for any pair of formulae p1 and p2, if both p1 -> p2 and p1 -> (not p2) are valid then p1 is not satisfiable. 

Prove by way of contradiction that this is true. 

My approach was assuming that p1 is satisfiable. So I proceeded to draw the truth tables for p1, p2, not p2, p1-> p2, p1-> not p2, p1->p2 and p1-> not p2. since the truth values did not come up as all true for the last row, then p1 is not valid since it is not a tautology.

Does that satisfy my prove?