Here is the corrected version of your proof.
Let A, B, and C be sets. If f(x) is a 1 - 1 and onto mapping from A --> B and g(x) is a 1-1 and onto mapping from B --> C, show
g(f(x)) is 1-1 and onto.
PF: Let x be an element of C. Then we know that since g(x) is 1 - 1 and onto then there is exactly one element, call it y, of B that
g(y) = x. Then we also know that since f(x) is 1 - 1 and onto there exists exactly one element z of A such that f(z) = y. Since we have an element in C that we can show is an element of A operated on by the two functions then g(f(x)) is 1-1 and onto.
Please tell me if I forgot to correct something. The answer to ln(2)=0 was that the series was conditionally convergent, and it's not allowed to rearrange the terms in such series.
My next "proof" is very simple (or it should be at least):