Explanation

(b)
[table] $p$ | $q$ | $p \land q$ | $\sim p$ | $\sim p \vee q$ | $(p \land q)$
$\rightarrow$
$( \sim p \vee q)$ | $\sim\lbrack(p \land q)$
$\rightarrow$
$( \sim p \vee q)\rbrack$ ; T
T
F
F | T
F
T
F | T
F
F
F | F
F
T
T | T
F
T
T | T
T
T
T | F
F
F
F [/table]
It is clear from the table that
$\sim\lbrack(p \land q) \rightarrow ( \sim p \vee q)\rbrack$
is a contradiction.