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Metamath Proof Explorer

Theorem rabeqbidv

Description: Equality of restricted class abstractions. (Contributed by Jeff Madsen, 1-Dec-2009)

Ref Expression
Hypotheses rabeqbidv.1 ⊢ ( 𝜑 → 𝐴 = 𝐵 )
rabeqbidv.2 ⊢ ( 𝜑 → ( 𝜓 ↔ 𝜒 ) )
Assertion rabeqbidv ( 𝜑 → { 𝑥 ∈ 𝐴 ∣ 𝜓 } = { 𝑥 ∈ 𝐵 ∣ 𝜒 } )

Proof

Step Hyp Ref Expression
1 rabeqbidv.1 ⊢ ( 𝜑 → 𝐴 = 𝐵 )
2 rabeqbidv.2 ⊢ ( 𝜑 → ( 𝜓 ↔ 𝜒 ) )
3 2 adantr ⊢ ( ( 𝜑 ∧ 𝑥 ∈ 𝐴 ) → ( 𝜓 ↔ 𝜒 ) )
4 1 3 rabeqbidva ⊢ ( 𝜑 → { 𝑥 ∈ 𝐴 ∣ 𝜓 } = { 𝑥 ∈ 𝐵 ∣ 𝜒 } )