Here is the logic equation and I am trying to prove the relation ($'$ stands for complement): $$_1_3' + _2'_3' +_1_3 +_2'_3 = _1'_2' + _1_2 + _1_2'$$
What I am doing: $$x_2' (x_1x_3' + x_1x_3 + x_3' + x_3)$$ $$x_2'(x_1(x_3' + x_3) + x_3' + x_3)$$ $$x_2'(x_1 + x_3' + x_3)$$
.. and I do not where to go further, $x_3' + x_3$ is a tautology but I think I cannot eliminate it when it is expression of ORs. It does not seem I am getting the expected result. Anyone can help? Thanks
After drawing some Karnaugh maps, we have: \begin{align*} x_1x_3' + x_2'x_3' + x_1x_3 + x_2'x_3 &= (x_1 + x_2')x_3' + (x_1 + x_2')x_3 \\ &= (x_1 + x_2')(x_3' + x_3) \\ &= (x_1 + x_2')(1) \\ &= x_1 + x_2' \end{align*} Likewise, we have: \begin{align*} x_1'x_2' + x_1x_2 + x_1x_2' &= x_1'x_2' + x_1x_2 + (x_1x_2' + x_1x_2') \\ &= (x_1x_2 + x_1x_2') + (x_1'x_2' + x_1x_2') \\ &= x_1(x_2 + x_2') + x_2'(x_1' + x_1) \\ &= x_1(1) + x_2'(1) \\ &= x_1 + x_2' \end{align*}
The first thing to do is to observe that both of these equations can be significantly simplified. $x_1x_2+x_1x_2'=x_1(x_2+x_2')=x_1$, so the right-hand side is equal to $x_1'x_2'+x_1=x_1'x_2'+x_1+x_1x_2'=x_1+x_2'$. Here I could add $x_1x_2'$ because $x_1x_2'\leq x_1$ so $x_1+x_1x_2'=x_1$. To reduce the left-hand side to this form is simple: just factor the $x_1$ and the $x_2'$ out of their respective terms.
