Question

is there any fast way of solving this?

Answer 1

\[p \wedge \lnot q \vee \lnot (p \wedge q)\]

Answer 2

one of possible truth table combination following the order of operation
\[T \wedge \lnot F \vee \lnot (T \wedge F)\]

Answer 3

demorgan laws :

\(\large \neg (p\land q) = \neg p \lor \neg q\)

that means the given boolean expression evaluates true when atleast one of (p, q) is F

Answer 4

for every two sets A and B
\[(A \cup B)^c = A^c \cap B^c; (A \cap B)^c = A^c \cup B^c \]