168 lines
4.2 KiB
Plaintext
168 lines
4.2 KiB
Plaintext
/-
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Chapter 3
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Propositions and Proofs
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-/
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-- ========================================
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-- Exercise 1
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--
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-- Prove the following identities.
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-- ========================================
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namespace ex1
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open or
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variable (p q r : Prop)
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-- Commutativity of ∧ and ∨
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example : p ∧ q ↔ q ∧ p :=
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Iff.intro
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(fun ⟨hp, hq⟩ => show q ∧ p from ⟨hq, hp⟩)
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(fun ⟨hq, hp⟩ => show p ∧ q from ⟨hp, hq⟩)
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example : p ∨ q ↔ q ∨ p :=
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Iff.intro
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(fun h => h.elim Or.inr Or.inl)
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(fun h => h.elim Or.inr Or.inl)
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-- Associativity of ∧ and ∨
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example : (p ∧ q) ∧ r ↔ p ∧ (q ∧ r) :=
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Iff.intro
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(fun ⟨⟨hp, hq⟩, hr⟩ => ⟨hp, hq, hr⟩)
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(fun ⟨hp, hq, hr⟩ => ⟨⟨hp, hq⟩, hr⟩)
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example : (p ∨ q) ∨ r ↔ p ∨ (q ∨ r) :=
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Iff.intro
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(fun h₁ => h₁.elim
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(fun h₂ => h₂.elim Or.inl (Or.inr ∘ Or.inl))
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(Or.inr ∘ Or.inr))
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(fun h₁ => h₁.elim
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(Or.inl ∘ Or.inl)
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(fun h₂ => h₂.elim (Or.inl ∘ Or.inr) Or.inr))
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-- Distributivity
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example : p ∧ (q ∨ r) ↔ (p ∧ q) ∨ (p ∧ r) :=
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Iff.intro
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(fun ⟨hp, hqr⟩ => hqr.elim (Or.inl ⟨hp, ·⟩) (Or.inr ⟨hp, ·⟩))
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(fun h₁ => h₁.elim
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(fun ⟨hp, hq⟩ => ⟨hp, Or.inl hq⟩)
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(fun ⟨hp, hr⟩ => ⟨hp, Or.inr hr⟩))
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example : p ∨ (q ∧ r) ↔ (p ∨ q) ∧ (p ∨ r) :=
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Iff.intro
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(fun h => h.elim
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(fun hp => ⟨Or.inl hp, Or.inl hp⟩)
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(fun ⟨hq, hr⟩ => ⟨Or.inr hq, Or.inr hr⟩))
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(fun ⟨h₁, h₂⟩ => h₁.elim
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Or.inl
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(fun hq => h₂.elim Or.inl (fun hr => Or.inr ⟨hq, hr⟩)))
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-- Other properties
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example : (p → (q → r)) ↔ (p ∧ q → r) :=
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Iff.intro
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(fun h ⟨hp, hq⟩ => h hp hq)
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(fun h hp hq => h ⟨hp, hq⟩)
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example : ((p ∨ q) → r) ↔ (p → r) ∧ (q → r) :=
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Iff.intro
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(fun h =>
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have h₁ : p → r := h ∘ Or.inl
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have h₂ : q → r := h ∘ Or.inr
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show (p → r) ∧ (q → r) from ⟨h₁, h₂⟩)
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(fun ⟨h₁, h₂⟩ h => h.elim h₁ h₂)
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example : ¬(p ∨ q) ↔ ¬p ∧ ¬q :=
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Iff.intro
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(fun h => ⟨h ∘ Or.inl, h ∘ Or.inr⟩)
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(fun h₁ h₂ => h₂.elim (absurd · h₁.left) (absurd · h₁.right))
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example : ¬p ∨ ¬q → ¬(p ∧ q) :=
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fun h₁ h₂ => h₁.elim (absurd h₂.left ·) (absurd h₂.right ·)
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example : ¬(p ∧ ¬p) :=
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fun h => absurd h.left h.right
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example : p ∧ ¬q → ¬(p → q) :=
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fun ⟨hp, nq⟩ hpq => absurd (hpq hp) nq
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example : ¬p → (p → q) :=
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fun np hp => absurd hp np
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example : (¬p ∨ q) → (p → q) :=
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fun npq hp => npq.elim (absurd hp ·) id
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example : p ∨ False ↔ p :=
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Iff.intro (fun hpf => hpf.elim id False.elim) Or.inl
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example : p ∧ False ↔ False :=
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Iff.intro (fun ⟨_, hf⟩ => hf) False.elim
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example : (p → q) → (¬q → ¬p) :=
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fun hpq nq hp => absurd (hpq hp) nq
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end ex1
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-- ========================================
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-- Exercise 2
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--
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-- Prove the following identities. These require classical reasoning.
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-- ========================================
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namespace ex2
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open Classical
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variable (p q r s : Prop)
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example (hp : p) : (p → r ∨ s) → ((p → r) ∨ (p → s)) :=
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fun h => (h hp).elim
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(fun hr => Or.inl (fun _ => hr))
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(fun hs => Or.inr (fun _ => hs))
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example : ¬(p ∧ q) → ¬p ∨ ¬q :=
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fun npq => (em p).elim
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(fun hp => (em q).elim
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(fun hq => False.elim (npq ⟨hp, hq⟩))
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Or.inr)
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Or.inl
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example : ¬(p → q) → p ∧ ¬q :=
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fun h =>
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have lhs : p := byContradiction
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fun np => h (fun (hp : p) => absurd hp np)
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⟨lhs, fun hq => h (fun _ => hq)⟩
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example : (p → q) → (¬p ∨ q) :=
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fun hpq => (em p).elim (fun hp => Or.inr (hpq hp)) Or.inl
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example : (¬q → ¬p) → (p → q) :=
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fun h hp => byContradiction
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fun nq => absurd hp (h nq)
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example : p ∨ ¬p := em p
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example : (((p → q) → p) → p) :=
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fun h => byContradiction
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fun np =>
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suffices hp : p from absurd hp np
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h (fun (hp : p) => absurd hp np)
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end ex2
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-- ========================================
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-- Exercise 3
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--
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-- Prove `¬(p ↔ ¬p)` without using classical logic.
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-- ========================================
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namespace ex3
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variable (p : Prop)
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example (hp : p) : ¬(p ↔ ¬p) :=
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fun h => absurd hp (Iff.mp h hp)
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end ex3
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