## PDNF and PCNF in Discrete MathematicsTo understand the PDNF, we have to first learn about the DNF (Disjunction normal form) and Min-terms. To understand the PCNF, we have to learn about the CNF (Conjunction normal form) and maxterms. ## DNFDNF stands for
## CNFCNF stands for For example: - (X ∨ Y) ∧ (Z ∨ U)
- (¬X ∨ Y ∨ Z) ∧ (U ∨ Z)
## Minterms using three variablesFor a given number of variables, the formula to calculate the minterms is described as follows: If there are two variables, X and Y, then minterm will contain 4 possible formulas that have conjunctions of X and Y or its negation will be: 2 X ∧ Y, ¬X ∧ Y, X ∧ ¬Y, ¬X ∧ ¬Y If there are there variables, then the available minterms for these variables will be 2 (X ∧ Y ∧ Z), (¬X ∧ Y ∧ Z), (X ∧ ¬Y ∧ Z), (X ∧ Y ∧ ¬Z), (¬X ∧ ¬Y ∧ Z), (X ∧ ¬Y ∧ ¬Z), (¬X ∧ Y ∧ ¬Z), and (¬X ∧ ¬Y ∧ ¬Z). Now we will show the truth table of minterms of X and Y, which is described as follows:
With the help of this table, we have cleared the following details: - There is no equivalence between two minterms.
- Each minterm has the truth value T for exactly one combination of the truth values of the variables X and Y.
## Maxterms using three variablesThe maxterms can be described as the duals of minterms. The formula to calculate the maxterms is described as follows: If there are two variables, X and Y, then maxterm will contain 4 possible formulas that have disjunctions of X and Y, or its negation will be: 2 X ∨ Y, ¬X ∨ Y, X ∨ ¬Y, ¬X ∨ ¬Y If there are there variables, then the available maxterms for these variables will be:
Each of the maxterm has the truth value F for exactly one combination of the truth values of the variables. ## PDNF
Here + is used to indicate that sum is the main operator. We always get confused while learning about the terms PDNF (Principal disjunction normal form) and DNF (Disjunction normal form), but there is a huge difference between both these terms. It is not necessary that the expression of DNF (Disjunction normal form) contains the same length of all the variables.
- The expression
**(X.Y'.Z) + (X'.Y.Z) + (X.Y)**can be known as an example of DNF, but this expression cannot be PDNF. - The expression
**(X.Y'.Z) + (X'.Y.Z) + (X.Y.Z')**can be known as an example of both DNF, and PDNF.
## Method to construct the PDNFThere are 2 methods by which we can obtain the PDNF of the given formula. We will discuss them one by one.
- In this step, we will first construct a truth table for the given formula.
- Now we will see the truth table, and for every truth value T, we will choose the minterm, which also has value T for the same combinations of T values of X and Y.
- In the last step, we will see the disjunction of these minterms, and it will be equivalent to the given formula.
In this table, only one value of X → Y is false. So PDNF of X → Y will be: (X ∧ Q) ∨ (¬X ∧ Q) ∨ (¬X ∧ ¬Q) Hence ∴ X → Y ⇔ (X ∧ Q) ∨ (¬X ∧ Q) ∨ (¬X ∧ ¬Q)
In this table, there are various false values in the expression (X ∧ Y) ∨ (¬X ∧ Z) ∨ (Y ∧ Z). For the PDNF, we will only select the true values. So PDNF of (X ∧ Y) ∨ (¬X ∧ Z) ∨ (Y ∧ Z) will be: (X ∧ Y ∧ Z) ∨ (X ∧ Y ∧ ¬Z) ∨ (¬X ∧ Y ∧ Z) ∨ (¬X ∧ ¬Y ∧ Z) ## Method 2: Without constructing the truth tableIn the following way, we can obtain the PDNF of the given formula. - In this step, we will replace → with their equivalent formula, which is used to have ¬, ∧, and ∨.
- Now, we will apply negation on the variables with the help of De Morgan's laws, which will follow the distributive law.
