Enderton. Formally verify theorem 3G.

finite-set-exercises
Joshua Potter 2023-06-26 07:08:25 -06:00
parent 3db21d7a1a
commit 931e617161
3 changed files with 56 additions and 8 deletions

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@ -3194,7 +3194,7 @@ For any one-to-one function $F$, $F^{-1}$ is also one-to-one.
\end{proof} \end{proof}
\subsection{\partial{Theorem 3G}}% \subsection{\verified{Theorem 3G}}%
\label{sub:theorem-3g} \label{sub:theorem-3g}
\begin{theorem}[3G] \begin{theorem}[3G]
@ -3207,6 +3207,14 @@ For any one-to-one function $F$, $F^{-1}$ is also one-to-one.
\begin{proof} \begin{proof}
\statementpadding
\lean*{Bookshelf/Enderton/Set/Chapter\_3}
{Enderton.Set.Chapter\_3.theorem\_3g\_i}
\lean{Bookshelf/Enderton/Set/Chapter\_3}
{Enderton.Set.Chapter\_3.theorem\_3g\_ii}
Suppose $F$ is a one-to-one \nameref{ref:function}. Suppose $F$ is a one-to-one \nameref{ref:function}.
Then \nameref{sub:lemma-1} indicates $F^{-1}$ is a one-to-one function with Then \nameref{sub:lemma-1} indicates $F^{-1}$ is a one-to-one function with
domain $\ran{F}$ and range $\dom{F}$. domain $\ran{F}$ and range $\dom{F}$.

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@ -436,18 +436,28 @@ theorem exercise_6_9_ii {A : Set (Set.Relation α)}
Assume that `F` is a one-to-one function. If `x ∈ dom F`, then `F⁻¹(F(x)) = x`. Assume that `F` is a one-to-one function. If `x ∈ dom F`, then `F⁻¹(F(x)) = x`.
-/ -/
theorem theorem_3g_i {F : Set.Relation α} theorem theorem_3g_i {α : Type _} {x y : α} {F : Set.Relation α}
(hF : Set.Relation.isOneToOne F) (hx : x ∈ F.dom) (hF : Set.Relation.isOneToOne F) (hx : x ∈ Set.Relation.dom F)
: ∃! y, (x, y) ∈ F ∧ (y, x) ∈ F.inv := by : ∃! y, (x, y) ∈ F ∧ (y, x) ∈ F.inv := by
sorry simp only [Set.Relation.mem_self_comm_mem_inv, and_self]
have ⟨y, hy⟩ := Set.Relation.dom_exists hx
refine ⟨y, hy, ?_⟩
intro y₁ hy₁
unfold Set.Relation.isOneToOne at hF
exact (Set.Relation.single_valued_eq_unique hF.left hy hy₁).symm
/-- ### Theorem 3G (ii) /-- ### Theorem 3G (ii)
Assume that `F` is a one-to-one function. If `y ∈ ran F`, then `F(F⁻¹(y)) = y`. Assume that `F` is a one-to-one function. If `y ∈ ran F`, then `F(F⁻¹(y)) = y`.
-/ -/
theorem theorem_3g_ii {F : Set.Relation α} theorem theorem_3g_ii {F : Set.Relation α}
(hF : Set.Relation.isOneToOne F) (hy : x ∈ F.ran) (hF : Set.Relation.isOneToOne F) (hy : y ∈ F.ran)
: ∃! x, (x, y) ∈ F ∧ (y, x) ∈ F.inv := by : ∃! x, (x, y) ∈ F ∧ (y, x) ∈ F.inv := by
sorry simp only [Set.Relation.mem_self_comm_mem_inv, and_self]
have ⟨x, hx⟩ := Set.Relation.ran_exists hy
refine ⟨x, hx, ?_⟩
intro x₁ hx₁
unfold Set.Relation.isOneToOne at hF
exact (Set.Relation.single_rooted_eq_unique hF.right hx hx₁).symm
end Enderton.Set.Chapter_3 end Enderton.Set.Chapter_3

