Theorem ceqsrexv 2856

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 2450	. . 3 |- ( E. x e. B ( x = A /\ ph ) <-> E. x ( x e. B /\ ( x = A /\ ph ) ) )
2		an12 551	. . . 4 |- ( ( x = A /\ ( x e. B /\ ph ) ) <-> ( x e. B /\ ( x = A /\ ph ) ) )
3	2	exbii 1593	. . 3 |- ( E. x ( x = A /\ ( x e. B /\ ph ) ) <-> E. x ( x e. B /\ ( x = A /\ ph ) ) )
4	1, 3	bitr4i 186	. 2 |- ( E. x e. B ( x = A /\ ph ) <-> E. x ( x = A /\ ( x e. B /\ ph ) ) )
5		eleq1 2229	. . . . 5 |- ( x = A -> ( x e. B <-> A e. B ) )
6		ceqsrexv.1	. . . . 5 |- ( x = A -> ( ph <-> ps ) )
7	5, 6	anbi12d 465	. . . 4 |- ( x = A -> ( ( x e. B /\ ph ) <-> ( A e. B /\ ps ) ) )
8	7	ceqsexgv 2855	. . 3 |- ( A e. B -> ( E. x ( x = A /\ ( x e. B /\ ph ) ) <-> ( A e. B /\ ps ) ) )
9	8	bianabs 601	. 2 |- ( A e. B -> ( E. x ( x = A /\ ( x e. B /\ ph ) ) <-> ps ) )
10	4, 9	syl5bb 191	1 |- ( A e. B -> ( E. x e. B ( x = A /\ ph ) <-> ps ) )

Syntax hints:   -> wi 4   /\ wa 103   <-> wb 104   = wceq 1343   E.wex 1480   e. wcel 2136   E.wrex 2445

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-io 699  ax-5 1435  ax-7 1436  ax-gen 1437  ax-ie1 1481  ax-ie2 1482  ax-8 1492  ax-10 1493  ax-11 1494  ax-i12 1495  ax-bndl 1497  ax-4 1498  ax-17 1514  ax-i9 1518  ax-ial 1522  ax-i5r 1523  ax-ext 2147

This theorem depends on definitions:  df-bi 116  df-tru 1346  df-nf 1449  df-sb 1751  df-clab 2152  df-cleq 2158  df-clel 2161  df-nfc 2297  df-rex 2450  df-v 2728

This theorem is referenced by:  ceqsrexbv  2857  ceqsrex2v  2858  f1oiso  5794  creur  8854  creui  8855