TITLE: Formal logic proof of absolute value formula
QUESTION [6 upvotes]: 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?
$(x>0 \implies z = x) \wedge (x < 0 \implies z = -x) \implies z \ge 0$

REPLY [4 votes]: Hint: What happens if x = 0?  Does that say anything about z?

REPLY [3 votes]: 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:
$(z < 0) \wedge ((x>0 \implies z = x) \wedge (x < 0 \implies z = -x))$
$\implies (((x \geq 0 \implies z = x) \wedge z < 0) \wedge ((x < 0 \implies z = -x) \wedge z < 0))$
$\implies ((x \geq 0 \implies x < 0) \wedge (x < 0 \implies -x < 0))$
$\implies ((F \wedge (x < 0 \implies -x < 0)) \vee ((x \geq 0 \implies x < 0) \wedge F))$
$\implies (F \vee F)$
$\implies F$
$\therefore \neg((z < 0) \wedge ((x \geq 0 \implies z = x) \wedge (x < 0 \implies z = -x)))$
$\therefore (x \geq 0 \implies z = x) \wedge (x < 0 \implies z = -x) \implies z \ge 0$