Truth values of $p \rightarrow r$ is $F$ and $p \leftrightarrow q$ is $F$. Then the truth values of $(\sim p…

Truth values of $p \rightarrow r$ is $F$ and $p \leftrightarrow q$ is $F$. Then the truth values of $(\sim p \vee q) \rightarrow(p \vee \sim q)$ and $(p \wedge \sim q) \rightarrow(\sim p \wedge q)$ are respectively
  1. $\mathrm{T}, \mathrm{F}$
  2. $\mathrm{F}, \mathrm{T}$
  3. $\mathrm{T}, \mathrm{T}$
  4. $\mathrm{F}, \mathrm{F}$

Solution

Truth values of $p \rightarrow r$ is $F$ and $p \leftrightarrow q$ is $F$ $\begin{aligned} \therefore \quad & p \equiv T, q \equiv F, r \equiv F \\ & (\sim p \vee q) \rightarrow(p \vee \sim q) \\ & \equiv(\sim T \vee F) \rightarrow(T \vee \sim F) \\ & \equiv(F \vee F) \rightarrow(T \vee T) \\ & \equiv F \rightarrow T \\ & \equiv T \\ & (p \wedge \sim q) \rightarrow(\sim p \wedge q) \\ & \equiv(T \wedge \sim F) \rightarrow(\sim T \wedge F) \\ & \equiv(T \wedge T) \rightarrow(F \wedge F) \\ & \equiv T \rightarrow F \\ & \equiv F \end{aligned}$

Asked in: MHT CET 2024 (16 May Shift 2)

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