Theorem Proving in Lean. Solutions to chapter 3 exercises.

src/theorem-proving-in-lean/exercises-3.lean

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+/- Exercises 3.7
+ -
+ - Avigad, Jeremy. ‘Theorem Proving in Lean’, n.d.
+-/
+
+-- Exercise 1
+--
+-- Prove the following identities, replacing the "sorry" placeholders with
+-- actual proofs.
+section ex_1
+
+open or
+
+variables p q r : Prop
+
+-- Commutativity of ∧ and ∨
+example : p ∧ q ↔ q ∧ p :=
+  iff.intro
+    (assume ⟨p, q⟩, ⟨q, p⟩)
+    (assume ⟨q, p⟩, ⟨p, q⟩)
+example : p ∨ q ↔ q ∨ p :=
+  iff.intro
+    (assume h, h.elim
+      (assume hp : p, inr hp)
+      (assume hq : q, inl hq))
+    (assume h, h.elim
+      (assume hq : q, inr hq)
+      (assume hp : p, inl hp))
+
+-- Associativity of ∧ and ∨
+example : (p ∧ q) ∧ r ↔ p ∧ (q ∧ r) :=
+  iff.intro
+    (assume ⟨⟨p, q⟩, r⟩, ⟨p, q, r⟩)
+    (assume ⟨p, q, r⟩, ⟨⟨p, q⟩, r⟩)
+example : (p ∨ q) ∨ r ↔ p ∨ (q ∨ r) :=
+  iff.intro
+    (assume h₁, h₁.elim
+      (assume h₂, h₂.elim (assume p, inl p) (assume q, inr (inl q)))
+      (assume r, inr (inr r)))
+    (assume h₁, h₁.elim
+      (assume p, inl (inl p))
+      (assume h₂, h₂.elim (assume q, inl (inr q)) (assume r, inr r)))
+
+-- Distributivity
+example : p ∧ (q ∨ r) ↔ (p ∧ q) ∨ (p ∧ r) :=
+  iff.intro
+    (assume ⟨hp, hqr⟩, hqr.elim
+      (assume hq, inl ⟨hp, hq⟩)
+      (assume hr, inr ⟨hp, hr⟩))
+    (assume h₁, h₁.elim
+      (assume ⟨hp, hq⟩, ⟨hp, inl hq⟩)
+      (assume ⟨hp, hr⟩, ⟨hp, inr hr⟩))
+example : p ∨ (q ∧ r) ↔ (p ∨ q) ∧ (p ∨ r) :=
+  iff.intro
+    (assume h, h.elim
+      (assume hp, ⟨inl hp, inl hp⟩)
+      (assume ⟨hq, hr⟩, ⟨inr hq, inr hr⟩))
+    (assume ⟨h₁, h₂⟩, h₁.elim
+      (assume hp, inl hp)
+      (assume hq, h₂.elim (assume hp, inl hp) (assume hr, inr ⟨hq, hr⟩)))
+
+-- Other properties
+example : (p → (q → r)) ↔ (p ∧ q → r) :=
+  iff.intro
+    (assume h ⟨hp, hq⟩, h hp hq)
+    (assume h hp hq, h ⟨hp, hq⟩)
+
+example : ((p ∨ q) → r) ↔ (p → r) ∧ (q → r) :=
+  iff.intro
+    (assume h,
+      have h₁ : p → r, from (assume hp, h (or.inl hp)),
+      have h₂ : q → r, from (assume hq, h (or.inr hq)),
+      ⟨h₁, h₂⟩)
+    (assume ⟨hpr, hqr⟩, assume hpq, hpq.elim hpr hqr)
+
+example : ¬(p ∨ q) ↔ ¬p ∧ ¬q :=
+  iff.intro
+    (assume h, and.intro
+      (assume hp, h (inl hp))
+      (assume hq, h (inr hq)))
+    (assume h₁ h₂, h₂.elim
+      (assume hp, have np : ¬p, from h₁.left, absurd hp np)
+      (assume hq, have nq : ¬q, from h₁.right, absurd hq nq))
+
+example : ¬p ∨ ¬q → ¬(p ∧ q) :=
+  assume h₁ h₂,
+  h₁.elim
+    (assume np : ¬p, have hp : p, from h₂.left, absurd hp np)
+    (assume nq : ¬q, have hq : q, from h₂.right, absurd hq nq)
+
+example : ¬(p ∧ ¬p) :=
+  assume h,
+  have hp : p, from h.left,
+  have np : ¬p, from h.right,
+  absurd hp np
+
+example : p ∧ ¬q → ¬(p → q) :=
+  assume ⟨hp, nq⟩ hpq,
+  absurd (hpq hp) nq
+
+example : ¬p → (p → q) :=
+  assume np hp,
+  absurd hp np
+
+example : (¬p ∨ q) → (p → q) :=
+  assume npq hp,
+  npq.elim
+    (assume np, absurd hp np)
+    (assume hq, hq)
+
+example : p ∨ false ↔ p :=
+  iff.intro
+    (assume hpf, hpf.elim
+      (assume hp, hp)
+      (assume hf, false.elim hf))
+    (assume hp, inl hp)
+
+example : p ∧ false ↔ false :=
+  iff.intro
+    (assume ⟨hp, hf⟩, hf)
+    (assume hf, false.elim hf)
+
+example : (p → q) → (¬q → ¬p) :=
+  assume hpq nq hp,
+  absurd (hpq hp) nq
+
+end ex_1
+
+-- Example 2
+--
+-- Prove the following identities, replacing the “sorry” placeholders with
+-- actual proofs. These require classical reasoning.
+section ex_2
+
+open classical
+
+variables p q r s : Prop
+
+example (hp : p) : (p → r ∨ s) → ((p → r) ∨ (p → s)) :=
+  assume h₁,
+  or.elim (h₁ hp)
+    (assume hr, or.inl (assume hp, hr))
+    (assume hs, or.inr (assume hp, hs))
+
+example : ¬(p ∧ q) → ¬p ∨ ¬q :=
+  assume npq,
+  or.elim (em p)
+    (assume hp, or.elim (em q)
+      (assume hq, false.elim (npq ⟨hp, hq⟩))
+      (assume nq, or.inr nq))
+    (assume np, or.inl np)
+
+example : ¬(p → q) → p ∧ ¬q :=
+  assume h₁,
+  and.intro
+    (by_contradiction (
+      assume np,
+      h₁ (λ (hp : p), absurd hp np)
+    ))
+    (assume hq, h₁ (λ (hp : p), hq))
+
+example : (p → q) → (¬p ∨ q) :=
+  assume hpq,
+  or.elim (em p)
+    (assume hp, or.inr (hpq hp))
+    (assume np, or.inl np)
+
+example : (¬q → ¬p) → (p → q) :=
+  assume h₁ hp,
+  by_contradiction (assume nq, absurd hp (h₁ nq))
+
+example : p ∨ ¬p := em p
+
+example : (((p → q) → p) → p) :=
+  assume h₁,
+  by_contradiction (
+    assume np,
+    suffices hp : p, from absurd hp np,
+    h₁ (λ (hp : p), absurd hp np)
+  )
+
+end ex_2
+
+-- Example 3
+--
+-- Prove `¬(p ↔ ¬p)` without using classical logic.
+section ex_3
+
+variable p : Prop
+
+example (hp : p) : ¬(p ↔ ¬p) :=
+  assume h,
+  absurd hp (iff.elim_left h (hp))
+
+end ex_3