Step 1- Injectivity. Step 2- Substitute $P(m,f(t)-f(m+2014))$ to get $f(f(t)-f(m+2014))=t-m$, for appropriate $t$. Step 3- Thus order the elements for integers $>2014$, using $k^{th}$ least value principle with induction for integers larger than 2014. Step 4- $P(k-f(n),n): f(k)-f(k+2014-f(n))=n$, for large enough $k$. Also notice that if $f(n) \leq 2014$, we get a direct contradiction, as by statement, $f(f(n)+m)-f(m+2014)=n$, and for large $m$, the order is ascending, thus $n<0$. Step 5- For large $k$, minimum difference between $f(k)$ and $f(k+2014-f(n))$ would be $f(n)-2014$, as our dey is ascending with integer steps. Thus $n \geq f(n)-2014=>n +2014 \geq f(n)$. But we know $f(n) \geq n+2014$ by ordering for $n>2014$, thus $f(n)=n+2014$ for $n>2014$. Step 6- $P(m,n)$, for large $m$ and $n<2014$: As for large inputs, we know the value of $f(m)$, thus $n=f(n)+m-m-2014 => f(n)=n+2014$ for all $n$.