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Try doing some basic maths questions in the Lean Theorem Prover. Functions, real numbers, equivalence relations and groups. Click on README.md and then on "Open in CoCalc with one click".

Project: Xena
Views: 18536
License: APACHE
/-
Copyright (c) 2016 Microsoft Corporation. All rights reserved.
Released under Apache 2.0 license as described in the file LICENSE.
Authors: Jeremy Avigad

Examples from the tutorial.
-/
import tactic.finish
open auto

section
variables p q r s : Prop

-- commutativity of ∧ and ∨
example : p ∧ q ↔ q ∧ p := by finish
example : p ∨ q ↔ q ∨ p := by finish

-- associativity of ∧ and ∨
example : (p ∧ q) ∧ r ↔ p ∧ (q ∧ r) := by finish
example : (p ∨ q) ∨ r ↔ p ∨ (q ∨ r) := by finish

-- distributivity
example : p ∧ (q ∨ r) ↔ (p ∧ q) ∨ (p ∧ r) := by finish [iff_def]
example : p ∨ (q ∧ r) ↔ (p ∨ q) ∧ (p ∨ r) := by finish [iff_def]

-- other properties
example : (p → (q → r)) ↔ (p ∧ q → r) := by finish [iff_def]
example : ((p ∨ q) → r) ↔ (p → r) ∧ (q → r) := by finish [iff_def]
example : ¬(p ∨ q) ↔ ¬p ∧ ¬q := by finish
example : ¬p ∨ ¬q → ¬(p ∧ q) := by finish
example : ¬(p ∧ ¬ p) := by finish
example : p ∧ ¬q → ¬(p → q) := by finish
example : ¬p → (p → q) := by finish
example : (¬p ∨ q) → (p → q) := by finish
example : p ∨ false ↔ p := by finish
example : p ∧ false ↔ false := by finish
example : ¬(p ↔ ¬p) := by finish
example : (p → q) → (¬q → ¬p) := by finish

-- these require classical reasoning
example : (p → r ∨ s) → ((p → r) ∨ (p → s)) := by finish
example : ¬(p ∧ q) → ¬p ∨ ¬q := by finish
example : ¬(p → q) → p ∧ ¬q := by finish
example : (p → q) → (¬p ∨ q) := by finish
example : (¬q → ¬p) → (p → q) := by finish
example : p ∨ ¬p := by finish
example : (((p → q) → p) → p) := by finish
end


section

variables (A : Type) (p q : A → Prop)
variable a : A
variable r : Prop

example : (∃ x : A, r) → r := by finish
-- TODO(Jeremy): can we get these automatically?
example (a : A) : r → (∃ x : A, r) := begin safe; apply a_2; assumption end
example : (∃ x, p x ∧ r) ↔ (∃ x, p x) ∧ r := by finish

theorem foo': (∃ x, p x ∨ q x) ↔ (∃ x, p x) ∨ (∃ x, q x) :=
by finish [iff_def]

example (h : ∀ x, ¬ ¬ p x) : p a := by finish
example (h : ∀ x, ¬ ¬ p x) : ∀ x, p x := by finish

example : (∀ x, p x) ↔ ¬ (∃ x, ¬ p x) := by finish

example : (∃ x, p x) ↔ ¬ (∀ x, ¬ p x) := by finish
example : (¬ ∃ x, p x) ↔ (∀ x, ¬ p x) := by finish
example : (¬ ∀ x, p x) ↔ (∃ x, ¬ p x) := by finish
example : (∃ x, ¬ p x) → (¬ ∀ x, p x) := by finish

example : (∀ x, p x → r) ↔ (∃ x, p x) → r := by finish [iff_def]
-- TODO(Jeremy): can we get these automatically?
example (a : A) : (∃ x, p x → r) ↔ (∀ x, p x) → r := begin safe [iff_def]; exact h a end
example (a : A) : (∃ x, r → p x) ↔ (r → ∃ x, p x) := begin safe [iff_def]; exact h a end

example : (∃ x, p x → r) → (∀ x, p x) → r := by finish
example : (∃ x, r → p x) → (r → ∃ x, p x) := by finish

end