Metamath Proof Explorer


Theorem bj-rabeqbid

Description: Version of rabeqbidv with two disjoint variable conditions removed and the third replaced by a nonfreeness hypothesis. (Contributed by BJ, 27-Apr-2019)

Ref Expression
Hypotheses bj-rabeqbid.nf ⊢ Ⅎ 𝑥 𝜑
bj-rabeqbid.1 ⊢ ( 𝜑 → 𝐴 = 𝐵 )
bj-rabeqbid.2 ⊢ ( 𝜑 → ( 𝜓 ↔ 𝜒 ) )
Assertion bj-rabeqbid ( 𝜑 → { 𝑥 ∈ 𝐴 ∣ 𝜓 } = { 𝑥 ∈ 𝐵 ∣ 𝜒 } )

Proof

Step Hyp Ref Expression
1 bj-rabeqbid.nf ⊢ Ⅎ 𝑥 𝜑
2 bj-rabeqbid.1 ⊢ ( 𝜑 → 𝐴 = 𝐵 )
3 bj-rabeqbid.2 ⊢ ( 𝜑 → ( 𝜓 ↔ 𝜒 ) )
4 1 2 rabeqd ⊢ ( 𝜑 → { 𝑥 ∈ 𝐴 ∣ 𝜓 } = { 𝑥 ∈ 𝐵 ∣ 𝜓 } )
5 1 3 rabbid ⊢ ( 𝜑 → { 𝑥 ∈ 𝐵 ∣ 𝜓 } = { 𝑥 ∈ 𝐵 ∣ 𝜒 } )
6 4 5 eqtrd ⊢ ( 𝜑 → { 𝑥 ∈ 𝐴 ∣ 𝜓 } = { 𝑥 ∈ 𝐵 ∣ 𝜒 } )