorem theorem_3k_b_ii {F : Set.HRelation α β} {𝓐 : Set (Set α)}
(hF : isSingleRooted F) (h𝓐 : Set.Nonempty 𝓐)
: image F (⋂₀ 𝓐) = ⋂₀ { image F A | A ∈ 𝓐} := by
rw [Set.Subset.antisymm_iff]
refine ⟨theorem_3k_b_i, ?_⟩
show ∀ v, v ∈ ⋂₀ {x | ∃ A, A ∈ 𝓐 ∧ image F A = x} → v ∈ image F (⋂₀ 𝓐)
intro v hv
simp only [
Set.mem_sInter,
Set.mem_setOf_eq,
forall_exists_index,
and_imp,
forall_apply_eq_imp_iff₂
] at hv
unfold image at hv
simp only [Set.mem_setOf_eq] at hv
have ⟨u, hu⟩ : ∃ u, (∀ (a : Set α), a ∈ 𝓐 → u ∈ a) ∧ (u, v) ∈ F := by
have ⟨A, hA⟩ := h𝓐
have ⟨_, ⟨_, hv'⟩⟩ := hv A hA
have ⟨u, hu⟩ := hF v (mem_pair_imp_snd_mem_ran hv')
simp only [and_imp] at hu
refine ⟨u, ?_, hu.left.right⟩
intro a ha
have ⟨u₁, hu₁⟩ := hv a ha
have := hu.right u₁ (mem_pair_imp_fst_mem_dom hu₁.right) hu₁.right
rw [← this]
exact hu₁.left
unfold image
simp only [Set.mem_sInter, Set.mem_setOf_eq]
exact ⟨u, hu⟩
/-! #### 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:
```
F⟦A⟧ - F⟦B⟧ ⊆ F⟦A - B⟧.
```
Equality holds if `F` is single-rooted.
-/
theorem theorem_3k_c_i {F : Set.HRelation α β} {A B : Set α}
: image F A \ image F B ⊆ image F (A \ B) := by
show ∀ v, v ∈ image F A \ image F B → v ∈ image F (A \ B)
intro v hv
have hv' : v ∈ image F A ∧ v ∉ image F B := hv
conv at hv' => arg 1; unfold image; simp only [Set.mem_setOf_eq, eq_iff_iff]
have ⟨u, hu⟩ := hv'.left
have hw : ∀ w ∈ B, (w, v) ∉ F := by
intro w hw nw
have nv := hv'.right
unfold image at nv
simp only [Set.mem_setOf_eq, not_exists, not_and] at nv
exact absurd nw (nv w hw)
have hu' : u ∉ B := by
by_contra nu
exact absurd hu.right (hw u nu)
unfold image
simp only [Set.mem_diff, Set.mem_setOf_eq]
exact ⟨u, ⟨hu.left, hu'⟩, hu.right⟩
theorem theorem_3k_c_ii {F : Set.HRelation α β} {A B : Set α}
(hF : isSingleRooted F)
: image F A \ image F B = image F (A \ B) := by
rw [Set.Subset.antisymm_iff]
refine ⟨theorem_3k_c_i, ?_⟩
show ∀ v, v ∈ image F (A \ B) → v ∈ image F A \ image F B
intro v hv
unfold image at hv
simp only [Set.mem_diff, Set.mem_setOf_eq] at hv
have ⟨u, hu⟩ := hv
have hv₁ : v ∈ image F A := by
unfold image
simp only [Set.mem_setOf_eq]
exact ⟨u, hu.left.left, hu.right⟩
have hv₂ : v ∉ image F B := by
intro nv
unfold image at nv
simp only [Set.mem_setOf_eq] at nv
have ⟨u₁, hu₁⟩ := nv
have := single_rooted_eq_unique hF hu.right hu₁.right
rw [← this] at hu₁
exact absurd hu₁.left hu.left.right
exact ⟨hv₁, hv₂⟩
/-! #### Corollary 3L
For any function `G` and sets `A`, `B`, and `𝓐`:
```
G⁻¹⟦ 𝓐⟧ = {G⁻¹⟦A⟧ | A ∈ 𝓐},
G⁻¹⟦𝓐⟧ = ⋂ {G⁻¹⟦A⟧ | A ∈ 𝓐} for 𝓐 ≠ ∅,
G⁻¹⟦A - B⟧ = G⁻¹⟦A⟧ - G⁻¹⟦B⟧.
```
-/
theorem corollary_3l_i {G : Set.HRelation β α} {𝓐 : Set (Set α)}
: image (inv G) (⋃₀ 𝓐) = ⋃₀ {image (inv G) A | A ∈ 𝓐} := theorem_3k_a
theorem corollary_3l_ii {G : Set.HRelation β α} {𝓐 : Set (Set α)}
(hG : isSingleValued G) (h𝓐 : Set.Nonempty 𝓐)
: image (inv G) (⋂₀ 𝓐) = ⋂₀ {image (inv G) A | A ∈ 𝓐} := by
have hG' : isSingleRooted (inv G) :=
single_valued_self_iff_single_rooted_inv.mp hG
exact theorem_3k_b_ii hG' h𝓐
theorem corollary_3l_iii {G : Set.HRelation β α} {A B : Set α}
(hG : isSingleValued G)
: image (inv G) (A \ B) = image (inv G) A \ image (inv G) B := by
have hG' : isSingleRooted (inv G) :=
single_valued_self_iff_single_rooted_inv.mp hG
exact (theorem_3k_c_ii hG').symm
/-- #### Exercise 3.12
Assume that `f` and `g` are functions and show that
```
f ⊆ g ↔ dom f ⊆ dom g ∧ (∀ x ∈ dom f) f(x) = g(x).
```
-/
theorem exercise_3_12 {f g : Set.HRelation α β}
(hf : isSingleValued f) (_ : isSingleValued g)
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: f ⊆ g ↔ dom f ⊆ dom g ∧
(∀ x ∈ dom f, ∃! y : β, (x, y) ∈ f ∧ (x, y) ∈ g) := by
apply Iff.intro
· intro h
apply And.intro
· show ∀ x, x ∈ dom f → x ∈ dom g
intro x hx
have ⟨y, hy⟩ := dom_exists hx
exact mem_pair_imp_fst_mem_dom (h hy)
· intro x hx
have ⟨y, hy⟩ := dom_exists hx
refine ⟨y, ⟨hy, h hy⟩, ?_⟩
intro y₁ hy₁
exact single_valued_eq_unique hf hy₁.left hy
· intro ⟨_, hx⟩
show ∀ p, p ∈ f → p ∈ g
intro (x, y) hp
have ⟨y₁, hy₁⟩ := hx x (mem_pair_imp_fst_mem_dom hp)
rw [single_valued_eq_unique hf hp hy₁.left.left]
exact hy₁.left.right
/-- #### Exercise 3.13
Assume that `f` and `g` are functions with `f ⊆ g` and `dom g ⊆ dom f`. Show
that `f = g`.
-/
theorem exercise_3_13 {f g : Set.HRelation α β}
(hf : isSingleValued f) (hg : isSingleValued g)
(h : f ⊆ g) (h₁ : dom g ⊆ dom f)
: f = g := by
have h₂ := (exercise_3_12 hf hg).mp h
have hfg := Set.Subset.antisymm_iff.mpr ⟨h₁, h₂.left⟩
ext p
have (a, b) := p
apply Iff.intro
· intro hp
have ⟨x, hx⟩ := h₂.right a (mem_pair_imp_fst_mem_dom hp)
rw [single_valued_eq_unique hf hp hx.left.left]
exact hx.left.right
· intro hp
rw [← hfg] at h₂
have ⟨x, hx⟩ := h₂.right a (mem_pair_imp_fst_mem_dom hp)
rw [single_valued_eq_unique hg hp hx.left.right]
exact hx.left.left
/-- #### Exercise 3.14 (a)
Assume that `f` and `g` are functions. Show that `f ∩ g` is a function.
