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More LaTeX proof embedding. Finish up "Set Difference Size".

fin-dom-ran
Joshua Potter 2023-11-12 11:21:00 -07:00
parent f215a3180a
commit b596478a36
3 changed files with 341 additions and 84 deletions
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29
Bookshelf/Enderton/Set.tex
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@ -9204,7 +9204,7 @@
Clearly, if $S' = S$, then $S'$ is finite.
Therefore suppose $S'$ is a proper subset of $S$.
By definition of finite set, $S$ is \nameref{ref:equinumerous} to some
By definition of a finite set, $S$ is \nameref{ref:equinumerous} to some
natural number $n$.
Let $f$ be a one-to-one correspondence between $S$ and $n$.
Then $f \restriction S'$ is a one-to-one correspondence between $S'$ and
@ -9299,20 +9299,18 @@
By \nameref{sub:theorem-6a}, $n \equin \ran{f}$.
\end{proof}
\subsection{\pending{Set Difference Size}}%
\subsection{\verified{Set Difference Size}}%
\hyperlabel{sub:set-difference-size}
\begin{lemma}
Let $A \equin m$ for some natural number $m$ and $B \subseteq A$.
Then there exists some $n \in \omega$ such that $B \equin n$ and
Then there exists some $n \in \omega$ such that $n \leq m$, $B \equin n$ and
$A - B \equin m - n$.
\end{lemma}
\code{Bookshelf/Enderton/Set/Chapter\_6}
{Enderton.Set.Chapter\_6.sdiff\_size}
TODO: Update IH and add additional lemmas to fully prove this out.
\begin{proof}
Let
@ -9320,7 +9318,7 @@
\hyperlabel{sub:set-difference-size-ih}
S = \{m \in \omega \mid
\forall A \equin m, \forall B \subseteq A, \exists n
\in \omega (B \equin n \land A - B \equin m - n)\}.
\in \omega (n \leq m \land B \equin n \land A - B \equin m - n)\}.
\end{equation}
We prove that (i) $0 \in S$ and (ii) if $n \in S$ then $n^+ \in S$.
Afterward we prove (iii) the lemma statement.
@ -9329,9 +9327,9 @@
\hyperlabel{par:set-difference-size-i}
Let $A \equin 0$ and $B \subseteq A$.
Then it follows $A = B = \emptyset$.
Therefore $B \equin 0$ and $$A - B = \emptyset \equin 0 = 0 - 0$$ as
expected.
Then it follows $A = B = \emptyset = 0$.
Since $0 \leq 0$, $B \equin 0$, and $A - B = \emptyset \equin 0 = 0 - 0$,
it follows $0 \in S$.
\paragraph{(ii)}%
\hyperlabel{par:set-difference-size-ii}
@ -9345,7 +9343,7 @@
Then $B - \{a\} \subseteq A - \{a\}$ and $f$ is a one-to-one
correspondence between $A - \{a\}$ and $m$.
By \ihref{sub:set-difference-size-ih}, there exists some $n \in \omega$
such that $B - \{a\} \equin n$ and
such that $n \leq m$, $B - \{a\} \equin n$ and
\begin{equation}
\hyperlabel{par:set-difference-size-ii-eq1}
(A - \{a\}) - (B - \{a\}) \equin m - n.
@ -9377,7 +9375,7 @@
& = \{x \mid x \in A - B \land x \neq a\} \nonumber \\
& = \{x \mid x \in A - B \land x \not\in \{a\}\} \nonumber \\
& = (A - B) - \{a\}.
\label{par:set-difference-size-ii-eq2}
\hyperlabel{par:set-difference-size-ii-eq2}
\end{align}
Since $a \in A$ and $a \in B$, $(A - B) - \{a\} = A - B$.
Thus
@ -9396,7 +9394,7 @@
\begin{align*}
(A - \{a\}) - (B - \{a\})
& = (A - \{a\}) - B \\
& \equin m - n & \eqref{par:set-difference-size-ii-eq1}.
