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