-/
theorem exercise_3_14_a {f g : Set.HRelation α β}
(hf : isSingleValued f) (_ : isSingleValued g)
: isSingleValued (f ∩ g) :=
single_valued_subset hf (Set.inter_subset_left f g)
/-- #### 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)`.
-/
theorem exercise_3_14_b {f g : Set.HRelation α β}
(hf : isSingleValued f) (hg : isSingleValued g)
: isSingleValued (f g) ↔
(∀ x ∈ dom f ∩ dom g, ∃! y, (x, y) ∈ f ∧ (x, y) ∈ g) := by
apply Iff.intro
· intro h x hx
have ⟨y₁, hy₁⟩ := hf x hx.left
have ⟨y₂, hy₂⟩ := hg x hx.right
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have : y₁ = y₂ := single_valued_eq_unique h
(Or.inl hy₁.left.right)
(Or.inr hy₂.left.right)
rw [← this] at hy₂
refine ⟨y₁, ⟨hy₁.left.right, hy₂.left.right⟩, ?_⟩
intro y₃ hfy₃
exact single_valued_eq_unique hf hfy₃.left hy₁.left.right
· intro h x hx
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by_cases hfx : x ∈ dom f <;>
by_cases hgx : x ∈ dom g
· -- `x ∈ dom f ∧ x ∈ dom g`
have ⟨y₁, hy₁⟩ := hf x hfx
have ⟨y₂, hy₂⟩ := hg x hgx
refine ⟨y₁, ⟨?_, Or.inl hy₁.left.right⟩, ?_⟩
· unfold ran
simp only [Set.mem_image, Set.mem_union, Prod.exists, exists_eq_right]
exact ⟨x, Or.inl hy₁.left.right⟩
· intro y₃ hy₃
apply Or.elim hy₃.right
· intro hxy
exact single_valued_eq_unique hf hxy hy₁.left.right
· refine fun hxy => single_valued_eq_unique hg hxy ?_
have : y₁ = y₂ := by
have ⟨z, ⟨hz, _⟩⟩ := h x ⟨hfx, hgx⟩
rw [
single_valued_eq_unique hf hy₁.left.right hz.left,
single_valued_eq_unique hg hy₂.left.right hz.right
]
rw [this]
exact hy₂.left.right
· -- `x ∈ dom f ∧ x ∉ dom g`
have ⟨y, hy⟩ := dom_exists hfx
have hxy : (x, y) ∈ f g := (Set.subset_union_left f g) hy
refine ⟨y, ⟨mem_pair_imp_snd_mem_ran hxy, hxy⟩, ?_⟩
intro y₁ hy₁
apply Or.elim hy₁.right
· intro hx'
exact single_valued_eq_unique hf hx' hy
· intro hx'
exact absurd (mem_pair_imp_fst_mem_dom hx') hgx
· -- `x ∉ dom f ∧ x ∈ dom g`
have ⟨y, hy⟩ := dom_exists hgx
have hxy : (x, y) ∈ f g := (Set.subset_union_right f g) hy
refine ⟨y, ⟨mem_pair_imp_snd_mem_ran hxy, hxy⟩, ?_⟩
intro y₁ hy₁
apply Or.elim hy₁.right
· intro hx'
exact absurd (mem_pair_imp_fst_mem_dom hx') hfx
· intro hx'
exact single_valued_eq_unique hg hx' hy
· -- `x ∉ dom f ∧ x ∉ dom g`
exfalso
unfold dom at hx
simp only [
Set.mem_image,
Set.mem_union,
Prod.exists,
exists_and_right,
exists_eq_right
] at hx
have ⟨_, hy⟩ := hx
apply Or.elim hy
· intro hz
exact absurd (mem_pair_imp_fst_mem_dom hz) hfx
· intro hz
exact absurd (mem_pair_imp_fst_mem_dom hz) hgx
/-- #### 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.
-/
theorem exercise_3_15 {𝓐 : Set (Set.HRelation α β)}
(h𝓐 : ∀ F ∈ 𝓐, isSingleValued F)
(h : ∀ F, ∀ G, F ∈ 𝓐 → G ∈ 𝓐 → F ⊆ G G ⊆ F)
: isSingleValued (⋃₀ 𝓐) := by
intro x hx
have ⟨y₁, hy₁⟩ := dom_exists hx
refine ⟨y₁, ⟨mem_pair_imp_snd_mem_ran hy₁, hy₁⟩, ?_⟩
intro y₂ hy₂
have ⟨f, hf⟩ : ∃ f : Set.HRelation α β, f ∈ 𝓐 ∧ (x, y₁) ∈ f := hy₁
have ⟨g, hg⟩ : ∃ g : Set.HRelation α β, g ∈ 𝓐 ∧ (x, y₂) ∈ g := hy₂.right
apply Or.elim (h f g hf.left hg.left)
· intro hf'
have := hf' hf.right
exact single_valued_eq_unique (h𝓐 g hg.left) hg.right this
· intro hg'
have := hg' hg.right
exact single_valued_eq_unique (h𝓐 f hf.left) this hf.right
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/-! #### 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.
-/
theorem exercise_3_17_i {F : Set.HRelation β γ} {G : Set.HRelation α β}
(hF : isSingleRooted F) (hG : isSingleRooted G)
: isSingleRooted (comp F G):= by
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intro v hv
have ⟨u₁, hu₁⟩ := ran_exists hv
have hu₁' := hu₁
unfold comp at hu₁'
simp only [Set.mem_setOf_eq] at hu₁'
have ⟨t₁, ht₁⟩ := hu₁'
unfold ExistsUnique
refine ⟨u₁, ⟨mem_pair_imp_fst_mem_dom hu₁, hu₁⟩, ?_⟩
intro u₂ hu₂
have hu₂' := hu₂
unfold comp at hu₂'
simp only [Set.mem_setOf_eq] at hu₂'
have ⟨_, ⟨t₂, ht₂⟩⟩ := hu₂'
have ht : t₁ = t₂ := single_rooted_eq_unique hF ht₁.right ht₂.right
rw [ht] at ht₁
exact single_rooted_eq_unique hG ht₂.left ht₁.left
theorem exercise_3_17_ii {F : Set.HRelation β γ} {G : Set.HRelation α β}
(hF : isOneToOne F) (hG : isOneToOne G)
: isOneToOne (comp F G) := And.intro
(single_valued_comp_is_single_valued hF.left hG.left)
(exercise_3_17_i hF.right hG.right)
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/-! #### Exercise 3.18
Let `R` be the set
```
{⟨0, 1⟩, ⟨0, 2⟩, ⟨0, 3⟩, ⟨1, 2⟩, ⟨1, 3⟩, ⟨2, 3⟩}
```
Evaluate the following: `R ∘ R`, `R ↾ {1}`, `R⁻¹ ↾ {1}`, `R⟦{1}⟧`, and
`R⁻¹⟦{1}⟧`.
