I am going through the following lecture notes on lambda calculus by Barendregt and Barendsen :


Here at page 12 , after introducing fixed point theorem a small exercise which is an attempt to prove the following has been posted : $ \exists G \forall X $ $GX =SGX $.

It first proceeds to the implication

$Gx$ = $SGx$

which implies : $G$ = $ \lambda x .SGx $

which implies : $G $ = $ (\lambda gx.Sgx)G$

and then it concludes that :

$G$ = $Y$ $( \lambda g x.Sgx)$ where $Y $ is the fixed point combinator .

I am unable to figure out how was G found out to be $Y$ $( \lambda g x.Sgx)$ in the last step ? How was this conclusion reached up to ?

  • $\begingroup$ Y satisfies the equation $Y f = f (Y f)$ for all $f$, and so $g=Yf$ solves $g = fg$ for all $f$. $\endgroup$ Commented Mar 4, 2018 at 2:13
  • $\begingroup$ @YuvalFilmus : And f here would be $( \lambda gx.Sgx) $ ? $\endgroup$ Commented Mar 4, 2018 at 7:02
  • $\begingroup$ Right, pattern matching would suggest that. $\endgroup$ Commented Mar 4, 2018 at 10:37
  • 1
    $\begingroup$ The direction of implication is wrong: taking $G=YF$ implies $G=FG$, not the other way around. So, taking $G=Y\ldots$ is only one way to solve the requirement on $G$ (other solutions exist), but that's enough for the goal. All your uses of "implies" should instead be "is implied by". $\endgroup$
    – chi
    Commented Mar 7, 2018 at 13:54
  • $\begingroup$ @Chi : Yes , sorry for my mistake $\endgroup$ Commented Mar 10, 2018 at 18:40

2 Answers 2


The Y combinator satisfies the equation $Y f = f(Y f)$. Hence $g = Yf$ solves the equation $g = f g$. In your case, the function $f$ is $\lambda g x.Sgx$.


$$ \begin{align} & G=Y(\lambda gx.Sgx) & (*)\\ & G=(\lambda gx.Sgx) Y (\lambda gx.Sgx) & \text{(by FPT (ii))} \\ & G=(\lambda gx.Sgx)G & \text{(by *)} \end{align} $$ Therefore $G=Y(\lambda gx.Sgx) \implies G=(\lambda gx.Sgx)G$


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