Theorem rabeqbidva 3440

Description: Equality of restricted class abstractions. (Contributed by Mario Carneiro, 26-Jan-2017.)

Hypotheses

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

Assertion

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

Distinct variable groups:   𝑥,𝐴   𝑥,𝐵   𝜑,𝑥

Allowed substitution hints:   𝜓(𝑥)   𝜒(𝑥)

Proof of Theorem rabeqbidva

Step	Hyp	Ref	Expression
1		rabeqbidva.2	. . 3 ⊢ ((𝜑 ∧ 𝑥 ∈ 𝐴) → (𝜓 ↔ 𝜒))
2	1	rabbidva 3431	. 2 ⊢ (𝜑 → {𝑥 ∈ 𝐴 ∣ 𝜓} = {𝑥 ∈ 𝐴 ∣ 𝜒})
3		rabeqbidva.1	. . 3 ⊢ (𝜑 → 𝐴 = 𝐵)
4	3	rabeqdv 3439	. 2 ⊢ (𝜑 → {𝑥 ∈ 𝐴 ∣ 𝜒} = {𝑥 ∈ 𝐵 ∣ 𝜒})
5	2, 4	eqtrd 2764	1 ⊢ (𝜑 → {𝑥 ∈ 𝐴 ∣ 𝜓} = {𝑥 ∈ 𝐵 ∣ 𝜒})

Syntax hints:   → wi 4   ↔ wb 205   ∧ wa 395   = wceq 1533   ∈ wcel 2098  {crab 3424

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1789  ax-4 1803  ax-5 1905  ax-6 1963  ax-7 2003  ax-8 2100  ax-9 2108  ax-ext 2695

This theorem depends on definitions:  df-bi 206  df-an 396  df-ex 1774  df-sb 2060  df-clab 2702  df-cleq 2716  df-clel 2802  df-rab 3425

This theorem is referenced by:  rabeqbidv  3441  natpropd  17931  gsumpropd2lem  18602  elntg  28711  rmfsupp2  32853  poimirlem28  37006  scotteqd  43485  domnmsuppn0  47234  eenglngeehlnm  47613