prove that $\Sigma \vdash \phi_1$ and $\Sigma \vdash \phi_2$ leads to $\Sigma \vdash \phi_1 \wedge \phi_2$.

Question

I try to prove that if $\Sigma \vdash \phi_1$ and $\Sigma \vdash \phi_2$ then $\Sigma \vdash \phi_1 \wedge \phi_2$.

Notice that, the ONLY rule of inference of the system is modes ponens and the set of logical axioms are all the tautologies.

I know that I can use completness theorem of propositional calculus to prove that easily but the point is , I was trying to prove completness theorem of propositional logic and to finish the proof I need to prove this as a claim which is I'm stuck with.

I know that I need to prove that $\Sigma \vdash \neg(\phi_1\rightarrow \neg\phi_2)$.

Any ideas or hints?

Answer 1

$\phi_1 \rightarrow (\phi_2 \rightarrow (\phi_1 \land \phi_2))$ is a tautology. Thus, apply modus ponens twice ...