# Simplifying boolean expressions

## Homework Statement

Simplify the following Boolean expressions to a minimum number of literals
(a+b+c')(a'b'+c)

2. The attempt at a solution
Whenever I tried this I made no progress in reducing the number of literals, I just reordered the expression.

(a+b+c')(a'b'+c)
=aa'b'+a'bb'+a'b'c'+ac+bc+cc'
=0b'+a'0+a'b'c'+ac+bc+0
=0+0+a'+b'+c'+ac+bc+0
=a'b'c'+ac+bc
=a'b'c'+c(a+b)

I began with 6 literals and ended with 6. What else can I try?

gabbagabbahey
Homework Helper
Gold Member

## Homework Statement

Simplify the following Boolean expressions to a minimum number of literals
(a+b+c')(a'b'+c)

2. The attempt at a solution
Whenever I tried this I made no progress in reducing the number of literals, I just reordered the expression.

(a+b+c')(a'b'+c)
=aa'b'+a'bb'+a'b'c'+ac+bc+cc'
=0b'+a'0+a'b'c'+ac+bc+0
=0+0+a'+b'+c'+ac+bc+0
=a'b'c'+ac+bc
=a'b'c'+c(a+b)

I began with 6 literals and ended with 6. What else can I try?

Are you allowed to use XOR?

It doesn't say we can't and we did cover it, however it doesn't show up in any of the other problems or examples so far

I tried to use X(N)OR and this is what I came up with:

From
=a'b'c'+ac+bc

I can multiply ac and bc by (1) in the form (x+x')

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

Get rid of one of the two abc (redundant)

=a'b'c'+abc+ab'c+a'bc

Factor

=bc(a+a')+b'(ac+a'c')
=bc(1)+b'(aXORc)'
=bc+b'(aXORc)'

Now I have 5 literals, but is that really the most simplified term? I did all this work just to eliminate 1 literal :(

NascentOxygen
Staff Emeritus