import GlimpseOfLean.Library.Basic

open Function

Conjunctions

In this file, we learn how to handle the conjunction ("logical and") operator and the existential quantifier.

In Lean the conjunction of two statements P and Q is denoted by P ∧ Q, read as "P and Q".

We can use a conjunction similarly to the ↔: * If h : P ∧ Q then h.1 : P and h.2 : Q. * To prove P ∧ Q use the constructor tactic.

Furthermore, we can decompose conjunction and equivalences. * If h : P ∧ Q, the tactic rcases h with ⟨hP, hQ⟩ gives two new assumptions hP : P and hQ : Q. * If h : P ↔ Q, the tactic rcases h with ⟨hPQ, hQP⟩ gives two new assumptions hPQ : P → Q and hQP : Q → P.

example (p q r s : Prop) (h : p → r) (h' : q → s) : p ∧ q → r ∧ s := by
  intro hpq
  rcases hpq with ⟨hp, hq⟩
  constructor
  · exact h hp
  · exact h' hq

One can also prove a conjunction without the constructor tactic by gathering both sides using the ⟨/⟩ brackets, so the above proof can be rewritten as.

example (p q r s : Prop) (h : p → r) (h' : q → s) : p ∧ q → r ∧ s := by
  intro hpq
  exact ⟨h hpq.1, h' hpq.2⟩

You can choose your own style in the next exercise.

example (p q r : Prop) : (p → (q → r)) ↔ p ∧ q → r := by
  sorry

Of course Lean doesn't need any help to prove this kind of logical tautologies. This is the job of the tauto tactic, which can prove true statements in propositional logic.

example (p q r : Prop) : (p → (q → r)) ↔ p ∧ q → r := by
  tauto

Extential quantifiers

In order to prove ∃ x, P x, we give some x₀ using tactic use x₀ and then prove P x₀. This x₀ can be an object from the local context or a more complicated expression. In the example below, the property to check after use is true by definition so the proof is over.

example : ∃ n : ℕ, 8 = 2*n := by
  use 4

In order to use h : ∃ x, P x, we use the rcases tactic to fix one x₀ that works.

Again h can come straight from the local context or can be a more complicated expression.

example (n : ℕ) (h : ∃ k : ℕ, n = k + 1) : n > 0 := by
  -- Let's fix k₀ such that n = k₀ + 1.
  rcases h with ⟨k₀, hk₀⟩
  -- It now suffices to prove k₀ + 1 > 0.
  rw [hk₀]
  -- and we have a lemma about this
  exact Nat.succ_pos k₀

The next exercises use divisibility in ℤ (beware the ∣ symbol which is not ASCII).

By definition, a ∣ b ↔ ∃ k, b = a*k, so you can prove a ∣ b using the use tactic.

example (a b c : ℤ) (h₁ : a ∣ b) (h₂ : b ∣ c) : a ∣ c := by
  sorry

We can now start combining quantifiers, using the definition

Surjective (f : X → Y) := ∀ y, ∃ x, f x = y

example (f g : ℝ → ℝ) (h : Surjective (g ∘ f)) : Surjective g := by
  sorry

This is the end of this file about ∃ and ∧. You've learned about tactics * rcases * tauto * use

This is the end of the Basics folder. We deliberately left out the logical or operator and everything around negation so that you could move as quickly as possible into actual mathematical content. You now get to choose one file from the Topics.

See the bottom of 03Forall for descriptions of the choices.