-/
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section Exercise_3_18
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variable {R : Set.Relation }
variable (hR : R = {(0, 1), (0, 2), (0, 3), (1, 2), (1, 3), (2, 3)})
theorem exercise_3_18_i
: comp R R = {(0, 2), (0, 3), (1, 3)} := by
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rw [hR]
unfold comp
simp only [Set.mem_singleton_iff, Set.mem_insert_iff, or_self, Prod.mk.injEq]
ext x
have (a, b) := x
apply Iff.intro
· simp only [Set.mem_setOf_eq, Set.mem_singleton_iff, Set.mem_insert_iff]
intro ⟨t, ht₁, ht₂⟩
casesm* _ _
all_goals case _ hl hr => first
| {rw [hl.right] at hr; simp at hr}
| {rw [hl.left] at hr; simp at hr}
| {rw [hl.left, hr.right]; simp}
· simp only [
Set.mem_singleton_iff,
Set.mem_insert_iff,
Prod.mk.injEq,
Set.mem_setOf_eq
]
intro h
casesm* _ _
· case _ h =>
refine ⟨1, Or.inl ⟨h.left, rfl⟩, ?_⟩
iterate 3 right
left
exact ⟨rfl, h.right⟩
· case _ h =>
refine ⟨1, Or.inl ⟨h.left, rfl⟩, ?_⟩
iterate 4 right
left
exact ⟨rfl, h.right⟩
· case _ h =>
refine ⟨2, ?_, ?_⟩
· iterate 3 right
left
exact ⟨h.left, rfl⟩
· iterate 5 right
exact ⟨rfl, h.right⟩
theorem exercise_3_18_ii
: restriction R {1} = {(1, 2), (1, 3)} := by
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rw [hR]
unfold restriction
ext p
have (a, b) := p
simp only [
Set.mem_singleton_iff,
Set.mem_insert_iff,
Set.mem_setOf_eq,
or_self
]
apply Iff.intro
· intro ⟨hp, ha⟩
rw [ha]
simp only [Prod.mk.injEq, true_and]
casesm* _ _
all_goals case _ h => first
| {rw [ha] at h; simp at h}
| {simp only [Prod.mk.injEq] at h; left; exact h.right}
| {simp only [Prod.mk.injEq] at h; right; exact h.right}
· intro h
apply Or.elim h
· intro hab
simp only [Prod.mk.injEq] at hab
refine ⟨?_, hab.left⟩
iterate 3 right
left
rw [hab.left, hab.right]
· intro hab
simp only [Prod.mk.injEq] at hab
refine ⟨?_, hab.left⟩
iterate 4 right
left
rw [hab.left, hab.right]
theorem exercise_3_18_iii
: restriction (inv R) {1} = {(1, 0)} := by
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rw [hR]
unfold inv restriction
ext p
have (a, b) := p
simp only [
Set.mem_singleton_iff,
Set.mem_insert_iff,
or_self,
exists_eq_or_imp,
exists_eq_left,
Set.mem_setOf_eq,
Prod.mk.injEq
]
apply Iff.intro
· intro ⟨hb, ha⟩
casesm* _ _
all_goals case _ hr => first
| exact ⟨ha, hr.right.symm⟩
| rw [ha] at hr; simp at hr
· intro ⟨ha, hb⟩
rw [ha, hb]
simp
theorem exercise_3_18_iv
: image R {1} = {2, 3} := by
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rw [hR]
unfold image
ext y
simp
theorem exercise_3_18_v
: image (inv R) {1} = {0} := by
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rw [hR]
unfold inv image
ext y
simp
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end Exercise_3_18
/-! #### Exercise 3.19
Let
```
A = {⟨∅, {∅, {∅}}⟩, ⟨{∅}, ∅⟩}.
```
Evaluate each of the following: `A(∅)`, `A⟦∅⟧`, `A⟦{∅}⟧`, `A⟦{∅, {∅}}⟧`,
`A⁻¹`, `A ∘ A`, `A ↾ ∅`, `A ↾ {∅, {∅}}`, ` A`.
-/
section Exercise_3_19
variable {A : Set.Relation (Set (Set (Set α)))}
variable (hA : A = {(∅, {∅, {∅}}), ({∅}, ∅)})
theorem exercise_3_19_i
: (∅, {∅, {∅}}) ∈ A := by
rw [hA]
simp
theorem exercise_3_19_ii
: image A ∅ = ∅ := by
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unfold image
simp
theorem exercise_3_19_iii
: image A {∅} = {{∅, {∅}}} := by
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unfold image
rw [hA]
ext x
simp only [
Set.mem_singleton_iff,
Prod.mk.injEq,
Set.mem_insert_iff,
exists_eq_left,
true_and
]
apply Iff.intro
· intro hx
simp at hx
apply Or.elim hx
· simp
· intro ⟨h, _⟩
exfalso
rw [Set.ext_iff] at h
have := h ∅
simp at this
· intro hx
rw [hx]
simp
theorem exercise_3_19_iv
: image A {∅, {∅}} = {{∅, {∅}}, ∅} := by
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unfold image
rw [hA]
ext x
simp only [
Set.mem_singleton_iff,
Set.mem_insert_iff,
Prod.mk.injEq,
exists_eq_or_imp,
true_and,
exists_eq_left,
Set.singleton_ne_empty,
false_and,
false_or,
Set.mem_setOf_eq
]
apply Iff.intro
· intro h
apply Or.elim h
· intro hx₁
apply Or.elim hx₁
· intro hx₂; left ; exact hx₂
· intro hx₂; right; exact hx₂.right
· intro hx₂
right
exact hx₂
· intro h
apply Or.elim h
· intro hx₁; iterate 2 left
exact hx₁
· intro hx₁; right; exact hx₁
theorem exercise_3_19_v
: inv A = {({∅, {∅}}, ∅), (∅, {∅})} := by
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unfold inv
rw [hA]
ext x
simp only [
Set.mem_singleton_iff,
Prod.mk.injEq,
Set.mem_insert_iff,
exists_eq_or_imp,
exists_eq_left,
Set.mem_setOf_eq
]
apply Iff.intro
· intro hx
apply Or.elim hx
· intro hx₁; left ; rw [← hx₁]
· intro hx₁; right; rw [← hx₁]
· intro hx
apply Or.elim hx
· intro hx₁; left ; rw [← hx₁]
· intro hx₁; right; rw [← hx₁]
theorem exercise_3_19_vi
: comp A A = {({∅}, {∅, {∅}})} := by
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unfold comp
rw [hA]
ext x
have (a, b) := x
simp only [
Set.mem_singleton_iff, Prod.mk.injEq, Set.mem_insert_iff, Set.mem_setOf_eq
]
apply Iff.intro
· intro ⟨t, ht₁, ht₂⟩
casesm* _ _
all_goals case _ hl hr => first
| {
rw [hl.right] at hr
have := hr.left
rw [Set.ext_iff] at this
simp at this
}
| exact ⟨hl.left, hr.right⟩
· intro ⟨ha, hb⟩
refine ⟨∅, ?_, ?_⟩
· right; rw [ha]; simp
· left ; rw [hb]; simp
theorem exercise_3_19_vii
: restriction A ∅ = ∅ := by
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unfold restriction
rw [hA]
simp
theorem exercise_3_19_viii
: restriction A {∅} = {(∅, {∅, {∅}})} := by
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unfold restriction
rw [hA]
ext x
have (a, b) := x
simp only [
Set.mem_singleton_iff, Prod.mk.injEq, Set.mem_insert_iff, Set.mem_setOf_eq
