Theorem ceqsrexv 1885

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 1531	. . . . 5 |- (x = A -> (x e. B <-> A e. B))
2		ceqsrexv.1	. . . . 5 |- (x = A -> (ph <-> ps))
3	1, 2	anbi12d 627	. . . 4 |- (x = A -> ((x e. B /\ ph) <-> (A e. B /\ ps)))
4	3	ceqsexgv 1884	. . 3 |- (A e. B -> (E.x(x = A /\ (x e. B /\ ph)) <-> (A e. B /\ ps)))
5	4	bianabs 652	. 2 |- (A e. B -> (E.x(x = A /\ (x e. B /\ ph)) <-> ps))
6		df-rex 1647	. . 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 1049	. . 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 531	1 |- (A e. B -> (E.x e. B (x = A /\ ph) <-> ps))

Syntax hints:   -> wi 3   <-> wb 146   /\ wa 223   = wceq 954   e. wcel 956  E.wex 978  E.wrex 1643

This theorem is referenced by:  ceqsrex2v 1886  reuxfr2 2898  f1oiso 3895  dfisum 7135

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-7 960  ax-gen 961  ax-8 962  ax-12 966  ax-17 969  ax-4 971  ax-5o 973  ax-6o 976  ax-9o 1121  ax-16 1208  ax-11o 1216  ax-ext 1457

This theorem depends on definitions:  df-bi 147  df-an 225  df-ex 979  df-sb 1170  df-clab 1462  df-cleq 1467  df-clel 1470  df-rex 1647  df-v 1808

Copyright terms: Public domain