does $B'C + BC' = 1$

Question

I'm studying Boolean algebra: I know this is a straightforward answer. I'm a bit confused about if $B'C + BC'$ would be opposites in this case and would follow the rule that $A + A' = 1$ Or how can that simplify?

Thanks so much!

Answer 1

There is no way to reduce $B'C + BC'$. However, you can see that $B'C + BC'$ is not equal to $1$. Try drawing a truth table for all values of $B$ and $C$.

$$\begin{array}{|c|c|c|}\hline B & C & B'C & BC' & B'C + BC' \\\hline 0 & 0 & 0 & 0 & 0 \\\hline 0 & 1 & 1 & 0 & 1\\\hline 1 & 0 & 0 & 1 & 1 \\\hline 1 & 1 & 0 & 0 & 0\\\hline\end{array}$$ Does this remind you of a specific gate?

Answer 2

If both $B$ and $C$ are true (or false) then the equation is false.

Indeed recall DeMorgan's law \begin{eqnarray*} (B^{'} C)^{'} =B + C' \neq BC'. \end{eqnarray*}

Answer 3

$B \,\textrm{XOR}\, C = B'C + BC'$ is known as EXCLUSIVE OR. This is true (equal to $1$) when only one of $B$ or $C$ are true. If both or neither are true, than false (equal to $0$).