Theorem rabeqbidva 3485

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 3477	. 2 ⊢ (𝜑 → {𝑥 ∈ 𝐴 ∣ 𝜓} = {𝑥 ∈ 𝐴 ∣ 𝜒})
3		rabeqbidva.1	. . 3 ⊢ (𝜑 → 𝐴 = 𝐵)
4	3	rabeqdv 3483	. 2 ⊢ (𝜑 → {𝑥 ∈ 𝐴 ∣ 𝜒} = {𝑥 ∈ 𝐵 ∣ 𝜒})
5	2, 4	eqtrd 2854	1 ⊢ (𝜑 → {𝑥 ∈ 𝐴 ∣ 𝜓} = {𝑥 ∈ 𝐵 ∣ 𝜒})

Syntax hints:   → wi 4   ↔ wb 208   ∧ wa 398   = wceq 1531   ∈ wcel 2108  {crab 3140

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1790  ax-4 1804  ax-5 1905  ax-6 1964  ax-7 2009  ax-8 2110  ax-9 2118  ax-ext 2791

This theorem depends on definitions:  df-bi 209  df-an 399  df-ex 1775  df-sb 2064  df-clab 2798  df-cleq 2812  df-clel 2891  df-rab 3145

This theorem is referenced by:  natpropd  17238  gsumpropd2lem  17881  elntg  26762  rmfsupp2  30859  poimirlem28  34912  scotteqd  40563  domnmsuppn0  44407  eenglngeehlnm  44716