Theorem rabeqbidva 2597

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 2592	. 2 ⊢ (𝜑 → {𝑥 ∈ 𝐴 ∣ 𝜓} = {𝑥 ∈ 𝐴 ∣ 𝜒})
3		rabeqbidva.1	. . 3 ⊢ (𝜑 → 𝐴 = 𝐵)
4		rabeq 2595	. . 3 ⊢ (𝐴 = 𝐵 → {𝑥 ∈ 𝐴 ∣ 𝜒} = {𝑥 ∈ 𝐵 ∣ 𝜒})
5	3, 4	syl 14	. 2 ⊢ (𝜑 → {𝑥 ∈ 𝐴 ∣ 𝜒} = {𝑥 ∈ 𝐵 ∣ 𝜒})
6	2, 5	eqtrd 2113	1 ⊢ (𝜑 → {𝑥 ∈ 𝐴 ∣ 𝜓} = {𝑥 ∈ 𝐵 ∣ 𝜒})

Syntax hints:   → wi 4   ∧ wa 102   ↔ wb 103   = wceq 1284   ∈ wcel 1433  {crab 2352

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-io 662  ax-5 1376  ax-7 1377  ax-gen 1378  ax-ie1 1422  ax-ie2 1423  ax-8 1435  ax-10 1436  ax-11 1437  ax-i12 1438  ax-bndl 1439  ax-4 1440  ax-17 1459  ax-i9 1463  ax-ial 1467  ax-i5r 1468  ax-ext 2063

This theorem depends on definitions:  df-bi 115  df-tru 1287  df-nf 1390  df-sb 1686  df-clab 2068  df-cleq 2074  df-clel 2077  df-nfc 2208  df-ral 2353  df-rab 2357

This theorem is referenced by: (None)