# Fixed Point Combinator Turing proof

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.

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.