prove that a wff is not satisfiable. 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?
 A: Assume that $p_1$ is satisfiable.
This amounts to saying that there is a truth assignment $v_1$ such that $v_1(p_1)=T$.
We have that $p_1 \rightarrow p_2$ is valid; this in turn means that for every truth assignment the formula is evaluated to $T$.
In particular, we have : $v_1(p_1 \rightarrow p_2)=T$.
Now we apply the truth table for $\rightarrow$ and conclude with $v_1(p_2)=T$
Also $p_1 \rightarrow (\lnot p_2)$ is valid, i.e. evaluated to $T$ by every truth assignments, and in particular by $v_1$.
Now, apply again the truth table for $\rightarrow$ : due to the fact that $v_1(p_1)=v_1(p_2)=T$ we have that $v_1(\lnot p_2)=F$, and thus $v_1(p_1 \rightarrow (\lnot p_2))=F$, contrary to the fact that the formula is valid.
Having reached a contradiction, we conclude that our initial assumption is not teneable, and thus that $p_1$ is not satisfiable.
A: There exists a very ancient rule of reasoning called modus ponendo ponens.  It says that "if (x->y) holds true, and x holds true, then y holds true."
Now, suppose that p1 is satisfiable.  Since (p1 -> p2) is valid, it follows that p2 is valid.  Since (p1 -> (not p2)) is valid, and p1 is valid, it follows that (not p2) is valid.  Thus, p2 and (not p2) are valid under the hypothesis of p1 as satisfiable.  This make for a contradiction, and thus p1 is unsatisfiable.
