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The expression ~(~p to q) is logically e...

The expression `~(~p to q)` is logically equivalent to

A

`p wedge ~q`

B

`p wedge q`

C

`~p wedge q`

D

`~p wedge ~q`

Text Solution

Verified by Experts

The correct Answer is:
D
Since, the expression
`p to q -= ~ p vee q`
So, ` ~p to q -= p ve q`
and therefore `~(~p to q)-= ~(p vee q)`
`-= (~p) wedge (~q) " " `[by De Morgan's law]
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