# PDNF and PCNF in Discrete Mathematics

To 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.

### DNF

DNF stands for Disjunction normal form. A logical formula will be known as the DNF if it is a disjunction of conjunction with every variable. In the DNF, every variable or its negation must indicate once in each conjunction. In other words, a compound statement will be known as the DNF if it is obtained by operating OR among variables that are connected by the ANDs, and the negation of variables is also included in this case.

For example:

### CNF

CNF stands for Conjunction normal form. A logical formula will be known as the CNF if it is a conjunction of disjunction with every variable. In the CNF, every variable or its negation must indicate once in each disjunction. In other words, a compound statement will be known as the CNF if it is obtained by operating AND among variables that are connected by the Ors, and the negation of variables is also included in this case.

For example:

1. (X ∨ Y) ∧ (Z ∨ U)
2. (¬X ∨ Y ∨ Z) ∧ (U ∨ Z)

### Minterms using three variables

For 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:

22 = 4

X ∧ Y, ¬X ∧ Y, X ∧ ¬Y, ¬X ∧ ¬Y

If there are there variables, then the available minterms for these variables will be

23 = 8

(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:

 X Y X ∧ Y ¬X ∧ Y X ∧ ¬Y ¬X ∧ ¬Y T T T F F F T F F T F F F T F F T F F F F F F T

With the help of this table, we have cleared the following details:

1. There is no equivalence between two minterms.
2. Each minterm has the truth value T for exactly one combination of the truth values of the variables X and Y.

### Maxterms using three variables

The 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:

22 = 4

X ∨ Y, ¬X ∨ Y, X ∨ ¬Y, ¬X ∨ ¬Y

If there are there variables, then the available maxterms for these variables will be:

(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).

Each of the maxterm has the truth value F for exactly one combination of the truth values of the variables.

## PDNF

PDNF is also known as the Principal Disjunction normal form. The PDNF is used to perform the sum of products (SOP). In other words, a formula ? will be known as the PDNF if ? is the sum of min terms. The expression of PDNF will be indicated in the following way:

For example: Suppose there are three variables X, Y, and Z. The expression of PDNF with the help of these variables will be (X.Y'.Z) + (X'.Y.Z) + (X.Y.Z').

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.

For example:

1. The expression (X.Y'.Z) + (X'.Y.Z) + (X.Y) can be known as an example of DNF, but this expression cannot be PDNF.
2. 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 PDNF

There are 2 methods by which we can obtain the PDNF of the given formula. We will discuss them one by one.

Method 1: By using Truth table

1. In this step, we will first construct a truth table for the given formula.
2. 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.
3. In the last step, we will see the disjunction of these minterms, and it will be equivalent to the given formula.

Example 1: In this example, we have an expression X → Y, and we have to determine the PDNF with the help of truth table.

Solution: The truth table of X → Y is described as follows:

X Y X → Y Minterm
T T T X ∧ Q
T F F X ∧ ¬Q
F T T ¬X ∧ Q
F F T ¬X ∧ ¬Q

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)

Example 2: In this example, we have an expression (X ∧ Y) ∨ (¬X ∧ Z) ∨ (Y ∧ Z), and we have to determine the PDNF with the help of truth table.

Solution: The truth table of (X ∧ Y) ∨ (¬X ∧ Z) ∨ (Y ∧ Z) is described as follows:

X Y Z Minterm X ∧ Z ¬X ∧ Z Y ∧ Z (X ∧ Y) ∨ (¬X ∧ Z) ∨ (Y ∧ Z)
T T T X ∧ Y ∧ Z T F T T
T T F X ∧ Y ∧ ¬Z T F F T
T F T X ∧ ¬Y ∧ Z F F F F
T F F X ∧ ¬Y ∧ ¬Z F F F F
F T T ¬X ∧ Y ∧ Z F T T T
F T F ¬X ∧ Y ∧ ¬Z F F F F
F F T ¬X ∧ ¬Y ∧ Z F T F T
F F F ¬X ∧ ¬Y ∧ ¬Z F F F F

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 table

In the following way, we can obtain the PDNF of the given formula.

