Answer
(a) and (b) are logically equivalent
Work Step by Step
let us assume 3 statements
p : x$\lt$2 (so ~p : x$\geq$2)
q : x$\gt$1 (so ~q : x$\leq$1)
r : x$\lt$3 (so ~r : $\geq$3)
so (a) converts to p ∨ ~(q ∧ r)
and (b) converts to ~q ∨ (p ∧ ~r)
so we have p ∨ ~(q ∧ r) ≡ p ∨ (~q ∨ ~r) [De Morgan’s law]
p ∨ (~q ∨ ~r) ≡ p ∨ (~r ∨ ~q) [commutative law]
p ∨ (~r ∨ ~q) ≡ (p ∨ ~r) ∨ ~q [Associative law]
(p ∨ ~r) ∨ ~q ≡ ~q ∨ (p ∨ ~r) [commutative law]
Hence we have (a) ≡ (b)
