Metamath Proof Explorer
Description: Equality deduction for restricted universal quantifier. (Contributed by NM, 16-Nov-1995) (Proof shortened by Steven Nguyen, 5-May-2023)
|
|
Ref |
Expression |
|
Hypothesis |
raleqbi1dv.1 |
⊢ ( 𝐴 = 𝐵 → ( 𝜑 ↔ 𝜓 ) ) |
|
Assertion |
raleqbi1dv |
⊢ ( 𝐴 = 𝐵 → ( ∀ 𝑥 ∈ 𝐴 𝜑 ↔ ∀ 𝑥 ∈ 𝐵 𝜓 ) ) |
Proof
Step |
Hyp |
Ref |
Expression |
1 |
|
raleqbi1dv.1 |
⊢ ( 𝐴 = 𝐵 → ( 𝜑 ↔ 𝜓 ) ) |
2 |
|
id |
⊢ ( 𝐴 = 𝐵 → 𝐴 = 𝐵 ) |
3 |
2 1
|
raleqbidvv |
⊢ ( 𝐴 = 𝐵 → ( ∀ 𝑥 ∈ 𝐴 𝜑 ↔ ∀ 𝑥 ∈ 𝐵 𝜓 ) ) |