Simpler answer from truth table?

Discussion in 'Feedback and Suggestions' started by calamari, Jan 28, 2004.

  1. calamari

    Thread Starter New Member

    Jan 28, 2004

    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:


    This should be able to be simplified to


    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.):


    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.