Simplify the boolean function below by using algebra laws.

Question

I've been stuck on this question for some time, if anyone happens to solve it please explain step by step.

$$(A +B ) \times ( A' + C ) \times ( B + C )$$

Answer 1

Since you are stuck, I will just indicate the first steps as hints

Step 1. Use distributivity to get a sum of products (instead of products of sums).

Step 2. Simplify using rules like $a + a = a$, $aa' = 0$ and $a + a' = 1$.

Answer 2

$(A +B ) \times ( A' + C ) \times ( B + C ) = [(A+B)\times A' + (A+B)\times C]\times (B+C) $

=$[A\times A'+B \times A' + A \times C + B \times C]\times (B+C) $

$A \times A' =0$

So, $B \times B \times A' +B \times A' \times C + A \times B \times C +A \times C \times C + B \times B \times C + B \times C \times C$

=$A' \times B + A' \times B \times C+ A \times B \times C+A \times C +B \times C = A' \times B + A \times C + B \times C$

write $B \times C$ as $A \times B \times C + A' \times B \times C$

=$ A' \times B + A \times C + A \times B \times C + A' \times B \times C$

=$A' \times B + A' \times B \times C+ A \times C + A \times B \times C$

=$A' \times B + A \times C$

Answer 3

Multiply it out to get a "flat" expression that's a sum of 8 terms $XYZ$, where each $X,Y,Z$ is one of $A,A',B,C$, possibly with more than one occurrence.

Then simplify, simplify, simplify. Don't forget the absorption law: $XY+Y = Y$ (and so $XYZ+XY = XY$). The whole thing reduces to something much simpler.

Answer 4

You need to dig out your textbook and do some work. There are lots of tricks, which only come from doing the work.

$$(A+B)×(\overline A+C)×(B+C)$$ $$(A+B)\ (\overline A+C)\ (B+C)$$ Less busy if you remove $×$. Think of it like addition and multiplication.

Consensus Law (13a): $(X + Y)\ (\overline X + Z)\ (Y + Z) = (X + Y)\ (X + Z)$ $$(A+B)\ (\overline A+C)$$ Distributive Law (8b): $(W + X)\ (Y + Z) = W Y + W Z + X Y + X Z$ $${A \overline A} + AC + \overline AB + CB$$ Complement Law (4a): $X • \overline X = 0$ $$AC + \overline AB + CB$$ Consensus Law (13b): $X Y + \overline X Z + Y Z = X Y + \overline X Z$ $$AC + \overline AB$$

Laws and Theorems of Boolean Algebra