& \equin m - n. & \eqref{par:set-difference-size-ii-eq1}
\end{align*}
The above implies that there exists a one-to-one correspondence $g$
between $(A - \{a\}) - B$ and $m - n$.
@ -9408,7 +9406,8 @@
\subparagraph{Subconclusion}%
The above two cases are exhaustive and both conclude the existence of
some $n \in \omega$ such that $B \equin n$ and $A - B \equin m^+ - n$.
some $n \in \omega$ such that $n \leq m^+$, $B \equin n$ and
$A - B \equin m^+ - n$.
Hence $m^+ \in S$.
\paragraph{(iii)}%
@ -9418,8 +9417,8 @@
\nameref{ref:inductive-set}.
Thus \nameref{sub:theorem-4b} implies $S = \omega$.
Hence, for all $A \equin m$ for some $m \in \omega$, if $B \subseteq A$,
then there exists some $n \in \omega$ such that $B \equin n$ and
$A - B \equin m - n$.
then there exists some $n \in \omega$ such that $n \leq m$,
$B \equin n$, and $A - B \equin m - n$.
\end{proof}

173
Bookshelf/Enderton/Set/Chapter_4.lean
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@ -299,9 +299,25 @@ m ∈ n ↔ m ⬝ p ∈ n ⬝ p.
theorem theorem_4n_i (m n p : ℕ)
: m < n ↔ m + p < n + p := by
/-
> Let `m` and `n` be natural numbers.
>
> ##### (⇒)
> Suppose `m ∈ n`. Let
>
> `S = {p ∈ ω | m + p ∈ n + p}`
-/
have hf : ∀ m n : ℕ, m < n → m + p < n + p := by
induction p with
/-
> It trivially follows that `0 ∈ S`.
-/
| zero => simp
/-
> Next, suppose `p ∈ S`. That is, suppose `m + p ∈ n + p`. By *Lemma 4L(a)*,
> this holds if and only if `(m + p)⁺ ∈ (n + p)⁺`. *Theorem 4I* then implies
> that `m + p⁺ ∈ n + p⁺` meaning `p⁺ ∈ S`.
-/
| succ p ih =>
intro m n hp
have := ih m n hp
@ -310,29 +326,76 @@ theorem theorem_4n_i (m n p : ℕ)
have h₂ : (n + p).succ = n + p.succ := rfl
rwa [← h₁, ← h₂]
apply Iff.intro
/-
> Thus `S` is an inductive set. Hence *Theorem 4B* implies `S = ω`. Therefore,
> for all `p ∈ ω`, `m ∈n` implies `m + p ∈ n + p`.
-/
· exact hf m n
/-
> ##### (⇐)
> Let `p` be a natural number and suppose `m + p ∈ n + p`. By the
> *Trichotomy Law for `ω`*, there are two cases to consider regarding how `m`
> and `n` relate to one another:
-/
· intro h
match @trichotomous ℕ LT.lt _ m n with
| Or.inl h₁ =>
exact h₁
| Or.inr (Or.inl h₁) =>
/-
> ###### Case 1
> Suppose `m = n`. Then `m + p ∈ n + p = m + p`. *Lemma 4L(b)* shows this is
> impossible.
-/
rw [← h₁] at h
exact absurd h (lemma_4l_b (m + p))
| Or.inr (Or.inr h₁) =>
/-
> ###### Case 2
> Suppose `n ∈ m`. Then *(⇒)* indicates `n + p ∈ m + p`. But this contradicts
> the *Trichotomy Law for `ω`* since, by hypothesis, `m + p ∈ n + p`.
-/
have := hf n m h₁
exact absurd this (Nat.lt_asymm h)
| Or.inl h₁ =>
/-
> ###### Conclusion
> By trichotomy, it follows `m ∈ n`.
-/
exact h₁
#check Nat.add_lt_add_iff_right
theorem theorem_4n_ii (m n p : ℕ)
: m < n ↔ m * p.succ < n * p.succ := by
/-
> Let `m` and `n` be natural numbers.
>
> ##### (⇒)
> Suppose `m ∈ n`. Let
>
> `S = {p ∈ ω | m ⬝ p⁺ ∈ n ⬝ p⁺}`.
