Mathematical Reasoning Question 5
Question 5 - 29 January - Shift 1
If $p, q$ and $r$ are three propositions, then which of the following combination of truth values of $p, q$ and $r$ makes the logical expression
${(p \vee q) \wedge((\sim p) \vee r)} \to((\sim q) \vee r)$ false ?
(1) $p=T, q=F, r=T$
(2) $p=T, q=T, r=F$
(3) $p=F, q=T, r=F$
(4) $p=T, q=F, r=F$
Show Answer
Answer: (3)
Solution:
| $p$ | $q$ | $r$ | $(p \vee q) \wedge((\sim p) \vee r)$ | $\sim q \vee r$ | |
|---|---|---|---|---|---|
| $(1)$ | $T$ | $F$ | $T$ | $T$ | $T$ |
| $(2)$ | $T$ | $T$ | $F$ | $F$ | $F$ |
| $(3)$ | $F$ | $T$ | $F$ | $T$ | $F$ |
| $(4)$ | $T$ | $F$ | $F$ | $F$ | $T$ |
Option (3) $(p \vee q) \wedge(\sim q \vee r) \to(\sim p \vee r)$ will be False.
