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is that proof correct

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In proving that :$x\leq y\wedge 0\leq z\Longrightarrow xz\leq yz$ we have the following proof:

Let ,:$x\leq y\wedge 0\leq z$ and

let ,$\neg(xz\leq yz)$.......................................................1

But from (1) and using the trichotomy law we have :yz<xz.And using the fact $0\leq z$ we have for 0= z, y0<x0 => 0<0 , a contradiction since ~(0<0)

Hence $xz\leq yz$

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No this "proof" is not valid! In particular, the negation of "$xz\le yz$" is NOT "$xz> yz$" for all values of x, y, z satisfying the conditions. It is that $xz> yz$ for some such values of x, y, and z.

Indeed, you cannot prove this unless you use some property of "inequality" such as "if $x\ge 0$ and $y\ge 0$ then $xy\ge 0$".

Edited by Country Boy

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