- Now we will drop those elementary products, which are a contradiction. After this, we can get minterms in disjunction with the help of introducing the missing factors. If there are some identical minterms in the disjunction, then we will delete them.
¬X ∨ Y ⇔ (¬X ∧ T) ∨ (Y ∧ T) [∵ A ∧ T ⇔ A] ⇔ (¬X ∧ (Y ∨ ¬Y)) ∨ (Y ∧ (X ∨ ¬X)) [∵ X ∨ ¬X ⇔ T] ⇔ (¬X ∧ Y) ∨ (¬X ∧ ¬Y) ∨ (Y ∧ X) ∨ (Y ∧ ¬X) [∵ X ∧ (Y ∨ Z) ⇔ (X ∧ Y) ∨ (X ∧ Z)] ⇔ (¬X ∧ Y) ∨ (¬X ∧ ¬Y) ∨ (X ∧ Y) [∵ X ∨ X ⇔ X] Hence the required PDNF is shown below: (X ∧ Y) ∨ (¬X ∧ Y) ∨ (¬X ∧ ¬Y)
⇔ (X ∧ Y ∧ T) ∨ (¬X ∧ Z ∧ T) ∨ (Y ∧ Z ∧ T) ⇔ (X ∧ Y ∧ (Z ∨ ¬Z)) ∨ (¬X ∧ Z ∧ (Y ∨ ¬Y)) ∨ (Y ∧ Z ∧ (X ∨ ¬X)) ⇔ (X ∧ Y ∧ Z) ∨ (X ∧ Y ∧ ¬Z) ∨ (¬X ∧ Z ∧ Y) ∨ (¬X ∧ Z ∧ ¬Y) ∨ (Y ∧ Z ∧ X) ∨ (Y ∧ Z ∧ ¬X) ⇔ (X ∧ Y ∧ Z) ∨ (X ∧ Y ∧ ¬Z) ∨ (¬X ∧ Y ∧ Z) ∨ (¬X ∧ ¬Y ∧ Z) Hence the required PDNF is shown below: (X ∧ Y ∧ Z) ∨ (X ∧ Y ∧ ¬Z) ∨ (¬X ∧ Y ∧ Z) ∨ (¬X ∧ ¬Y ∧ Z)
→ ((X → Y) ∧ ¬(¬Y ∨ ¬X)) ⇔ ¬X ∨ ((¬X ∨ Y) ∧ (Y ∧ X)) ⇔ ¬X ∨ ((¬X ∧ Y ∧ X) ∨ (Y ∧ Y ∧ X)) ⇔ ¬X ∨ F ∨ (X ∧ Y) ⇔ ¬X ∨ (X ∧ Y) ⇔ (¬X ∧ T) ∨ (X ∧ Y) ⇔ (¬X ∧ (Y ∨ ¬Y)) ∨ (X ∧ Y) ⇔ (¬X ∧ Y) ∨ (¬X ∧ ¬Y) ∨ (X ∧ Y) Hence the required PDNF is shown below: (X ∧ Y) ∨ (¬X ∧ Y) ∨ (¬X ∧ ¬Y) ## PCNFPCNF is also known as the
Here '.' is used to indicate that product is the main operator. We always get confused while learning about the terms PCNF (Principal conjunction normal form) and CNF (Conjunction normal form), but there is a difference between both these terms. It is not necessary that the expression of CNF (Conjunction normal form) contains the same length of all the variables.
- The expression
**(X + Y' + Z) . (X' + Y + Z) . (X + Y)**can be known as an example of CNF, but this expression cannot be PCNF. - The expression
**(X + Y' + Z) . (X' + Y + Z) . (X + Y + Z')**can be known as an example of both CNF, and PCNF.
The method through which we can get the PCNF for a given formula is the same as the one, which we have previously described.
