Enderton. Exercises 3.16 - 3.18.
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@ -4137,7 +4137,7 @@ Show that there is no set to which every function belongs.
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\end{proof}
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\subsection{\pending{Exercise 3.17}}%
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\subsection{\verified{Exercise 3.17}}%
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\label{sub:exercise-3.17}
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Show that the composition of two single-rooted sets is again single-rooted.
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@ -4145,6 +4145,14 @@ Conclude that the composition of two one-to-one functions is again one-to-one.
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\begin{proof}
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\statementpadding
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\lean*{Bookshelf/Enderton/Set/Chapter\_3}
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{Enderton.Set.Chapter\_3.exercise\_3\_17\_i}
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\lean{Bookshelf/Enderton/Set/Chapter\_3}
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{Enderton.Set.Chapter\_3.exercise\_3\_17\_ii}
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Let $F$ and $G$ be two single-rooted sets.
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Consider $F \circ G$.
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By definition of the \nameref{ref:composition} of sets,
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@ -4174,7 +4182,7 @@ Conclude that the composition of two one-to-one functions is again one-to-one.
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\end{proof}
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\subsection{\pending{Exercise 3.18}}%
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\subsection{\verified{Exercise 3.18}}%
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\label{sub:exercise-3.18}
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Let $R$ be the set
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@ -4185,11 +4193,27 @@ Evaluate the following: $R \circ R$, $R \restriction \{1\}$,
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\begin{proof}
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\statementpadding
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\lean*{Bookshelf/Enderton/Set/Chapter\_3}
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{Enderton.Set.Chapter\_3.exercise\_3\_18\_i}
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\lean*{Bookshelf/Enderton/Set/Chapter\_3}
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{Enderton.Set.Chapter\_3.exercise\_3\_18\_ii}
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\lean*{Bookshelf/Enderton/Set/Chapter\_3}
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{Enderton.Set.Chapter\_3.exercise\_3\_18\_iii}
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\lean*{Bookshelf/Enderton/Set/Chapter\_3}
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{Enderton.Set.Chapter\_3.exercise\_3\_18\_iv}
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\lean{Bookshelf/Enderton/Set/Chapter\_3}
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{Enderton.Set.Chapter\_3.exercise\_3\_18\_v}
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\begin{enumerate}[(i)]
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\item $R \circ R = \{
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\left< 0, 2 \right>,
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\left< 0, 3 \right>,
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\left< 0, 3 \right>,
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\left< 1, 3 \right>
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\}$.
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\item $R \restriction \{1\} = \{
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@ -1,6 +1,7 @@
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import Bookshelf.Enderton.Set.Chapter_2
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import Bookshelf.Enderton.Set.OrderedPair
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import Bookshelf.Enderton.Set.Relation
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import Mathlib.Tactic.CasesM
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/-! # Enderton.Set.Chapter_3
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@ -469,7 +470,7 @@ dom (F ∘ G) = {x ∈ dom G | G(x) ∈ dom F}.
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-/
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theorem theorem_3h_dom {F G : Set.Relation α}
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(_ : F.isSingleValued) (hG : G.isSingleValued)
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: dom (F.comp G) = {x ∈ dom G | ∃! y, (x, y) ∈ G ∧ y ∈ dom F } := by
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: dom (F.comp G) = {x ∈ dom G | ∃! y, (x, y) ∈ G ∧ y ∈ dom F} := by
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let rhs := {x ∈ dom G | ∃! y, (x, y) ∈ G ∧ y ∈ dom F }
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rw [Set.Subset.antisymm_iff]
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apply And.intro
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@ -720,7 +721,8 @@ f ⊆ g ↔ dom f ⊆ dom g ∧ (∀ x ∈ dom f) f(x) = g(x).
