$p \Rightarrow q$ is true

Which is the logic value of $(p \vee r) \Rightarrow (q \vee r)$ ?

I did at this point and I can not siplify it more:

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

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

Could someone help me please and explain how should I simplify it?

  • 1
    $\begingroup$ Hint: Either $r$ is true or it is false ... $\endgroup$ – Henning Makholm Oct 23 '16 at 16:15
  • $\begingroup$ Well, there is also a way to see this without assuming excluded middle ;) $\endgroup$ – Oles Wohnzimmer Oct 23 '16 at 16:20
  • $\begingroup$ because of associativity you can drop parentheses around $q \vee r$. And then combine the $r$ with the $\neg p \wedge \neg r$ $\endgroup$ – Bram28 Oct 23 '16 at 16:23
  • $\begingroup$ Thank you @HenningMakholm , but when can I suppose that a letter is true or false? $\endgroup$ – Francisca Sousa Oct 23 '16 at 16:37
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    $\begingroup$ @FranciscaSousa: In classical logic you can always assume that anything you can write down (such as, for example, $r$) is either true or false. $\endgroup$ – Henning Makholm Oct 23 '16 at 16:45

If truth value of $r$ is $T$, then $p\lor T\implies q\lor T\equiv (T\implies T)\equiv T$.

If truth value of $r$ is $F$, then $p\lor F\implies q\lor F\equiv (p\implies q)\equiv T$.


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