Statement-1: sim (p Leftrightarrow sim q) is | vedclass

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(A) Statement-$1$: $\sim (p \Leftrightarrow \sim q)$ is equivalent to $p \Leftrightarrow q$.Statement-$2$: $\sim (p \Leftrightarrow \sim q)$ is a taut

Vedclass · Accepted Answer

Statement-$1$ is true. Statement-$2$ is false.

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(A) Statement-$1$: $\sim (p \Leftrightarrow \sim q)$ is equivalent to $p \Leftrightarrow q$.Statement-$2$: $\sim (p \Leftrightarrow \sim q)$ is a tautology.$p$$q$$\sim q$$(p \Leftrightarrow q)$$(p \Leftrightarrow \sim q)$$\sim (p \Leftrightarrow \sim q)$$T$$T$$F$$T$$F$$T$$T$$F$$T$$F$$T$$F$$F$$T$$F$$F$$T$$F$$F$$F$$T$$T$$F$$T$From the truth table,we can see that the column for $\sim (p \Leftrightarrow \sim q)$ is identical to the column for $(p \Leftrightarrow q)$,so Statement-$1$ is true.However,the column for $\sim (p \Leftrightarrow \sim q)$ contains both $T$ and $F$ values,so it is not a tautology. Thus,Statement-$2$ is false.

Statement-$1$: $\sim (p \Leftrightarrow \sim q)$ is equivalent to $p \Leftrightarrow q$.
Statement-$2$: $\sim (p \Leftrightarrow \sim q)$ is a tautology.

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Statement $-1 :$ $\sim (p \leftrightarrow \sim q)$ is equivalent to $p \leftrightarrow q$.
Statement $-2 :$ $\sim (p \leftrightarrow \sim q)$ is a tautology.

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Statement-$1$: $\sim (p \Leftrightarrow \sim q)$ is equivalent to $p \Leftrightarrow q$.Statement-$2$: $\sim (p \Leftrightarrow \sim q)$ is a tautology.