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

Bookshelf/Enderton/Set.tex

@@ -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 $A = B = \emptyset$.
+      Then it follows $A = B = \emptyset = 0$.
-      Therefore $B \equin 0$ and $$A - B = \emptyset \equin 0 = 0 - 0$$ as
+      Since $0 \leq 0$, $B \equin 0$, and $A - B = \emptyset \equin 0 = 0 - 0$,
-        expected.
+        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
+        then there exists some $n \in \omega$ such that $n \leq m$,
-        $A - B \equin m - n$.
+        $B \equin n$, and $A - B \equin m - n$.
 
   \end{proof}
 

Bookshelf/Enderton/Set/Chapter_4.lean

@@ -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

Bookshelf/Enderton/Set/Chapter_6.lean

@@ -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 ht₂ : f t ∈ Set.Iio n := hf.left ht₁.left
-        have ht₃ : F t ∈ R := nr (F t) ht₂
+      have ht₃ : f t ∈ R := nr (f t) ht₂
       exact absurd ⟨t, ht₁, rfl⟩ ht₃.right
 
-    suffices Set.BijOn F S' R by
+  have : 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⟩
     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`
+> Since `f` is one-to-one and onto, there exists a unique value `a ∈ A` such
-    -- such that `f(a) = m`.
+> 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)