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Truth tabes

  1. Sep 16, 2009 #1
    Hi, I got the following truth table...

    P1 P2 P3 P(P1,P2,P3)

    T T T T
    T T F T
    T F T F
    F T T F
    T F F T
    F T F T
    F F T F
    F F F T

    I need the conjunctive and disjunctive normal form for this truth table. I have no idea how to do this simple problem:frown: I hope somebody can help me!
  2. jcsd
  3. Sep 16, 2009 #2
    Disjunctive Normal Form (DNF) is the sum of all fundamental row products for which the value is 1.

    Conjunction Normal Form (CNF) is the product of all fundamental row sums for which the value is 0.


    [tex]\begin{array}{c|c|c} p & q & pq' + p'q \\
    1 & 1 & 0 \\
    1 & 0 & 1 \\
    0 & 1 & 1 \\
    0 & 0 & 0

    DNF = pq' + p'q

    CNF = (p + q)(p' + q')

    I hope this is making sense. Now try it with your table.

  4. Sep 16, 2009 #3
    No..sorry, dosen't make much sense to me. DNF would be T and CNF would be F, right?
    No idea, ugh.
  5. Sep 16, 2009 #4
    DNF and CNF are expressions. For example (truth tables omitted)

    Given [itex]p \leftrightarrow q[/itex]

    The DNF is the expression pq + p'q' and the CNF is the expression (p + q')(p' + q).

    You can prove (using truth tables among other things) that for any Boolean expression, it is equivalent to both its DNF and CNF.

  6. Sep 17, 2009 #5
    (1) Consider the lines where P(P1,P2,P3) has the value T.
    For example the second line: P1=T, P2=T, P3=F
    From these values you form the conjunctive term [itex](P1 \wedge P2 \wedge \neg \bar{P3})[/itex]
    Here, only P3 is negated because P3=F.

    (2) For each line where P(P1,P2,P3) has the value T form the
    corresponding conjunctive term.

    (3) To finally get your DNF form the disjunction of all the conjunctive terms from (2).
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