475 lines
10 KiB
Plaintext
475 lines
10 KiB
Plaintext
/-
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Chapter 5
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Tactics
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-/
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-- ========================================
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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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-- ========================================
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namespace ex1
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-- ----------------------------------------
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-- Exercises 3.1
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-- ----------------------------------------
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section ex3_1
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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 := by
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apply Iff.intro
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· intro ⟨hp, hq⟩
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exact ⟨hq, hp⟩
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· intro ⟨hq, hp⟩
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exact ⟨hp, hq⟩
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example : p ∨ q ↔ q ∨ p := by
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apply Iff.intro
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· intro
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| Or.inl hp => exact Or.inr hp
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| Or.inr hq => exact Or.inl hq
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· intro
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| Or.inl hq => exact Or.inr hq
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| Or.inr hp => exact Or.inl hp
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-- Associativity of ∧ and ∨
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example : (p ∧ q) ∧ r ↔ p ∧ (q ∧ r) := by
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apply Iff.intro
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· intro ⟨⟨hp, hq⟩, hr⟩
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exact ⟨hp, hq, hr⟩
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· intro ⟨hp, hq, hr⟩
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exact ⟨⟨hp, hq⟩, hr⟩
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example : (p ∨ q) ∨ r ↔ p ∨ (q ∨ r) := by
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apply Iff.intro
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· intro
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| Or.inl (Or.inl hp) => exact Or.inl hp
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| Or.inl (Or.inr hq) => exact Or.inr (Or.inl hq)
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| Or.inr hr => exact Or.inr (Or.inr hr)
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· intro
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| Or.inl hp => exact Or.inl (Or.inl hp)
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| Or.inr (Or.inl hq) => exact Or.inl (Or.inr hq)
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| Or.inr (Or.inr hr) => exact Or.inr hr
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-- Distributivity
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example : p ∧ (q ∨ r) ↔ (p ∧ q) ∨ (p ∧ r) := by
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apply Iff.intro
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· intro
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| ⟨hp, Or.inl hq⟩ => exact Or.inl ⟨hp, hq⟩
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| ⟨hp, Or.inr hr⟩ => exact Or.inr ⟨hp, hr⟩
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· intro
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| Or.inl ⟨hp, hq⟩ => exact ⟨hp, Or.inl hq⟩
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| Or.inr ⟨hp, hr⟩ => exact ⟨hp, Or.inr hr⟩
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example : p ∨ (q ∧ r) ↔ (p ∨ q) ∧ (p ∨ r) := by
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apply Iff.intro
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· intro
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| Or.inl hp => exact ⟨Or.inl hp, Or.inl hp⟩
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| Or.inr ⟨hq, hr⟩ => exact ⟨Or.inr hq, Or.inr hr⟩
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· intro
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| ⟨Or.inl hp, _⟩ => exact Or.inl hp
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| ⟨Or.inr _, Or.inl hp⟩ => exact Or.inl hp
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| ⟨Or.inr hq, Or.inr hr⟩ => exact Or.inr ⟨hq, hr⟩
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-- Other properties
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example : (p → (q → r)) ↔ (p ∧ q → r) := by
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apply Iff.intro
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· intro h ⟨hp, hq⟩
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exact h hp hq
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· intro h hp hq
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exact h ⟨hp, hq⟩
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example : ((p ∨ q) → r) ↔ (p → r) ∧ (q → r) := by
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apply Iff.intro
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· intro h
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apply And.intro
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· intro hp
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exact h (Or.inl hp)
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· intro hq
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exact h (Or.inr hq)
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· intro ⟨hpr, hqr⟩ h
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apply Or.elim h
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· intro hp
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exact hpr hp
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· intro hq
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exact hqr hq
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example : ¬(p ∨ q) ↔ ¬p ∧ ¬q := by
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apply Iff.intro
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· intro h
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apply And.intro
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· intro hp
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exact h (Or.inl hp)
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· intro hq
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exact h (Or.inr hq)
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· intro ⟨np, nq⟩
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intro
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| Or.inl hp => exact absurd hp np
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| Or.inr hq => exact absurd hq nq
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example : ¬p ∨ ¬q → ¬(p ∧ q) := by
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intro
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| Or.inl np => intro h; exact absurd h.left np
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| Or.inr nq => intro h; exact absurd h.right nq
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example : ¬(p ∧ ¬p) := by
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intro ⟨hp, np⟩
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exact absurd hp np
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example : p ∧ ¬q → ¬(p → q) := by
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intro ⟨hp, nq⟩ h
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exact absurd (h hp) nq
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example : ¬p → (p → q) := by
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intro np hp
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exact absurd hp np
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example : (¬p ∨ q) → (p → q) := by
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intro
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| Or.inl np => intro hp; exact absurd hp np
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| Or.inr hq => exact fun _ => hq
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example : p ∨ False ↔ p := by
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apply Iff.intro
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· intro
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| Or.inl hp => exact hp
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| Or.inr ff => exact False.elim ff
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· intro hp
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exact Or.inl hp
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example : p ∧ False ↔ False := by
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apply Iff.intro
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· intro ⟨_, ff⟩
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exact ff
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· intro ff
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exact False.elim ff
