# Help with proving boolean algebra identity

#### pghaffari

Joined Oct 1, 2011
13
Prove: a' + a(a'b + b'c)' = a' + b + c'

This is what I get so far for the LHS

a' + a(ab'bc') --> a' + ab'bc' .. So we know b'b is always 0 so does that mean we can just ignore it? ---> a' + ac' --> a' + c' .. but i am missing a b somewhere..

also problem 2:

(a'b' + c)(b+a)(b' + ac) = a'bc

my question is.. can you foil boolean algebra out like in regular algebra? like for this example can i do:

a'b'b + a'ab' + cb + ac ... etc? or no??

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#### djsfantasi

Joined Apr 11, 2010
5,687
bb' is always 0 so you can ignore it in an OR function
I would perform a different set of operations on your left-hand side first, then apply DeMorgans from there. Note that c = c'' by the way!

#### pghaffari

Joined Oct 1, 2011
13
What else can i do for problem 1?

a' + a(a'b + b'c)' --> can I use this law: (a + a'b ) = a+b , so that it becomes

a' + (a'b + b'c)' ? even if i did it this way, I get same thing:

a' + (ab'bc') --> which makes the inside always 0 and since its an AND function the whole thing is 0? so its just a' .. what am i doing wrong?

#### pghaffari

Joined Oct 1, 2011
13
Ok I got the problems up. Thanks.

wxy + w'x(yz+yz') + x'(zw+zy') + z(x'w' + y'x) = xy+z ... lol so confusing!!!

this is what i have at the end..

xy + zw' + zx' + y'x' + y'z

How do i reduce this any further??

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#### Georacer

Joined Nov 25, 2009
5,181
Since you haven't posted your full operation progression and I have come to a different conclusion, I will assume that you have done something wrong in between.

#### pghaffari

Joined Oct 1, 2011
13
wxy + w'x(yz+yz') + x'(zw+zy') + z(x'w'+y'x)
wxy + w'x(zy+y) + x'(zw+y') + z(xy' + w')
w'xyz + wx'z + x'y' + xy'z + w'z
xy + z(w'(xy+q) + wx' + xy') + x'y'
xy + z(w' + wx' + xy') + x'y' -> xy + w'z + wx'z + wy'z + x'y'
xy + w'z + wx'z + y' (xz+x') -> xy + w'z + wx'z +y'(x'+z)
xy + z(x'w+w') + y'(x'+z) -> xy + z(w'+x') + y'(x'+z)
xy + zw' + zx' +y'x' + y'z
xy + z' + zw' + y'x' + y'z
xy+z+w'z+x'y' + z
xy+z+wz'+x'y'
xy+z+w+y'

stuck here

#### Georacer

Joined Nov 25, 2009
5,181
You might have understood something wrong, but all of your parentheses in the second line are wrong.

yz+yz'=y(z+z')=y*1=y
Similarly
zw+zy'=z(w+y')
and
x'w'+y'x
cannot be simplified.

Care to take another look into your theory and give it another shot?

#### pghaffari

Joined Oct 1, 2011
13
wxy + w'x(y) + x'(z(w+y')) + z(x'w' + y'x)
xy(w+w') + x'wz + x'y'z + x'w'z + xy'z
xy + z(x'y' + xy') + z(x'w + y'w')
xy + z(y'(x'+x)) + z(x'(w+w'))
xy + y'z + x'z
xy + z(x' + y')

im so close.. where di i go wrong? i have to get final result xy + z....

AH i got it nevermind!! it just becomes xy + z(xy)' ==> xy + z !!

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