⇔ [¬(¬X ) ∨ Z] ∧ [(Y → X) ∧ (X → Y)] ⇔ (X ∨ Z) ∧ [(¬Y ∨ X) ∧ (¬X ∨ Y)] ⇔ (X ∨ Z ∨ F) ∧ [(¬Y ∨ X ∨ F) ∧ (¬X ∨ Y ∨ F)] ⇔ [(X ∨ Z) ∨ (Y ∧ ¬Y)] ∧ [¬Y ∨ X) ∨ (Z ∧ ¬Z)] ∧ [(¬X ∨ Y) ∨ (Z ∧ ¬Z)] ⇔ (X ∨ Z ∨ Y) ∧ (X ∨ Z ∨ ¬Y) ∧ (X ∨ ¬Y ∨ Z) ∧ (X ∨ ¬Y ∨ ¬Z) ∧ (¬X ∨ Y ∨ Z) ∧ (¬X ∨ Y ∨ ¬Z) ⇔ (X ∨ Y ∨ Z) ∧ (X ∨ ¬Y ∨ Z) ∧ (X ∨ ¬Y ∨ ¬Z) ∧ (¬X ∨ Y ∨ Z) ∧ (¬X ∨ Y ∨ ¬Z) Hence the required PCNF is shown below: (X ∨ Y ∨ Z) ∧ (X ∨ ¬Y ∨ Z) ∧ (X ∨ ¬Y ∨ ¬Z) ∧ (¬X ∨ Y ∨ Z) ∧ (¬X ∨ Y ∨ ¬Z) ## Examples of PDNF and PCNFThere are various examples of PCNF and PDNF, and some of them are shown below:
Suppose S ⇔ (X ∧ Y) ∨ (¬X ∧ Z) ⇔ (X ∧ Y ∧ T) ∨ (¬X ∧ Z ∧ T) ⇔ (X ∧ Y ∧ (Z ∨ ¬Z)) ∨ (¬X ∧ Z ∧ (Y ∨ ¬Y)) ⇔ (X ∧ Y ∧ Z) ∨ (X ∧ Y ∧ ¬Z) ∨ (¬X ∧ Z ∧ Y) ∨ (¬X ∧ Z ∧ ¬Y) ⇔ (X ∧ Y ∧ Z) ∨ (X ∧ Y ∧ ¬Z) ∨ (¬X ∧ Y ∧ Z) ∨ (¬X ∧ ¬Y ∧ Z) This equation is the sum of minterms. Hence, we can say that it shows the PDNF. Now we will determine the PCNF by collecting the remaining minterms in S like this: ¬S ⇔ (X ∧ ¬Y ∧ Z) ∨ (X ∧ ¬Y ∧ ¬Z) ∨ (¬X ∧ Y ∧ ¬Z) ∨ (¬X ∧ ¬Y ∧ ¬Z) ¬(¬S) ⇔ (¬X ∨ Y ∨ ¬Z) ∧ (¬X ∨ Y ∨ Z) ∧ (X ∨ ¬Y ∨ Z) ∧ (X ∨ Y ∨ Z) S ⇔ (¬X ∨ Y ∨ ¬Z) ∧ (¬X ∨ Y ∨ Z) ∧ (X ∨ ¬Y ∨ Z) ∧ (X ∨ Y ∨ Z) This equation is the product of maxterms. Hence, we can say that it shows the PCNF.
