Truth Table Analysis: P1, P2, P3

In summary: Example:P1=P2=T, P3=F, P(P1,P2,P3)=TP1=T, P2=T, P3=FSo, the conjunctive term would be (P1 \wedge P2 \wedge \neg P3)Now, for each line where P(P1,P2,P3) has the value T you would form the corresponding conjunctive term.In summary, the DNF for the given truth table is (P1 \wedge P2 \wedge \neg P3) \vee (P1 \wedge P2 \wedge P3) \vee (P1 \wedge \neg P2 \wedge \
  • #1
XodoX
203
0
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!
 
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  • #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.

e.g.

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

DNF = pq' + p'q

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

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

--Elucidus
 
  • #3
No..sorry, dosen't make much sense to me. DNF would be T and CNF would be F, right?
No idea, ugh.
 
  • #4
XodoX said:
No..sorry, dosen't make much sense to me. DNF would be T and CNF would be F, right?
No idea, ugh.

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.

--Elucidus
 
  • #5
XodoX said:
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!

(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).
 

FAQ: Truth Table Analysis: P1, P2, P3

What is a truth table analysis?

A truth table analysis is a method used in logic and computer science to systematically determine the truth or falsity of a compound logical statement based on the truth values of its individual components.

What is P1, P2, and P3 in a truth table analysis?

P1, P2, and P3 are variables that represent the individual components of a compound logical statement. They can be either true or false, and their combination determines the overall truth value of the statement.

How do you create a truth table for P1, P2, and P3?

To create a truth table for P1, P2, and P3, you need to list out all possible combinations of truth values for each variable and determine the truth value of the compound statement for each combination. This can be done systematically by starting with all variables being true and changing one variable at a time to false, or vice versa.

What is the purpose of a truth table analysis?

The purpose of a truth table analysis is to provide a comprehensive visual representation of all possible truth values for a compound logical statement. This allows for a better understanding of the logical relationships between the individual components and the overall truth value of the statement.

What are some common applications of truth table analysis?

Truth table analysis is commonly used in logic, computer science, and mathematics to analyze logical statements and determine their validity. It is also used in the design and testing of digital circuits and in creating decision trees for problem-solving.

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