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I was reading about the principle of explosion and there is one thing I don't get.

Based on my understanding, the principle of explosion goes as follows: Given two statements $P$ and $Q$, and suppose $P$ and $\neg P$ are both true. Then we assert that $P\lor Q$ is true since $P$ is true. Finally, since $\neg P$ is true, it follows that $Q$ must be true for $P\lor Q$ to be true.

How does this form of deduction make sense? What I don't get is that if both $P$ and $\neg P$ are true, then it is a contradiction. At this point, we cannot conclude whether $P$ is true or false. So, it would be meaningless to determine whether $P\lor Q$ is true.

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    $\begingroup$ The point is from a contradiction you can deduce any $Q$ in your formal proof system. $\endgroup$
    – user10354138
    Oct 8 at 2:42
  • $\begingroup$ If $P$ and $\lnot P$ have both been derived at a point, then you can conclude that $P$ is true - because, you have done so. Also that it is false; simultaneously. $\endgroup$
    – Graham Kemp
    Oct 8 at 2:48

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There are many ways to see why the principle of explosion "should" be true. Here's one which I think should be particularly approachable (even if it's ultimately not "the right way" to see why explosion should be true):

Surely you agree that from $P$ and $\lnot P$ we can derive false, $\bot$. But now for any statement $Q$ at all, $\bot \to Q$ is a tautology. If you like, this is because

$$\bot \to Q \equiv \lnot \bot \lor q \equiv \top \lor q \equiv \top$$

So if we have $P$ and $\lnot P$ we can get $\bot$, and from $\bot$ we can get any $Q$ we like.

I hope this helps ^_^

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  • $\begingroup$ Thanks, what do you mean by getting $\bot$ from $P$ and $\neg P$? Contradiction implies $P$ is false? $\endgroup$
    – Jimmy Yang
    Oct 8 at 3:02
  • $\begingroup$ You can think of $\lnot P$ as an abbreviation for $P \to \bot$. So from $P$ and $P \to \bot$ we can derive $\bot$. $\endgroup$
    – HallaSurvivor
    Oct 8 at 3:47
  • $\begingroup$ I don't understand the point of deriving $\bot$... $\endgroup$
    – Jimmy Yang
    Oct 8 at 5:39
  • $\begingroup$ The point is that, for any $Q$, $\bot \to Q \equiv \top$ (by the proof I gave in the body of the answer). So as soon as we've derived $\bot$, we know that we can derive any $Q$ we want! $\endgroup$
    – HallaSurvivor
    Oct 8 at 6:17

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