Lambda Calculus - Fixed Point Theorem

Thu, 06 Feb 2025 00:00:00 +0000

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 $X \stackrel{\beta}{=} \lambda y . X$, because $FX \equiv (\lambda x . xy)(\lambda y . X) \stackrel{\beta}{\rightarrow} (\lambda y . X) y \stackrel{\beta}{\rightarrow} X$.
What if $F = \lambda x . xx$ then? We have $X \stackrel{\beta}{=} \lambda x . x$ or $X \stackrel{\eta}{=} I$ (the identity abstraction). Then $FX = FI = II = I$.

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Fixed-Point-Combinator on Siddharth Mishra

Lambda Calculus - Fixed Point Theorem

Fixed Point Theorem