Recall that a ⇒ b ≡ ¬a ∨ b.
• r ⇒ (p ∧ q) ≡ ¬r ∨ (p ∧ q)
In row C, q is false so p ∧ q false, and r is true so ¬r is false.
¬r ∨ (p ∧ q) ≡ false ∨ false ≡ false
• r ⇒ (p ∨ q) ≡ ¬r ∨ (p ∨ q) = p ∨ q ∨ ¬r
In each of rows A, C, and E, at least one of p or q is true, so
p ∨ q ∨ ¬r = true
• (q ∧ r) ⇒ p ≡ ¬(q ∧ r) ∨ p ≡ (¬q ∨ ¬ r) ∨ p = p ∨ ¬q ∨ ¬r
In row E, p is false and both q and r are true, so ¬q and ¬r are both false.
false ∨ false ∨ false = false
• (q ∨ r) ⇒ p ≡ ¬(q ∨ r) ∨ p ≡ (¬q ∧ ¬r) ∨ p
In row E, p is false and both q and r are true, so both ¬q and ¬r are false.
(¬q ∧ ¬r) ∨ p ≡ (false ∧ false) ∨ false ≡ false ∨ false ≡ false