]
apply Iff.intro
· intro ⟨h, ha⟩
apply Or.elim h
· simp
· intro ⟨ha', _⟩
exfalso
rw [ha', Set.ext_iff] at ha
simp at ha
· intro ⟨ha, hb⟩
rw [ha, hb]
simp
theorem exercise_3_19_ix
: restriction A {∅, {∅}} = A := by
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unfold restriction
rw [hA]
ext x
have (a, b) := x
simp only [
Set.mem_singleton_iff,
Prod.mk.injEq,
Set.mem_insert_iff,
Set.mem_setOf_eq
]
apply Iff.intro
· intro ⟨h₁, h₂⟩
casesm* _ _
· case _ hl _ => left ; exact hl
· case _ hl _ => left ; exact hl
· case _ hl _ => right; exact hl
· case _ hl _ => right; exact hl
· intro h₁
apply Or.elim h₁ <;>
· intro ⟨ha, hb⟩
rw [ha, hb]
simp
theorem exercise_3_19_x
: ⋃₀ ⋃₀ A.toOrderedPairs = {∅, {∅}, {∅, {∅}}} := by
unfold toOrderedPairs OrderedPair Set.sUnion sSup Set.instSupSetSet
rw [hA]
ext x
simp only [
Set.mem_singleton_iff,
Prod.mk.injEq,
Set.mem_image,
Set.mem_insert_iff,
exists_eq_or_imp,
exists_eq_left,
Set.singleton_ne_empty,
Set.mem_setOf_eq
]
apply Iff.intro
· intro ⟨a, ⟨t, ht₁, ht₂⟩, hx⟩
apply Or.elim ht₁
· intro ht
rw [← ht] at ht₂
simp only [Set.mem_singleton_iff, Set.mem_insert_iff] at ht₂
apply Or.elim ht₂
· intro ha
rw [ha] at hx
simp only [Set.mem_singleton_iff] at hx
left
exact hx
· intro ha
rw [ha] at hx
simp at hx
apply Or.elim hx <;>
· intro hx'; rw [hx']; simp
· intro ht
rw [← ht] at ht₂
simp only [
Set.mem_singleton_iff, Set.singleton_ne_empty, Set.mem_insert_iff
] at ht₂
apply Or.elim ht₂
· intro ha
rw [ha] at hx
simp only [Set.mem_singleton_iff] at hx
rw [hx]
simp
· intro ha
rw [ha] at hx
simp only [
Set.mem_singleton_iff, Set.singleton_ne_empty, Set.mem_insert_iff
] at hx
apply Or.elim hx <;>
· intro hx'; rw [hx']; simp
· intro hx
apply Or.elim hx
· intro hx₁
rw [hx₁]
refine ⟨{{∅}, ∅}, ⟨{{{∅}}, {{∅}, ∅}}, ?_⟩, ?_⟩ <;> simp
· intro hx₁
apply Or.elim hx₁
· intro hx₂
rw [hx₂]
refine ⟨{{∅}, ∅}, ⟨{{{∅}}, {{∅}, ∅}}, ?_⟩, ?_⟩ <;> simp
· intro hx₂
rw [hx₂]
refine ⟨{∅, {∅, {∅}}}, ⟨{{∅}, {∅, {∅, {∅}}}}, ?_⟩, ?_⟩ <;> simp
end Exercise_3_19
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/-- #### Exercise 3.20
Show that `F ↾ A = F ∩ (A × ran F)`.
-/
theorem exercise_3_20 {F : Set.HRelation α β} {A : Set α}
: restriction F A = F ∩ (Set.prod A (ran F)) := by
calc restriction F A
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_ = {p | p ∈ F ∧ p.fst ∈ A} := rfl
_ = {p | p ∈ F ∧ p.fst ∈ A ∧ p.snd ∈ ran F} := by
ext x
have (a, b) := x
simp only [
Set.mem_setOf_eq, Set.sep_and, Set.mem_inter_iff, iff_self_and, and_imp
]
intro hF _
exact ⟨hF, mem_pair_imp_snd_mem_ran hF⟩
_ = {p | p ∈ 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
/-- #### Exercise 3.22 (a)
Show that the following is correct for any sets.
```
A ⊆ B → F⟦A⟧ ⊆ F⟦B⟧
```
-/
theorem exercise_3_22_a {A B : Set α} {F : Set.HRelation α β} (h : A ⊆ B)
: image F A ⊆ image F B := by
show ∀ x, x ∈ image F A → x ∈ image F B
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intro x hx
have ⟨u, hu⟩ := hx
have := h hu.left
exact ⟨u, this, hu.right⟩
/-- #### Exercise 3.22 (b)
Show that the following is correct for any sets.
```
(F ∘ G)⟦A⟧ = F⟦G⟦A⟧⟧
```
-/
theorem exercise_3_22_b {A B : Set α} {F : Set.HRelation α β}
: image (comp F G) A = image F (image G A) := by
calc image (comp F G) A
_ = { v | ∃ u ∈ A, (u, v) ∈ comp F G } := rfl
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_ = { v | ∃ u ∈ A, ∃ a, (u, a) ∈ G ∧ (a, v) ∈ F } := rfl
_ = { v | ∃ a, ∃ u ∈ A, (u, a) ∈ G ∧ (a, v) ∈ F } := by
ext p
apply Iff.intro
· intro ⟨u, hu, a, ha⟩
exact ⟨a, u, hu, ha⟩
· intro ⟨a, u, hu, ha⟩
exact ⟨u, hu, a, ha⟩
_ = { v | ∃ a, (∃ u ∈ A, (u, a) ∈ G) ∧ (a, v) ∈ F } := by
ext p
apply Iff.intro
· intro ⟨a, u, h⟩
exact ⟨a, ⟨u, h.left, h.right.left⟩, h.right.right⟩
· intro ⟨a, ⟨u, hu⟩, ha⟩
exact ⟨a, u, hu.left, hu.right, ha⟩
_ = { v | ∃ a ∈ { w | ∃ u ∈ A, (u, w) ∈ G }, (a, v) ∈ F } := rfl
_ = { v | ∃ a ∈ image G A, (a, v) ∈ F } := rfl
_ = image F (image G A) := rfl
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/-- #### Exercise 3.22 (c)
Show that the following is correct for any sets.
```
Q ↾ (A B) = (Q ↾ A) (Q ↾ B)
```
-/
theorem exercise_3_22_c {A B : Set α} {Q : Set.Relation α}
: restriction Q (A B) = (restriction Q A) (restriction Q B) := by
calc restriction Q (A B)
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_ = { p | p ∈ Q ∧ p.1 ∈ A B } := rfl
_ = { p | p ∈ Q ∧ (p.1 ∈ A p.1 ∈ B) } := rfl
_ = { p | (p ∈ Q ∧ p.1 ∈ A) (p ∈ Q ∧ p.1 ∈ B) } := by
ext p
simp only [Set.sep_or, Set.mem_union, Set.mem_setOf_eq]
_ = { p | p ∈ Q ∧ p.1 ∈ A} { p | p ∈ Q ∧ p.1 ∈ B } := rfl
_ = (restriction Q A) (restriction Q B) := rfl
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/-- #### 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`.
-/
theorem exercise_3_23_i {A : Set α} {B : Set.HRelation α β} {I : Set.Relation α}
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(hI : I = { p | p.1 ∈ A ∧ p.1 = p.2 })
: comp B I = restriction B A := by
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rw [Set.Subset.antisymm_iff]
apply And.intro
· show ∀ p, p ∈ comp B I → p ∈ restriction B A
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intro (x, y) hp
have ⟨t, ht⟩ := hp
rw [hI] at ht
simp only [Set.mem_setOf_eq] at ht
show (x, y) ∈ B ∧ x ∈ A
rw [← ht.left.right] at ht
exact ⟨ht.right, ht.left.left⟩
· show ∀ p, p ∈ restriction B A → p ∈ comp B I
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unfold restriction comp
rw [hI]
intro (x, y) hp
refine ⟨x, ⟨hp.right, rfl⟩, hp.left⟩
/-- #### 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`.