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-/
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theorem exercise_3_12 {f g : Set.Relation α}
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(hf : f.isSingleValued) (_ : g.isSingleValued)
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: f ⊆ g ↔ dom f ⊆ dom g ∧ (∀ x ∈ dom f, ∃! y : α, (x, y) ∈ f ∧ (x, y) ∈ g) := by
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: f ⊆ g ↔ dom f ⊆ dom g ∧
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(∀ x ∈ dom f, ∃! y : α, (x, y) ∈ f ∧ (x, y) ∈ g) := by
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apply Iff.intro
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· intro h
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apply And.intro
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@ -786,17 +788,16 @@ theorem exercise_3_14_b {f g : Set.Relation α}
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· intro h x hx
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have ⟨y₁, hy₁⟩ := hf x hx.left
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have ⟨y₂, hy₂⟩ := hg x hx.right
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have : y₁ = y₂ := by
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have hf' : (x, y₁) ∈ f ∪ g := Or.inl hy₁.left.right
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have hg' : (x, y₂) ∈ f ∪ g := Or.inr hy₂.left.right
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exact single_valued_eq_unique h hf' hg'
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have : y₁ = y₂ := single_valued_eq_unique h
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(Or.inl hy₁.left.right)
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(Or.inr hy₂.left.right)
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rw [← this] at hy₂
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refine ⟨y₁, ⟨hy₁.left.right, hy₂.left.right⟩, ?_⟩
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intro y₃ hfy₃
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exact single_valued_eq_unique hf hfy₃.left hy₁.left.right
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· intro h x hx
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by_cases hfx : x ∈ dom f
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· by_cases hgx : x ∈ dom g
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by_cases hfx : x ∈ dom f <;>
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by_cases hgx : x ∈ dom g
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· -- `x ∈ dom f ∧ x ∈ dom g`
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have ⟨y₁, hy₁⟩ := hf x hfx
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have ⟨y₂, hy₂⟩ := hg x hgx
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@ -827,7 +828,6 @@ theorem exercise_3_14_b {f g : Set.Relation α}
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exact single_valued_eq_unique hf hx' hy
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· intro hx'
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exact absurd (mem_pair_imp_fst_mem_dom hx') hgx
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· by_cases hgx : x ∈ dom g
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· -- `x ∉ dom f ∧ x ∈ dom g`
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have ⟨y, hy⟩ := dom_exists hgx
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have hxy : (x, y) ∈ f ∪ g := (Set.subset_union_right f g) hy
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@ -878,6 +878,174 @@ theorem exercise_3_15 {𝓐 : Set (Set.Relation α)}
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have := hg' hg.right
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exact single_valued_eq_unique (h𝓐 f hf.left) this hf.right
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/-! #### Exercise 3.17
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Show that the composition of two single-rooted sets is again single-rooted.
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Conclude that the composition of two one-to-one functions is again one-to-one.
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-/
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theorem exercise_3_17_i {F G : Set.Relation α}
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(hF : F.isSingleRooted) (hG : G.isSingleRooted)
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: (F.comp G).isSingleRooted := by
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intro v hv
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have ⟨u₁, hu₁⟩ := ran_exists hv
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have hu₁' := hu₁
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unfold comp at hu₁'
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simp only [Set.mem_setOf_eq] at hu₁'
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have ⟨t₁, ht₁⟩ := hu₁'
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unfold ExistsUnique
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refine ⟨u₁, ⟨mem_pair_imp_fst_mem_dom hu₁, hu₁⟩, ?_⟩
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intro u₂ hu₂
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have hu₂' := hu₂
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unfold comp at hu₂'
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simp only [Set.mem_setOf_eq] at hu₂'
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have ⟨_, ⟨t₂, ht₂⟩⟩ := hu₂'
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have ht : t₁ = t₂ := single_rooted_eq_unique hF ht₁.right ht₂.right
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rw [ht] at ht₁
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exact single_rooted_eq_unique hG ht₂.left ht₁.left
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theorem exercise_3_17_ii {F G : Set.Relation α}
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(hF : F.isOneToOne) (hG : G.isOneToOne)
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: (F.comp G).isOneToOne := And.intro
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(single_valued_comp_is_single_valued hF.left hG.left)
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(exercise_3_17_i hF.right hG.right)
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/-! #### Exercise 3.18
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Let `R` be the set
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```
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{⟨0, 1⟩, ⟨0, 2⟩, ⟨0, 3⟩, ⟨1, 2⟩, ⟨1, 3⟩, ⟨2, 3⟩}
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```
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Evaluate the following: `R ∘ R`, `R ↾ {1}`, `R⁻¹ ↾ {1}`, `R⟦{1}⟧`, and
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`R⁻¹⟦{1}⟧`.