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example : (p → q) → (¬q → ¬p) := by
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intro hpq nq hp
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exact absurd (hpq hp) nq
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end ex3_1
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-- ----------------------------------------
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-- Exercises 3.2
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-- ----------------------------------------
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section ex3_2
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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)) := by
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intro h
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apply (h hp).elim
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· intro hr
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exact Or.inl (fun _ => hr)
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· intro hs
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exact Or.inr (fun _ => hs)
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example : ¬(p ∧ q) → ¬p ∨ ¬q := by
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intro h
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apply (em p).elim
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· intro hp
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apply (em q).elim
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· intro hq
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exact False.elim (h ⟨hp, hq⟩)
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· intro nq
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exact Or.inr nq
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· intro np
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exact Or.inl np
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example : ¬(p → q) → p ∧ ¬q := by
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intro h
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apply And.intro
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· apply byContradiction
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intro np
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apply h
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intro hp
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exact absurd hp np
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· intro hq
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apply h
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intro _
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exact hq
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example : (p → q) → (¬p ∨ q) := by
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intro hpq
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apply (em p).elim
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· intro hp
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exact Or.inr (hpq hp)
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· intro np
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exact Or.inl np
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example : (¬q → ¬p) → (p → q) := by
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intro hqp hp
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apply byContradiction
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intro nq
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exact absurd hp (hqp nq)
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example : p ∨ ¬p := by apply em
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example : (((p → q) → p) → p) := by
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intro h
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apply (em p).elim
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· intro hp
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exact hp
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· intro np
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apply h
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intro hp
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exact absurd hp np
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end ex3_2
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-- ----------------------------------------
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-- Exercises 3.3
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-- ----------------------------------------
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section ex3_3
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variable (p : Prop)
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example (hp : p) : ¬(p ↔ ¬p) := by
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intro h
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exact absurd hp (h.mp hp)
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end ex3_3
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-- ----------------------------------------
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-- Exercises 4.1
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-- ----------------------------------------
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section ex4_1
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variable (α : Type _)
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variable (p q : α → Prop)
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example : (∀ x, p x ∧ q x) ↔ (∀ x, p x) ∧ (∀ x, q x) := by
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apply Iff.intro
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· intro h
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apply And.intro
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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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· intro h hx
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have lhs : ∀ (x : α), p x := And.left h
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have rhs : ∀ (x : α), q x := And.right h
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exact ⟨lhs hx, rhs hx⟩
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example : (∀ x, p x → q x) → (∀ x, p x) → (∀ x, q x) := by
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intro h₁ h₂ hx
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exact h₁ hx (h₂ hx)
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example : (∀ x, p x) ∨ (∀ x, q x) → ∀ x, p x ∨ q x := by
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intro
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| Or.inl h => intro hx; exact Or.inl (h hx)
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| Or.inr h => intro hx; exact Or.inr (h hx)
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end ex4_1
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-- ----------------------------------------
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-- Exercises 4.2
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-- ----------------------------------------
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section ex4_2
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variable (α : Type _)
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variable (p q : α → Prop)
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variable (r : Prop)
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example : α → ((∀ _ : α, r) ↔ r) := by
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intro ha
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apply Iff.intro
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· intro har
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apply har
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exact ha
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· intro hr _
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exact hr
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section
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open Classical
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example : (∀ x, p x ∨ r) ↔ (∀ x, p x) ∨ r := by
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apply Iff.intro
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· intro h
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apply (em r).elim
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· intro hr
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exact Or.inr hr
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· intro nr
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apply Or.inl
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· intro hx
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apply (h hx).elim
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· exact id
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· intro hr
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exact absurd hr nr
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· intro h₁ hx
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apply h₁.elim
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· intro h₂
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exact Or.inl (h₂ hx)
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· intro hr
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exact Or.inr hr
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end
