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I've been trying to prove this for a long time now, anyone willing to offer some help or get me pointed in the right direction?

$(x>0 \implies z = x) \wedge (x < 0 \implies z = -x) \implies z \ge 0$

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Re-tagged as "logic" instead of "higher-order-logic", as this is first-order logic. –  Alex Becker Mar 21 '11 at 0:13
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2 Answers

up vote 4 down vote accepted

Hint: What happens if x = 0? Does that say anything about z?

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As Robert's answer indicates, the statement is false. However, changing the first $>$ to $\geq$ gives a true statement which I would use a proof by contradiction, that is assume $z < 0$ and $(x \geq 0 \implies z = x) \wedge (x < 0 \implies z = -x)$ and derive a contradiction. This can be done as follows:

$(z < 0) \wedge ((x>0 \implies z = x) \wedge (x < 0 \implies z = -x))$ $\implies (((x \geq 0 \implies z = x) \wedge z < 0) \wedge ((x < 0 \implies z = -x) \wedge z < 0))$ $\implies ((x \geq 0 \implies x < 0) \wedge (x < 0 \implies -x < 0))$ $\implies ((F \wedge (x < 0 \implies -x < 0)) \vee ((x \geq 0 \implies x < 0) \wedge F))$ $\implies (F \vee F)$ $\implies F$ $\therefore \neg((z < 0) \wedge ((x \geq 0 \implies z = x) \wedge (x < 0 \implies z = -x)))$ $\therefore (x \geq 0 \implies z = x) \wedge (x < 0 \implies z = -x) \implies z \ge 0$

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As @Robert's answer indicates, the statement is actually false, since $x$ need not be either $<0$ or $>0$; it could be $0$ (if it's implicitly assumed to be a real number; else it could also be $2+3\mathrm{i}$ or something else entirely). It follows that this proof must be flawed. The error is in replacing $(x>0\implies x<0)$ by $F$, since this implication is actually true if $x>0$ is false. –  joriki Mar 24 '11 at 6:13
    
@Joriki: Thanks, I missed that. Fixed. –  Alex Becker Mar 24 '11 at 6:36
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