Theorem ceqsrexv 1889

Description: Elimination of a restricted existential quantifier, using implicit substitution.

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

Proof of Theorem ceqsrexv

Step	Hyp	Ref	Expression
1		eleq1 1534	. . . . 5 |- (x = A -> (x e. B <-> A e. B))
2		ceqsrexv.1	. . . . 5 |- (x = A -> (ph <-> ps))
3	1, 2	anbi12d 628	. . . 4 |- (x = A -> ((x e. B /\ ph) <-> (A e. B /\ ps)))
4	3	ceqsexgv 1888	. . 3 |- (A e. B -> (E.x(x = A /\ (x e. B /\ ph)) <-> (A e. B /\ ps)))
5	4	bianabs 653	. 2 |- (A e. B -> (E.x(x = A /\ (x e. B /\ ph)) <-> ps))
6		df-rex 1650	. . 3 |- (E.x e. B (x = A /\ ph) <-> E.x(x e. B /\ (x = A /\ ph)))
7		an12 484	. . . 4 |- ((x = A /\ (x e. B /\ ph)) <-> (x e. B /\ (x = A /\ ph)))
8	7	exbii 1051	. . 3 |- (E.x(x = A /\ (x e. B /\ ph)) <-> E.x(x e. B /\ (x = A /\ ph)))
9	6, 8	bitr4 176	. 2 |- (E.x e. B (x = A /\ ph) <-> E.x(x = A /\ (x e. B /\ ph)))
10	5, 9	syl5bb 532	1 |- (A e. B -> (E.x e. B (x = A /\ ph) <-> ps))

Syntax hints:   -> wi 3   <-> wb 146   /\ wa 223   = wceq 956   e. wcel 958  E.wex 980  E.wrex 1646

This theorem is referenced by:  ceqsrex2v 1890  reuxfr2 2903  f1oiso 3904  dfisum 7191

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-7 962  ax-gen 963  ax-8 964  ax-12 968  ax-17 971  ax-4 973  ax-5o 975  ax-6o 978  ax-9o 1123  ax-16 1210  ax-11o 1218  ax-ext 1459

This theorem depends on definitions:  df-bi 147  df-an 225  df-ex 981  df-sb 1172  df-clab 1464  df-cleq 1469  df-clel 1472  df-rex 1650  df-v 1812

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