Which of the following is a tautology?

We recall that a statement is called a tautology when it remains true for every possible truth-value assignment of its component propositions. We shall examine each option one by one, reducing the compound statement with standard logical identities and looking for a possible counter-example. If even a single assignment makes the statement false, then it is not a tautology.

We begin with Option A:

$$ (\sim p)\;\wedge\;(p \vee q)\;\to\; q $$

First we simplify the antecedent "$$(\sim p)\wedge (p\vee q)$$". Using the distributive law

$$ a\wedge(b\vee c)\;=\;(a\wedge b)\;\vee\;(a\wedge c), $$

with $$a=\sim p,\; b=p,\; c=q,$$ we obtain

$$ (\sim p)\wedge(p\vee q)\;=\;[(\sim p)\wedge p] \;\vee\;[(\sim p)\wedge q]. $$

Now $$ (\sim p)\wedge p $$ is a contradiction, i.e. it is always false, so it may be dropped from the disjunction:

$$ (\sim p)\wedge(p\vee q)\;=\;(\sim p)\wedge q. $$

Therefore Option A becomes

$$ [(\sim p)\wedge q]\;\to\;q. $$

Next we recall the definition of implication:

$$ a\to b\;\equiv\;\sim a\;\vee\;b. $$

Replacing $$a$$ with $$ (\sim p)\wedge q $$ and $$b$$ with $$q$$ gives

$$ [(\sim p)\wedge q]\;\to\;q\;\equiv\;\sim[(\sim p)\wedge q]\;\vee\;q. $$

Apply De Morgan’s law to the negation inside:

$$ \sim[(\sim p)\wedge q]\;=\;\sim(\sim p)\;\vee\;\sim q\;=\;p\;\vee\;\sim q. $$

So the whole expression is

$$ (p\;\vee\;\sim q)\;\vee\;q. $$

The associative and commutative laws for disjunction permit us to regroup and obtain

$$ p\;\vee\;(q\;\vee\;\sim q). $$

Within the parentheses, $$ q\;\vee\;\sim q $$ is the law of excluded middle, which is always true, i.e. a tautology $$T$$. Hence we have

$$ p\;\vee\;T\;=\;T. $$

Thus Option A simplifies to a statement that is invariably true, making it a tautology.

Now we test Option B:

$$ (q\to p)\;\vee\;\sim(p\to q). $$

First expand each implication using $$a\to b \equiv \sim a\;\vee\;b$$:

$$ (q\to p) = (\sim q)\;\vee\;p, \qquad (p\to q) = (\sim p)\;\vee\;q.$$ Hence

$$ \sim(p\to q) = \sim[(\sim p)\vee q] = p\wedge\sim q $$ by De Morgan’s law. Therefore Option B becomes

$$ (\sim q\;\vee\;p)\;\vee\;(p\wedge\sim q). $$

We look for a truth-value combination that makes the whole disjunction false. A disjunction is false only when each of its components is false. So we need

$$ \sim q\;\vee\;p = F, \quad\text{and}\quad p\wedge\sim q = F. $$

Take $$p = F$$ and $$q = T$$. Then

$$\sim q = F,\quad p = F,\quad \therefore \sim q\;\vee\;p = F.$$ Also $$p\wedge\sim q = F\wedge F = F.$$ Both parts are false, so the entire statement is false for this assignment. Hence Option B is not a tautology.

Next we inspect Option C:

$$ (\sim q)\;\vee\;(p\wedge q)\;\to\;q. $$

Choose $$q = F$$ (i.e. $$q$$ is false). Then $$\sim q = T$$, so the antecedent $$(\sim q)\;\vee\;(p\wedge q)$$ is true irrespective of $$p$$. The consequent $$q$$, however, is false. Therefore the implication becomes $$T\to F,$$ which is false. So Option C is also not a tautology.

Finally Option D:

$$ (p\to q)\;\wedge\;(q\to p). $$

This is logically equivalent to the biconditional $$p\leftrightarrow q,$$ which is true only when $$p$$ and $$q$$ share the same truth value. Take $$p = T,\, q = F$$. Then $$p\to q = F$$, so the whole conjunction is false. Consequently Option D is not a tautology.

Only Option A survives every test and is always true.

Hence, the correct answer is Option A.