Archangel Macsika
Show that these compound propositions are tautologies: (1) (¬q ∧ (p → q)) → ¬p (2) ((p ∨ q) ∧ ¬p) → q
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1. Solution for (¬q ∧ (p → q)) → ¬p
(¬q ∧ (p → q)) → ¬p
≡ (¬q ∧ (¬ p ∨ q)) → ¬p
≡ ¬((¬q ∧ (¬ p ∨ q))) ∨ ¬p
≡ ((¬(¬q)) ∨ (¬(¬ p ∨ q))) ∨ ¬p
≡ ((q) ∨ (¬(¬ p)) ∧ (¬ q)) ∨ ¬p
≡ (q ∨ (p ∧ ¬ q)) ∨ ¬p
≡ ((q ∨ p) ∧ (q ∨ ¬ q)) ∨ ¬p
≡ ((q ∨ p) ∧ T) ∨ ¬p
≡ ((q ∨ p) ∨ ¬p
≡ q ∨ (p ∨ ¬p)
≡ q ∨ T
≡ T
2. Solution for ((p ∨ q) ∧ ¬p) → q
((p ∨ q) ∧ ¬p) → q
≡(¬p ∧ (p ∨ q)) ∨ q
≡ ¬ (¬p ∧ (p ∨ q)) ∨ q
≡ (¬ (¬p) ∨ ¬(p ∨ q)) ∨ q
≡ (p ∨ ¬(p ∨ q)) ∨ q
≡ (p ∨ (¬p ∧ ¬q)) ∨ q
≡ ((p ∨ ¬p) ∧ (p ∨ ¬q)) ∨ q
≡ (T ∧ (p ∨ ¬q)) ∨ q
≡ ((p ∨ ¬q) ∧ T) ∨ q
≡ (p ∨ ¬q) ∨ q
≡ p ∨ (¬q ∨ q)
≡ p ∨ (q ∨ ¬q)
≡ p ∨ T
≡ T