Equivalence of Formula in Discrete mathematicsSuppose 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:
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:
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:
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 ProcessIn 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:
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:
Solution: The negation of all the statements with the help of De Morgan's law is described one by one like this:
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:
Solution: The negation of all the statements is described one by one like this:
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.
In this truth table, we can see that the columns of (X → Y) → Z and X → (Y → Z) do not contain identical values. |