Theorem Proving in Lean. Finish exercises 5.
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@ -149,41 +149,38 @@ example : (∃ x : α, r) → r :=
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example (a : α) : r → (∃ x : α, r) :=
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assume hr,
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exists.intro a hr
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⟨a, hr⟩
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example : (∃ x, p x ∧ r) ↔ (∃ x, p x) ∧ r :=
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iff.intro
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(assume ⟨hx, ⟨hp, hr⟩⟩, ⟨exists.intro hx hp, hr⟩)
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(assume ⟨⟨hx, hp⟩, hr⟩, exists.intro hx ⟨hp, hr⟩)
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(assume ⟨hx, ⟨hp, hr⟩⟩, ⟨⟨hx, hp⟩, hr⟩)
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(assume ⟨⟨hx, hp⟩, hr⟩, ⟨hx, ⟨hp, hr⟩⟩)
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example : (∃ x, p x ∨ q x) ↔ (∃ x, p x) ∨ (∃ x, q x) :=
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iff.intro
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(assume ⟨hx, hpq⟩, hpq.elim
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(assume hp, or.inl (exists.intro hx hp))
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(assume hq, or.inr (exists.intro hx hq)))
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(assume hp, or.inl (⟨hx, hp⟩))
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(assume hq, or.inr (⟨hx, hq⟩)))
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(assume h, h.elim
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(assume ⟨hx, hp⟩, exists.intro hx (or.inl hp))
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(assume ⟨hx, hq⟩, exists.intro hx (or.inr hq)))
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(assume ⟨hx, hp⟩, ⟨hx, or.inl hp⟩)
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(assume ⟨hx, hq⟩, ⟨hx, or.inr hq⟩))
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example : (∀ x, p x) ↔ ¬ (∃ x, ¬ p x) :=
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iff.intro
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(assume h ⟨hx, np⟩, np (h hx))
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(assume h hx, classical.by_contradiction (
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assume np,
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h (exists.intro hx np)
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))
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(assume h hx, classical.by_contradiction (assume np, h ⟨hx, np⟩))
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example : (∃ x, p x) ↔ ¬ (∀ x, ¬ p x) :=
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iff.intro
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(assume ⟨hx, hp⟩ h, absurd hp (h hx))
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(assume h, classical.by_contradiction (
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assume h',
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h (λ (x : α), assume hp, h' (exists.intro x hp))
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h (λ (x : α), assume hp, h' ⟨x, hp⟩)
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))
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example : (¬ ∃ x, p x) ↔ (∀ x, ¬ p x) :=
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iff.intro
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(assume h hx hp, h (exists.intro hx hp))
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(assume h hx hp, h ⟨hx, hp⟩)
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(assume h ⟨hx, hp⟩, absurd hp (h hx))
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lemma forall_negation : (¬ ∀ x, p x) ↔ (∃ x, ¬ p x) :=
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@ -192,7 +189,7 @@ lemma forall_negation : (¬ ∀ x, p x) ↔ (∃ x, ¬ p x) :=
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assume h',
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h (λ (x : α), classical.by_contradiction (
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assume np,
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h' (exists.intro x np)
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h' ⟨x, np⟩
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))
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))
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(assume ⟨hx, np⟩ h, absurd (h hx) np)
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@ -209,21 +206,21 @@ example (a : α) : (∃ x, p x → r) ↔ (∀ x, p x) → r :=
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iff.intro
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(assume ⟨hx, hp⟩ h, hp (h hx))
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(assume h₁, or.elim (classical.em (∀ x, p x))
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(assume h₂, exists.intro a (assume hp, h₁ h₂))
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(assume h₂, ⟨a, assume hp, h₁ h₂⟩)
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(assume h₂,
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have h₃ : (∃ x, ¬p x), from iff.elim_left (forall_negation α p) h₂,
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match h₃ with
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⟨hx, hp⟩ := exists.intro hx (assume hp', absurd hp' hp)
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⟨hx, hp⟩ := ⟨hx, (assume hp', absurd hp' hp)⟩
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end))
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example (a : α) : (∃ x, r → p x) ↔ (r → ∃ x, p x) :=
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iff.intro
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(assume ⟨hx, hrp⟩ hr, exists.intro hx (hrp hr))
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(assume ⟨hx, hrp⟩ hr, ⟨hx, hrp hr⟩)
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(assume h, or.elim (classical.em r)
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(assume hr, match h hr with
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⟨hx, hp⟩ := exists.intro hx (assume hr, hp)
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⟨hx, hp⟩ := ⟨hx, (assume hr, hp)⟩
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end)
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(assume nr, exists.intro a (assume hr, absurd hr nr)))
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(assume nr, ⟨a, (assume hr, absurd hr nr)⟩))
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end ex_5
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@ -0,0 +1,499 @@
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/- Exercises 5.8
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-
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- Avigad, Jeremy. ‘Theorem Proving in Lean’, n.d.
