Boolean algebra simplification hw

Question

I'm given the equation $F=(a+b+c)(a'+b')(a+b'+c)$ and it's supposed to simplify into a sum of two product terms, each with two literals. I know the answer is $ab'+a'c$, but I'm unsure how to get there. I know the laws and postulates that I need to use. As much as I hate to admit it, I only know how to handle sums of products.

Answer 1

$$(a+b+c)(a'+b')(a+b'+c) = [aa'+ab'+a'b+bb'+a'c+b'c](a+b'+c) \\ = [0+ab'+a'b +0+a'c+b'c](a+b'+c) \\ = (ab'+a'b+a'c+b'c)(a+b'+c) \\ = aab' + ab'b'+ab'c + a'ab+a'bb'+a'bc +a'ac+a'b'c+a'cc+ab'c+b'b'c+b'cc \\= ab'+ab'+ab'c+0+0+a'bc+0+a'b'c+a'c+ab'c+b'c+b'c \\ = (ab'+ab') + ab'c+a'c(b+b')+a'c+b'c(a+1+1) \\ = ab'+ab'c+a'c(1)+a'c+b'c \\ = ab'(1+c) + a'c+b'c \\ = ab'+a'c+b'c$$

I've tried to simplify this in a few different ways but I always come out with the same answer which involves 3 terms. I suppose you could simplify the last line into $ab'+c(ab)'$ using DeMorgan's rule but I don't see how you obtain only $ab'+a'c$ in the end. I am fairly certain an error in my expansion/simplification will become noticeable if someone else would be kind enough to check it.

Answer 2

$$ F = (a + b + c) ( \bar a + \bar{b}) (a + \bar{b} + c) $$ I would rearrange this to put the outer two terms together $$= (a + b + c) (a + \bar{b} + c)( \bar a + \bar{b}) $$ since you can distribute with those two and cancel out a lot of things $$ = (a + a \bar{b} + ac + c) ( \bar a + \bar{b}) $$ For example $aa$ became $a$ and things like $a \bar{a}$ are just dropped since you can't have both $a$ and $\bar{a}$ at the same time. Then if you distribute the right hand term into the left you get $$ = \bar{a} c + a \bar{b} c + \bar{b} c $$ where again we had many terms with $a \bar{a}$ or $b \bar{b}$ that we can just drop. For the very last step you can simplify the last two terms :)

Answer 3

$$(a +b+c)(a'+b')(a +b'+c)= \text{ (Adjacency)}$$

$$(a+c)(a'+b')= \text{ (Distribution)}$$

$$aa'+ab'+ca'+cb'=\text{ (Complement)}$$

$$0+ab'+ca'+cb'=\text{ (Identity)}$$

$$ab'+ca'+cb'=\text{ (Consensus)}$$

$$ab'+ca'= \text{ (Commutation)}$$

$$ab'+a'c$$

The Consensus Law says that:

Consensus

$XY +Y'Z=XY +Y'Z +XZ$