Lattice disjoint

For any x in X, let ${\displaystyle x^{+}:=\sup \left\{x,0\right\}}$ and ${\displaystyle x^{-}:=\sup \left\{-x,0\right\}}$, where note that both of these elements are ${\displaystyle \geq 0}$ and ${\displaystyle x=x^{+}-x^{-}}$ with ${\displaystyle |x|=x^{+}+x^{-}}$. Then ${\displaystyle x^{+}}$ and ${\displaystyle x^{-}}$ are disjoint, and ${\displaystyle x=x^{+}-x^{-}}$ is the unique representation of x as the difference of disjoint elements that are ${\displaystyle \geq 0}$.[2] For all x and y in X, ${\displaystyle \left|x^{+}-y^{+}\right|\leq |x-y|}$ and ${\displaystyle x+y=\sup\{x,y\}+\inf\{x,y\}}$.[3] If y ≥ 0 and x ≤ y then x⁺ ≤ y. Moreover, ${\displaystyle x\leq y}$ if and only if ${\displaystyle x^{+}\leq y^{+}}$ and ${\displaystyle x^{-}\leq x^{-1}}$.[2]