260 lines
7.5 KiB
Plaintext
260 lines
7.5 KiB
Plaintext
/-
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Chapter 4
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Quantifiers and Equality
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-/
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-- ========================================
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-- Exercise 1
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--
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-- Prove these equivalences. You should also try to understand why the reverse
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-- implication is not derivable in the last example.
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-- ========================================
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namespace ex1
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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) :=
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Iff.intro
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(fun h => ⟨fun x => And.left (h x), fun x => And.right (h x)⟩)
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(fun ⟨h₁, h₂⟩ x => ⟨h₁ x, h₂ x⟩)
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example : (∀ x, p x → q x) → (∀ x, p x) → (∀ x, q x) :=
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fun h₁ h₂ x =>
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have px : p x := h₂ x
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h₁ x px
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example : (∀ x, p x) ∨ (∀ x, q x) → ∀ x, p x ∨ q x :=
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fun h₁ x => h₁.elim
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(fun h₂ => Or.inl (h₂ x))
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(fun h₂ => Or.inr (h₂ x))
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-- The implication in the above example cannot be proven in the other direction
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-- because it may be the case predicate `p x` holds for certain values of `x`
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-- but not others that `q x` may hold for (and vice versa).
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end ex1
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-- ========================================
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-- Exercise 2
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--
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-- It is often possible to bring a component of a formula outside a universal
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-- quantifier, when it does not depend on the quantified variable. Try proving
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-- these (one direction of the second of these requires classical logic).
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-- ========================================
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namespace ex2
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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) :=
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fun a => Iff.intro (fun h => h a) (fun hr _ => 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 :=
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Iff.intro
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(fun h₁ => (em r).elim
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Or.inr
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(fun nr => Or.inl (fun x => (h₁ x).elim id (absurd · nr))))
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(fun h₁ => h₁.elim
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(fun h₂ x => Or.inl (h₂ x))
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(fun hr _ => Or.inr hr))
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end
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example : (∀ x, r → p x) ↔ (r → ∀ x, p x) :=
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Iff.intro
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(fun h hr hx => h hx hr)
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(fun h hx hr => h hr hx)
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end ex2
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-- ========================================
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-- Exercise 3
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--
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-- Consider the "barber paradox," that is, the claim that in a certain town
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-- there is a (male) barber that shaves all and only the men who do not shave
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-- themselves. Prove that this is a contradiction.
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-- ========================================
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namespace ex3
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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 :=
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have b : shaves barber barber ↔ ¬shaves barber barber := h barber
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(em (shaves barber barber)).elim
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(fun b' => absurd b' (Iff.mp b b'))
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(fun b' => absurd (Iff.mpr b b') b')
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end ex3
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-- ========================================
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-- Exercise 4
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--
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-- Remember that, without any parameters, an expression of type `Prop` is just
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-- an assertion. Fill in the definitions of `prime` and `Fermat_prime` below,
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-- and construct each of the given assertions. For example, you can say that
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-- there are infinitely many primes by asserting that for every natural number
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-- `n`, there is a prime number greater than `n.` Goldbach’s weak conjecture
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-- states that every odd number greater than `5` is the sum of three primes.
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-- Look up the definition of a Fermat prime or any of the other statements, if
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-- necessary.
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-- ========================================
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namespace ex4
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def even (a : Nat) := ∃ b, a = 2 * b
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def odd (a : Nat) := ¬even a
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def prime (n : Nat) : Prop :=
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n > 1 ∧ ∀ (m : Nat), (1 < m ∧ m < n) → n % m ≠ 0
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def infinitelyManyPrimes : Prop :=
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∀ (n : Nat), (∃ (m : Nat), m > n ∧ prime m)
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def FermatPrime (n : Nat) : Prop :=
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∃ (m : Nat), n = 2^(2^m) + 1
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def infinitelyManyFermatPrimes : Prop :=
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∀ (n : Nat), (∃ (m : Nat), m > n ∧ FermatPrime m)
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def GoldbachConjecture : Prop :=
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∀ (n : Nat), even n ∧ n > 2 →
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∃ (x y : Nat), prime x ∧ prime y ∧ x + y = n
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def Goldbach'sWeakConjecture : Prop :=
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∀ (n : Nat), odd n ∧ n > 5 →
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∃ (x y z : Nat), prime x ∧ prime y ∧ prime z ∧ x + y + z = n
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def Fermat'sLastTheorem : Prop :=
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∀ (n : Nat), n > 2 → (∀ (a b c : Nat), a^n + b^n ≠ c^n)
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end ex4
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-- ========================================
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-- Exercise 5
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--
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-- Prove as many of the identities listed in Section 4.4 as you can.
