Simplifying boolean expressions

[magnifik]

i have two boolean expressions that I'm simply stuck on. (any variable with the ' mark means NOT). i used http://hopper.unco.edu/KARNAUGH1.1/Function.html" to check my answers for both..apparently they are both supposed to equal 1. i have tried solving the expressions multiple times, and each time i try i get stuck

(A + B)(A' + B + C)(A' + B + C')
= AA' + AB + AC + A'B + BB + BC(A' + B + C')
= AB + AC + A'B + B + BC(A' + B + C')
= B(A + A') + AC + B + BC(A' + B + C')
= B + AC + B + BC(A' + B + C')
= (B + AC + BC)(A' + B + C')
= A'B + BB + BC' + A'AC + BAC + ACC' + A'BC + BBC + BCC'
= A'B + B + BC' + ABC + A'BC + BBC
= A'B + B + BC' + ABC + A'BC + BC
= A'B + B + BC(A + A') + B(C + C')
= A'B + B + BC + B
= B(A' + 1 + C + 1)
= B(A' + C + 1) = A'B + BC + B
need help after this

A + ABC + A'BC + A'B + DA + DA'
= A + BC(A + A') + A'B + D(A + A') by distributive property
= A + BC + A'B + D by complement (A + A' = 1)
i don't know what to do after this...

 

[Zryn]

For the first one (assuming all your math is correct), keep in mind that from the truth table for an OR, 'X OR 1 = 1' (where X = anything), hence if you had BC + B = B(C+1), C OR 1 = 1, hence BC + B = 1B = B (since 1 AND X = X). You could extend this to say that B(A'+C+1) is comprised of A'+C being the aforementioned 'anything' and thus (A'+C) + 1 = 1, so you are left with B ... which is not 1.

*For the second one, what do you get if you simplify (A+ABC) and (A'BC + A'B) and (DA+DA') instead of the choices you made?

 

[Zryn]

If you use that website, be a little careful.

I put (A + B)(A' + B + C)(A' + B + C') (original equation) in and it said the answer was 1.
I put (A + B)(A' + B + C) in and it said the answer was 1.
I expanded this out to AA'+AB+AC+A'B+BB+BC and it said the answer was A'+B+C.
I simplified this to AB+AC+A'B+B+BC and it said the answer was A'+B+C.
I simplified this to AC + B(A+A'+1+C) and it said the answer was 1.
I simplified this to AC + B and it said the answer was A'+B+C.

I put in A+B and it said the answer was 1.
I put in A+B+C and it said the answer was 1.

Hopefully by now you can see the pattern ... its gone bonkers and shouldn't be trusted.

I went over your first equation and did my own version and I agree with your result.

 

[Quinzio]

Is this what you have to simplify ?
(A + B)(A' + B + C)(A' + B + C')

If so, you can solve it mentally.
Try to spot in the formula some regularity, some "weakness".

I will go on with the solution in the spoiler, if you want to work it yourself, don't open it.

Spoiler

(A + B)(A' + B + C)(A' + B + C')

Notice (A' + B + C)(A' + B + C')
let's compute it for all values of C
C = 0
(A' + B + 0)(A' + B + 1) =
(A' + B ) (1) =
(A' + B )C = 1
the same... the expression is simmetrical toward C
let's do it...
(A' + B + 1)(A' + B + 0) =
(1 ) (A' + B) =
(A' + B )So...
(A' + B + 0)(A' + B + 1) = (A' + B )

Let's put in the rest
(A + B)(A' + B )

same thing as before
A = 0
(0 + B)(1 + B ) = B(1) = B

A = 1 the same as before...Solution is
(A + B)(A' + B + C)(A' + B + C') = B

Make a truth table to convince yourself the the expression depends only on B and it's indifferent respect A and C

 

[magnifik]

For the first one (assuming all your math is correct), keep in mind that from the truth table for an OR, 'X OR 1 = 1' (where X = anything), hence if you had BC + B = B(C+1), C OR 1 = 1, hence BC + B = 1B = B (since 1 AND X = X). You could extend this to say that B(A'+C+1) is comprised of A'+C being the aforementioned 'anything' and thus (A'+C) + 1 = 1, so you are left with B ... which is not 1.

*For the second one, what do you get if you simplify (A+ABC) and (A'BC + A'B) and (DA+DA') instead of the choices you made?

for the second one, if i simplify it by that, i get
(A + ABC) + (A'BC + A'B) + (DA + DA')
= A(1+BC) + A'(BC + B) + D(A+A')
= A + A'(B(C+1)) + D
= A + A'B + D
= (A + A'B) + D
= A + B + D
is that really it's simplest form?