Theorem ceqsrexv 2890

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 2478	. . 3 |- ( E. x e. B ( x = A /\ ph ) <-> E. x ( x e. B /\ ( x = A /\ ph ) ) )
2		an12 561	. . . 4 |- ( ( x = A /\ ( x e. B /\ ph ) ) <-> ( x e. B /\ ( x = A /\ ph ) ) )
3	2	exbii 1616	. . 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 2256	. . . . 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 2889	. . 3 |- ( A e. B -> ( E. x ( x = A /\ ( x e. B /\ ph ) ) <-> ( A e. B /\ ps ) ) )
9	8	bianabs 611	. 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 1364   E.wex 1503   e. wcel 2164   E.wrex 2473

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 710  ax-5 1458  ax-7 1459  ax-gen 1460  ax-ie1 1504  ax-ie2 1505  ax-8 1515  ax-10 1516  ax-11 1517  ax-i12 1518  ax-bndl 1520  ax-4 1521  ax-17 1537  ax-i9 1541  ax-ial 1545  ax-i5r 1546  ax-ext 2175

This theorem depends on definitions:  df-bi 117  df-tru 1367  df-nf 1472  df-sb 1774  df-clab 2180  df-cleq 2186  df-clel 2189  df-nfc 2325  df-rex 2478  df-v 2762

This theorem is referenced by:  ceqsrexbv  2891  ceqsrex2v  2892  f1oiso  5869  creur  8978  creui  8979