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The book Discrete Mathematics by Kenneth A. Ross says: "Let's verify the logic implication $(p\wedge q) \implies p$. For that, we need to consider only just the case when $p\wedge q$ is true; i.e., both, $p$ and $q$, are true. This gives us the truncated table:"

(pwedge q) rightarrow p "

Why the author considers that portion of the truth table? What reasoning I must understand to follow that validation technique?

N.B.: The same technique is applied to $(p \rightarrow q) \Rightarrow [(p \vee r) \rightarrow (q \vee r)]$:

$(p \rightarrow q) \Rightarrow [(p \vee r) \rightarrow (q \vee r)]$

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    $\begingroup$ It may help your Readers to cite which "book" and author your Question is based on. $\endgroup$ – hardmath Jan 29 '16 at 14:52
  • $\begingroup$ Kenneth A. Ross - Discrete Mathematics. $\endgroup$ – InfZero Jan 29 '16 at 14:53
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Recall that an implication $R\implies S$ is "automatically true" in cases where the hypothesis $R$ is false. So the only interesting cases to check are those for which the hypothesis is true.

In this case the hypothesis is $p \wedge q$, so you can proceed just with analyzing what would be implied by that being true.

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The relation $p\rightarrow r$ is logically equivalent to $\neg p\vee r$. So for the case where $p$ is false, $\neg p$ is true and the outcome is then true. So the statement would be false only if $p$ is true and $r$ is false. Now replace this $p$ with $p\wedge q$ and $r$ by $p$ and you have a tautology.

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