-/
theorem exercise_3_23_ii {A C : Set α} {I : Set.Relation α}
(hI : I = { p | p.1 ∈ A ∧ p.1 = p.2 })
: image I C = A ∩ C := by
calc image I C
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_ = { v | ∃ u ∈ C, (u, v) ∈ I } := rfl
_ = { v | ∃ u ∈ C, u ∈ A ∧ u = v } := by
ext v
apply Iff.intro
· intro ⟨u, h₁, h₂⟩
rw [hI] at h₂
exact ⟨u, h₁, h₂⟩
· intro ⟨u, h₁, h₂⟩
refine ⟨u, h₁, ?_⟩
· rw [hI]
exact h₂
_ = { v | v ∈ C ∧ v ∈ A } := by
ext v
simp only [Set.mem_setOf_eq, Set.sep_mem_eq, Set.mem_inter_iff]
apply Iff.intro
· intro ⟨u, hC, hA, hv⟩
rw [← hv]
exact ⟨hC, hA⟩
· intro ⟨hC, hA⟩
exact ⟨v, hC, hA, rfl⟩
_ = C ∩ A := rfl
_ = A ∩ C := Set.inter_comm C A
/-- #### Exercise 3.24
Show that for a function `F`, `F⁻¹⟦A⟧ = { x ∈ dom F | F(x) ∈ A }`.
-/
theorem exercise_3_24 {F : Set.HRelation α β} {A : Set β}
(hF : isSingleValued F)
: image (inv F) A = { x ∈ dom F | ∃! y : β, (x, y) ∈ F ∧ y ∈ A } := by
calc image (inv F) A
_ = { x | ∃ y ∈ A, (y, x) ∈ inv F } := rfl
_ = { x | ∃ y ∈ A, (x, y) ∈ F } := by simp only [mem_self_comm_mem_inv]
_ = { x | x ∈ dom F ∧ (∃ y : β, (x, y) ∈ F ∧ y ∈ A) } := by
ext x
apply Iff.intro
· intro ⟨y, hy, hyx⟩
exact ⟨mem_pair_imp_fst_mem_dom hyx, y, hyx, hy⟩
· intro ⟨_, y, hxy, hy⟩
exact ⟨y, hy, hxy⟩
_ = { x ∈ dom F | ∃ y : β, (x, y) ∈ F ∧ y ∈ A } := rfl
_ = { x ∈ dom F | ∃! y : β, (x, y) ∈ F ∧ y ∈ A } := by
ext x
simp only [Set.mem_setOf_eq, and_congr_right_iff]
intro _
apply Iff.intro
· intro ⟨y, hy⟩
refine ⟨y, hy, ?_⟩
intro y₁ hy₁
exact single_valued_eq_unique hF hy₁.left hy.left
· intro ⟨y, hy⟩
exact ⟨y, hy.left⟩
/-- #### Exercise 3.25 (b)
Show that the result of part (a) holds for any function `G`, not necessarily
one-to-one.
-/
theorem exercise_3_25_b {G : Set.HRelation α β} (hG : isSingleValued G)
: comp G (inv G) = { p | p.1 ∈ ran G ∧ p.1 = p.2 } := by
ext p
have (x, y) := p
apply Iff.intro
· unfold comp inv
intro h
simp only [Prod.exists, Set.mem_setOf_eq, Prod.mk.injEq] at h
have ⟨t, ⟨a, b, ⟨hab, hb, ha⟩⟩, ht⟩ := h
rw [hb, ha] at hab
exact ⟨mem_pair_imp_snd_mem_ran hab, single_valued_eq_unique hG hab ht⟩
· intro h
simp only [Set.mem_setOf_eq] at h
unfold comp inv
simp only [Prod.exists, Set.mem_setOf_eq, Prod.mk.injEq]
have ⟨t, ht⟩ := ran_exists h.left
exact ⟨t, ⟨t, x, ht, rfl, rfl⟩, by rwa [← h.right]⟩
/-- #### 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`.
-/
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
Show that `dom (F ∘ G) = G⁻¹⟦dom F⟧` for any sets `F` and `G`. (`F` and `G` need
not be functions.)
-/
theorem exercise_3_27 {F : Set.HRelation β γ} {G : Set.HRelation α β}
: dom (comp F G) = image (inv G) (dom F) := by
rw [Set.Subset.antisymm_iff]
apply And.intro
· show ∀ x, x ∈ dom (comp F G) → x ∈ image (inv G) (dom F)
intro x hx
have ⟨y, t, ht⟩ := dom_exists hx
have htF : t ∈ dom F := mem_pair_imp_fst_mem_dom ht.right
unfold image inv
simp only [Prod.exists, Set.mem_setOf_eq, Prod.mk.injEq]
exact ⟨t, htF, x, t, ht.left, rfl, rfl⟩
· show ∀ x, x ∈ image (inv G) (dom F) → x ∈ dom (comp F G)
intro x hx
unfold image at hx
simp only [mem_self_comm_mem_inv, Set.mem_setOf_eq] at hx
have ⟨u, hu⟩ := hx
have ⟨t, ht⟩ := dom_exists hu.left
unfold dom comp
simp only [
Set.mem_image,
Set.mem_setOf_eq,
Prod.exists,
exists_and_right,
exists_eq_right
]
exact ⟨t, u, hu.right, ht⟩
/-- #### 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`
maps `𝒫 A` one-to-one into `𝒫 B`.
