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



Answer to a math question 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

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

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 rp (modus tollens):
\neg r\neg p
Thus, \neg p\neg r
From q \vee r and \neg p, we get:
Therefore, p → r is true.

Now let's prove r → p:
Assume r.
Since from rp, we have rp.
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}

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