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When using the word "or" in proofs, what if one of the statements is true and provides the correct justification that finishes the proof, and the other thing you state is either false or true but not always true? Would this mean that the proof is incorrect? For example if you arrive at a contradiction and say this is false because otherwise $p$ or $q$ would happen leading to a contradiction, but $p$ is the statement that actually leads to the correct contradiction, whereas $q$ is not necessarily true, would this mean that the proof is incorrect?

I would think it would be incorrect because the statements $P \implies Q \vee R$ is logically equivalent to $P \wedge \sim Q \implies R$.

kevin firme
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  • In $Q\lor R$, $Q$ and $R$ play a symmetrical role. –  Feb 19 '21 at 17:16
  • If $(P\lor Q)$ lead to a contradiction you can't claim that it is $P$ and not $Q$ that led to the contradiction because $P\lor Q$ would be true when $P$ is false and $Q$ is true. $P$ can't contribute alone to the contradiction in that case because $P$ isn't actually true in that case. In that case $Q$ IS contributing to the contradiction. – fleablood Feb 19 '21 at 18:02
  • If "25 is prime or My name is moo cow" leads to a contradiction I have to consider "My name is moo cow and 25 is not prime" $\implies$ "25 is prime or my name is moo cow" leads to a contradiction. So "my name is moo cow and 25 is not prime" can not both be true. But $25$ isn't prime so that didn't lead to the contradiction. SO "my name is moo cow" had to be the thing that lead to the contradiction. In an OR statement one statement alone can't be the only thing leading to the contradiction because neither statement alone must be true. – fleablood Feb 19 '21 at 18:08

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You are asking two different questions.

In a direct proof: Prove $P$ or $Q$. Pf: Bunch of stuff...... therefore $P$ is true. And if $P$ is true then $P$ or $Q$ is true. SO $P$ or $Q$.

That's an utterly valid and correct proof.

In a proof: by contradiction: Prove "$P$ or $Q$" is not true. Pf: If we assume $P$ we get a contradiction $R$ that we know is not true. So as $P \implies (P\lor Q)$. And $P \implies R$. We have a contradiction is completely invalid. and wrong.

It's wrong because $P \implies (P\lor Q)$. But $(P\lor Q) \not \implies P$. So we haven't proven anything and we have not reached the contradiction yet. $Q\lor P$ was a contradiction but $P$ is not the only way to get $Q\lor P$.

On the other hand if $(P\lor Q)$ does lead to a contradiction we can conclude that $P$ is false, because $P$ would imply $(P\lor Q)$ which implies the contradiction. That is valid.

.....

But you seem to be arguing that if assuming $p$ or $q$ leads to a contradiction but it's because $p$ and not $q$ that lead to the contradiction that that wouldn't be valid. That's not logical because $p$ only can't lead to a contradiction because $p$ need not be true for $p$ and $q$ to be true.

If I assume "Either The moon is made of blue cheese or $7$ is prime" I won't be able to reach a contradiction based on "The moon is made of blue cheese" alone.

To have "Either the moon is made of blue cheese or $7$ is prime" to lead to a contradiction we must have all three of these statements lead to a contradiction:

The moon is made of blue cheese and $7$ is prime must lead to a contradiction (it could because the moon is not made of blue cheese)

The moon is made of blue cheese and $7$ is not prime must lead to a contradiction (it could because the moon is not made of blue cheese and $7$ is prime).

The moon is not made of blue cheese and $7$ is prime must lead to a contradiction (IMPOSSIBLE.... The moon isn't made of blue cheese and $7$ is prime so this can't ever lead to a contradiction!)

So "The moon is made of blue cheese or 7 is prime" can't lead to a contradiction because it IS true.

fleablood
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For "$p$ or $q$" to be true, it suffices that one of $p$, $q$ be true. Both can also be true.

E.g.

  • $n$ is an arbitrary integer iff $n$ is even or $n$ is odd.

  • the natural $n$ is different of $2$ iff $n$ is odd or $n$ is composite.

  • if $n$ is even then $n=2m$ or $n=2m+1$.

For a slightly less trivial example, $n$ is a perfect number implies that $n$ is even or $n$ is odd.

  • So when writing a proof, you can state something correct or something that is complete nonsense, and the proof will remain correct, as if you never used the "or" in the first place? – kevin firme Feb 19 '21 at 17:20
  • @kevinfirme What do you mean by 'you [can] state'? Is the verb 'to state' equivalent to 'assume' here, or to 'conclude' or 'infer' or something else? – CiaPan Feb 19 '21 at 17:35
  • @CiaPan I would say conclude. Suppose you arrive at a contradiction and justify the contradiction by saying " the contradiction appears because of a or b" but a is the real reason for the contradiction and b is nonsense, does the fact that b is nonsense nullify the correctness of the proof? – kevin firme Feb 19 '21 at 17:38
  • @kevinfirme: it is not a complete nonsense, it is just a truth. The same holds with the definition of "implies". False implies both false and true. In common speech, this sounds illogical. In formal logics, it is a truth like any other. –  Feb 19 '21 at 17:41
  • In that case you won't be able to get a contradiction. Take the follow argument: "Ether the moon is made of blue cheese or 7 is a prime number" now we reach a contradiction and therefore I conclude 7 is not a prime number... you are claiming that would be invalid. But it's utterly valid (but moot) because you will NEVER be able to get a contradiction. Because "Either the moon is made of blue cheese and 7 is a prime number" is a true statement and can't lead to a contradiction. – fleablood Feb 19 '21 at 17:44