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The truth table for $(\sim p\lor q)$ is as follows:

$p$ $q$ $\sim p$ $(\sim p\lor q)$
T T F T
T F F F
F T T T
F F T T

from the above, it is clearly visible that it is $(p\to q)$  truth table.

$(p\to q)$ is a conditional statement. the truth table is shown below:

$p$ $q$ $(p\to q)$
T T T
T F F
F T T
F F T

Another approach:

$\because (p\lor q) \equiv (\sim p\to q)$

using above property we can rewrite the given expression:

$(\sim p\lor q) \ = [\sim(\sim p)\to q]$

$(\sim p\lor q) \equiv (p\to q)$

So option ($A)$ is correct.

edited by
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From the truth table of implication ( p-> q) see below img.

It Clearly States that Option A.  ∼p∨q

is ans.

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