Exercise logic.propositional.proof.unicode
Description
Prove two propositions equivalent (unicode support)
Examples
easy (7)
¬(p ∧ q) ∨ s ∨ ¬r == (p ∧ q) → (r → s)
¬(p ∧ (q ∨ r)) == ¬p ∨ (¬q ∧ ¬r)
medium (14)
¬((p → q) → (p ∧ q)) == (p → q) ∧ (¬p ∨ ¬q)
¬((p ↔ q) → (p ∨ (p ↔ q))) == F
(p → q) → (p → s) == (¬q → ¬p) → (¬s → ¬p)
(p → ¬q) → q == (s ∨ (s → (q ∨ p))) ∧ q
(p → q) ∧ (r → q) == (p ∨ r) → q
p ∧ (q ∨ s) == (q ∧ ¬s ∧ p) ∨ (p ∧ s)
difficult (10)
(p ∧ ¬q) ∨ (q ∧ ¬p) == (p ∨ q) ∧ ¬(p ∧ q)
(p ∧ q ∧ r) ∨ (¬p ∧ q) == (¬p ∧ q ∧ ¬r) ∨ (q ∧ r)
(q → (¬p → q)) → p == ¬p → (q ∧ p ∧ q ∧ q)
p → (q → r) == (p → q) → (p → r)
¬((p → q) → ¬(q → p)) == p ↔ q
(p → q) ↔ (p → r) == (p → (q ∧ r)) ∨ ¬(p → (q ∨ r))
p ∨ (q ∧ r) == (p ∧ ¬q) ∨ (p ∧ ¬r) ∨ (q ∧ r)
(p ∧ q) ∨ (¬q ∧ r) == (p ∧ r) ∨ (p ∧ q ∧ ¬r) ∨ (¬p ∧ ¬q ∧ r)