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-/
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import tactic.rcases
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-- Exercise 1
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--
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-- Go back to the exercises in Chapter 3 and Chapter 4 and redo as many as you
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-- can now with tactic proofs, using also `rw` and `simp` as appropriate.
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section ex_1_3_1
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variables p q r : Prop
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-- Commutativity of ∧ and ∨
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example : p ∧ q ↔ q ∧ p :=
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begin
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split,
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{ rintro ⟨p, q⟩, exact ⟨q, p⟩ },
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{ rintro ⟨q, p⟩, exact ⟨p, q⟩ },
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end
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example : p ∨ q ↔ q ∨ p :=
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begin
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split,
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{ rintro (p | q),
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{ exact or.inr p },
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{ exact or.inl q },
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},
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{ rintro (q | p),
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{ exact or.inr q },
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{ exact or.inl p },
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},
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end
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-- Associativity of ∧ and ∨
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example : (p ∧ q) ∧ r ↔ p ∧ (q ∧ r) :=
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begin
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split,
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{ rintro ⟨⟨p, q⟩, r⟩, exact ⟨p, q, r⟩ },
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{ rintro ⟨p, q, r⟩, exact ⟨⟨p, q⟩, r⟩ },
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end
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example : (p ∨ q) ∨ r ↔ p ∨ (q ∨ r) :=
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begin
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split,
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{ rintro ((p | q) | r),
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{ exact or.inl p },
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{ exact or.inr (or.inl q) },
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{ exact or.inr (or.inr r) },
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},
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{ rintro (p | q | r),
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{ exact or.inl (or.inl p) },
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{ exact or.inl (or.inr q) },
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{ exact or.inr r },
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},
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end
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-- Distributivity
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example : p ∧ (q ∨ r) ↔ (p ∧ q) ∨ (p ∧ r) :=
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begin
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split,
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{ rintro ⟨p, (q | r)⟩,
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{ exact or.inl ⟨p, q⟩ },
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{ exact or.inr ⟨p, r⟩ },
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},
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{ rintro (⟨p, q⟩ | ⟨p, r⟩),
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{ exact ⟨p, or.inl q⟩ },
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{ exact ⟨p, or.inr r⟩ },
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},
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end
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example : p ∨ (q ∧ r) ↔ (p ∨ q) ∧ (p ∨ r) :=
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begin
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split,
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{ rintro (p | ⟨q, r⟩),
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{ exact ⟨or.inl p, or.inl p⟩ },
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{ exact ⟨or.inr q, or.inr r⟩ },
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},
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{ rintro ⟨(p | q), (p | r)⟩;
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exact or.inl p <|> exact or.inr ⟨q, r⟩,
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},
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end
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-- Other properties
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example : (p → (q → r)) ↔ (p ∧ q → r) :=
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begin
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split,
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{ rintros h ⟨hp, hq⟩, exact h hp hq },
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{ intros h hp hq, exact h ⟨hp, hq⟩ },
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end
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example : ((p ∨ q) → r) ↔ (p → r) ∧ (q → r) :=
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begin
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split,
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{ intros h₁,
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split,
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{ intro p, exact h₁ (or.inl p) },
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{ intro q, exact h₁ (or.inr q) },
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},
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{ rintros ⟨hp, hq⟩ h,
