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Can we say that the conclusion in this argument: (P v Q), P |- Q breaks "The Law of Excluded Middle"? And that is the reason why argument is invalid?

I recently studied "The Law of Excluded Middle":

In logic, the law of excluded middle (or the principle of excluded middle) states that for any proposition, either that proposition is true or its negation is true. Wiki

$$\begin{array}{|c|c|c|} \hline p&q&p∨ q\\ \hline T&T&T\\ T&F&T\\ F&T&T\\ F&F&F\\\hline \end{array}$$

"The following argument: (P v Q), P |- Q is invalid. Because,

Premise 1: there are three instance in truth table where (P v Q) is True (1st three in above table),

Premise 2: there are two instance in truth table where (P) is True for (P v Q) to be true at the same time (1st two in above table),

Conclusion: In this scenario Q is both True and False for (P) and (P v Q) to be true, right? and that is the reason why this argument is invalid.

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  • $\begingroup$ Your judgement (P v Q), P |- Q holds neither in classical logic nor in intuitionistic logic. $\endgroup$ – Ilya Vlasov Feb 14 '19 at 15:23
  • $\begingroup$ @IlyaVlasov well, that is what I have already stated. My question is different :) $\endgroup$ – Ubi hatt Feb 14 '19 at 15:24
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The law of the excluded middle has nothing to do with why this argument is false. The law of the excluded middle says that $P \lor \lnot P$ is always true, but even in logics that do not have that law, you cannot conclude $Q$ from $P$ and $P \lor Q$.

Using your truth table interpretation of $\vdash$, the reason $P, P \lor Q \vdash Q$ is false is that there is a line in the truth table where $P$ and $P \lor Q$ are true, but $Q$ is false, namely, the second one.

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  • $\begingroup$ Thanks! I am new to logic and trying to understand core concepts :) $\endgroup$ – Ubi hatt Feb 14 '19 at 15:25

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