It should be ||🍎(x) -🍌(x)| +|🍌(x) - 🍇(x)|| >=|🍎(x) -🍌(x) +🍌(x) - 🍇(x)| = |🍎(x) - 🍇(x)| I missed the abs that I added in the previous step.
let me make the variables less annoying:
||x-y|+|y-z|| >= |x-y+y-z| = |x-z|
we are getting rid of the abs around |x-y| and |y-z| so the 2 y’s can cancel out. We can do this because |x-y| >= x-y because |q| >= q
I think this could use a bit more elaboration, since if x-y+y-z < -(|x-y|+|y-z|), then ||x-y|+|y-z|| >= |x-y+y-z| wouldnt be true. This is impossible though since q >= -|q|
I’m confused about this step in the final condition’s proof:
|🍎(x) -🍌(x)| +|🍌(x) - 🍇(x)| >=|🍎(x) -🍌(x) +🍌(x) - 🍇(x)| = |🍎(x) - 🍇(x)| since |q| >= q forall q
I can see how it’s true by proving that |p| + |q| >= |p + q|, but that’s not stated anywhere and I can’t figure out how |q| >= q forall q is relevant.
Also, thanks a lot for making/showing a proof :D
It should be ||🍎(x) -🍌(x)| +|🍌(x) - 🍇(x)|| >=|🍎(x) -🍌(x) +🍌(x) - 🍇(x)| = |🍎(x) - 🍇(x)| I missed the abs that I added in the previous step.
let me make the variables less annoying:
||x-y|+|y-z|| >= |x-y+y-z| = |x-z| we are getting rid of the abs around |x-y| and |y-z| so the 2 y’s can cancel out. We can do this because |x-y| >= x-y because |q| >= q
I think this could use a bit more elaboration, since if x-y+y-z < -(|x-y|+|y-z|), then ||x-y|+|y-z|| >= |x-y+y-z| wouldnt be true. This is impossible though since q >= -|q|
I see, thanks! :3