Proving a theorem of the form p implies (q or r)

Question

I wanted to know, if I encountered a theorem of the form

p implies (q or r)

Is it sufficient to show that p implies q (this is my attempt). Since we only require one of the two to be true in an or statement, for the whole implication to be true.

Answer 1

Yes: if you manage to prove $p\to q$, then you will also have a proof of $p\to(q\lor r)$ (or so close to a proof as not to matter in everyday mathematics).

However, this is somewhat unlikely. If someone took the effort to phrase the theorem as $p\to(q\lor r)$ instead of $p\to q$ (which would be a more useful theorem), then it must have been because they didn't expect that $p\to q$ can be proved. They may be wrong about that, of course -- but in many classroom settings, it would probably be more likely that you're misunderstanding something when you think you can prove $p\to q$.

Answer 2

It is true that $p\to q$ implies $p\to (q\lor r)$. A proof of this:

1. Hypothesis: $p\to q$
2. Assume $p$.
3. By 1-2, $q$ is true.
4. By 3, $q\lor r$ is true.
5. By 2-4, we have $p\to (q\lor r)$.

Therefore, as you asked, it is sufficient to prove $p\to q$, but it is not very likely that you will be able to do that in actual theorems, since usually it is not the case that $p\to q$. If $p\to q$, one would just state $p\to q$ instead $p\to (q\lor r)$, which would be redundant in that case.

Answer 3

Yes; if you’ve shown $p \implies q$ then p implies q or r is true this can easily be verified using truth tables. However, if you want to show $p \implies q$ or $r$, it suffices to show that p and not q implies r, as they are logically equivalent. Meaning that when one statement is true or false the other statement is true or false, respectively.

Answer 4

Please note that $p \to (q \lor r) \Leftrightarrow p \to (\neg r \to q) \Leftrightarrow (p \land \neg r) \to q$

So: if you can show that assuming both $p$ and $\neg r$ leads to $q$, then you are done. This is much safer then trying to shows that $p$ all by itself implies $q$.

Now, if it turns out that you never have to use #$\neg r$, ok, fine, that works. But having $\neg r$ as an extra assumption may well be necessary, and even if it isn't, it could still make it easier to get to $q$.