-/
have hf : ∀ m n : ℕ, m < n → m * p.succ < n * p.succ := by
intro m n hp₁
induction p with
| zero =>
/-
> `0 ∈ S` by *Right Multiplicative Identity*.
-/
simp only [Nat.mul_one]
exact hp₁
| succ p ih =>
/-
> Next, suppose `p ∈ S`. That is, `m ⬝ p⁺ ∈ n ⬝ p⁺`. Then
>
> `m ⬝ p⁺⁺ = m ⬝ p⁺ + m` *Theorem 4J*
> ` ∈ n ⬝ p⁺ + m` *(i)*
> ` = m + n ⬝ p⁺` *Theorem 4K-2*
> ` ∈ n + n ⬝ p⁺` *(i)*
> ` = n ⬝p⁺ + n` *Theorem 4K-2*
> ` n ⬝ p⁺⁺` *Theorem 4J*
>
> Therefore `p⁺ ∈ S`.
-/
have hp₂ : m * p.succ < n * p.succ := by
by_cases hp₃ : p = 0
· rw [hp₃] at *
@ -347,17 +410,43 @@ theorem theorem_4n_ii (m n p : ℕ)
_ = n * p.succ + n := by rw [theorem_4k_2]
_ = n * p.succ.succ := rfl
apply Iff.intro
/-
> Thus `S` is an inductive set. Hence *Theorem 4B* implies `S = ω`. By
> *Theorem 4C*, every natural number except `0` is the successor of some natural
> number. Therefore, for all `p ∈ ω` such that `p ≠ 0`, `m ∈ n` implies
`m ⬝ p ∈ n ⬝ p`.
-/
· exact hf m n
· intro hp
/-
> ##### (⇐)
> Let `p ≠ 0` be a natural number and suppose `m ⬝ p ∈ n ⬝ p`. By the
> *Trichotomy Law for `ω`*, there are two cases to consider regarding how `m`
> and `n` relate to one another.
-/
intro hp
match @trichotomous ℕ LT.lt _ m n with
| Or.inl h₁ =>
exact h₁
| Or.inr (Or.inl h₁) =>
/-
> ###### Case 1
> Suppose `m = n`. Then `m ⬝ p ∈ n ⬝ p = m ⬝ p`. *Lemma 4L(b)* shows this is
> impossible.
-/
rw [← h₁] at hp
exact absurd hp (lemma_4l_b (m * p.succ))
| Or.inr (Or.inr h₁) =>
/-
> ###### Case 2
> Suppose `n ∈ m`. Then *(⇒)* indicates `n ⬝ p ∈ m ⬝ p`. But this contradicts
> *Trichotomy Law for `ω`* since, by hypothesis, `m ⬝ p ∈ n ⬝ p`.
-/
have := hf n m h₁
exact absurd this (Nat.lt_asymm hp)
| Or.inl h₁ =>
/-
> ###### Conclusion
> By trichotomy, it follows `m ∈ n`.
-/
exact h₁
#check Nat.mul_lt_mul_of_pos_right
@ -372,15 +461,30 @@ m ⬝ p = n ⬝ p ∧ p ≠ 0 ⇒ m = n.
theorem corollary_4p_i (m n p : ℕ) (h : m + p = n + p)
: m = n := by
/-
> Suppose `m + p = n + p`. By the *Trichotomy Law for `ω`*, there are two cases
> to consider regarding how `m` and `n` relate to one another.
-/
match @trichotomous ℕ LT.lt _ m n with
| Or.inl h₁ =>
/-
> If `m ∈n`, then *Theorem 4N* implies `m + p ∈ n + p`.
-/
rw [theorem_4n_i m n p, h] at h₁
exact absurd h₁ (lemma_4l_b (n + p))
| Or.inr (Or.inl h₁) =>
exact h₁
| Or.inr (Or.inr h₁) =>
/-
> If `n ∈ m`, then *Theorem 4N* implies `n + p ∈ m + p`.
