Theorem sbc5g 1450

Description: An equivalence for class substitution.

Assertion

Ref	Expression
sbc5g	|- (A e. B -> ([A / x]ph <-> E.x(x = A /\ ph)))

Distinct variable group(s):   x,A

Proof of Theorem sbc5g

Step	Hyp	Ref	Expression
1		ax-17 925	. . . . . 6 |- (y e. A -> A.x y e. A)
2	1	hbsbc 1446	. . . . 5 |- ((A e. V -> [A / x]ph) -> A.x(A e. V -> [A / x]ph))
3		sbceq1 1443	. . . . . 6 |- (x = A -> (ph <-> [A / x]ph))
4	3	imbi2d 464	. . . . 5 |- (x = A -> ((A e. V -> ph) <-> (A e. V -> [A / x]ph)))
5	2, 4	ceqsexg 1411	. . . 4 |- (A e. V -> (E.x(x = A /\ (A e. V -> ph)) <-> (A e. V -> [A / x]ph)))
6		biimt 549	. . . . . . 7 |- (A e. V -> (ph <-> (A e. V -> ph)))
7	6	anbi2d 468	. . . . . 6 |- (A e. V -> ((x = A /\ ph) <-> (x = A /\ (A e. V -> ph))))
8	7	biexdv 936	. . . . 5 |- (A e. V -> (E.x(x = A /\ ph) <-> E.x(x = A /\ (A e. V -> ph))))
9		biimt 549	. . . . 5 |- (A e. V -> (E.x(x = A /\ ph) <-> (A e. V -> E.x(x = A /\ ph))))
10	8, 9	bitr3d 408	. . . 4 |- (A e. V -> (E.x(x = A /\ (A e. V -> ph)) <-> (A e. V -> E.x(x = A /\ ph))))
11	5, 10	bitr3d 408	. . 3 |- (A e. V -> ((A e. V -> [A / x]ph) <-> (A e. V -> E.x(x = A /\ ph))))
12	11	pm5.74rd 446	. 2 |- (A e. V -> (A e. V -> ([A / x]ph <-> E.x(x = A /\ ph))))
13		elisset 1354	. 2 |- (A e. B -> A e. V)
14	12, 13, 13	sylc 62	1 |- (A e. B -> ([A / x]ph <-> E.x(x = A /\ ph)))

Syntax hints:   -> wi 2   <-> wb 127   /\ wa 196  E.wex 678   = wceq 1091   e. wcel 1092  Vcvv 1348  [wsbc 1440

This theorem is referenced by:  sbc5 1452  sbc2or 1454  sbcgf 1469

This theorem was proved from axioms:  ax-1 3  ax-2 4  ax-3 5  ax-mp 6  ax-4 673  ax-5 674  ax-6 675  ax-7 676  ax-gen 677  ax-8 798  ax-9 799  ax-10 800  ax-11 801  ax-12 802  ax-16 922  ax-17 925  ax-ext 1074

This theorem depends on definitions:  df-bi 128  df-an 198  df-ex 679  df-sb 853  df-clab 1093  df-cleq 1097  df-clel 1099  df-v 1349  df-sbc 1441

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