Suppose S ⇔ (X ∨ Z) ∧ (X ∨ ¬Y) ⇔ ((X ∨ Z) ∨ F) ∧ ((X ∨ ¬Y) ∨ F) ⇔ ((X ∨ Z) ∨ (Y ∨ ¬Y)) ∧ ((X ∨ ¬Y) ∧ (Z ∨ ¬Z)) ⇔ ((X ∨ Z ∨ Y) ∧ (X ∨ Z ∨ ¬Y) ∧ (X ∨ ¬Y ∨ Z) ∧ (X ∨ ¬Y ∨ ¬Z) ⇔ ((X ∨ Y ∨ Z) ∧ (X ∨ ¬Y ∨ Z) ∧ (X ∨ ¬Y ∨ ¬Z) This equation is the product of maxterms. Hence, we can say that it shows the PCNF. Now we will determine the PDNF by collecting the remaining maxterms in S like this: ¬S ⇔ (¬X ∨ Y ∨ Z) ∧ (X ∨ Y ∨ ¬Z) ∧ (¬X ∨ ¬Y ∨ Z) ∧ (¬X ∨ Y ∨ ¬Z) ∧ (¬X ∨ ¬Y ∨ ¬Z) ¬(¬S) ⇔ (X ∧ ¬Y ∧ ¬Z) ∨ (¬X ∧ ¬Y ∧ Z) ∨ (X ∧ Y ∧ ¬Z) ∨ (X ∧ ¬Y ∧ Z) ∨ (X ∧ Y ∧ Z) S ⇔ (X ∧ ¬Y ∧ ¬Z) ∨ (¬X ∧ ¬Y ∧ Z) ∨ (X ∧ Y ∧ ¬Z) ∨ (X ∧ ¬Y ∧ Z) ∨ (X ∧ Y ∧ Z) This equation is the sum of minterms. Hence, we can say that it shows the PDNF.
Suppose S: X → (Y ∧ X) ∧ (¬X → (¬Y ∧ ¬Z)) ⇔ ¬X ∨ (Y ∧ X) ∧ (X ∨ (¬Y ∧ ¬Z)) ⇔ (¬X ∨ Y) ∧ (¬X ∨ X) ∧ (X ∨ ¬Y) ∧ (X ∨ ¬Z) ⇔ (¬X ∨ Y) ∧ T ∧ (X ∨ ¬Y) ∧ (X ∨ ¬Z) ⇔ (¬X ∨ Y) ∧ (X ∨ ¬Y) ∧ (X ∨ ¬Z) ⇔ (¬X ∨ Y ∨ F) ∧ (X ∨ ¬Y ∨ F) ∧ (X ∨ ¬Z ∨ F) ⇔ (¬X ∨ Y ∨ (Z ∧ ¬Z)) ∧ (X ∨ ¬Y ∨ (Z ∧ ¬Z)) ∧ (X ∨ ¬Z ∨ (Y ∧ ¬Y)) ⇔ (¬X ∨ Y ∨ Z) ∧ (¬X ∨ Y ∨ ¬Z)) ∧ (X ∨ ¬Y ∨ Z) ∧ (X ∨ ¬Y ∨ ¬Z) ∧ (X ∨ ¬Z ∨ Y) ∧ (X ∨ ¬Z ∨ ¬Y) ⇔ (¬X ∨ Y ∨ Z) ∧ (¬X ∨ Y ∨ ¬Z)) ∧ (X ∨ ¬Y ∨ Z) ∧ (X ∨ ¬Y ∨ ¬Z) ∧ (X ∨ Y ∨ ¬Z) This equation is the product of maxterms. Hence, we can say that it shows the PCNF. Now we will determine the PDNF by collecting the remaining maxterms in S like this: ¬S: (X ∨ Y ∨ Z) ∧ (¬X ∨ ¬Y ∨ Z) ∧ (¬X ∨ ¬Y ∨ ¬Z) Now we will take negation on both sides of this equation, and then we will get the following: ¬(¬S): (¬X ∧ ¬Y ∧ ¬Z) ∨ (X ∧ Y ∧ ¬Z) ∨ (X ∧ Y ∧ Z) S ⇔ (¬X ∧ ¬Y ∧ ¬Z) ∨ (X ∧ Y ∧ ¬Z) ∨ (X ∧ Y ∧ Z) This equation is the sum of minterms. Hence, we can say that it shows the PDNF.
