TITLE: The logical structure of a certain proof.
QUESTION [1 upvotes]: Let $x,y,z$ be real numbers with $0<x<y<z<1$. Prove that at least two of the numbers $x,y,z$ are within $\frac{1}{2}$ unit from one another.

The proof I'm referring to is this

$0<x<y\wedge y-x>\frac12\implies y>\frac12$
$y<z<1\wedge z-y>\frac12\implies y<\frac12$
Contradiction
Therefore ....... 

However the statement we ought to prove has the following logical form
$$(\forall x)(\forall y)(\forall z)(0<x<y<z<1\implies([|x-y|\le\frac12\wedge|z-y|\le\frac12]\vee[|z-y|\le\frac12\wedge|x-z|\le\frac12]\vee[|x-z|\le\frac12\wedge|x-y|\le\frac12]))$$
So to prove it by contradiction one would assume $0<x<y<z<1$ and then assume the negation of the consequent which is $$\neg([|x-y|\le\frac12\wedge|z-y|\le\frac12]\vee[|z-y|\le\frac12\wedge|x-z|\le\frac12]\vee[|x-z|\le\frac12\wedge|x-y|\le\frac12]))\iff([|x-y|>\frac12\vee|z-y|>\frac12]\wedge[|z-y|>\frac12\vee|x-z|>\frac12]\wedge[|x-z|>\frac12\vee|x-y|>\frac12]))$$
So to prove it wrong we all need to take just one piece (because of the $\wedge$) for example $[|x-y|>\frac12\vee|z-y|>\frac12]$. 
So we assume that $[|x-y|>\frac12\vee|z-y|>\frac12]$, but in the proof I was shown they assumed that $[|x-y|>\frac12\wedge|z-y|>\frac12]$
Does that mean that the proof is wrong?

REPLY [2 votes]: You are making some confusion. What you have to prove is that
$$\forall x,y,z ((0<x<y<z<1) \Rightarrow((|x-y| < 1/2) \vee (|x-z| < 1/2) \vee (|y-z| < 1/2))$$
So arguing by contradiction you have
$$\exists x,y,z  ((0<x<y<z<1) \wedge (|x-y| \ge 1/2) \wedge (|x-z| \ge 1/2) \wedge (|y-z| \ge 1/2) )$$
widh leads to the contradiction in the proof.
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