I've obtained a formula through cannonical representation, which is:
$$A\cdot \overline{B\cdot S}+A\cdot B\cdot \overline{S}+\overline{A}\cdot B\cdot S+A\cdot B \cdot S$$
And I'm trying to simplify it to:
$$( A \cdot \overline{S}) + (B \cdot S)\tag{1}$$
Following the laws here, I've done this:
$$A\cdot B\cdot \overline{S}+\overline{A}\cdot B\cdot S+(A\cdot B \cdot S+A\cdot \overline{B\cdot S})$$
$$A\cdot B\cdot \overline{S}+\overline{A}\cdot B\cdot S+A\cdot (BS+\overline{BS})\tag{Complementation 2}$$
$$A\cdot B\cdot \overline{S}+\overline{A}\cdot B\cdot S+A\cdot (1)\tag{Identity for ∧}$$
$$A\cdot B\cdot \overline{S}+\overline{A}\cdot B\cdot S+A$$
I'm having problem at this step, I'm not sure how to proceed, I did the following:
$$B\cdot (A\cdot \overline{S}+\overline{A}\cdot S)+A$$
Trying to obtain some insight, but from here, I have no idea on how to proceed to the form $(1)$.
Here's a simple solution, assuming I've correctly interpreted the question: $$ AB'S' + ABS' + A'BS + ABS = \\ A(B' + B)S' + (A' + A)BS = \\ AS' + BS $$ as desired. Here, $A'$ means "not $A$". Note that the notation $\overline{B\cdot S}$ is unclear since $(BS)' \neq B'S'$. For a good way of making such simplifications in general, look into Karnaugh maps.