Boolean Simplification - Q=A.B.(~B+C)+B.C+B

Question

I've been struggling with boolean simplification in class, and took it to practice some more at home. I found a list of questions, but they don't have any answers or workings. This one I'm stuck on, if you could answer clearly showing each step I'd much appreciate:

Q=A.B.(~B+C)+B.C+B

I tried looking for a calculator to give me the answer and then to work out how to get to that, but I'm lost

(I'm new to this)

Edit: ~B = NOT B

@AdiC I meant instead of B' - but what I actually wanted to say is, that the whole equation is somehow ambiguous. sorry, joking attempt failed — Lovis, Mar 01 '18 at 11:32
@Lovis Oh :) I've never done this so I wouldn't have understood :) — Adi219, Mar 01 '18 at 11:33
@J.McGill It's fine :) It's just that all the online sites I've found use B' so I assumed that's what you meant. — Adi219, Mar 01 '18 at 11:35
@AdiC I've seen that too, just the most recent one I saw was that format so I put that — J. McGill, Mar 01 '18 at 11:38
Since I've never done this, I have a question. Is there any boolean algebraic equation linking `A`, `B`, and `C`? Maybe something like `A = B + C`? Or are they completely independent of each other? — Adi219, Mar 01 '18 at 11:43
I'm voting to close this question as off-topic because it is about boolean algebra and [math.se] instead of programming or software development. — Pang, Mar 02 '18 at 01:27

miku · Answer 1 · 2018-03-01T12:00:50.733

Let's be lazy and use sympy, a Python library for symbolic computation.

>>> from sympy import *
>>> from sympy.logic import simplify_logic
>>> a, b, c = symbols('a, b, c')
>>> expr = a & b & (~b | c) | b & c | b # A.B.(~B+C)+B.C+B
>>> simplify_logic(expr)
b

There are two ways to go about such a formula:

Applying simplifications,
Brute force

Let's look at brute force first. The following is a dense truth table (for a better looking table, look at Wα), enumerating all possible value for a, b and c, alongside the values of the expression.

a b c -- a & b & (~b | c) | b & c | b = Q
0 0 0    0   0    10 1 0    0 0 0   0 = 0
0 0 1    0   0    10 1 1    0 0 1   0 = 0
0 1 0    0   1    01 0 0    1 0 0   1 = 1
0 1 1    0   1    01 1 1    1 1 1   1 = 1
1 0 0    1   0    10 1 0    0 0 0   0 = 0
1 0 1    1   0    10 1 1    0 0 1   0 = 0
1 1 0    1   1    01 1 0    1 0 0   1 = 1
1 1 1    1   1    01 1 1    1 1 1   1 = 1

You can also think of the expression as a tree, which will depend on the precedence rules (e.g. usually AND binds stronger than OR, see also this question on math.se).

So the expression:

a & b & (~b | c) | b & c | b

is a disjunction of three terms:

a & b & (~b | c)
b & c
b

You can try to reason about the individual terms, knowing that only one has to be true (as this is a disjunction).

The last two will be true, if and only if b is true. For the first, this a bit harder to see, but if you look closely: you have now a conjunction (terms concatenated by AND): All of them must be true, for the whole expression to be true, so a and b must be true. Especially b must be true.

In summary: For the whole expression to be true, in all three top-level cases, b must be true (and it will be false, if b is false). So it simplifies to just b.

Explore more on Wolfram Alpha:

https://www.wolframalpha.com/input/?i=a+%26+b+%26+(~b+%7C+c)+%7C+b+%26+c+%7C+b

It's good and all to have the answer, but I won't exactly be having a python program in my exam to use. Really I need to understand how to get the answer as well — J. McGill, Mar 01 '18 at 11:35
@J.McGill, *if in doubt, use brute force*. I updated my answer. — miku, Mar 01 '18 at 11:39

score 0 · Answer 2 · answered Mar 01 '18 at 11:32

I've never done this, so I'm using this site to help me.

A.B.(B' + C) = A.(B.B' + B.C) = A.(0 + B.C) = A.(B.C)

So the expression is now A.(B.C) + B.C + B.

Not sure about this, but I'm guessing A.(B.C) + (B.C) = (A + 1).(B.C). This equals A.(B.C).

So the expression is now A.(B.C) + B.

As A.(B + C) = B.(A.C), the expression is now B.(A.C) + B, which equals (B + 1).(A.C) = B.(A.C).

NOTE: This isn't complete yet, so please avoid downvoting as I'm not finished yet (posted this to help the OP understand the first part).

score 0 · Answer 3 · answered Mar 04 '18 at 19:09

A.B.(~B+C) + B.C + B = A.B.~B + A.B.C + B.C + B    ; Distribution
                     = A.B.C + B.C + B             ; Because B.~B = 0
                     = B.C + B                     ; Because A.B.C <= B.C
                     = B                           ; Because B.C <= B

Boolean Simplification - Q=A.B.(~B+C)+B.C+B

3 Answers3