Q. $( S 1)(p \Rightarrow q) \vee(p \wedge(\sim q))$ is a tautology $(S2) ((\sim p) \Rightarrow(\sim q)) \wedge((\sim p) \vee q)$ is a contradiction. Then
Solution:
$p$
$q$
$p \Rightarrow q$
$\sim q$
$p \wedge \sim q$
$(p \Rightarrow q) \vee(p \wedge \sim q)$
T
T
T
F
F
T
T
F
F
T
T
T
F
T
T
F
F
T
F
F
T
T
F
T
$\sim p$
$\sim q$
$\sim p \Rightarrow \sim q$
$\sim p \vee q$
$((\sim p) \Rightarrow(\sim q)) \wedge(\sim p) \vee q)$
F
F
T
T T
F
T
T
F F
T
F
F
T F
T
T
T
T T
| $p$ | $q$ | $p \Rightarrow q$ | $\sim q$ | $p \wedge \sim q$ | $(p \Rightarrow q) \vee(p \wedge \sim q)$ |
|---|---|---|---|---|---|
| T | T | T | F | F | T |
| T | F | F | T | T | T |
| F | T | T | F | F | T |
| F | F | T | T | F | T |
| $\sim p$ | $\sim q$ | $\sim p \Rightarrow \sim q$ | $\sim p \vee q$ | $((\sim p) \Rightarrow(\sim q)) \wedge(\sim p) \vee q)$ |
|---|---|---|---|---|
| F | F | T | T | T |
| F | T | T | F | F |
| T | F | F | T | F |
| T | T | T | T | T |