Theorem rabeqbidva 2855

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 2850	. 2 ⊢ (φ → {x ∈ A ∣ ψ} = {x ∈ A ∣ χ})
3		rabeqbidva.1	. . 3 ⊢ (φ → A = B)
4		rabeq 2853	. . 3 ⊢ (A = B → {x ∈ A ∣ χ} = {x ∈ B ∣ χ})
5	3, 4	syl 15	. 2 ⊢ (φ → {x ∈ A ∣ χ} = {x ∈ B ∣ χ})
6	2, 5	eqtrd 2385	1 ⊢ (φ → {x ∈ A ∣ ψ} = {x ∈ B ∣ χ})

Syntax hints:   → wi 4   ↔ wb 176   ∧ wa 358   = wceq 1642   ∈ wcel 1710  {crab 2618

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-6 1729  ax-7 1734  ax-11 1746  ax-12 1925  ax-ext 2334

This theorem depends on definitions:  df-bi 177  df-or 359  df-an 360  df-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649  df-clab 2340  df-cleq 2346  df-clel 2349  df-nfc 2478  df-ral 2619  df-rab 2623

This theorem is referenced by: (None)