Question

The following statement $(p\to q)\to \left[\right(~p\to q)\to q]$ is

• A. equivalent to $~p\to q$
• B. equivalent to $p\to ~q$
• C. a fallacy
• D. a tautology

Answer 1

The correct option is D

a tautology

Explanation for the correct option:

Explain the negation in tautology for the given statement

Given, $(p\to q)\to \left[\right(~p\to q)\to q]$

Tautology: A tautology is a logical statement in which the conclusion is equivalent to the premise. More colloquially, it is formula in propositional calculus which is always true.

Using this table we can clearly able to see that the given statement is a tautology.

p	q	$p\to q$	$(~p\to q)$	$\left(~p\to q\right)\to q$	$(p\to q)\to \left[\left(~p\to q\right)\to q\right]$
T	T	T	T	T	T
T	F	F	T	F	T
F	T	T	T	T	T
F	F	T	F	T	T

Here , option $\left(D\right)$is correct because $(p\to q)\to \left[\right(~p\to q)\to q]$ is equivalent to T ( true ) and all the values in the last column are true. So , $(p\to q)\to \left[\right(~p\to q)\to q]$ is a tautology.

Explanation for the in-correct option:

Since all the values in the last column are true, hence the given statement is a tautology.

Here , option $\left(A\right)$is discarded because $(p\to q)\to \left[\right(~p\to q)\to q]$ is equivalent to T ( true ) . So , $(p\to q)\to \left[\right(~p\to q)\to q]$ is not equivalent to $~p\to q$.

Here , option $\left(B\right)$is discarded because $(p\to q)\to \left[\right(~p\to q)\to q]$ is equivalent to T ( true ) . So , $(p\to q)\to \left[\right(~p\to q)\to q]$ is not equivalent to $p\to ~q$.

Here , option $\left(C\right)$is discarded because $(p\to q)\to \left[\right(~p\to q)\to q]$ is equivalent to T ( true ) and all the values in the last column are true. So , $(p\to q)\to \left[\right(~p\to q)\to q]$ is not a fallacy.

Hence , the given statement $\mathbf{(}\mathbf{p}\mathbf{}\mathbf{\to }\mathbf{}\mathbf{q}\mathbf{)}\mathbf{}\mathbf{\to }\mathbf{}\mathbf{\left[}\mathbf{\right(}\mathbf{~}\mathbf{p}\mathbf{}\mathbf{\to }\mathbf{}\mathbf{q}\mathbf{)}\mathbf{}\mathbf{\to }\mathbf{}\mathbf{q}\mathbf{]}$ is a tautology.