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

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?

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1 Answer 1

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 $(r\lor s)\land t\to r\land t$.

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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

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