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example : (∀ x, r → p x) ↔ (r → ∀ x, p x) := by
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apply Iff.intro
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· intro h hr hx
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exact h hx hr
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· intro h hx hr
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exact h hr hx
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end ex4_2
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-- ----------------------------------------
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-- Exercises 4.3
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-- ----------------------------------------
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section ex4_3
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open Classical
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variable (men : Type _)
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variable (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) : False := by
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apply (em (shaves barber barber)).elim
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· intro hb
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exact absurd hb ((h barber).mp hb)
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· intro nb
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exact absurd ((h barber).mpr nb) nb
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end ex4_3
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-- ----------------------------------------
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-- Exercises 4.5
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-- ----------------------------------------
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section ex4_5
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open Classical
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variable (α : Type _)
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variable (p q : α → Prop)
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variable (r s : Prop)
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example : (∃ _ : α, r) → r := by
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intro ⟨_, hr⟩
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exact hr
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example (a : α) : r → (∃ _ : α, r) := by
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intro hr
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exact ⟨a, hr⟩
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example : (∃ x, p x ∧ r) ↔ (∃ x, p x) ∧ r := by
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apply Iff.intro
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· intro ⟨hx, hp, hr⟩
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exact ⟨⟨hx, hp⟩, hr⟩
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· intro ⟨⟨hx, hp⟩, hr⟩
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exact ⟨hx, hp, hr⟩
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example : (∃ x, p x ∨ q x) ↔ (∃ x, p x) ∨ (∃ x, q x) := by
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apply Iff.intro
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· intro
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| ⟨hx, Or.inl hp⟩ => exact Or.inl ⟨hx, hp⟩
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| ⟨hx, Or.inr hq⟩ => exact Or.inr ⟨hx, hq⟩
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· intro
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| Or.inl ⟨hx, hp⟩ => exact ⟨hx, Or.inl hp⟩
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| Or.inr ⟨hx, hq⟩ => exact ⟨hx, Or.inr hq⟩
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example : (∀ x, p x) ↔ ¬(∃ x, ¬p x) := by
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apply Iff.intro
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· intro ha ⟨hx, np⟩
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exact absurd (ha hx) np
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· intro he hx
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apply byContradiction
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intro np
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exact he ⟨hx, np⟩
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example : (∃ x, p x) ↔ ¬(∀ x, ¬p x) := by
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apply Iff.intro
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· intro ⟨hx, hp⟩ h
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exact absurd hp (h hx)
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· intro h₁
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apply byContradiction
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intro h₂
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apply h₁
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intro hx hp
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exact h₂ ⟨hx, hp⟩
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example : (¬∃ x, p x) ↔ (∀ x, ¬p x) := by
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apply Iff.intro
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· intro h hx hp
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exact h ⟨hx, hp⟩
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· intro h ⟨hx, hp⟩
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exact absurd hp (h hx)
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theorem forall_negation : (¬∀ x, p x) ↔ (∃ x, ¬p x) := by
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apply Iff.intro
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· intro h₁
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apply byContradiction
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intro h₂
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exact h₁ (fun (x : α) => by
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apply byContradiction
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intro np
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exact h₂ ⟨x, np⟩)
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· intro ⟨hx, np⟩ h
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exact absurd (h hx) np
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example : (¬∀ x, p x) ↔ (∃ x, ¬p x) := forall_negation α p
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example : (∀ x, p x → r) ↔ (∃ x, p x) → r := by
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apply Iff.intro
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· intro h ⟨hx, hp⟩
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exact h hx hp
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· intro h hx hp
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exact h ⟨hx, hp⟩
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example (a : α) : (∃ x, p x → r) ↔ (∀ x, p x) → r := by
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apply Iff.intro
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· intro ⟨hx, hp⟩ h
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apply hp
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exact h hx
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· intro h₁
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apply (em (∀ x, p x)).elim
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· intro h₂
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exact ⟨a, fun _ => h₁ h₂⟩
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· intro h₂
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have ⟨hx, np⟩ : (∃ x, ¬p x) := (forall_negation α p).mp h₂
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exact ⟨hx, fun hp => absurd hp np⟩
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example (a : α) : (∃ x, r → p x) ↔ (r → ∃ x, p x) := by
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apply Iff.intro
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· intro ⟨hx, h⟩ hr
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exact ⟨hx, h hr⟩
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· intro h
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apply (em r).elim
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· intro hr
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have ⟨hx, hp⟩ := h hr
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exact ⟨hx, fun _ => hp⟩
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· intro nr
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exact ⟨a, fun hr => absurd hr nr⟩
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end ex4_5
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end ex1
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-- ========================================
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-- Exercise 2
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--
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-- Use tactic combinators to obtain a one line proof of the following:
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-- ========================================
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namespace ex2
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example (p q r : Prop) (hp : p) : (p ∨ q ∨ r) ∧ (q ∨ p ∨ r) ∧ (q ∨ r ∨ p) :=
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by simp [*]
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end ex2
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