# Simpler answer from truth table?

#### calamari

Joined Jan 28, 2004
1
Hi,

First of all, let me thank you guys for an excellent site! Here's a minor improvement I've come up with for the second "Converting truth tables into Boolean expressions" example.

In that second example the truth table is converted into the Product-Of-Sums expression:

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

This should be able to be simplified to

(a(+)b)+(a(+)c)+(b(+)c)

where (+) represents exclusive-or.

Here is my proof:
(a+b+c)(a'b'c') Given
=(a+b+c)a'+(a+b+c)b'+(a+b+c)c' Distributive
=aa'+ba'+ca'+ab'+bb'+cb'+ac'+bc'+cc' Distributive
=0+ba'+ca'+ab'+0+cb'+ac'+bc'+0 Identity or Complement
=ba'+ca'+ab'+cb'+ac'+bc' Identity
=ab'+ba'+ac'+ca'+bc'+cb' Commutative
=ab'+a'b+ac'+a'c+bc'+b'c Commutative
=(ab'+a'b)+(ac'+a'c)+(bc'+b'c) Associative
=(a(+)b)+(a(+)c)+(b(+)c) Exclusive-or

Drawing this out saves a couple circuit components (4 vs 6).
I also compared the truth tables (to double check) and interestingly enough, in the process I found that only two of the three exclusive-or gates are necessary to generate the correct bit pattern (i.e.):

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

This brings the number of circuit components down to 3 vs 6. I don't have a proof of that except the truth table. It also seems that a'bc+ab'c'=a(+)b+a(+)c. Some proofs would be great.

calamari