Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

$(p \vee q) \wedge (p \rightarrow m) \wedge (q \rightarrow m) \rightarrow m$

Do I have to do a proof by case if I want to prove this directly?

If m is true, then the right hand side and the left hand side is true? That's it? How do you pick the right variable when there are like 10 variables though?

share|cite|improve this question
up vote 1 down vote accepted

You can argue like this:

$$\begin{align*} (p\lor q)\land(p\to m)\land(q\to m)&\leftrightarrow\Big((p\lor q)\land(p\to m)\Big)\land\Big((p\lor q)\land(q\to m)\Big)\\ &\leftrightarrow\left(\Big(p\land(p\to m)\Big)\lor\Big(q\land(p\to m)\Big)\right)\\ &\qquad\quad\land\left(\Big(p\land(q\to m)\Big)\lor\Big(q\land(q\to m)\Big)\right)\\ &\overset{*}\to\Big(p\land(p\to m)\Big)\land\Big(q\land(q\to m)\Big)\\ &\to m\land m\\ &\leftrightarrow m \end{align*}$$

The starred step follows from two applications of $(r\lor s)\land t\to r\land t$.

share|cite|improve this answer
is that a direct proof though? aren't we supposed to assume the left side is true? – Gladstone Asder Oct 31 '12 at 2:40
@Gladstone: It starts with the hypothesis and grinds through to the conclusion; it doesn’t use contradiction, or prove the contrapositive. In my book that makes it a direct proof. The only problem might be if you have some very specific definition of direct proof that you’re using, rather than the usual notion. – Brian M. Scott Oct 31 '12 at 2:42

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.