Fixed Point Theorem $ \forall F \quad \exists X \mid FX = X $ For all lambda expression $F$, there exists another lambda expression $X$, such that $FX = X$, meaning when $F$ is applied over $X$ we get $X$ (itself). Let’s take $F = \lambda x . x$ (the identity abstraction), Then any lambda expression can take place of $X$. Now consider $F = \lambda x . y$, then we have $X \equiv y$, because $(\lambda x . y)y = y$. If $F = \lambda x . xy$, then? Then can use $X \stackrel{\beta}{=} \lambda y ....