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-- ========================================
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namespace ex5
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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 :=
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fun ⟨_, hr⟩ => hr
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example (a : α) : r → (∃ _ : α, r) :=
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fun hr => ⟨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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(fun ⟨hx, ⟨hp, hr⟩⟩ => ⟨⟨hx, hp⟩, hr⟩)
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(fun ⟨⟨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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(fun ⟨hx, hpq⟩ => hpq.elim
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(fun hp => Or.inl ⟨hx, hp⟩)
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(fun hq => Or.inr ⟨hx, hq⟩))
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(fun h => h.elim
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(fun ⟨hx, hp⟩ => ⟨hx, Or.inl hp⟩)
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(fun ⟨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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(fun h ⟨hx, np⟩ => np (h hx))
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(fun h hx => byContradiction
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fun 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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(fun ⟨hx, hp⟩ h => absurd hp (h hx))
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(fun h => byContradiction
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fun h' => h (fun (x : α) hp => h' ⟨x, hp⟩))
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example : (¬∃ x, p x) ↔ (∀ x, ¬p x) :=
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Iff.intro
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(fun h hx hp => h ⟨hx, hp⟩)
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(fun h ⟨hx, hp⟩ => absurd hp (h hx))
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theorem forall_negation : (¬∀ x, p x) ↔ (∃ x, ¬p x) :=
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Iff.intro
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(fun h => byContradiction
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fun h' => h (fun (x : α) => byContradiction
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fun np => h' ⟨x, np⟩))
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(fun ⟨hx, np⟩ h => absurd (h hx) np)
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example : (¬∀ x, p x) ↔ (∃ x, ¬p x) :=
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forall_negation α p
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example : (∀ x, p x → r) ↔ (∃ x, p x) → r :=
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Iff.intro
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(fun h ⟨hx, hp⟩ => h hx hp)
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(fun h hx hp => h ⟨hx, hp⟩)
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example (a : α) : (∃ x, p x → r) ↔ (∀ x, p x) → r :=
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Iff.intro
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(fun ⟨hx, hp⟩ h => hp (h hx))
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(fun h₁ => (em (∀ x, p x)).elim
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(fun h₂ => ⟨a, fun _ => h₁ h₂⟩)
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(fun h₂ =>
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have h₃ : (∃ x, ¬p x) := Iff.mp (forall_negation α p) h₂
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match h₃ with
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| ⟨hx, hp⟩ => ⟨hx, fun hp' => absurd hp' hp⟩))
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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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(fun ⟨hx, hrp⟩ hr => ⟨hx, hrp hr⟩)
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(fun h => (em r).elim
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(fun hr => match h hr with
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| ⟨hx, hp⟩ => ⟨hx, fun _ => hp⟩)
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(fun nr => ⟨a, fun hr => absurd hr nr⟩))
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end ex5
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-- ========================================
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-- Exercise 6
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--
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-- Give a calculational proof of the theorem `log_mul` below.
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-- ========================================
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namespace ex6
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variable (log exp : Float → Float)
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variable (log_exp_eq : ∀ x, log (exp x) = x)
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variable (exp_log_eq : ∀ {x}, x > 0 → exp (log x) = x)
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variable (exp_pos : ∀ x, exp x > 0)
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variable (exp_add : ∀ x y, exp (x + y) = exp x * exp y)
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example (x y z : Float) : exp (x + y + z) = exp x * exp y * exp z :=
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by rw [exp_add, exp_add]
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example (y : Float) (h : y > 0) : exp (log y) = y := exp_log_eq h
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theorem log_mul {x y : Float} (hx : x > 0) (hy : y > 0) :
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log (x * y) = log x + log y :=
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calc log (x * y) = log (x * exp (log y)) := by rw [exp_log_eq hy]
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_ = log (exp (log x) * exp (log y)) := by rw [exp_log_eq hx]
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_ = log (exp (log x + log y)) := by rw [exp_add]
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_ = log x + log y := by rw [log_exp_eq]
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end ex6
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