Simplifying boolean expressions

[Pi Face]

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 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?

 

[Pi Face]

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

 

[Pi Face]

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]

a'b'c'+ac+bc

= c'·a'b' + c(a+b)

= c'·a'b' + c·(a'b')'

= ...