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apply or.elim h,
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{ intro p, exact hp p },
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{ intro q, exact hq q },
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},
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end
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example : ¬(p ∨ q) ↔ ¬p ∧ ¬q :=
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begin
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split,
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{ intro h,
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split,
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{ intro hp, apply h, exact or.inl hp },
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{ intro hq, apply h, exact or.inr hq },
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},
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{ rintros ⟨np, nq⟩ (p | q),
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{ apply np, assumption },
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{ apply nq, assumption },
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},
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end
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example : ¬p ∨ ¬q → ¬(p ∧ q) :=
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begin
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rintro (np | nq) ⟨hp, hq⟩,
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{ apply np, assumption },
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{ apply nq, assumption },
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end
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example : ¬(p ∧ ¬p) :=
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begin
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rintro ⟨hp, np⟩,
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exact absurd hp np
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end
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example : p ∧ ¬q → ¬(p → q) :=
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begin
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rintro ⟨hp, nq⟩ hpq,
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apply nq,
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exact hpq hp
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end
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example : ¬p → (p → q) :=
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begin
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intros np hp,
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exact absurd hp np,
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end
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example : (¬p ∨ q) → (p → q) :=
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begin
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rintros (np | hq) hp,
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{ exact absurd hp np },
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{ exact hq },
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end
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example : p ∨ false ↔ p :=
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begin
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split,
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{ rintro (hp | hf),
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{ exact hp },
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{ exact false.elim hf },
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},
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{ intro hp,
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exact or.inl hp,
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},
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end
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example : p ∧ false ↔ false :=
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begin
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split,
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{ rintro ⟨hp, hf⟩, assumption },
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{ intro hf, exact false.elim hf },
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end
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example : (p → q) → (¬q → ¬p) :=
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begin
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rintro hpq nq hp,
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apply nq,
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exact absurd (hpq hp) nq,
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end
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end ex_1_3_1
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section ex_1_3_2
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open classical
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variables p q r s : Prop
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example (hp : p) : (p → r ∨ s) → ((p → r) ∨ (p → s)) :=
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begin
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intro h,
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apply or.elim (h hp),
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{ intro hr, exact or.inl (assume hp, hr) },
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{ intro hs, exact or.inr (assume hp, hs) }
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end
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example : ¬(p ∧ q) → ¬p ∨ ¬q :=
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begin
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intro h₁,
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apply or.elim (classical.em p),
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{ intro hp,
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apply or.elim (classical.em q),
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{ assume hq, exact false.elim (h₁ ⟨hp, hq⟩) },
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{ assume nq, exact or.inr nq },
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},
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{ intro np,
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exact or.inl np,
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},
