Theorem sbc5 2877

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 2862	. 2 |- ( [. A / x ]. ph -> A e. _V )
2		exsimpl 1560	. . 3 |- ( E. x ( x = A /\ ph ) -> E. x x = A )
3		isset 2639	. . 3 |- ( A e. _V <-> E. x x = A )
4	2, 3	sylibr 133	. 2 |- ( E. x ( x = A /\ ph ) -> A e. _V )
5		dfsbcq2 2857	. . 3 |- ( y = A -> ( [ y / x ] ph <-> [. A / x ]. ph ) )
6		eqeq2 2104	. . . . 5 |- ( y = A -> ( x = y <-> x = A ) )
7	6	anbi1d 454	. . . 4 |- ( y = A -> ( ( x = y /\ ph ) <-> ( x = A /\ ph ) ) )
8	7	exbidv 1760	. . 3 |- ( y = A -> ( E. x ( x = y /\ ph ) <-> E. x ( x = A /\ ph ) ) )
9		sb5 1822	. . 3 |- ( [ y / x ] ph <-> E. x ( x = y /\ ph ) )
10	5, 8, 9	vtoclbg 2694	. 2 |- ( A e. _V -> ( [. A / x ]. ph <-> E. x ( x = A /\ ph ) ) )
11	1, 4, 10	pm5.21nii 658	1 |- ( [. A / x ]. ph <-> E. x ( x = A /\ ph ) )

Syntax hints:   /\ wa 103   <-> wb 104   = wceq 1296   E.wex 1433   e. wcel 1445   [wsb 1699   _Vcvv 2633   [.wsbc 2854

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 668  ax-5 1388  ax-7 1389  ax-gen 1390  ax-ie1 1434  ax-ie2 1435  ax-8 1447  ax-10 1448  ax-11 1449  ax-i12 1450  ax-bndl 1451  ax-4 1452  ax-17 1471  ax-i9 1475  ax-ial 1479  ax-i5r 1480  ax-ext 2077

This theorem depends on definitions:  df-bi 116  df-tru 1299  df-nf 1402  df-sb 1700  df-clab 2082  df-cleq 2088  df-clel 2091  df-nfc 2224  df-v 2635  df-sbc 2855

This theorem is referenced by:  sbc6g  2878  sbc7  2880  sbciegft  2883  sbccomlem  2927  csb2  2949  rexsns  3502