Use the Proof method to determine whether the following argument form is valid. 1. (~p v q) & ~(q & ~r). Given. 2. r → p . Given SHOW: p ↔ r

To show p \leftrightarrow r, we need to prove both p \rightarrow r and r \rightarrow p.

Given:
1. (\neg p \vee q) \wedge \neg(q \wedge \neg r)
2. r \rightarrow p

Let's start by proving p \rightarrow r:
Assume p.
From line 1, we know:
(\neg p \vee q) \wedge \neg(q \wedge \neg r)
Since we have assumed p, we get:
(\neg p \vee q) and \neg(q \wedge \neg r)
From \neg(q \wedge \neg r), we get:
\neg q \vee r
Since (\neg p \vee q), we can say:
q \vee r
Now, we can prove r by exhaustion using r → p (modus tollens):
\neg r → \neg p
Thus, \neg p → \neg r
From q \vee r and \neg p, we get:
r
Therefore, p → r is true.

Now let's prove r → p:
Assume r.
Since from r → p, we have r → p.
Therefore, r → p is true.

Since we have proved both p \rightarrow r and r \rightarrow p, we can conclude that:
p \leftrightarrow r, is valid.

\boxed{p \leftrightarrow r}