# Check my Boolean Algebra!! I think it's right...?

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1. Oct 19, 2015

### Master0fN0thing

1. The problem statement, all variables and given/known data
Am I doing this right? The question is...
(ab) + (a' + b') = 1

2. Relevant equations
(a) Commutative a · b = b · a a + b = b + a
(b) Associative (a · b) · c = a · (b · c) (a + b) + c = a + (b + c)
(c) Distributive a · (b + c) = (a · b) + (a · c) a + (b · c) = (a + b) · (a + c)
(d) Identity a · 1 = a a + 0 = 0
(e) Negation a + a' = 1 a · a' = 0
(f) Double negative (a')' = a
(g) Idempotent a · a = a a + a = a
(h) DeMorgan’s laws (a · b)' = a' + b' (a + b)' = a' · b'
(i) Universal bound a + 1 = a a · 0 = 0
(j) Absorption a · (a + b) = a a + (a · b) = a
(k) Complement of 1 and 0 1' = 0 0' = 1

3. The attempt at a solution
My Steps....
(ab) + (a' + b') = (ab + a') + b' [Associative]
= (a' + ab) + b' [Commutative]
= (a' + a)(a' + b) + b' [Distributive]
= 1(a' + b) + b' [Negation]
= (a' + b) + b' [Identity]
= a' + (b + b') [Associative]
= a' + 1 [Negation]
= 1 [Absorbtion]

2. Oct 19, 2015

### andrewkirk

How are you justifying the third step? It doesn't look like a straightforward application of the Distributive law to me.

3. Oct 19, 2015

### Master0fN0thing

I'm sorry. I probably should of spaced the relevant equations better. According to the properties in my textbook, one way the distributive property can be written as is...
a + (bc) = (a+b)(a +c)
so I'm treating a as my 1st part and bc as my 2nd part.
Now to my work... in the previous step, I have (a' + ab) + b' and I think I can use the distributive property stated to say
(a' + ab) + b' = (a' + a)(a' + b) + b'
if I consider a' as my first part and ab as my 2nd part. I hope that makes sense...

4. Oct 19, 2015

### andrewkirk

Yes, that looks OK.

5. Oct 19, 2015

### Master0fN0thing

Ok, cool. Thanks!