Theorem rabeqbida 3440

Description: Version of rabeqbidva 3428 with two disjoint variable conditions removed and the third replaced by a nonfreeness hypothesis. (Contributed by BJ, 27-Apr-2019.)

Hypotheses

Ref	Expression
rabeqbida.nf	⊢ Ⅎ𝑥𝜑
rabeqbida.1	⊢ (𝜑 → 𝐴 = 𝐵)
rabeqbida.2	⊢ ((𝜑 ∧ 𝑥 ∈ 𝐴) → (𝜓 ↔ 𝜒))

Assertion

Ref	Expression
rabeqbida	⊢ (𝜑 → {𝑥 ∈ 𝐴 ∣ 𝜓} = {𝑥 ∈ 𝐵 ∣ 𝜒})

Proof of Theorem rabeqbida

Step	Hyp	Ref	Expression
1		rabeqbida.nf	. . 3 ⊢ Ⅎ𝑥𝜑
2		rabeqbida.2	. . 3 ⊢ ((𝜑 ∧ 𝑥 ∈ 𝐴) → (𝜓 ↔ 𝜒))
3	1, 2	rabbida 3437	. 2 ⊢ (𝜑 → {𝑥 ∈ 𝐴 ∣ 𝜓} = {𝑥 ∈ 𝐴 ∣ 𝜒})
4		rabeqbida.1	. . 3 ⊢ (𝜑 → 𝐴 = 𝐵)
5	1, 4	rabeqd 3439	. 2 ⊢ (𝜑 → {𝑥 ∈ 𝐴 ∣ 𝜒} = {𝑥 ∈ 𝐵 ∣ 𝜒})
6	3, 5	eqtrd 2795	1 ⊢ (𝜑 → {𝑥 ∈ 𝐴 ∣ 𝜓} = {𝑥 ∈ 𝐵 ∣ 𝜒})

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 401   = wceq 1569  Ⅎwnf 1815   ∈ wcel 2144  {crab 3412

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1942  ax-6 1999  ax-7 2040  ax-8 2146  ax-9 2154  ax-12 2212  ax-ext 2732

This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1812  df-nf 1816  df-sb 2099  df-clab 2739  df-cleq 2752  df-clel 2835  df-rab 3413

This theorem is used by: (None)