How to show $\vdash(\phi_1\wedge\phi_2)\rightarrow(\phi_2\wedge\phi_1)$
Am I allowed to replace $(\phi_1\wedge\phi_2)=(\phi_2\wedge\phi_1)=\phi$ then the problem is reduced to show $\phi\rightarrow\phi$ or do I have to use deduction theorem, assuming the $(\phi_1\wedge\phi_2)$ is true and using an axiom $(\phi_1\wedge\phi_2)\rightarrow\phi_2$ twice $(\phi_1\wedge\phi_1)\rightarrow \phi_2$ obtain $\phi_2\wedge\phi_1$
It is forbidden to apply completeness theorem (still cannot find this in my notes, we had only deduction theorem)