If $(p \wedge \sim q) \wedge(p \wedge r) \rightarrow \sim p \vee q$ is false, then the truth values of $p,…
If $(p \wedge \sim q) \wedge(p \wedge r) \rightarrow \sim p \vee q$ is false, then the truth values of $p, q$ and $r$ are respectively
- $\mathrm{T}, \mathrm{T}, \mathrm{T}$
- F, F, F
- T, F, T
- $F, T, F$
Solution
$\begin{array}{ll} & (p \wedge \sim q) \wedge(p \wedge r) \rightarrow \sim p \vee q \equiv F \\ \therefore & \sim p \vee q \equiv F \\ \therefore & \sim p \equiv F \text { and } q \equiv F \\ \therefore & p \equiv T \text { and } q=F \\ \therefore & \text { Option }(3) \text { is correct. }\end{array}$
Asked in: MHT CET 2024 (04 May Shift 1)
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