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end
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example : ¬(p → q) → p ∧ ¬q :=
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begin
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intro h₁,
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split,
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{ by_contradiction np, apply h₁, intro hp, exact absurd hp np },
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{ intro hq, apply h₁, intro hp, exact hq },
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end
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example : (p → q) → (¬p ∨ q) :=
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begin
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intro hpq,
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apply or.elim (classical.em p),
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{ assume hp, exact or.inr (hpq hp) },
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{ assume np, exact or.inl np },
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end
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example : (¬q → ¬p) → (p → q) :=
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begin
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intros hqp hp,
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by_contradiction nq,
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exact absurd hp (hqp nq),
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end
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example : p ∨ ¬p :=
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by apply classical.em
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example : (((p → q) → p) → p) :=
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begin
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intro h,
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apply or.elim (classical.em p),
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{ intro hp, assumption },
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{ intro np, apply h, intro hp, exact absurd hp np },
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end
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end ex_1_3_2
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section ex_1_3_3
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variable p : Prop
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example (hp : p) : ¬(p ↔ ¬p) :=
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begin
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intro h₁,
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cases h₁ with h₂,
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exact absurd hp (h₂ hp),
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end
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end ex_1_3_3
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section ex_1_4_1
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variables (α : Type*) (p q : α → Prop)
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example : (∀ x, p x ∧ q x) ↔ (∀ x, p x) ∧ (∀ x, q x) :=
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begin
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split,
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{ intro h,
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split,
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{ intro hx, exact and.left (h hx) },
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{ intro hx, exact and.right (h hx) },
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},
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{ intros h hx,
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have hl : ∀ (x : α), p x, from and.left h,
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have hr : ∀ (x : α), q x, from and.right h,
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exact ⟨hl hx, hr hx⟩
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},
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end
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example : (∀ x, p x → q x) → (∀ x, p x) → (∀ x, q x) :=
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begin
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intros h₁ h₂ hx,
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exact h₁ hx (h₂ hx)
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end
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example : (∀ x, p x) ∨ (∀ x, q x) → ∀ x, p x ∨ q x :=
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begin
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rintro (h₁ | h₂),
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{ intro hx, exact or.inl (h₁ hx) },
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{ intro hx, exact or.inr (h₂ hx) },
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end
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end ex_1_4_1
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section ex_1_4_2
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variables (α : Type*) (p q : α → Prop)
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variable r : Prop
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example : α → ((∀ x : α, r) ↔ r) :=
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begin
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intro hα,
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split,
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{ intro hαr, apply hαr, exact hα },
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{ intros hr hα, exact hr },
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end
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section
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open classical
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example : (∀ x, p x ∨ r) ↔ (∀ x, p x) ∨ r :=
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begin
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split,
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{ intro h₁,
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apply or.elim (classical.em r),
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{ intro hr, exact or.inr hr },
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{ intro nr,
