I would start here: Link
Q1) Let m and n be a positive integers. For 1<x<n and 1<y<m the cell (x,y) is said to be a winning first move in the mxn chomp for red if there exists a winning strategy for in which (x,y) is the first move
a) Assume that (x,y) is a winning first move of red in mxn chomp for some positive integers m and n. PROVE THAT (x+2,y+5) is NOT A WINNING first move STRATEGY FOR RED in mxn chomp. (you can assume that x+2<n and y_5<m)
Does anyone know how to tackle this problem?
Then, I would suggest considering what would happen if player A starts in position (x+2,y+5), then player B takes position (x,y). The configuration for A's second turn is exactly the configuration of B's first turn if A starts in position (x,y). Hence, B can now play the winning strategy, and A will lose.