-/
rw [theorem_4n_i n m p, h] at h₁
exact absurd h₁ (lemma_4l_b (n + p))
/-
> Both of these contradict the *Trichotomy Law for `ω`* of `m + p` and `n + p`.
> Thus `m = n` is the only remaining possibility.
-/
| Or.inr (Or.inl h₁) =>
exact h₁
#check Nat.add_right_cancel
@ -391,25 +495,53 @@ Let `A` be a nonempty subset of `ω`. Then there is some `m ∈ A` such that
-/
theorem well_ordering_nat {A : Set ℕ} (hA : Set.Nonempty A)
: ∃ m ∈ A, ∀ n, n ∈ A → m ≤ n := by
-- Assume `A` does not have a least element.
/-
> Let `A` be a nonempty subset of `ω`. For the sake of contradiciton, suppose
> `A` does not have a least element.
-/
by_contra nh
simp only [not_exists, not_and, not_forall, not_le, exists_prop] at nh
-- If we show the complement of `A` is `ω`, then `A = ∅`, a contradiction.
/-
> It then suffices to prove that the complement of `A` equals `ω`. If we do so,
> then `A = ∅`, a contradiction.
-/
suffices A.compl = Set.univ by
have h := Set.univ_diff_compl_eq_self 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.
/-
> Define
>
> `S = {n ∈ ω | (∀ m ∈ n)m ∉ A}`.
>
> We prove `S` is an inductive set by showing that (i) `0 ∈ S` and (ii) if
> `n ∈ S`, then `n⁺ ∈ S`. Afterward we show that `ω - A = ω`, completing the
> proof.
-/
have : ∀ n : ℕ, (∀ m, m < n → m ∈ A.compl) := by
intro n
induction n with
| zero =>
/-
> #### (i)
> It vacuously holds that `0 ∈ S`.
-/
intro m hm
exact False.elim (Nat.not_lt_zero m hm)
| succ n ih =>
/-
> #### (ii)
> Suppose `n ∈ S`. We want to prove that
>
> `∀ m, m ∈ n⁺ ⇒ m ∉ A`.
>
> To this end, let `m ∈ ω` such that `m ∈ n⁺`. By definition of the successor,
> `m ∈ n` or `m = n`. If the former, `n ∈ S` implies `m ∉ A`. If the latter, it
> isn't possible for `n ∈ A` since the *Trichotomy Law for `ω`* would otherwise
> imply `n` is the least element of `A`, which is assumed to not exist. Hence
> `n⁺ ∈ S`.
-/
intro m hm
have hm' : m < n ∨ m = n := by
rw [Nat.lt_succ] at hm
@ -429,7 +561,12 @@ theorem well_ordering_nat {A : Set ℕ} (hA : Set.Nonempty A)
exact absurd hp.left (ih p hp.right)
· rw [h]
exact hn
/-
> #### Conclusion
> By *(i)* and *(ii)*, `S` is an inductive set. Since `S ⊆ ω`, *Theorem 4B*
> implies `S = ω`. Bu this immediately implies `ω = ω - A` meaning `A` is the
> empty set.
-/
ext x
simp only [Set.mem_univ, iff_true]
by_contra nh'
@ -451,11 +588,17 @@ theorem strong_induction_principle_nat (A : Set ℕ)
rw [this] at h'
simp only [Set.diff_empty] at h'
exact h'.symm
/-
> For the sake of contradiction, suppose `ω - A` is a nonempty set. By
> *Well Ordering of `ω`*, there exists a least element `m ∈ ω - A`.
-/
by_contra nh
have ⟨m, hm⟩ := well_ordering_nat (Set.nmem_singleton_empty.mp nh)
refine absurd (h m ?_) hm.left
/-
> Then every number less than `m` is in `A`. But then *(4.23)* implies `m ∈ A`,
> a contradiction. Thus `ω - A` is an empty set meaning `A = ω`.
-/
-- Show that every number less than `m` is in `A`.
intro x hx
by_contra nx

183
Bookshelf/Enderton/Set/Chapter_6.lean
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@ -763,15 +763,33 @@ Any subset of a finite set is finite.
