(WIP) mid of exercises on quantifiers_and_equality
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quantifiers_and_equality.lean
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quantifiers_and_equality.lean
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--
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-- 4. Quantifiers and Equality
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--
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-- The Universal Quantifier
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example (α : Type) (p q : α → Prop) : (∀ x : α, p x ∧ q x) → ∀ y : α, p y :=
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fun h : ∀ x : α, p x ∧ q x =>
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fun y : α =>
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show p y from (h y).left
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variable (α : Type) (r : α → α → Prop)
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variable (trans_r : ∀ {x y z}, r x y → r y z → r x z)
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variable (a b c : α)
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variable (hab : r a b) (hbc : r b c)
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#check trans_r
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#check @trans_r
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#check trans_r hab
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#check trans_r hab hbc
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-- Equality
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#check Eq.refl
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#check Eq.symm
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#check Eq.trans
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universe u
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#check @Eq.refl.{u}
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#check @Eq.symm.{u}
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#check @Eq.trans.{u}
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variable (a b c d : α)
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variable (hab : a = b) (hcb : c = b) (hcd : c = d)
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example : a = d :=
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Eq.trans (Eq.trans hab (Eq.symm hcb)) hcd
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example : a = d := (hab.trans hcb.symm).trans hcd
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variable (α β : Type)
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example (f : α → β) (a : α) : (fun x => f x) a = f a := Eq.refl _
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example (f : α → β) (a : α) : (fun x => f x) a = f a := rfl
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example (α : Type) (a b : α) (p : α → Prop)
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(h1 : a = b) (h2 : p a) : p b :=
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Eq.subst h1 h2
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example (α : Type) (a b : α) (p : α → Prop)
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(h1 : a = b) (h2 : p a) : p b :=
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h1 ▸ h2
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variable (α : Type)
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variable (a b : α)
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variable (f g : α → Nat)
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variable (h₁ : a = b)
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variable (h₂ : f = g)
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example : f a = f b := congrArg f h₁
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example : f a = g a := congrFun h₂ a
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variable (a : Nat) in
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example : a + 0 = a := Nat.add_zero a
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-- Calculational Proofs
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variable (a b c d : Nat) in
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variable (h1 : a = b) in
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variable (h2 : b = c + 1) in
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variable (h3 : c = d) in
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variable (h4 : e = 1 + d) in
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theorem T : a = e :=
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calc
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a = b := h1
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_ = c + 1 := h2
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_ = d + 1 := congrArg Nat.succ h3
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_ = 1 + d := Nat.add_comm d 1
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_ = e := h4.symm
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-- The Existential Quantifier
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example : ∃ x : Nat, x > 0 :=
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have h : 1 > 0 := Nat.zero_lt_succ 0
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Exists.intro 1 h
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variable (α : Type) (p q : α → Prop) in
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example (h : ∃ x, p x ∧ q x) : ∃ x, q x ∧ p x :=
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Exists.elim h
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fun w =>
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fun hw : p w ∧ q w =>
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Exists.intro w ⟨hw.right, hw.left⟩
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#check @Exists.elim.{u}
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#check Sigma
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variable (α : Type) (p q : α → Prop) in
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example (h : ∃ x, p x ∧ q x) : ∃ x, q x ∧ p x :=
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match h with
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-- More on the Proof Language
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variable (f : Nat → Nat)
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variable (h : ∀ x : Nat, f x ≤ f (x + 1))
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example : f 0 ≤ f 3 :=
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have : f 0 ≤ f 1 := h 0
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have : f 0 ≤ f 2 := Nat.le_trans (by assumption) (h 1)
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show f 0 ≤ f 3 from Nat.le_trans (by assumption) (h 2)
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example : f 0 ≥ f 1 → f 1 ≥ f 2 → f 0 = f 2 :=
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fun _ : f 0 ≥ f 1 =>
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fun _ : f 1 ≥ f 2 =>
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have : f 0 ≥ f 2 := Nat.le_trans ‹f 1 ≥ f 2› ‹f 0 ≥ f 1›
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have : f 0 ≤ f 2 := Nat.le_trans (h 0) (h 1)
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show f 0 = f 2 from Nat.le_antisymm this ‹f 0 ≥ f 2›
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-- Exercises
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variable (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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have h1 : (∀ (x : α), p x ∧ q x) → (∀ (x : α), p x) ∧ ∀ (x : α), q x :=
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fun h =>
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have h11 : ∀ (x : α), p x :=
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fun x : α => (h x).left
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have h12 : ∀ (x : α), q x :=
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fun x : α => (h x).right
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And.intro h11 h12
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have h2 : ((∀ (x : α), p x) ∧ ∀ (x : α), q x) → ∀ (x : α), p x ∧ q x :=
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fun h =>
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fun x : α =>
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⟨h.left x, h.right x⟩
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Iff.intro h1 h2
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example : (∀ x, p x → q x) → (∀ x, p x) → (∀ x, q x) :=
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fun h1 =>
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fun h2 =>
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fun x : α =>
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show q x from (h1 x) (h2 x)
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example : (∀ x, p x) ∨ (∀ x, q x) → ∀ x, p x ∨ q x :=
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fun h =>
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h.elim
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(fun hp =>
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fun x => Or.inl (hp x))
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(fun hq =>
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fun x => Or.inr (hq x))
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9
tactics.lean
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9
tactics.lean
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--
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-- 5. Tactics
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--
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--
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theorem test (p q : Prop) (hp : p) (hq : q) : p ∧ q ∧ p := by
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apply And.intro
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exact hp
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exact { left := hq, right := hp }
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