Theorem sbc5 2864

Description: An equivalence for class substitution. (Contributed by NM, 23-Aug-1993.) (Revised by Mario Carneiro, 12-Oct-2016.)

Assertion

Ref	Expression
sbc5	|- ( [. A / x ]. ph <-> E. x ( x = A /\ ph ) )

Distinct variable group:   x, A

Allowed substitution hint:   ph( x)

Proof of Theorem sbc5

Dummy variable y is distinct from all other variables.

Step	Hyp	Ref	Expression
1		sbcex 2849	. 2 |- ( [. A / x ]. ph -> A e. _V )
2		exsimpl 1554	. . 3 |- ( E. x ( x = A /\ ph ) -> E. x x = A )
3		isset 2626	. . 3 |- ( A e. _V <-> E. x x = A )
4	2, 3	sylibr 133	. 2 |- ( E. x ( x = A /\ ph ) -> A e. _V )
5		dfsbcq2 2844	. . 3 |- ( y = A -> ( [ y / x ] ph <-> [. A / x ]. ph ) )
6		eqeq2 2098	. . . . 5 |- ( y = A -> ( x = y <-> x = A ) )
7	6	anbi1d 454	. . . 4 |- ( y = A -> ( ( x = y /\ ph ) <-> ( x = A /\ ph ) ) )
8	7	exbidv 1754	. . 3 |- ( y = A -> ( E. x ( x = y /\ ph ) <-> E. x ( x = A /\ ph ) ) )
9		sb5 1816	. . 3 |- ( [ y / x ] ph <-> E. x ( x = y /\ ph ) )
10	5, 8, 9	vtoclbg 2681	. 2 |- ( A e. _V -> ( [. A / x ]. ph <-> E. x ( x = A /\ ph ) ) )
11	1, 4, 10	pm5.21nii 656	1 |- ( [. A / x ]. ph <-> E. x ( x = A /\ ph ) )

Syntax hints:   /\ wa 103   <-> wb 104   = wceq 1290   E.wex 1427   e. wcel 1439   [wsb 1693   _Vcvv 2620   [.wsbc 2841

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-io 666  ax-5 1382  ax-7 1383  ax-gen 1384  ax-ie1 1428  ax-ie2 1429  ax-8 1441  ax-10 1442  ax-11 1443  ax-i12 1444  ax-bndl 1445  ax-4 1446  ax-17 1465  ax-i9 1469  ax-ial 1473  ax-i5r 1474  ax-ext 2071

This theorem depends on definitions:  df-bi 116  df-tru 1293  df-nf 1396  df-sb 1694  df-clab 2076  df-cleq 2082  df-clel 2085  df-nfc 2218  df-v 2622  df-sbc 2842

This theorem is referenced by:  sbc6g  2865  sbc7  2867  sbciegft  2870  sbccomlem  2914  csb2  2936  rexsns  3486