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

  1. #1
    Super Member bigwave's Avatar
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    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
    prob14 (p \to q) \wedge (q \to p)\-xke.png
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  2. #2
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    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.
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  3. #3
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    Re: prob14 (p \to q) \wedge (q \to p)\\

    Quote Originally Posted by bigwave View Post
    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:$\begin{align*}&q\\
    \therefore p &\vee q
    \end{align*}$
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  4. #4
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    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]
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  5. #5
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    Re: prob14 (p \to q) \wedge (q \to p)\\

    Quote Originally Posted by bigwave View Post
    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 $p\wedge q$
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