Oh wow. The only solution is $f(A)$ is the smallest positive integer that dosent appear in $A$. Figuring out this solution is in fact the main difficulty of the problem. Verifying the conditions is not trivial but its a straightforward summation. To prove that it is the only solution, note that if $A\subset B$ then $f(A)\le f(B)$. Now taking $n=1$, we must have $f(\emptyset)=1$ and $f(\{1\})=2$. We can prove by strong induction that $f([1,n])=n+1$. Base cases are done above. Now if $f([1,m])=m+1$ for all $m\le n-1$, then for any set $A$ that contains $[1,m-1]$ but not $m$, $P(A, [1,m])$ implies that $f(A)=m$. Now using the second condition for $n$, we know the value of all $f(X)$ except $[1,n]$. Hence, this uniquely determines $f([1,n])$ to be $n+1$. Thus the induction is complete. As shown above, for any set $A$ that contains $[1,m-1]$ and not $m$, $P(A, [1,m])$ implies that $f(A)=m$, as desired.