-/
theorem corollary_6g {S S' : Set α} (hS : Set.Finite S) (hS' : S' ⊆ S)
: Set.Finite S' := by
rw [subset_iff_ssubset_or_eq] at hS'
/-
> Let `S` be a finite set and `S' ⊆ S`.
-/
rw [subset_iff_ssubset_or_eq, or_comm] at hS'
apply Or.elim hS'
/-
> Clearly, if `S' = S`, then `S'` is finite.
-/
· intro h
rwa [h]
/-
> Therefore suppose `S'` is a proper subset of `S`.
-/
intro h
/-
> By definition of a finite set, `S` is equinumerous to some natural number `n`.
> Let `f` be a one-to-one correspondence between `S` and `n`.
-/
rw [Set.finite_iff_equinumerous_nat] at hS
have ⟨n, F, hF⟩ := hS
have ⟨n, f, hf⟩ := hS
/-
> Then `f ↾ S'` is a one-to-one correspondence between `S'` and some proper
> subset of `n`.
-/
-- Mirrors logic found in `corollary_6c`.
let T := S \ S'
let R := (Set.Iio n) \ (F '' T)
let R := (Set.Iio n) \ (f '' T)
have hR : R ⊂ Set.Iio n := by
rw [Set.ssubset_def]
apply And.intro
@ -781,36 +799,33 @@ theorem corollary_6g {S S' : Set α} (hS : Set.Finite S) (hS' : S' ⊆ S)
· show ¬ ∀ x, x ∈ Set.Iio n → x ∈ R
intro nr
have ⟨t, ht₁⟩ : Set.Nonempty T := Set.diff_ssubset_nonempty h
have ht₂ : F t ∈ Set.Iio n := hF.left ht₁.left
have ht₃ : F t ∈ R := nr (F t) ht₂
have ht₂ : f t ∈ Set.Iio n := hf.left ht₁.left
have ht₃ : f t ∈ R := nr (f t) ht₂
exact absurd ⟨t, ht₁, rfl⟩ ht₃.right
suffices Set.BijOn F S' R by
have ⟨m, hm⟩ := lemma_6f hR
have := Set.equinumerous_trans ⟨F, this⟩ hm.right
exact Set.finite_iff_equinumerous_nat.mpr ⟨m, this⟩
have : Set.BijOn f S' R := by
refine ⟨?_, ?_, ?_⟩
· -- `Set.MapsTo f S' R`
intro x hx
dsimp only
simp only [Set.mem_diff, Set.mem_Iio, Set.mem_image, not_exists, not_and]
apply And.intro
· exact hF.left (subset_of_ssubset h hx)
· exact hf.left (subset_of_ssubset h hx)
· intro y hy
by_contra nf
have := hF.right.left (subset_of_ssubset h hx) hy.left nf.symm
have := hf.right.left (subset_of_ssubset h hx) hy.left nf.symm
rw [this] at hx
exact absurd hx hy.right
· -- `Set.InjOn f S'`
intro x₁ hx₁ x₂ hx₂ hf
intro x₁ hx₁ x₂ hx₂ hf'
have h₁ : x₁ ∈ S := subset_of_ssubset h hx₁
have h₂ : x₂ ∈ S := subset_of_ssubset h hx₂
exact hF.right.left h₁ h₂ hf
exact hf.right.left h₁ h₂ hf'
· -- `Set.SurjOn f S' R`
show ∀ x, x ∈ R → x ∈ F '' S'
show ∀ x, x ∈ R → x ∈ f '' S'
intro x hx
have h₁ := hF.right.right
have h₁ := hf.right.right
unfold Set.SurjOn at h₁
rw [Set.subset_def] at h₁
have ⟨y, hy⟩ := h₁ x hx.left
@ -820,9 +835,16 @@ theorem corollary_6g {S S' : Set α} (hS : Set.Finite S) (hS' : S' ⊆ S)
simp only [Set.mem_image, Set.mem_diff, not_exists, not_and] at hx
by_contra ny
exact (hx.right y ⟨hy.left, ny⟩) rfl
/-
> By *Lemma 6f*, `ran (f ↾ S')` is equinumerous to some `m < n`.
