In proving that :[latex]x\leq y\wedge 0\leq z\Longrightarrow xz\leq yz [/latex] we have the following proof:

 

Let ,:[latex]x\leq y\wedge 0\leq z[/latex] and

let ,[latex]\neg(xz\leq yz) [/latex].......................................................1

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

 

Hence [latex] xz\leq yz[/latex]

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

 

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

Edited August 18, 20169 yr by Country Boy