Amir Hossein wrote: Find all strictly ascending functions $f$ such that for all $x\in \mathbb R$, \[f(1-x)=1-f(f(x)).\] Let $P(x)$ be the assertion $f(1-x)=1-f(f(x))$ $P(1-f(x))$ $\implies$ $f(f(x))=1-f(f(1-f(x)))$ $P(x)$ $\implies$ $f(f(x))=1-f(1-x)$ So $f(f(1-f(x)))=f(1-x)$ $\implies$ (since increasing, and so injective) $f(1-f(x))=1-x$ $P(f(x))$ $\implies$ $f(1-f(x))=1-f(f(f(x)))$ and so (using previous conclusion) $f(f(f(x)))=x$ $\forall x$ Then, if $f(a)>a$ for some $a$, we get $f(f(a))>f(a)>a$ and $f(f(f(a)))>f(f(a))>f(a)>a$, impossible Same, if $f(a)<a$ for some $a$, we get $f(f(a))<f(a)<a$ and $f(f(f(a)))<f(f(a))<f(a)<a$, impossible Hence the unique solution $\boxed{f(x)=x}$ $\forall x$ which indeed is a solution.