Negation of inverse of the following statement pattern $(p \wedge q) \rightarrow(p \vee \sim q)$ is

Negation of inverse of the following statement pattern $(p \wedge q) \rightarrow(p \vee \sim q)$ is
  1. $\mathrm{p}$
  2. $\sim q$
  3. $\sim p$
  4. $\mathrm{q}$

Solution

Inverse of $(p \wedge q) \rightarrow(p \vee \sim q)$ is $\begin{aligned} & \sim(p \wedge q) \rightarrow \sim(p \vee \sim q) \\ & \equiv \sim[\sim(p \wedge q)] \vee \sim(p \vee \sim q) \ldots[p \rightarrow q \equiv \sim p \vee q] \end{aligned}$ $\begin{aligned} & \equiv(p \wedge q) \vee(\sim p \wedge q)...[De Morgan's law]\\ & \equiv(q \wedge p) \vee(q \wedge \sim p)...[Commutative law] \\ & \equiv q \wedge(p \vee \sim p)...[Distributive law] \\ & \equiv q \wedge T...[Complement law] \\ & \equiv q...[Identity law] \end{aligned}$ $\therefore \quad$ Negation of inverse of $(p \wedge q) \rightarrow(p \vee \sim q)$ is $\sim q$

Asked in: MHT CET 2023 (13 May Shift 1)

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