# Can we say that conclusion in this argument: (P v Q), P |- Q breaks “The Law of Excluded Middle”?

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.

• Your judgement (P v Q), P |- Q holds neither in classical logic nor in intuitionistic logic. – Ilya Vlasov Feb 14 '19 at 15:23
• @IlyaVlasov well, that is what I have already stated. My question is different :) – Ubi hatt Feb 14 '19 at 15:24

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.