Theorem rabeqbidva 2855

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

Hypotheses

Ref	Expression
rabeqbidva.1	|- (ph -> A = B)
rabeqbidva.2	|- ((ph /\ x e. A) -> (ps <-> ch))

Assertion

Ref	Expression
rabeqbidva	|- (ph -> {x e. A | ps} = {x e. B | ch})

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

Allowed substitution hints:   ps(x)   ch(x)

Proof of Theorem rabeqbidva

Step	Hyp	Ref	Expression
1		rabeqbidva.2	. . 3 |- ((ph /\ x e. A) -> (ps <-> ch))
2	1	rabbidva 2850	. 2 |- (ph -> {x e. A | ps} = {x e. A | ch})
3		rabeqbidva.1	. . 3 |- (ph -> A = B)
4		rabeq 2853	. . 3 |- (A = B -> {x e. A | ch} = {x e. B | ch})
5	3, 4	syl 15	. 2 |- (ph -> {x e. A | ch} = {x e. B | ch})
6	2, 5	eqtrd 2385	1 |- (ph -> {x e. A | ps} = {x e. B | ch})

Syntax hints:   -> wi 4   <-> wb 176   /\ wa 358   = wceq 1642   e. 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)