# tautology or not

• Dec 7th 2012, 04:21 PM
psolaki
tautology or not
In the following formula :

$|x+y| = |x|+|y|\Longleftrightarrow (x+y\geq 0 \Longrightarrow x+y=|x|+|y|)\wedge(x+y<0\Longrightarrow x+y=-|x|-|y|)$

I substituted :

1)|x+y|=|x|+|y| with a

2) $x+y\geq 0$ with b

3)x+y=|x|+|y| with c

4) x+y<0 with d

5) x+y = -|x|-|y| with e

Then

I tried to write the truth table of :

$a\Longleftrightarrow (b\Longrightarrow c)\wedge (d\Longrightarrow e)$

expecting to get a tautology.

But unfortunately i did not get a tautology. Why?

Does that mean that the above formula in not provable??
• Dec 7th 2012, 05:15 PM
Plato
Re: tautology or not
Quote:

Originally Posted by psolaki
In the following formula :
$|x+y| = |x|+|y| \longleftrightarrow (xy)\ge 0$

I suggest you try to prove the above.
• Dec 8th 2012, 02:43 AM
psolaki
Re: tautology or not
Quote:

Originally Posted by Plato
I suggest you try to prove the above.

To prove :

$|x+y|=|x|+|y|\Longleftrightarrow xy\geq 0$ is easy, but i do not see what that has to do with my problem