Theorem ceqsrexv 2972

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 2620	. . 3 |- (E.x e. B (x = A /\ ph) <-> E.x(x e. B /\ (x = A /\ ph)))
2		an12 772	. . . 4 |- ((x = A /\ (x e. B /\ ph)) <-> (x e. B /\ (x = A /\ ph)))
3	2	exbii 1582	. . 3 |- (E.x(x = A /\ (x e. B /\ ph)) <-> E.x(x e. B /\ (x = A /\ ph)))
4	1, 3	bitr4i 243	. 2 |- (E.x e. B (x = A /\ ph) <-> E.x(x = A /\ (x e. B /\ ph)))
5		eleq1 2413	. . . . 5 |- (x = A -> (x e. B <-> A e. B))
6		ceqsrexv.1	. . . . 5 |- (x = A -> (ph <-> ps))
7	5, 6	anbi12d 691	. . . 4 |- (x = A -> ((x e. B /\ ph) <-> (A e. B /\ ps)))
8	7	ceqsexgv 2971	. . 3 |- (A e. B -> (E.x(x = A /\ (x e. B /\ ph)) <-> (A e. B /\ ps)))
9	8	bianabs 850	. 2 |- (A e. B -> (E.x(x = A /\ (x e. B /\ ph)) <-> ps))
10	4, 9	syl5bb 248	1 |- (A e. B -> (E.x e. B (x = A /\ ph) <-> ps))

Syntax hints:   -> wi 4   <-> wb 176   /\ wa 358  E.wex 1541   = wceq 1642   e. wcel 1710  E.wrex 2615

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-rex 2620  df-v 2861

This theorem is referenced by:  ceqsrexbv  2973  ceqsrex2v  2974  fnasrn  5417  f1oiso  5499