Question

Which of the following is a tautology?

1. $P \rightarrow P \wedge Q$
2. $P \rightarrow Q \wedge V$
3. $(P \rightarrow Q) \wedge (Q \rightarrow R) \rightarrow P \rightarrow R$
4. $(P \rightarrow Q) \leftrightarrow (\sim Q \rightarrow \sim P)$

Answer 1

$P\to Q = \neg P \vee Q$
$\neg Q \to \neg P = Q \vee \neg P$

Hence, D is correct.

C choice is also correct as $\to$ is usually taken as right associative (See here: http://math.stackexchange.com/questions/12223/associativity-of-logical-connectives ). It'll be interpreted as

$\left(\left(P\to Q\right) \wedge \left(Q\to R\right)\right) \to \left(P \to R\right)$

This is a tautology.

Comments

Actually I got c by mistake.

If  → is right associative  then getting c is as tautology.

As we mentioned earlier, AND and OR are associative and commutative; so is .

We shall assume that they group from the left if it is necessary to be specific. The
other binary operators listed above are not associative. We shall generally show
parentheses around them explicitly to avoid ambiguity, but each of the operators
->, NAND, and NOR will be grouped from the left in strings of two or more of the
same operator. 

but this sentence making confusion.

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Yes. That is telling just the opposite. I'm not able to find a concrete place for the convention for associativity for $\to$, may be there is none.

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Sir Is there any proof or english translation so that I can get that Implication is right associative?

Answer 2

actually both (c) and (d) are correct.

d is called the contrapostive.

as for the third option

now let's say that the third option is incorrect then we can have (p -> r) false even when the left hand side is true.

but p -> r false means p = True and R = False. Now for the left hand side to be true Q has to be false and true at the same time. Since (p -> q) demands q to be false whereas (q -> r) demands q to be true. which is never possible hence that is also a tutology.

Answer 3

C is Correct.