# How to solve system of equations (boolean algebra)?

I have the following boolean algebra equations. They are supposed to be equivalent for $$ab=FF, FT, TF$$. I can show it with truth tables, but how can I prove it algebraically? $$\begin{cases} x_1 = \neg (b \lor y_1) \\ \tag 1 y_1 = \neg (a \lor x_1) \\ \end{cases}$$ $$\begin{cases} x_2 = \neg (\neg a \land y_2) \\ \tag 2 y_2 =\neg (\neg b \land x_2) \end{cases}$$ Starting with $$x_1=x_2$$: \begin{align} \neg (b \lor y_1)&= \neg (\neg a \land y_2) \\ \neg b \land \neg y_1&= a\lor\neg(y_2)\\ \end{align} I'm stuck here.

For $$y_1=y_2$$ I have: \begin{align} \neg (a \lor x_1) &= \neg (\neg b \land x_2) \\ \neg a \land \neg x_1 &= b \lor \neg x_2 \\ \end{align} I'm also stuck here, how to proceed?

• What do you mean by them being "equivalent for ab = ....". What is equivalent to what and in what sense? Apr 20 '20 at 16:32

I would just plug in those different truth-value combination for $$a$$ and $$b$$.
So, for example, when $$ab=FF$$, you get:
$$\begin{cases} x_1 = \neg (F \lor y_1) = \neg y_1\\ \tag 1 y_1 = \neg (F \lor x_1) = \neg x_1 \\ \end{cases}$$ $$\begin{cases} x_2 = \neg (\neg F \land y_2) = \neg (T \land y_2) = \neg y_2\\ \tag 2 y_2 =\neg (\neg F \land x_2) = \neg (T \land x_2) = \neg x_2\\ \end{cases}$$
You can see these two sets of equations are equivalent, in that you can set $$x_1=x_2$$ and $$y_1 = y_2$$ to get the same equations.
Do the same for the other truth-value pairs ... and finally show that you do not get an equivalence when $$ab=TT$$.
I am not sure how to just take the equations and solve for $$a$$ and $$b$$ ... also because it is not clear that you always need to have that $$x_1=x_2$$ and $$y_1=y_2$$ in order to get equivalence: in the above example you can also set $$x_1=y_2$$ and $$y_1=x_2$$ to demonstrate equivalence.