# Simplify the following Boolean expression to a minimum number of literals: (a + b + c') (a' b' + c)?

#### vlin2

Joined Feb 28, 2015
6
(a+b+c')(a'b'+c)

=aa' ab' ac + ba bb' bc + c'a' c'b' c'c

=1 ab' ac + ba 1 bc + c'a' c'b' 1

=1 + 1 + a' + b' + c' + ac + bc + 1

=a'b'c' + ac + bc

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

Did I do this problem correctly? I'm having trouble reducing the literals any further.

#### WBahn

Joined Mar 31, 2012
26,301
(a+b+c')(a'b'+c)

=aa' ab' ac + ba bb' bc + c'a' c'b' c'c

=1 ab' ac + ba 1 bc + c'a' c'b' 1

=1 + 1 + a' + b' + c' + ac + bc + 1

=a'b'c' + ac + bc

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

Did I do this problem correctly? I'm having trouble reducing the literals any further.
Your final result is correct (in that it is equal to the starting expression), but I don't follow your work. For instance:

=1 + 1 + a' + b' + c' + ac + bc + 1

Is simply 1 since 1 + anything is 1. So it would appear that you are using a very unusual notation.

Also, aa' is 0, not 1. That is assuming that you are using the normal convention of 0 for False, 1 for True, + for OR and * for AND.

#### vlin2

Joined Feb 28, 2015
6
Your final result is correct (in that it is equal to the starting expression), but I don't follow your work. For instance:

=1 + 1 + a' + b' + c' + ac + bc + 1

Is simply 1 since 1 + anything is 1. So it would appear that you are using a very unusual notation.

Also, aa' is 0, not 1. That is assuming that you are using the normal convention of 0 for False, 1 for True, + for OR and * for AND.
redid the problem

aa'b' + ac + ba'b' + bc + c'a'b' + c'c

= 0b' + ac + a'b + bc + a'b'c' + 0

= 0 + 0 + a' + b' + c' +ac + bc + 0 <- for this part why did the 0b' and the a'b turn to 0?

= a'b'c' + ac +bc + 0

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

#### WBahn

Joined Mar 31, 2012
26,301
redid the problem

aa'b' + ac + ba'b' + bc + c'a'b' + c'c

= 0b' + ac + a'b + bc + a'b'c' + 0

= 0 + 0 + a' + b' + c' +ac + bc + 0 <- for this part why did the 0b' and the a'b turn to 0?

= a'b'c' + ac +bc + 0

= a'b'c' + c(a+b)
Who's work is this? It would appear that it is not yours.

#### vlin2

Joined Feb 28, 2015
6
Got the answer online, trying to figure out how to get to it now.

Last edited:

#### vlin2

Joined Feb 28, 2015
6
Got the answer online, trying to figure out how to get to it now.
aa'b' + ac + ba'b' + bc + c'a'b' + c'c

aa' goes to 0
bb' goes to 0
c'c goes to 0

(0)b' +ac + (0)a' + bc + a'b'c' + 0

a'b'c' +c(a+b) after the zeros drop

#### WBahn

Joined Mar 31, 2012
26,301
Much better.

Now, as to whether it is the minimum number of literals, that depends on the rules. Notice that the original expression contains six literals. So does your solution. So by this metric neither is better than the other. If you are expected to use SOP or POS form, then neither is acceptable regardless of how many literals they have. If you are allowed to use the exclusive-OR function, you can cut that number in half.

#### vlin2

Joined Feb 28, 2015
6
Much better.

Now, as to whether it is the minimum number of literals, that depends on the rules. Notice that the original expression contains six literals. So does your solution. So by this metric neither is better than the other. If you are expected to use SOP or POS form, then neither is acceptable regardless of how many literals they have. If you are allowed to use the exclusive-OR function, you can cut that number in half.
Yeah I learned about XOR but the professor said to only use AND, and that XOR wouldn't be on the first exam.

#### WBahn

Joined Mar 31, 2012
26,301
Yeah I learned about XOR but the professor said to only use AND, and that XOR wouldn't be on the first exam.
Okay, so XOR is off the table. But if you are only allowed to use AND, then how are you going to do it since all of those expressions also use OR? They also use NOT, but that can be considered buried in the literals since a literal can either be the straight or the complemented version of a signal.