I am going through the following introduction to lambda calculus :
At page 12 , the following has been asked to prove
$ \exists $ $ G $ $ \forall X $ such that
$GX$ = $GG$
This means to find a function which takes a function as an argument and returns the same expression irrespective of the function .
The book posits the solution to be $ G $ = $Y$ $( \lambda gx. gg) $
On trying to verify this I first evaluated the L.H.S to be
$ GX $ = $ Y ( \lambda gx.gg ) $ $X$.
Now $X$ being the argument will be picked up by the first bound variable so .
$L.H.S $ = $ Y ( \lambda x . XX) $ = $( \lambda x . XX)$ $ Y ( \lambda x . XX) $ = $ Y ( \lambda x .$ $ Y ( \lambda x . XX)$ $ Y ( \lambda x . XX)$)
Here due to the presence of the form $ YF $ , I can carry on the reduction continuously by replacinf $YF$ with $FYF$
$R.H.S$ = $Y$ $( \lambda gx. gg) $ $Y$ $( \lambda gx. gg) $ = $Y$$ \lambda x.$ $( Y( \lambda gx. gg) $ $Y( \lambda gx. gg)) $
Now similarly the reduction can be carried on further by again consuming the first $g$ that comes in the expression with $Y( \lambda gx. gg)) $ and this can on perpetually and it will look like $ Y ( \lambda x .$ $ Y ( \lambda x . XX)$ $ Y ( \lambda x . XX)$)..... So I couldn't completely ascertain that they ( L.H.S and R.H.S) will turn out to be equal but I can see that L.H.S and R.H.S are going to be infinitely long and will resemble each other more and more as the number of reductions go on since the $ g$ 's in the R.H.S will continually get replaced .
Is this logic correct ? If not , where did I err ?
P.S. : On a general and on a lighter note , how are this lambda calculus equations solved ? (They look analogous to differential equations ).