-/
theorem exercise_3_28 {A : Set α} {B : Set β}
{f : Set.HRelation α β} {G : Set.HRelation (Set α) (Set β)}
(hf : isOneToOne f ∧ mapsInto f A B)
(hG : G = { p | p.1 ∈ 𝒫 A ∧ p.2 = image f p.1 })
: isOneToOne G ∧ mapsInto G (𝒫 A) (𝒫 B) := by
have dG : dom G = 𝒫 A := by
rw [hG]
ext p
unfold dom Prod.fst
simp
have hG₁ : isSingleValued G := by
intro x hx
have ⟨y, hy⟩ := dom_exists hx
refine ⟨y, ⟨mem_pair_imp_snd_mem_ran hy, hy⟩, ?_⟩
intro y₁ hy₁
rw [hG, Set.mem_setOf_eq] at hy
conv at hy₁ => rhs; rw [hG, Set.mem_setOf_eq]
simp only at *
rw [hy.right, hy₁.right.right]
apply And.intro
· show isOneToOne G
refine ⟨hG₁, ?_⟩
intro y hy
have ⟨X₁, hX₁⟩ := ran_exists hy
refine ⟨X₁, ⟨mem_pair_imp_fst_mem_dom hX₁, hX₁⟩, ?_⟩
intro X₂ hX₂
have hX₁' : y = image f X₁ := by
rw [hG] at hX₁
simp only [Set.mem_powerset_iff, Set.mem_setOf_eq] at hX₁
exact hX₁.right
have hX₂' : y = image f X₂ := by
have := hX₂.right
rw [hG] at this
simp only [Set.mem_powerset_iff, Set.mem_setOf_eq] at this
exact this.right
ext t
apply Iff.intro
· intro ht
rw [dG] at hX₂
simp only [Set.mem_powerset_iff] at hX₂
have ht' := hX₂.left ht
rw [← hf.right.dom_eq] at ht'
have ⟨ft, hft⟩ := dom_exists ht'
have hft' : ft ∈ image f X₂ := ⟨t, ht, hft⟩
rw [← hX₂', hX₁'] at hft'
have ⟨t₁, ht₁⟩ := hft'
rw [single_rooted_eq_unique hf.left.right hft ht₁.right]
exact ht₁.left
· intro ht
have hX₁sub := mem_pair_imp_fst_mem_dom hX₁
rw [dG] at hX₁sub
simp only [Set.mem_powerset_iff] at hX₁sub
have ht' := hX₁sub ht
rw [← hf.right.dom_eq] at ht'
have ⟨ft, hft⟩ := dom_exists ht'
have hft' : ft ∈ image f X₁ := ⟨t, ht, hft⟩
rw [← hX₁', hX₂'] at hft'
have ⟨t₁, ht₁⟩ := hft'
rw [single_rooted_eq_unique hf.left.right hft ht₁.right]
exact ht₁.left
· show mapsInto G (𝒫 A) (𝒫 B)
refine ⟨hG₁, dG, ?_⟩
show ∀ x, x ∈ ran G → x ∈ 𝒫 B
intro x hx
rw [hG] at hx
unfold ran Prod.snd at hx
simp only [
Set.mem_powerset_iff,
Set.mem_image,
Set.mem_setOf_eq,
Prod.exists,
exists_eq_right
] at hx
have ⟨a, ha⟩ := hx
rw [ha.right]
show ∀ y, y ∈ image f a → y ∈ B
intro y ⟨b, hb⟩
have hz := mem_pair_imp_snd_mem_ran hb.right
exact hf.right.ran_ss hz
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/-- #### Exercise 3.29
Assume that `f : A → B` and define a function `G : B → 𝒫 A` by
```
G(b) = {x ∈ A | f(x) = b}
```
Show that if `f` maps `A` *onto* `B`, then `G` is one-to-one. Does the converse
hold?
-/
theorem exercise_3_29 {f : Set.HRelation α β} {G : Set.HRelation β (Set α)}
{A : Set α} {B : Set β} (hf : mapsOnto f A B)
(hG : mapsInto G B (𝒫 A) ∧ G = { p | p.1 ∈ B ∧ p.2 = {x ∈ A | (x, p.1) ∈ f} })
: isOneToOne G := by
unfold isOneToOne
refine ⟨hG.left.is_func, ?_⟩
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intro y hy
have ⟨x₁, hx₁⟩ := ran_exists hy
refine ⟨x₁, ⟨mem_pair_imp_fst_mem_dom hx₁, hx₁⟩, ?_⟩
intro x₂ hx₂
have hG₁ : (x₁, {x ∈ A | (x, x₁) ∈ f}) ∈ G := by
rw [hG.right, ← hG.left.dom_eq]
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simp only [Set.mem_setOf_eq, and_true]
exact mem_pair_imp_fst_mem_dom hx₁
have hG₂ : (x₂, {x ∈ A | (x, x₂) ∈ f}) ∈ G := by
rw [hG.right, ← hG.left.dom_eq]
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simp only [Set.mem_setOf_eq, and_true]
exact hx₂.left
have heq : {x ∈ A | (x, x₁) ∈ f} = {x ∈ A | (x, x₂) ∈ f} := by
have h₁ := single_valued_eq_unique hG.left.is_func hx₁ hG₁
have h₂ := single_valued_eq_unique hG.left.is_func hx₂.right hG₂
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rw [← h₁, ← h₂]
rw [hG.right, ← hf.ran_eq] at hG₁
simp only [Set.mem_setOf_eq, and_true] at hG₁
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have ⟨t, ht⟩ := ran_exists hG₁
have : t ∈ {x ∈ A | (x, x₁) ∈ f} := by
refine ⟨?_, ht⟩
rw [← hf.dom_eq]
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exact mem_pair_imp_fst_mem_dom ht
rw [heq] at this
exact single_valued_eq_unique hf.is_func this.right ht
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/-- #### Theorem 3M
If `R` is a symmetric and transitive relation, then `R` is an equivalence
relation on `fld R`.
-/
theorem theorem_3m {R : Set.Relation α}
(hS : R.isSymmetric) (hT : R.isTransitive)
: R.isEquivalence (fld R) := by
refine ⟨Eq.subset rfl, ?_, hS, hT⟩
intro x hx
apply Or.elim hx
· intro h
have ⟨y, hy⟩ := dom_exists h
have := hS hy
exact hT hy this
· intro h
have ⟨t, ht⟩ := ran_exists h
have := hS ht
exact hT this ht
/-- #### Exercise 3.32 (a)
Show that `R` is symmetric **iff** `R⁻¹ ⊆ R`.
-/
theorem exercise_3_32_a {R : Set.Relation α}
: isSymmetric R ↔ inv R ⊆ R := by
apply Iff.intro
· intro hR
show ∀ p, p ∈ inv R → p ∈ R
intro (x, y) hp
simp only [mem_self_comm_mem_inv] at hp
exact hR hp
· intro hR
unfold isSymmetric
intro x y hp
rw [← mem_self_comm_mem_inv] at hp
exact hR hp
/-- #### Exercise 3.32 (b)
Show that `R` is transitive **iff** `R ∘ R ⊆ R`.
-/
theorem exercise_3_32_b {R : Set.Relation α}
: isTransitive R ↔ comp R R ⊆ R := by
apply Iff.intro
· intro hR
show ∀ p, p ∈ comp R R → p ∈ R
intro (x, y) hp
have ⟨t, ht⟩ := hp
exact hR ht.left ht.right
· intro hR
intro x y z hx hz
have : (x, z) ∈ comp R R := ⟨y, hx, hz⟩
exact hR this
/-- #### Exercise 3.33
Show that `R` is a symmetric and transitive relation **iff** `R = R⁻¹ ∘ R`.
-/
theorem exercise_3_33 {R : Set.Relation α}
: isSymmetric R ∧ isTransitive R ↔ R = comp (inv R) R := by
have hR : comp (inv R) R = { p | ∃ t, (p.1, t) ∈ R ∧ (p.2, t) ∈ R } := by
ext p
unfold comp inv
simp only [Prod.exists, Set.mem_setOf_eq, Prod.mk.injEq]
apply Iff.intro
· intro ⟨t, ht, a, b, h⟩
refine ⟨t, ht, ?_⟩
rw [← h.right.right, ← h.right.left]
exact h.left
· intro ⟨t, ht⟩
exact ⟨t, ht.left, p.snd, t, ht.right, rfl, rfl⟩
apply Iff.intro
· intro h
rw [Set.Subset.antisymm_iff]
apply And.intro
· show ∀ p, p ∈ R → p ∈ comp (inv R) R
intro (x, y) hp
have hy := h.left hp
have hx := h.right hp hy
rw [hR]
exact ⟨x, hx, hy⟩
· show ∀ p, p ∈ comp (inv R) R → p ∈ R
intro (x, y) hp
rw [hR] at hp
have ⟨_, ht⟩ := hp
have := h.left ht.right
exact h.right ht.left this
· intro h
have hS : isSymmetric R := by
intro x y hp
have : inv R = R := by
calc inv R
_ = inv (comp (inv R) R) := by conv => lhs; rw [h]
_ = comp (inv R) (inv (inv R)) := by rw [comp_inv_eq_inv_comp_inv]
_ = comp (inv R) R := by rw [inv_inv_eq_self]
_ = R := h.symm
rwa [← this, mem_self_comm_mem_inv]
refine ⟨hS, ?_⟩
intro x y z hx hy
have : (z, y) ∈ R := hS hy
rw [h, hR]
exact ⟨y, hx, this⟩
/-- #### 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?
