Metamath Proof Explorer


Theorem rabeq12f

Description: Equality deduction for restricted class abstraction. (Contributed by Giovanni Mascellani, 10-Apr-2018)

Ref Expression
Hypotheses rabeq12f.1 ⊢ Ⅎ 𝑥 𝐴
rabeq12f.2 ⊢ Ⅎ 𝑥 𝐵
Assertion rabeq12f ( ( 𝐴 = 𝐵 ∧ ∀ 𝑥 ∈ 𝐴 ( 𝜑 ↔ 𝜓 ) ) → { 𝑥 ∈ 𝐴 ∣ 𝜑 } = { 𝑥 ∈ 𝐵 ∣ 𝜓 } )

Proof

Step Hyp Ref Expression
1 rabeq12f.1 ⊢ Ⅎ 𝑥 𝐴
2 rabeq12f.2 ⊢ Ⅎ 𝑥 𝐵
3 rabbi ⊢ ( ∀ 𝑥 ∈ 𝐴 ( 𝜑 ↔ 𝜓 ) ↔ { 𝑥 ∈ 𝐴 ∣ 𝜑 } = { 𝑥 ∈ 𝐴 ∣ 𝜓 } )
4 3 biimpi ⊢ ( ∀ 𝑥 ∈ 𝐴 ( 𝜑 ↔ 𝜓 ) → { 𝑥 ∈ 𝐴 ∣ 𝜑 } = { 𝑥 ∈ 𝐴 ∣ 𝜓 } )
5 1 2 rabeqf ⊢ ( 𝐴 = 𝐵 → { 𝑥 ∈ 𝐴 ∣ 𝜓 } = { 𝑥 ∈ 𝐵 ∣ 𝜓 } )
6 4 5 sylan9eqr ⊢ ( ( 𝐴 = 𝐵 ∧ ∀ 𝑥 ∈ 𝐴 ( 𝜑 ↔ 𝜓 ) ) → { 𝑥 ∈ 𝐴 ∣ 𝜑 } = { 𝑥 ∈ 𝐵 ∣ 𝜓 } )