-/
have ⟨m, hm⟩ := lemma_6f hR
/-
> Then *Theorem 6A* indicates `S' ≈ m`. Hence `S'` is a finite set.
-/
have := Set.equinumerous_trans ⟨f, this⟩ hm.right
exact Set.finite_iff_equinumerous_nat.mpr ⟨m, this⟩
· intro h
rwa [h]
/-- ### Subset Size
@ -832,10 +854,18 @@ such that `B ≈ m`, `A ≈ n`, and `m ≤ n`.
lemma subset_size [DecidableEq α] [Nonempty α] {A B : Set α}
(hBA : B ⊆ A) (hA : Set.Finite A)
: ∃ m n : ℕ, B ≈ Set.Iio m ∧ A ≈ Set.Iio n ∧ m ≤ n := by
/-
> Let `A` be a finite set and `B` be a subset of `A`. By *Corollary 6G*, `B`
> must be finite. By definition of a finite set, there exists natural numbers
> `m, n ∈ ω` such that `B ≈ m` and `A ≈ n`.
-/
have ⟨n, hn⟩ := Set.finite_iff_equinumerous_nat.mp hA
have ⟨m, hm⟩ := Set.finite_iff_equinumerous_nat.mp (corollary_6g hA hBA)
refine ⟨m, n, hm, hn, ?_⟩
/-
> By the *Trichotomy Law for `ω`*, it suffices to prove that `m > n` is not
> possible for then either `m < n` or `m = n`.
-/
suffices ¬ m > n by
match @trichotomous ℕ LT.lt _ m n with
| Or.inr (Or.inl hr) => -- m = n
@ -844,11 +874,22 @@ lemma subset_size [DecidableEq α] [Nonempty α] {A B : Set α}
exact absurd hr this
| Or.inl hr => -- m < n
exact Nat.le_of_lt hr
/-
> For the sake of contradiction, assume `m > n`. By definition of equinumerous,
> there exists a one-to-one correspondence between `B` and `m`. *Theorem 6A*
> indicates there then exists a one-to-one correspondence `f` between `m` and
> `B`. Likewise, there exists a one-to-one correspondence `g` between `A` and
> `n`.
-/
by_contra nr
have ⟨f, hf⟩ := Set.equinumerous_symm hm
have ⟨g, hg⟩ := hn
/-
> Define `h : A → B` as `h(x) = f(g(x))` for all `x ∈ A`. Since `n ⊂ m` by
> *Corollary 4M*, `h` is well-defined. By *One-to-One Composition*, `h` must be
> one-to-one. thus `h` is a one-to-one correspondence between `A` and `ran h`,
> i.e. `A ≈ ran h`.
-/
let h x := f (g x)
have hh : Set.BijOn h A (h '' A) := by
refine ⟨?_, ?_, Eq.subset rfl⟩
@ -860,7 +901,11 @@ lemma subset_size [DecidableEq α] [Nonempty α] {A B : Set α}
refine Set.InjOn.comp hf.right.left hg.right.left ?_
intro x hx
exact Nat.lt_trans (hg.left hx) nr
/-
> But `n < m` meaning `ran h ⊂ B` which in turn is a proper subset of `A` by
> hypothesis. *Corollary 6C* states no finite set is equinumerous to a proper
> subset of itself, a contradiction.
-/
have : h '' A ⊂ A := by
rw [Set.ssubset_def]
apply And.intro
@ -918,8 +963,21 @@ lemma sdiff_size_aux [DecidableEq α] [Nonempty α]
: ∀ A : Set α, A ≈ Set.Iio m →
∀ B, B ⊆ A →
∃ n : ℕ, n ≤ m ∧ B ≈ Set.Iio n ∧ A \ B ≈ (Set.Iio m) \ (Set.Iio n) := by
/-
> `Let
>
> `S = {m ∈ ω | ∀ A ≈ m, ∀ B ⊆ A, ∃ n ∈ ω(n ≤ m ∧ B ≈ n ∧ A - B ≈ m - n) }`.