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exact or.inl begin
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intro hx,
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have h₂ : p hx ∨ r, from h₁ hx,
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apply or.elim h₂,
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{ intro hp, exact hp },
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{ intro hr, exact absurd hr nr },
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end
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},
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},
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{ assume h₁ hx,
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apply or.elim h₁,
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{ assume h₂, exact or.inl (h₂ hx) },
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{ assume hr, exact or.inr hr },
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},
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end
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end
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example : (∀ x, r → p x) ↔ (r → ∀ x, p x) :=
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begin
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split,
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{ intros h hr hx, exact h hx hr },
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{ intros h hx hr, exact h hr hx },
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end
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end ex_1_4_2
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section ex_1_4_3
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open classical
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variables (men : Type*) (barber : men)
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variable (shaves : men → men → Prop)
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example (h : ∀ x : men, shaves barber x ↔ ¬ shaves x x) :
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false :=
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begin
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apply or.elim (classical.em (shaves barber barber)),
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{ intro hb,
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apply iff.elim_left (h barber) hb,
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exact hb,
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},
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{ intro nb,
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have s, from iff.elim_right (h barber) nb,
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exact absurd s nb
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},
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end
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end ex_1_4_3
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section ex_1_4_5
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open classical
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variables (α : Type*) (p q : α → Prop)
|
||||
variables r s : Prop
|
||||
|
||||
example : (∃ x : α, r) → r :=
|
||||
begin
|
||||
rintro ⟨hx, hr⟩,
|
||||
exact hr,
|
||||
end
|
||||
|
||||
example (a : α) : r → (∃ x : α, r) :=
|
||||
begin
|
||||
intro hr,
|
||||
exact ⟨a, hr⟩
|
||||
end
|
||||
|
||||
example : (∃ x, p x ∧ r) ↔ (∃ x, p x) ∧ r :=
|
||||
begin
|
||||
split,
|
||||
{ rintro ⟨hx, ⟨hp, hr⟩⟩, exact ⟨⟨hx, hp⟩, hr⟩ },
|
||||
{ rintro ⟨⟨hx, hp⟩, hr⟩, exact ⟨hx, ⟨hp, hr⟩⟩ },
|
||||
end
|
||||
|
||||
example : (∃ x, p x ∨ q x) ↔ (∃ x, p x) ∨ (∃ x, q x) :=
|
||||
begin
|
||||
split,
|
||||
{ rintro ⟨hx, (hp | hq)⟩,
|
||||
{ exact or.inl ⟨hx, hp⟩ },
|
||||
{ exact or.inr ⟨hx, hq⟩ },
|
||||
},
|
||||
{ rintro (⟨hx, hp⟩ | ⟨hx, hq⟩),
|
||||
{ exact ⟨hx, or.inl hp⟩ },
|
||||
{ exact ⟨hx, or.inr hq⟩ },
|
||||
},
|
||||
end
|
||||
|
||||
example : (∀ x, p x) ↔ ¬ (∃ x, ¬ p x) :=
|
||||
begin
|
||||
split,
|
||||
{ rintros ha ⟨hx, np⟩, exact absurd (ha hx) np },
|
||||
{ intros he hx, by_contradiction np, exact he ⟨hx, np⟩ },
|
||||
end
|
||||
|
||||
example : (∃ x, p x) ↔ ¬ (∀ x, ¬ p x) :=
|
||||
begin
|
||||
split,
|
||||
{ rintro ⟨hx, hp⟩ h, exact absurd hp (h hx) },
|
||||
{ intro h₁,
|
||||
by_contradiction h₂,
|
||||
apply h₁,
|
||||
intros hx hp,
|
||||
exact h₂ ⟨hx, hp⟩,
|
||||
},
|
||||
end
|
||||
|
||||
example : (¬ ∃ x, p x) ↔ (∀ x, ¬ p x) :=
|
||||
begin
|
||||
split,
|
||||
{ intros h hx hp,
|
||||
exact h ⟨hx, hp⟩
|
||||
},
|
||||
{ rintros h ⟨hx, hp⟩,
|
||||
exact absurd hp (h hx)
|
||||
},
|
||||
end
|
||||
|
||||
lemma forall_negation : (¬ ∀ x, p x) ↔ (∃ x, ¬ p x) :=
|
||||
begin
|
||||
split,
|
||||
{ intro h₁,
|
||||
by_contradiction h₂,
|
||||
exact h₁ (λ (x : α), begin
|
||||
by_contradiction np,
|
||||
exact h₂ ⟨x, np⟩
|
||||
end)
|
||||
},
|
||||
{ rintros ⟨hx, np⟩ h, exact absurd (h hx) np },
|
||||
end
|
||||
|
||||
example : (¬ ∀ x, p x) ↔ (∃ x, ¬ p x) := forall_negation α p
|
||||
|
||||
example : (∀ x, p x → r) ↔ (∃ x, p x) → r :=
|
||||
begin
|
||||
split,
|
||||
{ rintros h ⟨hx, hp⟩, exact h hx hp },
|
||||
{ intros h hx hp,
|
||||
exact h ⟨hx, hp⟩
|
||||
},
|
||||
end
|
||||
|
||||
example (a : α) : (∃ x, p x → r) ↔ (∀ x, p x) → r :=
|
||||
begin
|
||||
split,
|
||||
{ rintros ⟨hx, hp⟩ h, apply hp, exact h hx },
|
||||
{ intro h₁,
|
||||
apply or.elim (classical.em (∀ x, p x)),
|
||||
{ intro h₂, exact ⟨a, (assume hp, h₁ h₂)⟩ },
|
||||
{ intro h₂,
|
||||
have h₃ : (∃ x, ¬p x), from iff.elim_left (forall_negation α p) h₂,
|
||||
exact let ⟨hx, hp⟩ := h₃ in ⟨hx, (assume hp', absurd hp' hp)⟩,
|
||||
},
|
||||
},
|
||||
end
|
||||
|
||||
example (a : α) : (∃ x, r → p x) ↔ (r → ∃ x, p x) :=
|
||||
begin
|
||||
split,
|
||||
{ rintros ⟨hx, h⟩ hr, exact ⟨hx, h hr⟩ },
|
||||
{ intro h,
|
||||
apply or.elim (classical.em r),
|
||||
{ intro hr,
|
||||
exact let ⟨hx, hp⟩ := h hr in ⟨hx, (assume hr, hp)⟩
|
||||
},
|
||||
{ intro nr, exact ⟨a, (assume hr, absurd hr nr)⟩ },
|
||||
},
|
||||
end
|
||||
|
||||
end ex_1_4_5
|
||||
|
||||
-- Exercise 2
|
||||
--
|
||||
-- Use tactic combinators to obtain a one line proof of the following:
|
||||
section ex_2
|
||||
|
||||
example (p q r : Prop) (hp : p) : (p ∨ q ∨ r) ∧ (q ∨ p ∨ r) ∧ (q ∨ r ∨ p) :=
|
||||
by simp *
|
||||
|
||||
end ex_2
|
||||
Loading…
Reference in New Issue