Using the word or in proofs 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 $P \wedge \sim Q \implies R$.
 A: 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.
A: 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.