>
> We prove that (i) `0 ∈ S` and (ii) if `n ∈ S` then `n⁺ ∈ S`. Afterward we
> prove (iii) the lemma statement.
-/
induction m with
| zero =>
/-
> #### (i)
> Let `A ≈ 0` and `B ⊆ A`. Then it follows `A = B = ∅ = 0`. Since `0 ≤ 0`,
> `B ≈ 0`, and `A - B = ∅ ≈ 0 = 0 - 0`, it follows `0 ∈ S`.
-/
intro A hA B hB
refine ⟨0, ?_⟩
simp only [
@ -935,17 +993,25 @@ lemma sdiff_size_aux [DecidableEq α] [Nonempty α]
exact hA
rw [this]
refine ⟨trivial, hB', Set.equinumerous_emptyset_emptyset⟩
| succ m ih =>
/-
> #### (ii)
> Suppose `m ∈ S` and consider `m⁺`. Let `A ≈ m⁺` and let `B ⊆ A`. By definition
> of equinumerous, there exists a one-to-one corerspondnece `f` between `A` and
> `m⁺`.
-/
intro A ⟨f, hf⟩ B hB
-- Since `f` is one-to-one and onto, there exists a unique value `a ∈ A`
-- such that `f(a) = m`.
/-
> Since `f` is one-to-one and onto, there exists a unique value `a ∈ A` such
> that `f(a) = m`.
-/
have hfa := hf.right.right
unfold Set.SurjOn at hfa
have ⟨a, ha₁, ha₂⟩ := (Set.subset_def ▸ hfa) m (by simp)
-- `f` is a one-to-one correspondence between `A - {a}` and `m`.
/-
> Then `B - {a} ⊆A - {a}` and `f` is a one-to-one correspondence between
> `A - {a}` and `m`.
-/
have hBA : B \ {a} ⊆ A \ {a} := Set.diff_subset_diff_left hB
have hfBA : Set.BijOn f (A \ {a}) (Set.Iio m) := by
refine ⟨?_, ?_, ?_⟩
@ -974,12 +1040,21 @@ lemma sdiff_size_aux [DecidableEq α] [Nonempty α]
by_contra nb
rw [← nb, hb.right] at ha₂
exact absurd ha₂ (Nat.ne_of_lt hx)
/-
> By *(IH)*, there exists some `n ∈ ω` such that `n ≤ m`, `B - {a} ≈ n` and
>
> `(A - {a}) - (B - {a}) ≈ m - n`. (6.4)
>
> There are two cases to consider:
-/
-- `(A - {a}) - (B - {a}) ≈ m - n`
have ⟨n, hn₁, hn₂, hn₃⟩ := ih (A \ {a}) ⟨f, hfBA⟩ (B \ {a}) hBA
by_cases hc : a ∈ B
· refine ⟨n.succ, ?_, ?_, ?_⟩
/-
> ##### Case 1
> Assume `a ∈ B`. Then `B ≈ n⁺`.
-/
· exact Nat.succ_le_succ hn₁
· -- `B ≈ Set.Iio n.succ`
have ⟨g, hg⟩ := hn₂
@ -1024,12 +1099,28 @@ lemma sdiff_size_aux [DecidableEq α] [Nonempty α]
rwa [if_neg hb₂]
· intro hx₁
exact absurd hx₁ hc₁
/-
> Furthermore, by definition of the set difference,
>
> `...`
-/
· have hA₁ : (A \ {a}) \ (B \ {a}) = (A \ B) \ {a} :=
Set.diff_mem_diff_mem_eq_diff_diff_mem
/-
> Since `a ∈ A` and `a ∈ B`, `(A - B) - {a} = A - B`.
