Mathematical Reasoning Question 9
Question 9 - 31 January - Shift 1
$(S 1)(p \Rightarrow q) \vee(p \wedge(\sim q))$ is a tautology
$(S 2)((\sim p) \Rightarrow(\sim q)) \wedge((\sim p) \vee q)$ is a Contradiction. Then
(1) only (S2) is correct
(2) both (S1) and (S2) are correct
(3) both (S1) and (S2) are wrong
(4) only (S1) is correct
Show Answer
Answer: (2)
Solution:
| $\mathbf{p}$ | $\mathbf{q}$ | $\mathbf{p} \Rightarrow \mathbf{q}$ | $\sim \mathbf{q}$ | $p \wedge \sim q$ | $(p \Rightarrow q) \vee(p \wedge \sim q)$ |
|---|---|---|---|---|---|
| $T$ | $T$ | $T$ | $F$ | $F$ | $T$ |
| $T$ | $F$ | $F$ | $T$ | $T$ | $T$ |
| $F$ | $T$ | $T$ | $F$ | $F$ | $T$ |
| $F$ | $F$ | $T$ | $T$ | $F$ | $T$ |
| $\sim \mathbf{p}$ | $\sim \mathbf{q}$ | $\sim \mathbf{p} \Rightarrow \sim \mathbf{q}$ | $\sim \mathbf{p} \vee \mathbf{q}$ | $((\sim \mathbf{p}) \Rightarrow(\sim \mathbf{q})) \wedge(\sim \mathbf{p}) \vee$ |
|---|---|---|---|---|
| $F$ | $F$ | $T$ | $T$ | $T$ |
| $F$ | $T$ | $T$ | $F$ | $F$ |
| $T$ | $F$ | $F$ | $T$ | $F$ |
| $T$ | $T$ | $T$ | $T$ | $T$ |
