1. ## tautology or not

In the following formula :

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

I substituted :

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

2) $\displaystyle x+y\geq 0$ 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 :

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

expecting to get a tautology.

But unfortunately i did not get a tautology. Why?

Does that mean that the above formula in not provable??

2. ## Re: tautology or not

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

I suggest you try to prove the above.

3. ## Re: tautology or not

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

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