# New easy theorem

I have added a new easy (but unnoticed before) theorem to my book:

Proposition $(\mathsf{RLD})_{\mathrm{out}} f\sqcup (\mathsf{RLD})_{\mathrm{out}} g = (\mathsf{RLD})_{\mathrm{out}}(f\sqcup g)$ for funcoids $f$, $g$.