Suppose S: (Y ∨ (X ∧ Z)) ∧ ¬((X ∨ Z) ∧ Y)) ⇔ (Y ∨ (X ∧ Z)) ∧ (¬(X ∨ Z) ∨ ¬Y) ⇔ (Y ∨ (X ∧ Z)) ∧ ((¬X ∧ ¬Z) ∨ ¬Y) ⇔ (Y ∨ X) ∧ (Y ∨ Z) ∧ (¬X ∨ ¬Y) ∧ (¬Z ∨ ¬Y) ⇔ (X ∨ Y) ∧ (Y ∨ Z) ∧ (¬X ∨ ¬Y) ∧ (¬Y ∨ ¬Z) ⇔ (X ∨ Y ∨ F) ∧ (Y ∨ Z ∨ F) ∧ (¬X ∨ ¬Y ∨ F) ∧ (¬Z ∨ ¬Y ∨ F) ⇔ (X ∨ Y ∨ (Z ∧ ¬Z)) ∧ (Y ∨ Z ∨ (X ∧ ¬X)) ∧ (¬X ∨ ¬Y ∨ (Z ∧ ¬Z)) ∧ (¬Z ∨ ¬Y ∨ (X ∧ ¬X)) ⇔ (X ∨ Y ∨ Z) ∧ (Y ∨ Y ∨ ¬Z) ∧ (Y ∨ Z ∨ X) ∧ (Y ∨ Z ∨ ¬X) ∧ (¬X ∨ ¬Y ∨ Z) ∧ (¬X ∨ ¬Y ∨ ¬Z) ∧ (¬Y ∨ ¬Z ∨ X) ∧ (¬Y ∨ ¬Z ∨ ¬X) ⇔ (X ∨ Y ∨ Z) ∧ (X ∨ Y ∨ ¬Z) ∧ (¬X ∨ Y ∨ Z) ∧ (¬X ∨ ¬Y ∨ Z) ∧ (X ∨ ¬Y ∨ ¬Z) ∧ (¬X ∨ ¬Y ∨ ¬Z) This equation is the product of maxterms. Hence, we can say that it shows the PCNF. Now we will determine the PDNF by collecting the remaining maxterms in S like this: ¬S: (X ∨ ¬Y ∨ Z) ∧ (¬X ∨ Y ∨ ¬Z) Now we will take negation on both sides of this equation, and then we will get the following: ¬(¬S): (¬X ∧ Y ∧ ¬Z) ∨ (X ∧ ¬Y ∧ Z) S: (¬X ∧ Y ∧ ¬Z) ∨ (X ∧ ¬Y ∧ Z) This equation is the sum of minterms. Hence, we can say that it shows the PDNF.
Suppose S: X ∨ (¬X → (Y ∨ (¬Y → Z))) ⇔ X ∨ (¬X → (Y ∨ (Y ∨ Z))) ⇔ X ∨ (X ∨ (Y ∨ (Y ∨ Z))) ⇔ X ∨ (X ∨ (Y ∨ Y) ∨ Z)) ⇔ X ∨ (X ∨ Y ∨ Z) ⇔ (X ∨ X) ∨ Y ∨ Z ⇔ X ∨ Y ∨ Z This equation is the product of maxterms. Hence, we can say that it shows the PCNF. Now we will determine the PDNF by collecting the remaining maxterms in S like this: ¬S: (¬X ∨ Y ∨ Z) ∧ (X ∨ ¬Y ∨ Z) ∧ (X ∨ Y ∨ ¬Z) ∧ (¬X ∨ ¬Y ∨ Z) ∧ (X ∨ ¬Y ∨ ¬Z) ∧ (¬X ∨ Y ∨ ¬Z) ∧ (¬X ∨ ¬Y ∨ ¬Z) Now we will take negation on both sides of this equation, and then we will get the following: ¬(¬S): (X ∧ ¬Y ∧ ¬Z) ∨ (¬X ∧ Y ∧ ¬Z) ∨ (¬X ∧ ¬Y ∧ Z) ∨ (X ∧ Y ∧ ¬Z) ∨ (¬X ∧ Y ∧ Z) ∨ (X ∧ ¬Y ∧ Z) ∨ (X ∧ Y ∧ Z) S: (X ∧ ¬Y ∧ ¬Z) ∨ (¬X ∧ Y ∧ ¬Z) ∨ (¬X ∧ ¬Y ∧ Z) ∨ (X ∧ Y ∧ ¬Z) ∨ (¬X ∧ Y ∧ Z) ∨ (X ∧ ¬Y ∧ Z) ∨ (X ∧ Y ∧ Z) This equation is the sum of minterms. Hence, we can say that it shows the PDNF.
