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 $P\implies (Q\lor R)$

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?

  • 1
    $\begingroup$ 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? $\endgroup$
    – 5xum
    Feb 27 at 4:32
  • $\begingroup$ 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. $\endgroup$
    – peterwhy
    Feb 27 at 4:42
  • $\begingroup$ You don't have to prove both $Q\land R$. But you have to prove $Q\lor R$. $\endgroup$
    – peterwhy
    Feb 27 at 4:44
  • $\begingroup$ @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? $\endgroup$
    – Alexis SM
    Feb 27 at 5:24
  • 1
    $\begingroup$ "We just need to show that x is even to proof xy is also even." But what if $x$ is odd? $\endgroup$
    – fleablood
    Feb 27 at 5:38

1 Answer 1


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.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .