Prove $(((p\land q)\to r)\land(p \to q))\to (p \to r)$ is satisfiable.

Question

I'm trying to learn how to apply shortcuts of a truth table, and was wondering if the following is correct:

Let $A=(p\land q)$

Let $B = (A \to r)$

Let $C=(p \to q)$

Let $D=(B\land C)$

Let $E=(p \to r)$

Let $F=(D\to E)$

Proof by contradiction, assume equation is not satisfiable, therefore $v(F)=f=v((D\to E))=F_{\to} (v(D),v(E))$, which means $v(D)=t$, and $v(E)=f$, by Definition Value of a Formula in a State $v$.

1. $v(E)=f=F_{\to} (v(p),v(r)), v(r)=f$ and $v(p)=t$, by Definition

2. $v(D)=t=v(B\land C)=F_{\land} (v(B),v(C)), v(B)=t$, and $v(C)=t$ by Definition.

3. $v(B)=t=F_{\to} (v(A),v(r)), v(A)=f$, since $v(r)=f$ in Step 1.

4. $v(C)=t=F_{\to} (v(p),v(q)),v(q)=t$, since $v(p)=t$ in Step 1.

5. $v(A)=f=F_{\land} (v(p),v(q))$ – not possible as $v(p),v(q)=t$ from (1), and (4) respectively.

Therefore The above equation is satisfiable by shortcut method. QED.

Answer 1

Actually you proved by contradiction that the negation of $F$ is not satisfiable. It follows that $F$ is a tautology, and hence satisfiable.

Answer 2

This statement is a tautology (natural deduction style like proof):

assume the left hand side

$((p\land q)\to r)\land(p \to q))$ holds and assume $p$. Then $p \to q$ tells us $q$ holds, so $p \land q$ holds.

So by $(p \land q) \to r$ we conclude $r$ and we have shown $p \to r$ and hence the total statement.

So any choice of truth values will make it true.