Let $$*, \square \in \{\wedge, \vee\}$$ be such that the Boolean expression $$(p * \sim q) \Rightarrow (p \square q)$$ is a tautology. Then:

We have to select the two connectives $$*$$ and $$\square$$ from the set $$\{\wedge ,\vee\}$$ in such a way that the statement

$$ (\,p * \sim q\,)\;\Rightarrow\;(p \;\square\; q) $$

is a tautology, that is, it must be true for every possible truth-value assignment of the propositional variables $$p$$ and $$q$$.

First recall the standard equivalence that rewrites an implication:

$$ a \Rightarrow b \;\equiv\; \lnot a \;\vee\; b. $$

Applying this formula to our expression we obtain

$$ (\,p * \sim q\,)\;\Rightarrow\;(p \square q) \;\equiv\; \lnot\,(p * \sim q) \;\vee\; (p \square q). $$

Because $$*$$ and $$\square$$ may each be either $$\wedge$$ (AND) or $$\vee$$ (OR), four distinct combinations are possible. We test them one by one, showing every algebraic step.

Case 1: $$*=\wedge,\;\square=\vee$$

Substituting these symbols gives

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

We now push the negation inside the parenthesis with De Morgan’s law:

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

Therefore the whole expression becomes

$$ \big((\lnot p) \;\vee\; q\big) \;\vee\; (p \;\vee\; q). $$

Using associativity and commutativity of $$\vee$$ we group the disjuncts that involve $$p$$:

$$ \big((\lnot p) \;\vee\; p\big) \;\vee\; q. $$

Since a proposition OR its negation is always true, we have

$$ (\lnot p) \;\vee\; p \;=\; \text{T} \quad\text{(a tautology).} $$

Thus the entire expression simplifies to

$$ \text{T} \;\vee\; q \;=\; \text{T}. $$

So for $$*=\wedge,\;\square=\vee$$ the implication is always true; hence this choice makes the given statement a tautology.

Case 2: $$*=\wedge,\;\square=\wedge$$

The expression now is

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

To check whether this is a tautology, take the assignment $$p=\text{T},\;q=\text{F}$$ (i.e. $$p=1,q=0$$).

Then we get

$$ \lnot p = 0,\quad q = 0,\quad p \wedge q = 1\wedge 0 = 0. $$

So the whole disjunction becomes $$0\;\vee\;0=0$$, which is false. Hence this case fails to be a tautology.

Case 3: $$*=\vee,\;\square=\vee$$

We have

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

Again by De Morgan,

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

Hence the whole statement is

$$ \big((\lnot p) \wedge q\big) \;\vee\; (p \vee q). $$

Choose $$p=\text{F},\;q=\text{F}$$. Then

$$ (\lnot p)\wedge q = 1\wedge 0 = 0,\quad p\vee q = 0\vee 0 = 0, $$

so the overall value is $$0\vee 0 = 0$$—not a tautology.

Case 4: $$*=\vee,\;\square=\wedge$$

The expression becomes

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

With the assignment $$p=\text{F},\;q=\text{F}$$ we get

$$ (\lnot p)\wedge q = 1\wedge 0 = 0,\quad p\wedge q = 0\wedge 0 = 0, $$

so the disjunction is again $$0\vee 0 = 0$$. Hence this case also fails to be a tautology.

Among all four possibilities, only Case 1—namely $$*=\wedge$$ and $$\square=\vee$$—produces a statement that is true under every possible truth assignment. All other choices admit at least one counter-example, so they are not tautologies.

Hence, the correct answer is Option C.