I'm trying to prove that $x = x^{+} - x^{-}$ for all $x$ in a vector lattice $E$.
The relevant definitions:
$E$ is a $\bf{\text{vector lattice}}$ if $E$ is a vector space, as well as a partially ordered set such that
(i) $x \leq y \Rightarrow x + z\leq y + z$ for all $x,y,z\in E$
(ii) $0\leq x \Rightarrow 0\leq tx$ for all $t>0$
(iii) $x,y\in E\Rightarrow x\vee y := sup(x,y)\in E$ and $x\wedge y:=inf(x,y)\in E$
For $x\in E$ we define $x^{+} = x\vee 0$ and $x^{-} = (-x)\wedge 0$.
I want to show that $x = x^{+} - x^{-}$, and I've already verified the following properties which are available:
1) $x + y = (x\vee y) + (x\wedge y)$
2) $x\vee y = -\left[(-x)\wedge (-y)\right]$
3) $(x\vee y)+z = (x+z)\vee (y+z)$
4) $(x\wedge y) + z = (x + z)\wedge(y+z)$.
This appears as Proposition 1.1.2 (ii) in these notes I am reading.
http://siba-ese.unisalento.it/index.php/quadmat/article/viewFile/8711/7967
Got it! Use the properties verified to get
$x^{+} - x^{-} = (x\vee 0) - ((-x)\vee 0) = (x \vee 0) + (x \wedge 0) = x + 0 = x$