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.