Theorem ceqsrexv 2725

Description: Elimination of a restricted existential quantifier, using implicit substitution. (Contributed by NM, 30-Apr-2004.)

Hypothesis

Ref	Expression
ceqsrexv.1	|- ( x = A -> ( ph <-> ps ) )

Assertion

Ref	Expression
ceqsrexv	|- ( A e. B -> ( E. x e. B ( x = A /\ ph ) <-> ps ) )

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

Allowed substitution hint:   ph( x)

Proof of Theorem ceqsrexv

Step	Hyp	Ref	Expression
1		df-rex 2354	. . 3 |- ( E. x e. B ( x = A /\ ph ) <-> E. x ( x e. B /\ ( x = A /\ ph ) ) )
2		an12 525	. . . 4 |- ( ( x = A /\ ( x e. B /\ ph ) ) <-> ( x e. B /\ ( x = A /\ ph ) ) )
3	2	exbii 1536	. . 3 |- ( E. x ( x = A /\ ( x e. B /\ ph ) ) <-> E. x ( x e. B /\ ( x = A /\ ph ) ) )
4	1, 3	bitr4i 185	. 2 |- ( E. x e. B ( x = A /\ ph ) <-> E. x ( x = A /\ ( x e. B /\ ph ) ) )
5		eleq1 2141	. . . . 5 |- ( x = A -> ( x e. B <-> A e. B ) )
6		ceqsrexv.1	. . . . 5 |- ( x = A -> ( ph <-> ps ) )
7	5, 6	anbi12d 456	. . . 4 |- ( x = A -> ( ( x e. B /\ ph ) <-> ( A e. B /\ ps ) ) )
8	7	ceqsexgv 2724	. . 3 |- ( A e. B -> ( E. x ( x = A /\ ( x e. B /\ ph ) ) <-> ( A e. B /\ ps ) ) )
9	8	bianabs 575	. 2 |- ( A e. B -> ( E. x ( x = A /\ ( x e. B /\ ph ) ) <-> ps ) )
10	4, 9	syl5bb 190	1 |- ( A e. B -> ( E. x e. B ( x = A /\ ph ) <-> ps ) )

Syntax hints:   -> wi 4   /\ wa 102   <-> wb 103   = wceq 1284   E.wex 1421   e. wcel 1433   E.wrex 2349

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-rex 2354  df-v 2603

This theorem is referenced by:  ceqsrexbv  2726  ceqsrex2v  2727  f1oiso  5485  creur  8036  creui  8037