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showing logical argument is valid or invalid

  1. $p\to q$
    $q\to (r \lor \lnot p)$
    $\lnot r$
    therefore, $\lnot p$

  2. $p\to (r \lor q)$
    $q\to\lnot r$
    therefore, $p$

I believe (1) is valid and (2) invalid. Am I correct?

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You are correct, and the full truth tables will demonstrate that. – Brian M. Scott Sep 18 '12 at 20:16
  1. If $\lnot r$ and $r \lor \lnot p$, then $\lnot p$; so if $q \to (r \lor \lnot p)$, then $q \to \lnot p$. But $p \to q$, so $p \to \lnot p$, and if both $p$ and $\lnot p$, then $\bot$, so $p \to \bot$, i.e. $\lnot p$. I leave the formal details to you, since you have not specified what formal system you are using.

  2. Suppose both $p$ and $q$ are false while $r$ is true. Then, vacuously, $p \to (r \lor q)$, $q \to \lnot r$, and $r$ are all satisfied while $p$ is not.

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The truth table has these eight lines: $$ \begin{array}{|c|c|c|} \hline p & q & r \\ \hline T & T & T \\ T & T & f \\ T & f & T \\ f & T & T \\ T & f & f \\ f & T & f \\ f & f & T \\ f & f & f \\ \hline \end{array} $$ In each line one can identify each of the premises of an argument as true or false, and likewise one can classify the conclusion as true or false: $$ \begin{array}{|c|c|c|c|c|} \hline p & q & r & p\to r\lor q & q\to\lnot r \\ \hline T & T & T & T & f \\ T & T & f & T & T \\ T & f & T & T & T \\ f & T & T & T & f \\ T & f & f & f & T \\ f & T & f & T & T \\ f & f & T & T & T \\ f & f & f & T & T \\ \hline {} \end{array} $$ In the seventh row of this table, all three premises of your second argument are true but the conclusion is false. Therefore the argument is not valid. An argument of this kind is valid precisely if the conclusion is true in every row of the true table in which the premises are true.

$p\to (r \lor q)$
$q\to\lnot r$
therefore, $p$

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