1. In this step, we will replace → with their equivalent formula, which is used to have ¬, ∧, and ∨.
2. Now, we will apply negation on the variables with the help of De Morgan's laws, which will follow the distributive law.
3. 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.

Example 1: In this example, we have an expression ¬X ∨ Y, and we have to determine the PDNF.

Solution:

¬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)

Example 2: In this example, we have an expression (X ∧ Y) ∨ (¬X ∧ Z) ∨ (Y ∧ Z), and we have to determine the PDNF.

Solution: (X ∧ Y) ∨ (¬X ∧ Z) ∨ (Y ∧ Z)

⇔ (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)

Example 3: In this example, we have an expression X → ((X → Y) ∧ ¬(¬Y ∨ ¬X)), and we have to determine the PDNF without constructing the truth table.

Solution: Here, we will use X → Y ⇔ ¬X ∨ Y and De Morgan's law, and then we will get the following:

→ ((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)

## PCNF

PCNF is also known as the Principal Conjunction normal form. The PCNF is used to perform the product of sums (POS). In other words, a formula ? will be known as the PCNF if ? is the product of max terms. The expression of PCNF will be indicated in the following way:

For example: Suppose there are three variables X, Y, and Z. The expression of PCNF with the help of these variables will be (X + Y' + Z) . (X' + Y + Z) . (X + Y + Z').

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.

For example:

1. The expression (X + Y' + Z) . (X' + Y + Z) . (X + Y) can be known as an example of CNF, but this expression cannot be PCNF.
2. 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.

Example: In this example, we have an expression (¬X → Z) ∧ (Y ↔ X), and we have to determine the PCNF.

Solution: (¬X → Z) ∧ (Y ↔ X)

⇔ [¬(¬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 PCNF

There are various examples of PCNF and PDNF, and some of them are shown below:

Example 1: In this example, we have an expression (X ∧ Y) ∨ (¬X ∧ Z). Now we have to obtain the PDNF (Principal disjunction normal form) and PCNF (Principal conjunction normal form).

Solution:

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.

Example 2: In this example, we have an expression (X ∨ Z) ∧ (X ∨ ¬Y). Now we have to obtain the PDNF (Principal disjunction normal form) and PCNF (Principal conjunction normal form).

Solution:

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.

Example 3: In this example, we have an expression X → (Y ∧ X) ∧ (¬X → (¬Y ∧ ¬Z)). Now we have to obtain the PDNF (Principal disjunction normal form) and PCNF (Principal conjunction normal form).

Solution:

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.

Example 4: In this example, we have an expression (Y ∨ (X ∧ Z)) ∧ ¬((X ∨ Z) ∧ Y)). Now we have to obtain the PDNF (Principal disjunction normal form) and PCNF (Principal conjunction normal form).

Solution:

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.

Example 5: In this example, we have an expression X ∨ (¬X → (Y ∨ (¬Y → Z))). Now we have to obtain the PDNF (Principal disjunction normal form) and PCNF (Principal conjunction normal form).

Solution:

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.

Example 6: In this example, we have an expression (¬X → Z) ∧ (Y ↔ X). Now we have to obtain the PDNF (Principal disjunction normal form) and PCNF (Principal conjunction normal form) with the help of equivalence.

Solution:

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.

Example 7: In this example, we have an expression Y ∨ (X ∧ Z) ∧ ¬((X ∨ Z) ∧ Y). Now we have to obtain the PDNF (Principal disjunction normal form).

Solution:

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)

Example 8: In this example, we have an expression (¬X ∨ ¬Y) → (X ↔ ¬Y). Now we have to obtain the PDNF (Principal disjunction normal form).

Solution: The truth table of (¬X ∨ ¬Y) → (X ↔ ¬Y) is described as follows:

 X Y ¬X ¬Y ¬X ∨ ¬Y X ↔ ¬Y S Minterm Maxterm T T F F F F T X∧ Y T F F T T T T X ∧ ¬Y F T T F T T T ¬X ∧ Y F F T T T F F X ∨ Y

Hence the PCNF is shown below like this:

X ∨ Y and

The PDNF is shown below like this:

(X ∧ Y) ∨ (X ∧ ¬Y) ∨ (¬X ∧ Y)

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