# Equivalence of Formula in Discrete mathematics

Suppose there are two formulas, X and Y. These formulas will be known as equivalence iff X ↔ Y is a tautology. If two formulas X ↔ Y is a tautology, then we can also write it as X ⇔ Y, and we can read this relation as X is equivalence to Y.

#### Note: There are some points which we should keep in mind while linear equivalence of formula, which are described as follows:

• ⇔ is used to indicate only symbol, but it is not connective.
• The truth value of X and Y will always be equal if X ↔ Y is a tautology.
• The equivalence relation contains two properties, i.e., symmetric and transitive.

### Method 1: Truth table method:

In this method, we will construct the truth tables of any two-statement formula and then check whether these statements are equivalent.

Example 1: In this example, we have to prove X ∨ Y ⇔ ¬(¬X ∧ ¬Y).

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

XYX ∨ Y¬X¬Y¬X ∧ ¬Y¬(¬X ∧ ¬Y)X ∨ Y ⇔ ¬(¬X ∧ ¬Y)
TTTFFFTT
TFTFTFTT
FTTTFFTT
FFFTTTFT

As we can see that X ∨ Y and ¬(¬X ∧ ¬Y) is a tautology. Hence X ∨ Y ⇔ ¬(¬X ∧ ¬Y).

Example 2: In this example, we have to prove (X → Y) ⇔ (¬X ∨ Y).

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

XYX → Y¬X¬X ∨ Y(X → Y) ⇔ (¬X ∨ Y)
TTTFTT
TFFFFT
FTTTTT
FFTTTT

As we can see that X → Y and (¬X ∨ Y) are a tautology. Hence (X → Y) ⇔ (¬X ∨ Y)

### Equivalence formula:

There are various laws that are used to prove the equivalence formula, which is described as follows:

Idempotent law: If there is one statement formula, then it will hold the following properties:

Associative law: If there are three statement formulas, then it will hold the following properties:

Commutative law: If there are two statement formulas, then it will hold the following properties:

Distributive law: If there are three statement formulas, then it will hold the following properties:

Identity law: If there is one statement formula, then it will hold the following properties:

Complement law: If there is one statement formula, then it will hold the following properties:

Absorption Law: If there are two statement formulas, then it will hold the following properties:

De Morgan's Law: If there are two statement formulas, then it will hold the following properties:

### Method 2: Replacement Process

In this method, we will assume a formula A : X → (Y → Z). The formula Y → Z can be known as the part of formula. If we replace this part of the formula, i.e., Y → Z, with the help of equivalence formula ¬Y ∨ Z in A, then we will get another formula, i.e., B : X → (¬Y ∨ Z). It is an easy process to verify whether the given formulas A and B are equivalent to each other or not. With the help of replacement process, we can get B from A.

Example 1: In this example, we have to prove that {X → (Y → Z) ⇔ X → (¬Y ∨ Z)} ⇔ (X ∧ Y) → Z.

Solution: Here, we will take the left side part and try to get the right side part.

Now we will use the Associative law like this:

Now we will use De Morgan's law like this:

Hence proved

Example 2: In this example, we have to prove that {(X → Y) ∧ (Z → Y)} ⇔ (X ∨ Z) → Y.

Solution: Here, we will take the left side part and try to get the right side part.

Hence proved

{(X → Y) ∧ (Z → Y)} ⇔ (X ∨ Z) → Y

Example 3: In this example, we have to prove that X → (Y → X) ⇔ ¬X → (X → Y).

Solution: Here, we will take the left side part and try to get the right side part.

Hence proved

Example 4: In this example, we have to prove that (¬X ∧ (¬Y ∧ Z)) ∨ (Y ∧ Z) ∨ (X ∧ Z) ⇔ Z.

Solution: Here, we will take the left side part and try to get the right side part.

Now we will use the Associative and Distributive laws like this:

Now we will use De Morgan's law like this:

Now we will use the Distributive law like this:

Hence proved

Example 5: In this example, we have to show that ((X ∨Y) ∧ ¬(¬X ∧ (¬Y ∨ ¬Z))) ∨ (¬X ∧ ¬Y) ∨ (¬X ∧ ¬Z) is a tautology.

Solution: Here, we will take little parts and solve them.

First, we will use De Morgan's law and get the following:

Therefore,

Also

Hence

Thus

Hence we can say that the given formula is a tautology.

Example 6: In this example, we have to show that (X ∧ Y) → (X ∨ Y) is a tautology.

Solution: (X ∧ Y) → (X ∨ Y)

Now we will use De Morgan's law like this:

Now we will use the Associative law and Commutative law like this:

Now we will use the Negation law like this:

Hence we can say that the given formula is a tautology.

Example 7: In this example, we have to write the negation of some statements, which are described as follows:

1. Marry will complete her education or accept the joining letter of XYZ Company.
2. Harry will go for a ride or run tomorrow.
3. If I get good marks, my cousin will be jealous.

Solution: First, we will solve the first statement like this:

1. Suppose X: Marry will complete her education.

Y: Accept the joining letter of XYZ Company.

We can use the following symbolic form to express this statement:

The negation of X ∨ Y is described as follows:

In conclusion, the negation of given statement will be:

2. Suppose X: Harry will go for a ride

Y: Harry will run tomorrow

We can use the following symbolic form to express this statement:

The negation of X ∨ Y is described as follows:

In conclusion, the negation of given statement will be:

3. Suppose X: If I get good marks.

Y: My cousin will be jealous.

We can use the following symbolic form to express this statement:

The negation of X → Y is described as follows:

In conclusion, the negation of given statement will be:

Example 8: In this example, we have to write the negation of some statements with the help of De Morgan's law. These statements are described as follows:

1. I need a diamond set and worth a gold ring.
2. You get a good job or you will not get a good partner.
3. I take a lot of work and I can't handle it.
4. My dog goes on a trip or it makes a mess in the house.

Solution: The negation of all the statements with the help of De Morgan's law is described one by one like this:

1. I don't need a diamond set or not worth a gold ring.
2. You cannot get a good job and you will get a good partner.
3. I do not take a lot of work or I can handle it.
4. My dog not goes on a trip and it does not make a mess in the house.

Example 9: In this example, we have some statements, and we have to write the negation of those statements. The statements are described as follows:

1. If it is raining, then the plan to go to the beach is cancelled.
2. If I study hard, then I will get good marks on the exam.
3. If I go to a late-night party, then I will get punishment by my father.
4. If you don't want to talk to me, then you have to block my number.

Solution: The negation of all the statements is described one by one like this:

1. If the plan to go to the beach is cancelled, then it is raining.
2. If I get good marks on the exam, then I study hard.
3. If I will get punishment by my father, then I go to a late-night party.
4. If you have to block my number, then you don't want to talk to me.

Example 10: In this example, we have to check whether (X → Y) → Z and X → (Y → Z) are logically equivalent or not. We have to justify our answer with the help of truth tables and with the help of rules of logic to simplify both expressions.

Solution: First, we will use method 1 to check whether (X → Y) → Z and X → (Y → Z) are logically equivalent, which is described as follows:

Method 1: Here, we will assume the following:

And

Method 2: Now, we will use the second method. In this method, we will use the truth table.

XYZX → Y(X → Y) → ZY → ZX → (Y → Z)
TTTTTTT
TTFTFFF
TFTFTTT
TFFFTTT
FTTTTTT
FTFTFFT
FFTTTTT
FFFTFTT

In this truth table, we can see that the columns of (X → Y) → Z and X → (Y → Z) do not contain identical values.