If A and B are logically equivalent we write A ≡ B. Alternatively, A is logically equivalent to B if and only if A ↔ B is a tautology.
We can use this fact about the biconditional to prove two statements are logically equivalent
Example: Prove that ∀x(P(x)⋀Q(x)) ≡∀xP(x) ⋀∀xQ(x).
Proof:
To show that the statements are logically equivalent we need to show ∀x(P(x)⋀Q(x)) ↔ ∀xP(x) ⋀∀xQ(x) is a tautology.
So, we need to show
∀x(P(x)⋀Q(x)) → ∀xP(x) ⋀∀xQ(x) is a tautology
and ∀xP(x) ⋀∀xQ(x) → ∀x(P(x)⋀Q(x)) is a tautology
First, we need to show∀x(P(x)⋀Q(x)) → ∀xP(x) ⋀∀xQ(x) is a tautology.
1. | ∀x(P(x)⋀Q(x)) | Premise |
2. | P(x)⋀Q(x) | Universal Instantiation |
3. | P(x) | Simplification |
4. | Q(x) | Simplification |
5. | ∀xP(x) | Universal Generalisation |
6. | ∀xQ(x) | Universal Generalisation |
7. | ∀xP(x)⋀∀xQ(x) | Conjunction |
Similarly, we need to show∀xP(x) ⋀∀xQ(x) → ∀x(P(x)⋀Q(x)) is a tautology.
1. | ∀xP(x) | Premise |
2. | ∀xQ(x) | Premise |
3. | P(x) | Universal Instantiation |
4. | Q(x) | Universal Instantiation |
5. | P(x)⋀Q(x) | Conjunction |
6. | ∀x(P(x)⋀Q(x)) | Universal Generalisation |
Therefore, ∀x(P(x)⋀Q(x)) → ∀xP(x) ⋀∀xQ(x) is a tautology.
Therefore, ∀x(P(x)⋀Q(x)) ↔ ∀xP(x) ⋀∀xQ(x) is a tautology i.e.∀x(P(x)⋀Q(x)) ≡∀xP(x) ⋀∀xQ(x).