Question
$\sim[\sim p \wedge(p \Leftrightarrow q)] \equiv$(1) $\mathrm{p} \vee \mathrm{q}$(2) $\mathrm{q} \wedge \mathrm{p}$(3) $\mathrm{T}$(4) $\mathrm{F}$
Step 1
This statement is a negation of two parts: $\sim p$ (not p) and $p \Leftrightarrow q$ (p if and only if q). Show more…
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$(p \vee \sim q) \wedge q$ is equivalent to (1) $\mathrm{p} \vee \mathrm{q}$ (2) $\mathrm{p} \wedge \mathrm{q}$ (3) $\mathrm{p} \vee \sim \mathrm{q}$ (4) $\mathrm{p} \wedge \sim \mathrm{q}$
$\sim[\sim p \vee(\sim p \Leftrightarrow q)]$ is equivalent to (1) $\mathrm{p} \wedge(\mathrm{p} \Leftrightarrow \mathrm{q})$ (2) $\mathrm{p} \wedge \mathrm{q}$ (3) Both (1) and (2) (4) None of these
$$[(p \vee q) \wedge(r \wedge s)] \wedge(t \vee \sim p)$$
Logic
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