-/
have hA₂ : (A \ B) \ {a} = A \ B := by
refine Set.not_mem_diff_eq_self ?_
by_contra na
exact absurd hc na.right
/-
> Thus
>
> `(A - {a} - (B - {a})) = (A - B) - {a}`
> ` = A - B`
> ` ≈ m - n` *(6.4)*
> ` ≈ m⁺ - n⁺` *(Theorem 6A)*
-/
rw [hA₁, hA₂] at hn₃
suffices (Set.Iio m) \ (Set.Iio n) ≈ (Set.Iio m.succ) \ (Set.Iio n.succ)
from Set.equinumerous_trans hn₃ this
@ -1063,13 +1154,24 @@ lemma sdiff_size_aux [DecidableEq α] [Nonempty α]
exact Nat.lt_succ.mp hx₂
· rw [hp] at hx₁
exact Nat.succ_lt_succ_iff.mp hx₁
/-
> ##### Case 2
> Assume `a ∉ B`. Then `B - {a} = B` (i.e. `B ≈ n`) and
>
> `(A - {a}) - (B - {a}) = (A - {a}) - B`
> ` ≈ m - n`. *(6.4)
-/
· have hB : B \ {a} = B := Set.not_mem_diff_eq_self hc
refine ⟨n, ?_, ?_, ?_⟩
· calc n
_ ≤ m := hn₁
_ ≤ m + 1 := by simp
· rwa [← hB]
/-
> The above implies that there exists a one-to-one correspondence `g` between
> `(A - {a}) - B` and `m - n`. Therefore `g ∪ {⟨a, m⟩}` is a one-to-one
> correspondence between `A - B` and `(m - n) ∪ {m}`.
-/
· rw [hB] at hn₃
have ⟨g, hg⟩ := hn₃
have hAB : A \ B ≈ (Set.Iio m) \ (Set.Iio n) ∪ {m} := by
@ -1104,7 +1206,7 @@ lemma sdiff_size_aux [DecidableEq α] [Nonempty α]
exact absurd h.symm (Nat.ne_of_lt this.right)
· rw [if_neg hc₁, if_pos hc₂] at h
have := hg.left ⟨⟨hx₁.left, hc₁⟩, hx₁.right⟩
simp at this
simp only [Set.Iio_diff_Iio, gt_iff_lt, not_lt, ge_iff_le, Set.mem_Ico] at this
exact absurd h (Nat.ne_of_lt this.right)
· rw [if_neg hc₁, if_neg hc₂] at h
exact hg.right.left ⟨⟨hx₁.left, hc₁⟩, hx₁.right⟩ ⟨⟨hx₂.left, hc₂⟩, hx₂.right⟩ h
@ -1120,10 +1222,13 @@ lemma sdiff_size_aux [DecidableEq α] [Nonempty α]
refine ⟨y, ?_, ?_⟩
· exact ⟨hy₁.left.left, hy₁.right⟩
· rwa [if_neg hy₁.left.right]
/-
> By *Theorem 6A*
>
> `A - B ≈ (m - n) ∪ {m} ≈ m⁺ - n`.
-/
suffices (Set.Iio m) \ (Set.Iio n) ∪ {m} ≈ (Set.Iio m.succ) \ (Set.Iio n)
from Set.equinumerous_trans hAB this
refine ⟨fun x => x, ?_, ?_, ?_⟩
· intro x hx
simp at hx ⊢
@ -1161,6 +1266,16 @@ lemma sdiff_size_aux [DecidableEq α] [Nonempty α]
· intro hx
left
exact hx
/-
> ##### Subconclusion
> The above two cases are exhaustive and both conclude the existence of some
> `n ∈ ω` such that `n ≤ m⁺`, `B ≈ n`, and `A - B ≈ m⁺ - n`. Hence `m⁺ ∈ S`.
>
> #### (iii)
> By *(i)* and *(ii)*, `S ⊆ ω` is an inductive set. Thus *Theorem 4B* implies
> `S = ω`. Hence, for all `A ≈ m` for some `m ∈ ω`, if `B ⊆ A`, then there
> exists some `n ∈ ω` such that `n ≤ m`, `B ≈ n`, and `A - B ≈ m - n`.
-/
lemma sdiff_size [DecidableEq α] [Nonempty α] {A B : Set α}
(hB : B ⊆ A) (hA : A ≈ Set.Iio m)