I had a thought while making my lunch. I'm reconsidering the line in the proof above which states:
and interpreting as because we can compare it with the equation . My thought was that we cannot simply compare the equations and say the inference is correct. So if we can't do the comparison then we can't prove that . (In other words we cannot solve the equation for as the inverse of h need not exist in a monoid.)
But this raises a different problem for me. Take a look at the problem of finding the inverse of an element ab where a and b are in a group G. We can do this: consider the element . I say this is equal to the inverse of the element of ab in G. For . Since we have
then by comparison.
So is the comparison a valid method or not?