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-/
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section
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variable {R : Set.Relation ℕ}
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variable (hR : R = {(0, 1), (0, 2), (0, 3), (1, 2), (1, 3), (2, 3)})
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theorem exercise_3_18_i
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: R.comp R = {(0, 2), (0, 3), (1, 3)} := by
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rw [hR]
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unfold comp
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simp only [Set.mem_singleton_iff, Set.mem_insert_iff, or_self, Prod.mk.injEq]
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ext x
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have (a, b) := x
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apply Iff.intro
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· simp only [Set.mem_setOf_eq, Set.mem_singleton_iff, Set.mem_insert_iff]
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intro ⟨t, ht₁, ht₂⟩
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casesm* _ ∨ _
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all_goals case _ hl hr => first
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| {rw [hl.right] at hr; simp at hr}
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| {rw [hl.left] at hr; simp at hr}
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| {rw [hl.left, hr.right]; simp}
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· simp only [
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Set.mem_singleton_iff,
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Set.mem_insert_iff,
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Prod.mk.injEq,
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Set.mem_setOf_eq
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]
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intro h
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casesm* _ ∨ _
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· case _ h =>
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refine ⟨1, Or.inl ⟨h.left, rfl⟩, ?_⟩
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iterate 3 right
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left
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exact ⟨rfl, h.right⟩
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· case _ h =>
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refine ⟨1, Or.inl ⟨h.left, rfl⟩, ?_⟩
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iterate 4 right
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left
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exact ⟨rfl, h.right⟩
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· case _ h =>
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refine ⟨2, ?_, ?_⟩
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· iterate 3 right
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left
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exact ⟨h.left, rfl⟩
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· iterate 5 right
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exact ⟨rfl, h.right⟩
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theorem exercise_3_18_ii
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: R.restriction {1} = {(1, 2), (1, 3)} := by
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rw [hR]
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unfold restriction
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ext p
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have (a, b) := p
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simp only [
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Set.mem_singleton_iff,
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Set.mem_insert_iff,
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Set.mem_setOf_eq,
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or_self
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]
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apply Iff.intro
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· intro ⟨hp, ha⟩
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rw [ha]
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simp only [Prod.mk.injEq, true_and]
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casesm* _ ∨ _
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all_goals case _ h => first
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| {rw [ha] at h; simp at h}
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| {simp only [Prod.mk.injEq] at h; left; exact h.right}
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| {simp only [Prod.mk.injEq] at h; right; exact h.right}
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· intro h
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apply Or.elim h
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· intro hab
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simp only [Prod.mk.injEq] at hab
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refine ⟨?_, hab.left⟩
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iterate 3 right
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left
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rw [hab.left, hab.right]
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· intro hab
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simp only [Prod.mk.injEq] at hab
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refine ⟨?_, hab.left⟩
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iterate 4 right
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left
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rw [hab.left, hab.right]
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theorem exercise_3_18_iii
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: R.inv.restriction {1} = {(1, 0)} := by
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rw [hR]
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unfold inv restriction
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ext p
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have (a, b) := p
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simp only [
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Set.mem_singleton_iff,
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Set.mem_insert_iff,
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or_self,
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exists_eq_or_imp,
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exists_eq_left,
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Set.mem_setOf_eq,
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Prod.mk.injEq
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]
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apply Iff.intro
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· intro ⟨hb, ha⟩
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casesm* _ ∨ _
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all_goals case _ hr => first
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| exact ⟨ha, hr.right.symm⟩
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| rw [ha] at hr; simp at hr
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· intro ⟨ha, hb⟩
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rw [ha, hb]
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simp
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theorem exercise_3_18_iv
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: R.image {1} = {2, 3} := by
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rw [hR]
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unfold image
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ext y
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simp
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theorem exercise_3_18_v
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: R.inv.image {1} = {0} := by
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rw [hR]
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unfold inv image
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ext y
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simp
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end
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end Relation
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end Enderton.Set.Chapter_3
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