I have to proof that Turing's combinator is a fixed point operator, but I can't get it. I tried this:

\begin{align*} Vg &= (UU)g = ((\lambda f.\lambda x.x(ffx)) (\lambda f.\lambda x.x(ffx)))g = (\lambda x.x((\lambda f.\lambda x.x(ffx))(\lambda f.\lambda x.x(ffx))x))g\\ &= g((\lambda f.\lambda x.x(ffx))(\lambda f.\lambda x.x(ffx))x)g \\&= g(UUx)g = gVxg \end{align*} The problem is that I'm getting an extra $x$ at the final. Did I make a mistake while doing beta reduction?

Any help will be appreciated.


1 Answer 1


Your second $\beta$ reduction is wrong.

After the first reduction, you (should) have $(\lambda x.x (U U x)) g$, so after the second you have $g (U U g)$ as required, but you haven't substituted the second $x$ correctly, leaving both it and the $g$ that the function is applied to, giving you $g (U U x) g$, though you have substituted the first $x$ correctly.


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