Metamath Proof Explorer


Theorem rexeqbidva

Description: Equality deduction for restricted universal quantifier. (Contributed by Mario Carneiro, 5-Jan-2017)

Ref Expression
Hypotheses raleqbidva.1 ⊢ ( 𝜑 → 𝐴 = 𝐵 )
raleqbidva.2 ⊢ ( ( 𝜑 ∧ 𝑥 ∈ 𝐴 ) → ( 𝜓 ↔ 𝜒 ) )
Assertion rexeqbidva ( 𝜑 → ( ∃ 𝑥 ∈ 𝐴 𝜓 ↔ ∃ 𝑥 ∈ 𝐵 𝜒 ) )

Proof

Step Hyp Ref Expression
1 raleqbidva.1 ⊢ ( 𝜑 → 𝐴 = 𝐵 )
2 raleqbidva.2 ⊢ ( ( 𝜑 ∧ 𝑥 ∈ 𝐴 ) → ( 𝜓 ↔ 𝜒 ) )
3 2 rexbidva ⊢ ( 𝜑 → ( ∃ 𝑥 ∈ 𝐴 𝜓 ↔ ∃ 𝑥 ∈ 𝐴 𝜒 ) )
4 1 rexeqdv ⊢ ( 𝜑 → ( ∃ 𝑥 ∈ 𝐴 𝜒 ↔ ∃ 𝑥 ∈ 𝐵 𝜒 ) )
5 3 4 bitrd ⊢ ( 𝜑 → ( ∃ 𝑥 ∈ 𝐴 𝜓 ↔ ∃ 𝑥 ∈ 𝐵 𝜒 ) )