For two statements $$p$$ and $$q$$, the logical statement $$(p \rightarrow q) \wedge (q \rightarrow \sim p)$$ is equivalent to

We begin with the compound statement

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

First, recall the fundamental implication formula

$$ p \rightarrow q \;\equiv\; \sim p \,\vee\, q, $$

which says “if p then q” is logically the same as “either not p or q.” Applying this rule to each implication separately, we obtain

$$ (p \rightarrow q)\;=\;\sim p \,\vee\, q, $$

and

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

Therefore our original statement becomes

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

Observe that both disjunctions contain the term $$\sim p$$, so it is natural to use the distributive law of Boolean algebra, which states

$$ (A \,\vee\, B)\;\wedge\;(A \,\vee\, C) \;\equiv\; A \,\vee\, (B \,\wedge\, C). $$

Here we identify

$$A = \sim p,\qquad B = q,\qquad C = \sim q.$$

Substituting these into the distributive formula, we have

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

Now, $$q \,\wedge\, \sim q$$ is a contradiction: one cannot have $$q$$ and “not $$q$$” true at the same time. Thus

$$ q \,\wedge\, \sim q \;\equiv\; \text{False}. $$

Using the identity $$A \,\vee\, \text{False} \equiv A$$, we finally arrive at

$$ \sim p \,\vee\, \text{False} \;\equiv\; \sim p. $$

Hence the entire logical statement simplifies to $$\sim p$$.

Looking at the given options, $$\sim p$$ corresponds to Option C.

Hence, the correct answer is Option C.