Anyone familiar with Boolean Algebra?

[prentila]

Hi guys, so I have been asked to simplify this boolean algebra B + A'C + AB' and I just don't know where to start.

Hi guys, so I have been asked to simplify this boolean algebra B + A'C + AB' and I just don't know where to start.

*boolean expression

 

[Dr.Peterson]

Hi guys, so I have been asked to simplify this boolean expression B + A'C + AB' and I just don't know where to start.

Have you been taught a particular procedure to follow? Then do whatever the first step is, and show us what you did. (Also show us a summary of your procedure, or an example you were given.)

On the other hand, if you were just given a list of properties (such as distribution and absorption) and set free to discover what you can do with them, then do so! This is the sort of thing that you only learn by trying, just as children learn to do things by playing with them. You must be willing to do something without being sure it is the "right" thing, and just see whether it accomplishes anything. If it does, do more; if it doesn't, back up and try something else.

In other words, the place to start is with anything you think of. Just try it.

If you were given examples, show those to us, and think about whether any of the things they do there will work for yours.

I tried looking for a site that teaches a procedure, and didn't find any to share with you. But I did find some useful examples. For example, try this or this.

The former says this: "When using the laws to simplify Boolean expressions, you are doing a pattern matching activity to look for parts of the expression that you can replace with a simpler version. To do this easily, you need to know all the Boolean laws; if it's possible to apply several laws, you need to consider which one will help most for the next step of the simplification."

 

[Dr.Peterson]

I tried looking for a site that teaches a procedure, and didn't find any to share with you. But I did find some useful examples. For example, try this or this.

In case you didn't think I gave you enough help, here is a specific example from the second link that is useful for your problem:



Do you see the similarity to yours?

The first link also has a similar example, which they simplified by a different technique that seems harder to me. But there are often many routes to the same result.

 

[nasi112]

You have three circles. Collect them. If a letter overlap with its complement, it cancels out.

 

[lex]

Or just use the associative law and commutative law to rearrange, and then use the distributive law.

[MATH]\qquad B+A^\prime C+AB^\prime[/MATH][MATH]\equiv\ A^\prime C+B+AB^\prime[/MATH][MATH]\equiv\ A^\prime C+(B+A)({B+B}^\prime)[/MATH] Distributive Law
[MATH]\equiv\ A^\prime C+B+A[/MATH]Now do a similar rearrangement and apply the distributive law, to get the final answer.

 

[prentila]

B

Or just use the associative law and commutative law to rearrange, and then use the distributive law.

[MATH]\qquad B+A^\prime C+AB^\prime[/MATH][MATH]\equiv\ A^\prime C+B+AB^\prime[/MATH][MATH]\equiv\ A^\prime C+(B+A)({B+B}^\prime)[/MATH] Distributive Law
[MATH]\equiv\ A^\prime C+B+A[/MATH]Now do a similar rearrangement and apply the distributive law, to get the final answer.

A'C + B + A
B + A + A'C
B + A + AC + A'C
B + A + C (A + A')
B + A + C(1)
B + A + C

Here's what I got, mine's a bit longer though. Thanks, everyone!

 

[lex]

I think your first line needs justification, but your final answer is grand. Job done.

My full solution was:

[MATH]\qquad B+A^\prime C+AB^\prime[/MATH][MATH]\equiv\ A^\prime C+B+AB^\prime[/MATH][MATH]\equiv\ A^\prime C+(B+A)({B+B}^\prime)[/MATH] Distributive Law
[MATH]\equiv\ A^\prime C+B+A[/MATH][MATH]\equiv\ B+A+A^\prime C[/MATH][MATH]\equiv\ B+\left(A+A^\prime\right)\left(A+C\right)[/MATH] Distributive Law
[MATH]\equiv\ A+B+C[/MATH]

 

[Dr.Peterson]

A'C + B + A
B + A + A'C
B + A + AC + A'C
B + A + C (A + A')
B + A + C(1)
B + A + C

Here's what I got, mine's a bit longer though. Thanks, everyone!

Here is mine, using the technique of the example I gave and showing most of the details, as you didn't:

B + A'C + AB'
(B + AB) + A'C + AB'
B + (AB + AB') + A'C
B + A(B + B') + A'C
B + A + A'C
B + (A + AC) + A'C
B + A + (AC + A'C)
B + A + (A + A')C
B + A + C

 

[lex]

@Dr.Peterson
Yes, a nice solution.

 

[prentila]

Here is mine, using the technique of the example I gave and showing most of the details, as you didn't:

B + A'C + AB'
(B + AB) + A'C + AB'
B + (AB + AB') + A'C
B + A(B + B') + A'C
B + A + A'C
B + (A + AC) + A'C
B + A + (AC + A'C)
B + A + (A + A')C
B + A + C

Here's another one, how about for this case? My initial answer is;

(X'+Y') (X'+Y) (X' + X)
(X'+Y') (X' + Y) (1)
(X' + Y') (X' + Y)
(XX)' + X'Y + (YX)' + Y'Y
X'(1 + Y) +(Y' + X') + 0
X' (1) + Y' + X'
X' + X' + Y'
X' + Y' or (XY)'

 

[lex]

@prentila
Note line 4: Y'X'[MATH]\not\equiv[/MATH](YX)'
Just keep it as Y'X' and your solution should then work.