Theorem ceqsrexv 2950

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 2528	. . 3 |- ( E. x e. B ( x = A /\ ph ) <-> E. x ( x e. B /\ ( x = A /\ ph ) ) )
2		an12 563	. . . 4 |- ( ( x = A /\ ( x e. B /\ ph ) ) <-> ( x e. B /\ ( x = A /\ ph ) ) )
3	2	exbii 1654	. . 3 |- ( E. x ( x = A /\ ( x e. B /\ ph ) ) <-> E. x ( x e. B /\ ( x = A /\ ph ) ) )
4	1, 3	bitr4i 187	. 2 |- ( E. x e. B ( x = A /\ ph ) <-> E. x ( x = A /\ ( x e. B /\ ph ) ) )
5		eleq1 2297	. . . . 5 |- ( x = A -> ( x e. B <-> A e. B ) )
6		ceqsrexv.1	. . . . 5 |- ( x = A -> ( ph <-> ps ) )
7	5, 6	anbi12d 473	. . . 4 |- ( x = A -> ( ( x e. B /\ ph ) <-> ( A e. B /\ ps ) ) )
8	7	ceqsexgv 2949	. . 3 |- ( A e. B -> ( E. x ( x = A /\ ( x e. B /\ ph ) ) <-> ( A e. B /\ ps ) ) )
9	8	bianabs 615	. 2 |- ( A e. B -> ( E. x ( x = A /\ ( x e. B /\ ph ) ) <-> ps ) )
10	4, 9	bitrid 192	1 |- ( A e. B -> ( E. x e. B ( x = A /\ ph ) <-> ps ) )

Syntax hints:   -> wi 4   /\ wa 104   <-> wb 105   = wceq 1398   E.wex 1541   e. wcel 2205   E.wrex 2523

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 717  ax-5 1496  ax-7 1497  ax-gen 1498  ax-ie1 1542  ax-ie2 1543  ax-8 1553  ax-10 1554  ax-11 1555  ax-i12 1556  ax-bndl 1558  ax-4 1559  ax-17 1575  ax-i9 1579  ax-ial 1583  ax-i5r 1584  ax-ext 2216

This theorem depends on definitions:  df-bi 117  df-tru 1401  df-nf 1510  df-sb 1812  df-clab 2221  df-cleq 2227  df-clel 2230  df-nfc 2375  df-rex 2528  df-v 2817

This theorem is referenced by:  ceqsrexbv  2951  ceqsrex2v  2952  f1oiso  6005  creur  9250  creui  9251