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@ -43,7 +43,7 @@ theorem mem_pair_imp_fst_mem_dom {R : Relation α} (h : (x, y) ∈ R)
If `x ∈ dom R`, there exists some `y` such that `⟨x, y⟩ ∈ R`. If `x ∈ dom R`, there exists some `y` such that `⟨x, y⟩ ∈ R`.
-/ -/
theorem dom_exists {R : Relation α} (hx : x ∈ R.dom) theorem dom_exists {R : Relation α} (hx : x ∈ R.dom)
: ∃ y, (x, y) ∈ R := by : ∃ y : α, (x, y) ∈ R := by
unfold dom at hx unfold dom at hx
simp only [mem_image, Prod.exists, exists_and_right, exists_eq_right] at hx simp only [mem_image, Prod.exists, exists_and_right, exists_eq_right] at hx
exact hx exact hx
@ -63,7 +63,7 @@ theorem mem_pair_imp_snd_mem_ran {R : Relation α} (h : (x, y) ∈ R)
If `x ∈ ran R`, there exists some `t` such that `⟨t, x⟩ ∈ R`. If `x ∈ ran R`, there exists some `t` such that `⟨t, x⟩ ∈ R`.
-/ -/
theorem ran_exists {R : Relation α} (hx : x ∈ R.ran) theorem ran_exists {R : Relation α} (hx : x ∈ R.ran)
: ∃ t, (t, x) ∈ R := by : ∃ t : α, (t, x) ∈ R := by
unfold ran at hx unfold ran at hx
simp only [mem_image, Prod.exists, exists_eq_right] at hx simp only [mem_image, Prod.exists, exists_eq_right] at hx
exact hx exact hx
@ -192,6 +192,21 @@ exists exactly one `x` such that `⟨x, y⟩ ∈ R`.
def isSingleRooted (R : Relation α) : Prop := def isSingleRooted (R : Relation α) : Prop :=
∀ y ∈ R.ran, ∃! x, x ∈ R.dom ∧ (x, y) ∈ R ∀ y ∈ R.ran, ∃! x, x ∈ R.dom ∧ (x, y) ∈ R
/--
A single-rooted `Relation` should map the same output to the same input.
-/
theorem single_rooted_eq_unique {R : Relation α} {x₁ x₂ y : α}
(hR : isSingleRooted R)
: (x₁, y) ∈ R → (x₂, y) ∈ R → x₁ = x₂ := by
intro hx₁ hx₂
unfold isSingleRooted at hR
have := hR y (mem_pair_imp_snd_mem_ran hx₁)
have ⟨y₁, hy₁⟩ := this
simp only [and_imp] at hy₁
have h₁ := hy₁.right x₁ (mem_pair_imp_fst_mem_dom hx₁) hx₁
have h₂ := hy₁.right x₂ (mem_pair_imp_fst_mem_dom hx₂) hx₂
rw [h₁, h₂]
/-- /--
A `Relation` `R` is said to be single-valued **iff** for all `x ∈ dom R`, there A `Relation` `R` is said to be single-valued **iff** for all `x ∈ dom R`, there
exists exactly one `y` such that `⟨x, y⟩ ∈ R`. exists exactly one `y` such that `⟨x, y⟩ ∈ R`.
@ -201,6 +216,21 @@ Notice, a `Relation` that is single-valued is a function.
def isSingleValued (R : Relation α) : Prop := def isSingleValued (R : Relation α) : Prop :=
∀ x ∈ R.dom, ∃! y, y ∈ R.ran ∧ (x, y) ∈ R ∀ x ∈ R.dom, ∃! y, y ∈ R.ran ∧ (x, y) ∈ R
/--
A single-valued `Relation` should map the same input to the same output.
-/
theorem single_valued_eq_unique {R : Relation α} {x y₁ y₂ : α}
(hR : isSingleValued R)
: (x, y₁) ∈ R → (x, y₂) ∈ R → y₁ = y₂ := by
intro hy₁ hy₂
unfold isSingleValued at hR
have := hR x (mem_pair_imp_fst_mem_dom hy₁)
have ⟨x₁, hx₁⟩ := this
simp only [and_imp] at hx₁
have h₁ := hx₁.right y₁ (mem_pair_imp_snd_mem_ran hy₁) hy₁
have h₂ := hx₁.right y₂ (mem_pair_imp_snd_mem_ran hy₂) hy₂
rw [h₁, h₂]
/-- /--
For a set `F`, `F⁻¹` is a function **iff** `F` is single-rooted. For a set `F`, `F⁻¹` is a function **iff** `F` is single-rooted.
-/ -/