Does the following given conjunction $$ x\lt 0\le y \wedge |x|\gt |y|$$ implies that either
- $|x+y|= x-y$
- $|x+y| = -x+y$
- $|x+y|=|x|-|y|$
- $|x+y|=-(x+y)$
or a combination of those proposition, is true?
I look at absolute values as being the magnitudes of vectors so geometrically the result that makes sense to me would be to say that $1$, $2$ are false and $3$, $4$ are true such that $$ x\lt 0\le y \wedge |x|\gt |y| \implies |x+y|=|x|-|y|\wedge \;|x+y|=-(x+y)\ $$ would be the correct implication. Is this the only true implication?
Your geometric intuition is good. Here is an arithmetic proof.
For convenience (to avoid having to write the same thing many times), let $P$ be the conjunction $x < 0 \leq y \land \lvert x \rvert > \lvert y \rvert.$
Suppose $P$ is true. Then:
Since $x < 0 \leq y,$ we know that $\lvert x \rvert = -x$ and $\lvert y \rvert = y.$
Since $\lvert x \rvert > \lvert y \rvert,$ we have $y < -x,$ so $x + y < 0$ and $\lvert x + y \rvert = -(x+y).$ This proves part 4.
Furthermore, $-(x+y) = -x - y = \lvert x \rvert - \lvert y \rvert,$ so $\lvert x + y \rvert = \lvert x \rvert - \lvert y \rvert.$ This proves part 3.
On the other hand, $\lvert x + y \rvert - (x - y) = -(x + y) - (x - y) = -2x > 0,$ so $\lvert x + y \rvert \neq (x - y),$ disproving part 1. (That is, $P$ implies statement 1 is false.)
Also, $\lvert x + y \rvert - (-x + y) = -(x + y) - (-x + y) = -2y.$ That is, $\lvert x + y \rvert = (-x + y) - 2y.$ Therefore $P$ implies that statement 2 is true if and only if $y = 0$; $P$ alone (without the additional condition $y=0$) do not imply statement 2.
So the conjunction $x < 0 \leq y \land \lvert x \rvert > \lvert y \rvert$ implies statements 3 and 4 but not statements 1 and 2.