1.

The statement `~(pharr~ q)`is(1) equivalent to`pharrq`(2) equivalent to`~ pharrq`(3) atautology(4) a fallacy

Answer» Here, we will represent `T` for True and `F` for False.
When `p=T` and `q = T`,
`~(p harr ~q) = T`, `~p harr q = F`, `p harr q = T`

When `p=T` and `q = F`,
`~(p harr ~q) = F`, `~p harr q = T`, `p harr q = F`

When `p=F` and `q = T`,
`~(p harr ~q) = F`, `~p harr q = T`, `p harr q = F`

When `p=F` and `q = F`,
`~(p harr ~q) = T`, `~p harr q = F`, `p harr q = T`


`:. ~(p harr ~q) ` is equivalent to `p harr q`.
Also, it is not a tautology as it contains false.
It is not a fallacy as it contains true also.
So, option `1` is the correct option.


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