0

Simplify $(\overline A+\overline{AB})\oplus B$.

Using Demorgan's Law that $A\oplus B = \overline AB+A\overline B$, I got an answer of $AB+\overline B$.

This looked as if it was in simplest terms, but it actually wasn't.

If $f(A,B)=AB+\overline B$, then $f(0,0)=1$, $f(1, 0)=1$, $f(0,1)=0$, and $f(1,1)=1$. All $4$ possible cases for $A$ and $B$ are handled.

But, the same results are acquired if $f(A,B)=A+\overline B$. And I would consider $A+\overline B$ to be more simpler than $AB+\overline B$.

How would you get from $AB+\overline B$ to $A+\overline B$?

Did
  • 279,727
  • 1
    $$A+\bar B=A(B+\bar B)+\bar B=AB+A\bar B+\bar B=AB+(A+1)\bar B=AB+\bar B$$ – Did Mar 26 '18 at 20:22
  • How did you spot that $\overline B=(A+1)\overline B$. What was the strategy there? –  Mar 26 '18 at 20:25
  • First "producing" $AB$ by force, then looking at the rest $A\bar B+\bar B$, obviously included in $\bar B$, and wondering if this was actually different from $\bar B$. – Did Mar 26 '18 at 20:35

1 Answers1

2

This is actually a very common occurrence when doing Boolean Algebra, so there is a known equivalence for that:

Reduction

$P + P'Q= P + Q$

Not all texts have reduction though (surprisingly, because again, this pattern often occurs), so you can derive it from more basic equivalences as follows:

$$P + P' Q \overset{Distribution}{=}$$

$$(P + P') (P + Q) \overset{Complement}{=}$$

$$1 (P + Q) \overset{Identity}{=}$$

$$P + Q$$

Bram28
  • 100,612
  • 6
  • 70
  • 118