If $a < b$ then $((f \wedge b) -a)^-=(f-a)^-$

Question

If $a < b$ then $$((f \wedge b) -a)^-=(f-a)^-$$ where $f$ is a real valued function. I can't figure out why this holds. I know that $f^-=-(f\wedge 0)$ so I tried showing this from definitions. I would greatly appreciate any help.

Answer 1

It's easy checking cases…

For $f \ge b > a$ it holds: $$((f \wedge b) -a)^- = (b - a)^- = 0 = (f - a)^-$$

For $f < b$ it's $$((f \wedge b) -a)^- = (f - a)^-$$

Nothing wild…