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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Brice

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62 Answers

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 provingp \rightarrow r :

Assumep .

From line 1, we know:

(\neg p \vee q) \wedge \neg(q \wedge \neg r)

Since we have assumedp , 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 prover by exhaustion using r → p (modus tollens):

\neg r → \neg p

Thus,\neg p → \neg r

Fromq \vee r and \neg p , we get:

r

Therefore,p → r is true.

Now let's prover → p :

Assumer .

Since fromr → p , we have r → p .

Therefore,r → p is true.

Since we have proved bothp \rightarrow r and r \rightarrow p , we can conclude that:

p \leftrightarrow r , is valid.

\boxed{p \leftrightarrow r}

Given:

1.

2.

Let's start by proving

Assume

From line 1, we know:

Since we have assumed

From

Since

Now, we can prove

Thus,

From

Therefore,

Now let's prove

Assume

Since from

Therefore,

Since we have proved both

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