functions - Proof verification: system of functional equation

A problem in Putnam Competition 1992(?). The question asked:

Prove that, the only solution of the system of functional equation with respect to $f:\mathbb Z\to\mathbb Z$: $$\begin{cases} f(f(n))=n\\ f(f(n+2)+2)=n\\ f(0)=1 \end{cases}$$
is $f(n)=1-n$.

I know that the usual way is to construct another solution, and show that the difference between the two solutions is zero. But I want to use equivalence condition this time. (i.e. Prove that the system is equivalent to say that $f(n)=1-n$).

Now, we have
$$f(f(n))=f(f(n+2)+2)$$
Take $f$ on both sides.
$$f(f(f(n)))=f(f(f(n+2)+2))$$

Again by the first equation,
$$f(n)=f(n+2)+2$$
And another useful thing is that,
$$f(1)=f(f(0))=0$$

So, the original system is equivalent to say the recurrent relation
$$\begin{cases} f(n)=f(n+2)+2\\ f(0)=1\\ f(1)=0 \end{cases}$$
The sequence-like function satisfying above is unique, trivially (as for all integers, we could find an unique image of it). So our proof is actually done already as $f(n)=1-n$ is simply a solution. Q.E.D.

In my proof, I used a lot of equivalence condition. I know that my proof is valid if the conditions are actually equivalent. I think so for myself, but I want peer reviews also. Thanks in advance.