# Why we have to proof both $Q$ and $R$ in $P\implies (Q\lor R)$

I'm studying proofs trying to use logic before starting with the proof.

A direct proof can be written as $$P\implies Q$$, by forcing $$P$$ to be true, we have to force $$Q$$ to be true so the statement stays true.

But in a case of the form $$P\implies (Q \lor R)$$, why do we have to prove both propositions to be true ($$Q, R$$) individually? True tables states 3 cases to be true when $$P$$ is true in this particular case

P Q R $$P\implies (Q\lor R)$$
T T T T
T T F T
T F T T

Suppose $$x\in\mathbb{R}$$. If $$xy$$ is even, then $$x$$ is even or $$y$$ in even.

In this case it also applies my reasoning through true tables. We just need to show that $$x$$ is even to proof $$xy$$ is also even. No matter which value $$y$$ have, it'll always be true $$xy$$ are even in $$x$$ is even.

What's the logic behind my wrong reasoning and why it is a must to proof both by cases, $$x$$ being even and $$y$$ being even?

• I don't understand the question. If $P\implies Q$ is true, then $P\implies (Q\lor R)$ is also true. Is that what you are asking?
– 5xum
Feb 27 at 4:32
• You just need to show that "$x$ is even" ($Q$) to prove "$x$ is even or $y$ is even" ($Q \lor R$). But given "$xy$ is even", $Q$ is not always true. Feb 27 at 4:42
• You don't have to prove both $Q\land R$. But you have to prove $Q\lor R$. Feb 27 at 4:44
• @5xum I'm asking why in direct proofs we have to proof both propositions $Q$ be true and $R$ be true? Why not just proof $Q$, which lead us to affirm $P\implies (Q\lor R)$ is true? Feb 27 at 5:24
• "We just need to show that x is even to proof xy is also even." But what if $x$ is odd? Feb 27 at 5:38

In general, proving that $$P\implies Q$$ is sufficient to prove that $$P\implies Q\lor R$$. Sufficient, but not necessary, and in fact, often times, proving $$P\implies Q$$ may be impossible!

Take, for example, the statement "If I have fewer than $$2$$ legs, then I either have one leg or I have no legs". Clearly, this is a true statement in which $$P$$ is "I have fewer than $$2$$ legs", $$Q$$ is "I have one leg" and $$R$$ is "I have no legs". So, in this case,

1. the statement $$P\implies Q\lor R$$ is true,
2. The statement $$P\implies Q$$ is false (since it equals "If I have fewer than two legs, I must have one leg", a statement contradicted by the existence of people with no legs)
3. The statement $$P\implies R$$ is false (since it equals "If I have fewer than two legs, I must have no legs", a statement contradicted by the existence of people with one leg)

In your case, if $$P$$ is "$$xy$$ is even", $$Q$$ is "$$x$$ is even" and $$R$$ is "$$y$$ is even", then proving $$P\implies Q$$ will be impossible, because the statement is false.

However, it is also not true, as you say, that you need to prove both propositions, i.e. both $$Q$$ and $$R$$. You only need to prove that $$Q\lor R$$ is true, i.e. that one of them is true. You don't have to prove that a specific single one of them is true, as that can be, as it is in your case, impossible.

What you can do, and what I advise you to do in this case, is to prove is that if $$P$$ is true, and $$Q$$ is false, then $$R$$ must be true.