Theorem sbc5 2787

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 2772	. 2 |- ( [. A / x ]. ph -> A e. _V )
2		exsimpl 1508	. . 3 |- ( E. x ( x = A /\ ph ) -> E. x x = A )
3		isset 2561	. . 3 |- ( A e. _V <-> E. x x = A )
4	2, 3	sylibr 137	. 2 |- ( E. x ( x = A /\ ph ) -> A e. _V )
5		dfsbcq2 2767	. . 3 |- ( y = A -> ( [ y / x ] ph <-> [. A / x ]. ph ) )
6		eqeq2 2049	. . . . 5 |- ( y = A -> ( x = y <-> x = A ) )
7	6	anbi1d 438	. . . 4 |- ( y = A -> ( ( x = y /\ ph ) <-> ( x = A /\ ph ) ) )
8	7	exbidv 1706	. . 3 |- ( y = A -> ( E. x ( x = y /\ ph ) <-> E. x ( x = A /\ ph ) ) )
9		sb5 1767	. . 3 |- ( [ y / x ] ph <-> E. x ( x = y /\ ph ) )
10	5, 8, 9	vtoclbg 2614	. 2 |- ( A e. _V -> ( [. A / x ]. ph <-> E. x ( x = A /\ ph ) ) )
11	1, 4, 10	pm5.21nii 620	1 |- ( [. A / x ]. ph <-> E. x ( x = A /\ ph ) )

Syntax hints:   /\ wa 97   <-> wb 98   = wceq 1243   E.wex 1381   e. wcel 1393   [wsb 1645   _Vcvv 2557   [.wsbc 2764

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-io 630  ax-5 1336  ax-7 1337  ax-gen 1338  ax-ie1 1382  ax-ie2 1383  ax-8 1395  ax-10 1396  ax-11 1397  ax-i12 1398  ax-bndl 1399  ax-4 1400  ax-17 1419  ax-i9 1423  ax-ial 1427  ax-i5r 1428  ax-ext 2022

This theorem depends on definitions:  df-bi 110  df-tru 1246  df-nf 1350  df-sb 1646  df-clab 2027  df-cleq 2033  df-clel 2036  df-nfc 2167  df-v 2559  df-sbc 2765

This theorem is referenced by:  sbc6g  2788  sbc7  2790  sbciegft  2793  sbccomlem  2832  csb2  2854  rexsns  3409  rexsnsOLD  3410