Suppose S: (¬X → Z) ∧ (Y ↔ X) ⇔ (X ∨ Z) ∧ ((Y → X) ∧ (X → Y)) ⇔ (X ∨ Z) ∧ (¬Y ∨ X) ∧ (¬X ∨ Y) ⇔ (X ∨ Z) ∧ (X ∨ ¬Y) ∧ (¬X ∨ Y) ⇔ (X ∨ Z ∨ F) ∧ (X ∨ ¬Y ∨ F) ∧ (¬X ∨ Y ∨ F) ⇔ (X ∨ Z ∨ (Y ∧ ¬Y)) ∧ (X ∨ ¬Y ∨ (Z ∧ ¬Z)) ∧ (¬X ∨ Y ∨ (Z ∧ ¬Z)) ⇔ (X ∨ Z ∨ Y) ∧ (X ∨ Z ∨ ¬Y) ∧ (X ∨ ¬Y ∨ Z) ∧ (X ∨ ¬Y ∨ ¬Z) ∧ (¬X ∨ Y ∨ Z) ∧ (¬X ∨ Y ∨ ¬Z) ⇔ (X ∨ Y ∨ Z) ∧ (¬X ∨ Y ∨ Z) ∧ (X ∨ ¬Y ∨ Z) ∧ (X ∨ ¬Y ∨ ¬Z) ∧ (¬X ∨ Y ∨ ¬Z) This equation is the product of maxterms. Hence, we can say that it shows the PCNF. Now we will determine the PDNF by collecting the remaining maxterms in S like this: ¬S: (X ∨ Y ∨ ¬Z) ∧ (¬X ∨ ¬Y ∨ Z) ∧ (¬X∨ ¬Y ∨ ¬Z) Now we will take negation on both sides of this equation, and then we will get the following: ¬(¬S): (¬X ∧ ¬Y ∧ Z) ∨ (X ∧ Y ∧ ¬Z) ∨ (X ∧ Y ∧ Z) S: (¬X ∧ ¬Y ∧ Z) ∨ (X ∧ Y ∧ ¬Z) ∨ (X ∧ Y ∧ Z) This equation is the sum of minterms. Hence, we can say that it shows the PDNF.
Y ∨ (X ∧ Z) ∧ ¬((X ∨ Z) ∧ Y) Now we will use De Morgan's law in this expression like this: ⇔ (Y ∨ (X ∧ Z)) ∧ (¬((X ∨ Z) <∧ Y)) Now we will again use De Morgan's law like this: ⇔ (Y ∨ (X ∧ Z)) ∧ ((¬X ∧ ¬Z) ∨ ¬Y)) Now we will use the Extended Distributive law like this: ⇔ (Y ∧ (¬X ∧ ¬Z)) ∨ (Y ∧ ¬Y) ∨ ((X ∧ Z) ∧ ¬X ∧ ¬Z) ((X ∧ Z) ∧ ¬Y) Now we will use the Negation law like this: ⇔ (¬X ∧ Y ∧ ¬Z) ∨ F ∨ (F ∧ Z ∧ ¬Z) ∨ (X ∧ ¬Y ∧ Z) Now we will again use the Negation law like this: ⇔ (¬X ∧ Y ∧ ¬Z) ∨ (X ∧ ¬Y ∧ Z)> Hence the required PDNF is shown below: (¬X ∧ Y ∧ ¬Z) ∨ (X ∧ ¬Y ∧ Z)
Hence the PCNF is shown below like this: X ∨ Y and The PDNF is shown below like this: (X ∧ Y) ∨ (X ∧ ¬Y) ∨ (¬X ∧ Y) |

For Videos Join Our Youtube Channel: Join Now

- Send your Feedback to [email protected]