§ Diagonal lemma for monotone functions

§ Proving that powering is continuous

(ff)n+1=(ff)(gg)n=g((ff)gn)=gf(fgn)=f(ffn)=f(fn+1) \begin{aligned} &(\sqcup_f f)^{n+1} \\ &= (\sqcup_f f) \circ (\sqcup_g g)^n \\ &= \sqcup_g ((\sqcup_f f) \circ g^n) \\ &= \sqcup_g \sqcup_f (f \circ g^n) \\ &= \sqcup_f (f \circ f^n) \\ &= \sqcup_f (f^{n+1}) \\ \end{aligned}