# Thread: prob14 (p \to q) \wedge (q \to p)\\

1. ## prob14 (p \to q) \wedge (q \to p)\\

Determine whether the symbolic form of the argument is valid or invalid
\begin{align*}\displaystyle
&(p \to q) \wedge (q \to p)\\
&q\\
&------\\
&\therefore p \lor q
\end{align*}

ok this should be easy just is its all new to me... and taking the class next spring

2. ## Re: prob14 (p \to q) \wedge (q \to p)\\

You can try a truth table so see if you get the same result. I find that the original statement is true when $(p\wedge q)\vee (\neg p \wedge \neg q)$ (XNOR), which is not the same as $p\vee q$. I have never done boolean algebra, so I am not sure what is expected.

3. ## Re: prob14 (p \to q) \wedge (q \to p)\\ Originally Posted by bigwave Determine whether the symbolic form of the argument is valid or invalid
\begin{align*}\displaystyle
&(p \to q) \wedge (q \to p)\\
&q\\
&------\\
&\therefore p \lor q
\end{align*}
ok this should be easy just is its all new to me... and taking the class next spring
Did you make this up? It is not a problem!
One of the basic argument forms is addition. That is:\begin{align*}&q\\ \therefore p &\vee q \end{align*}

4. ## Re: prob14 (p \to q) \wedge (q \to p)\\

You could also use logic rules. The result should be stronger than p v q, as an argument can be built from q alone? I constructed a truth table to convince myself afterward. Basically if q is true and p is false, then you'll see q -> p is false.

q [Assumption]
p v q [Disjunction Introduction]

Argument for conclusion p ^ q, and hence valid.

(p -> q) ^ (q -> p) [Assumption]
(p -> q) [ ^ Elimination]
(q -> p) [ ^ Elimination]
q [Assumption]
p [Implication]
-----
p ^ q [Conjunction Introduction]

5. ## Re: prob14 (p \to q) \wedge (q \to p)\\ Originally Posted by bigwave Determine whether the symbolic form of the argument is valid or invalid
\begin{align*}\displaystyle
&(p \to q) \wedge (q \to p)\\
&q\\
&------\\
&\therefore p \lor q
\end{align*}
ok this should be easy just is its all new to me... and taking the class next spring
Here is a much better question.
\begin{align*}\displaystyle &(p \to q) \wedge (q \to p)\\ &q\\ &------\\ &\therefore p \wedge q\end{align*}
From $(p \to q) \wedge (q \to p)$ using simplification we can get $(q\to p)$.
Now from $q~\&~(q\to p)$ using modus ponens we get $p$.
Then using conjunction it gives $p\wedge q$