# Proofs with Hypotheses Containing "or" statement

This is a meta-question about proofwriting. Suppose we were asked to prove the following dummy statement:

"If p or q, then r"

How should I go about approaching a direct proof? My best guess is the following: p or q means that p is true, q is true, or they are both true. As such, three cases need to be considered. In the first case, I would assume p is true and show that r is true. In the second I would assume q is true and show that r is true. Finally, I would assume p and q are true and show that r is true.

Given that my statement lacks particulars, I understand that it is likely unnecessary to consider all three cases in some contexts where there is potential for WLOG. Still, would it ever be wrong to consider all three? In particular, if we show the first two cases, is it ever necessary to show the third?

• The third case always follows from the first two. You show case 1 ($p \Rightarrow r$), then case 2 ($q \Rightarrow r$), and then case three follows from the observation that if $p$ and $q$ are true, then in particular $p$ is true, so you can use case 1. Commented Jul 20 at 17:47
• @BenSteffan that wouldn't be $p\land q\to r$? Or which cases would you use to prove it? Commented Jul 20 at 17:49
• @manooooh If you know that $p \to r$ (which is case 1), then case 3 follows from $p \wedge q \to p$ and case 1. Commented Jul 20 at 17:55
• $\lor$ behaves like a coproduct: $p \lor q \to r$ is equivalent to $(p \to r) \land (q \to r)$. Dually, $\land$ behaves like a product: $r \to p \land q$ is equivalent to $(r \to p) \land (r \to q)$. Commented Jul 20 at 17:58
• @manooooh I replied in my answer below. Commented Jul 21 at 5:52

You only need to check the first two cases, i.e. "if $$p$$ is true, then so is $$r$$" and "if $$q$$ is true, then so is $$r$$." Why? Because one of these cases is already enough to show case 3 "if $$p$$ and $$q$$ are true, then so is $$r$$:" If you know that $$p$$ and $$q$$ are true, then you in particular know that $$p$$ is true, and you've already shown for case 1 that if $$p$$ is true, then so is $$r$$ (or replace $$p$$ with $$q$$ and case 1 with 2 for the same result).

In the first case, I would assume p is true and show that r is true.

In the second case, I would assume q is true and show that r is true.

In the third case, I would assume p and q are true and show that r is true.

if we show the first two cases, is it ever necessary to show the third?

Yes, we don't need to consider Case 3, since

$$(P→ R) ∧ (Q→ R)\quad\implies\quad (P ∨ Q)→ R.\tag1$$

@manooooh:

that wouldn't be (P∧Q)→R ?

Yes too.

$$(P ∨ Q)→ R \quad\implies\quad (P ∧ Q)→ R.\tag2$$

Combining $$(1)$$ and $$(2):$$ $$(P→ R) ∧ (Q→ R)\quad\implies\quad (P ∧ Q)→ R.\tag3$$

Since implication $$(1)$$ is stronger than implication $$(3),$$ it is more useful.