-/
theorem exercise_3_34_a {𝓐 : Set (Set.Relation α)}
(_ : Set.Nonempty 𝓐) (h𝓐 : ∀ A ∈ 𝓐, isTransitive A)
: isTransitive (⋂₀ 𝓐) := by
intro x y z hx hy
simp only [Set.mem_sInter] at *
intro A hA
have hx' := hx A hA
have hy' := hy A hA
exact h𝓐 A hA hx' hy'
/-- #### Exercise 3.34 (b)
Assume that `𝓐` is a nonempty set, every member of which is a transitive
relation. Is ` 𝓐` a transitive relation?
-/
theorem exercise_3_34_b {𝓐 : Set (Set.Relation )}
(_ : Set.Nonempty 𝓐) (h𝓐 : 𝓐 = {{(1, 2), (2, 3), (1, 3)}, {(2, 1)}})
: (∀ A ∈ 𝓐, isTransitive A) ∧ ¬ isTransitive (⋃₀ 𝓐) := by
apply And.intro
· intro A hA
rw [h𝓐] at hA
simp only [Set.mem_singleton_iff, Set.mem_insert_iff] at hA
apply Or.elim hA
· intro hA₁
rw [hA₁]
intro x y z hx hy
simp only [Set.mem_singleton_iff, Set.mem_insert_iff, Prod.mk.injEq] at *
casesm* _ _
all_goals case _ hl hr => first
| {rw [hl.right] at hr; simp at hr}
| {rw [hl.left] at hr; simp at hr}
| {right; right; exact ⟨hl.left, hr.right⟩}
· intro hA₁
rw [hA₁]
intro x y z hx hy
simp only [Set.mem_singleton_iff, Set.mem_insert_iff, Prod.mk.injEq] at *
rw [hx.right] at hy
simp at hy
· intro h
have h₁ : (1, 2) ∈ ⋃₀ 𝓐 := by
simp only [Set.mem_sUnion]
exact ⟨{(1, 2), (2, 3), (1, 3)}, by rw [h𝓐]; simp, by simp⟩
have h₂ : (2, 1) ∈ ⋃₀ 𝓐 := by
simp only [Set.mem_sUnion]
exact ⟨{(2, 1)}, by rw [h𝓐]; simp, by simp⟩
have h₃ : (1, 1) ∉ ⋃₀ 𝓐 := by
simp only [Set.mem_sUnion]
rw [h𝓐]
intro ⟨t, ht⟩
simp only [Set.mem_singleton_iff, Set.mem_insert_iff] at ht
have := ht.right
apply Or.elim ht.left <;>
· intro ht₁
rw [ht₁] at this
simp at this
exact absurd (h h₁ h₂) h₃
/-- #### Exercise 3.35
Show that for any `R` and `x`, we have `[x]_R = R⟦{x}⟧`.
-/
theorem exercise_3_35 {R : Set.Relation α} {x : α}
: neighborhood R x = image R {x} := by
calc neighborhood R x
_ = { t | (x, t) ∈ R } := rfl
_ = { t | ∃ u ∈ ({x} : Set α), (u, t) ∈ R } := by simp
_ = image R {x} := rfl
/-- #### 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
equivalence relation on `A`.
-/
theorem exercise_3_36 {f : Set.HRelation α β}
{Q : Set.Relation α} {R : Set.Relation β} {A : Set α} {B : Set β}
(hf : mapsInto f A B) (hR : isEquivalence R B)
(hQ : Q = { p | ∃ fx fy : β, (p.1, fx) ∈ f ∧ (p.2, fy) ∈ f ∧ (fx, fy) ∈ R })
: isEquivalence Q A := by
refine ⟨?_, ?_, ?_, ?_⟩
· -- `fld Q ⊆ A`
show ∀ x, x ∈ fld Q → x ∈ A
intro x hx
rw [hQ] at hx
unfold fld dom ran at hx
simp only [
exists_and_left,
Set.mem_union,
Set.mem_image,
Set.mem_setOf_eq,
Prod.exists,
exists_and_right,
exists_eq_right
] at hx
apply Or.elim hx
· intro ⟨_, _, hx₁⟩
rw [← hf.dom_eq]
exact mem_pair_imp_fst_mem_dom hx₁.left
· intro ⟨_, _, _, _, hx₂⟩
rw [← hf.dom_eq]
exact mem_pair_imp_fst_mem_dom hx₂.left
· -- `isReflexive Q A`
intro x hx
rw [← hf.dom_eq] at hx
have ⟨fx, hfx⟩ := dom_exists hx
have := hR.refl fx (hf.ran_ss $ mem_pair_imp_snd_mem_ran hfx)
rw [hQ]
simp only [exists_and_left, Set.mem_setOf_eq]
exact ⟨fx, hfx, fx, hfx, this⟩
· -- `isSymmetric Q`
intro x y h
rw [hQ] at h
simp only [exists_and_left, Set.mem_setOf_eq] at h
have ⟨fx, hfx, fy, hfy, h'⟩ := h
have := hR.symm h'
rw [hQ]
simp only [exists_and_left, Set.mem_setOf_eq]
exact ⟨fy, hfy, fx, hfx, this⟩
· -- `isTransitive Q`
intro x y z hx hy
rw [hQ] at hx hy
simp only [exists_and_left, Set.mem_setOf_eq] at hx hy
have ⟨fx, hfx, fy, hfy, h₁⟩ := hx
have ⟨fy₁, hfy₁, fz, hfz, h₂⟩ := hy
have hfy' : fy = fy₁ := single_valued_eq_unique hf.is_func hfy hfy₁
rw [hfy'] at h₁
rw [hQ]
simp only [exists_and_left, Set.mem_setOf_eq]
exact ⟨fx, hfx, fz, hfz, hR.trans h₁ h₂⟩
/-- #### Exercise 3.37
Assume that `P` is a partition of a set `A`. Define the relation `Rₚ` as
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follows:
```
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xRₚy ↔ (∃ B ∈ Π)(x ∈ B ∧ y ∈ B).
```
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Show that `Rₚ` is an equivalence relation on `A`. (This is a formalized version
of the discussion at the beginning of this section.)
-/
theorem exercise_3_37 {P : Set (Set α)} {A : Set α}
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(hP : isPartition P A) (Rₚ : Set.Relation α)
(hRₚ : ∀ x y, (x, y) ∈ Rₚ ↔ ∃ B ∈ P, x ∈ B ∧ y ∈ B)
: isEquivalence Rₚ A := by
have hRₚ_eq : Rₚ = { p | ∃ B ∈ P, p.1 ∈ B ∧ p.2 ∈ B } := by
ext p
have (x, y) := p
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exact hRₚ x y
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refine ⟨?_, ?_, ?_, ?_⟩
· -- `fld Rₚ ⊆ A`
show ∀ x, x ∈ fld Rₚ → x ∈ A
rw [hRₚ_eq]
intro x hx
unfold fld dom ran at hx
simp only [
Set.mem_union,
Set.mem_image,
Set.mem_setOf_eq,
Prod.exists,
exists_and_right,
exists_eq_right
] at hx
apply Or.elim hx
· intro ⟨t, B, hB⟩
have := hP.p_subset B hB.left
exact this hB.right.left
· intro ⟨a, B, hB⟩
have := hP.p_subset B hB.left
exact this hB.right.right
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· -- `isReflexive Rₚ A`
intro x hx
rw [hRₚ_eq]
simp only [Set.mem_setOf_eq, and_self]
exact hP.exhaustive x hx
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· -- `isSymmetric Rₚ`
intro x y h
rw [hRₚ_eq] at h
simp only [Set.mem_setOf_eq] at h
have ⟨B, hB⟩ := h
rw [hRₚ_eq]
simp only [Set.mem_setOf_eq]
conv at hB => right; rw [and_comm]
exact ⟨B, hB⟩
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· -- `isTransitive Rₚ`
intro x y z hx hy
rw [hRₚ_eq] at hx hy
simp only [Set.mem_setOf_eq] at hx hy
have ⟨B₁, hB₁⟩ := hx
have ⟨B₂, hB₂⟩ := hy
have hB : B₁ = B₂ := by
have hy₁ : y ∈ B₁ := hB₁.right.right
have hy₂ : y ∈ B₂ := hB₂.right.left
have hy := hP.disjoint B₁ hB₁.left B₂ hB₂.left
rw [contraposition] at hy
simp at hy
suffices B₁ ∩ B₂ ≠ ∅ from hy this
intro h'
rw [Set.ext_iff] at h'
exact (h' y).mp ⟨hy₁, hy₂⟩
rw [hRₚ_eq]
simp only [Set.mem_setOf_eq]
exact ⟨B₁, hB₁.left, hB₁.right.left, by rw [hB]; exact hB₂.right.right⟩
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/-- #### 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
`Rₚ` of the preceding exercise, then the partition `A / Rₚ` is just `P`.
-/
theorem exercise_3_38 {P : Set (Set α)} {A : Set α}
(hP : isPartition P A) (Rₚ : Set.Relation α)
(hRₚ : ∀ x y, (x, y) ∈ Rₚ ↔ ∃ B ∈ P, x ∈ B ∧ y ∈ B)
: modEquiv (exercise_3_37 hP Rₚ hRₚ) = P := by
have hRₚ_eq : Rₚ = { p | ∃ B ∈ P, p.1 ∈ B ∧ p.2 ∈ B } := by
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ext p
have (x, y) := p
exact hRₚ x y
ext B
apply Iff.intro
· intro ⟨x, hx⟩
have ⟨B', hB'⟩ := partition_mem_mem_eq hP hx.left
simp only at hB'
suffices B = B' by
rw [this]
exact hB'.left.left
ext y
apply Iff.intro
· intro hy
rw [← hx.right, hRₚ_eq] at hy
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have ⟨B₁, hB₁⟩ := hy
have := hB'.right B₁ ⟨hB₁.left, hB₁.right.left⟩
rw [← this]
exact hB₁.right.right
· intro hy
rw [← hx.right, hRₚ_eq]
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exact ⟨B', hB'.left.left, hB'.left.right, hy⟩
· intro hB
have ⟨x, hx⟩ := hP.nonempty B hB
have hx' : x ∈ A := hP.p_subset B hB hx
refine ⟨x, hx', Eq.symm ?_⟩
calc B
_ = {t | x ∈ B ∧ t ∈ B} := by
ext y
apply Iff.intro
· intro hy
exact ⟨hx, hy⟩
· intro hy
exact hy.right
_ = {t | ∃ B₁ ∈ P, x ∈ B₁ ∧ t ∈ B₁} := by
ext y
apply Iff.intro
· intro hy
exact ⟨B, hB, hy⟩
· intro hy
have ⟨B₁, hB₁⟩ := hy
have ⟨B', hB'⟩ := partition_mem_mem_eq hP hx'
simp only [Set.mem_setOf_eq] at *
have : B = B₁ := by
have lhs := hB'.right B ⟨hB, hx⟩
have rhs := hB'.right B₁ ⟨hB₁.left, hB₁.right.left⟩
rw [lhs, rhs]
rw [this]
exact hB₁.right
_ = {t | (x, t) ∈ Rₚ } := by
rw [hRₚ_eq]
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simp only [Set.mem_setOf_eq]
_ = neighborhood Rₚ x := rfl
/-- #### 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
`R`.
-/
theorem exercise_3_39 {P : Set (Set α)} {R Rₚ : Set.Relation α} {A : Set α}
(hR : isEquivalence R A) (hP : P = modEquiv hR)
(hRₚ : ∀ x y, (x, y) ∈ Rₚ ↔ ∃ B ∈ P, x ∈ B ∧ y ∈ B)
: Rₚ = R := by
have hRₚ_eq : Rₚ = { p | ∃ B ∈ P, p.1 ∈ B ∧ p.2 ∈ B } := by
ext p
have (x, y) := p
exact hRₚ x y
ext p
have (x, y) := p
apply Iff.intro
· intro hp
rw [hRₚ_eq] at hp
have ⟨B, hB, hx, hy⟩ := hp
rw [hP] at hB
have ⟨z, hz⟩ := hB
rw [← hz.right] at hx hy
exact neighborhood_mem_imp_relate hR hx hy
· intro hp
have hxA : x ∈ A := hR.b_on (Or.inl (mem_pair_imp_fst_mem_dom hp))
have hyA : y ∈ A := hR.b_on (Or.inr (mem_pair_imp_snd_mem_ran hp))
rw [hRₚ_eq]
have hx : x ∈ neighborhood R x := neighborhood_self_mem hR hxA
have hy : y ∈ neighborhood R x := by
rw [← neighborhood_eq_iff_mem_relate hR hxA hyA] at hp
rw [hp]
exact neighborhood_self_mem hR hyA
refine ⟨neighborhood R x, ?_, ⟨hx, hy⟩⟩
rw [hP]
exact ⟨x, hxA, rfl⟩
/-- #### Exercise 3.41 (a)
Let `` be the set of real numbers and define the realtion `Q` on ` × ` by
`⟨u, v⟩ Q ⟨x, y⟩` **iff** `u + y = x + v`. Show that `Q` is an equivalence
relation on ` × `.
-/
theorem exercise_3_41_a {Q : Set.Relation ( × )}
(hQ : ∀ u v x y, ((u, v), (x, y)) ∈ Q ↔ u + y = x + v)
: isEquivalence Q (Set.univ : Set ( × )) := by
have hQ_eq : Q = {p | p.1.1 + p.2.2 = p.2.1 + p.1.2} := by
ext p
apply Iff.intro <;>
· intro hp
rwa [hQ] at *
refine ⟨?_, ?_, ?_, ?_⟩
· -- `fld Q ⊆ Set.univ`
show ∀ p, p ∈ fld Q → p ∈ Set.univ
intro _ _
simp only [Set.mem_univ]
· -- `isReflexive Q Set.univ`
intro (x, y) _
rw [hQ_eq]
simp
· -- `isSymmetric Q`
intro (u, v) (x, y) hp
rw [hQ_eq] at *
exact Eq.symm hp
· -- `isTransitive Q`
unfold isTransitive
intro (u, v) (x, y) (a, b) h₁ h₂
rw [hQ_eq] at *
simp at h₁ h₂
simp
have h₁' : u - v = x - y := by
have := sub_eq_of_eq_add h₁
rw [add_sub_right_comm] at this
exact eq_sub_of_add_eq this
have h₂' : x - y = a - b := by
have := sub_eq_of_eq_add h₂
rw [add_sub_right_comm] at this
exact eq_sub_of_add_eq this
rw [h₂'] at h₁'
have := eq_add_of_sub_eq' h₁'
rw [← add_sub_assoc] at this
have := add_eq_of_eq_sub this
conv => right; rw [add_comm]
exact this
end Relation
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end Enderton.Set.Chapter_3