Theorem sbc5 2932

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 2917	. 2 |- ( [. A / x ]. ph -> A e. _V )
2		exsimpl 1596	. . 3 |- ( E. x ( x = A /\ ph ) -> E. x x = A )
3		isset 2692	. . 3 |- ( A e. _V <-> E. x x = A )
4	2, 3	sylibr 133	. 2 |- ( E. x ( x = A /\ ph ) -> A e. _V )
5		dfsbcq2 2912	. . 3 |- ( y = A -> ( [ y / x ] ph <-> [. A / x ]. ph ) )
6		eqeq2 2149	. . . . 5 |- ( y = A -> ( x = y <-> x = A ) )
7	6	anbi1d 460	. . . 4 |- ( y = A -> ( ( x = y /\ ph ) <-> ( x = A /\ ph ) ) )
8	7	exbidv 1797	. . 3 |- ( y = A -> ( E. x ( x = y /\ ph ) <-> E. x ( x = A /\ ph ) ) )
9		sb5 1859	. . 3 |- ( [ y / x ] ph <-> E. x ( x = y /\ ph ) )
10	5, 8, 9	vtoclbg 2747	. 2 |- ( A e. _V -> ( [. A / x ]. ph <-> E. x ( x = A /\ ph ) ) )
11	1, 4, 10	pm5.21nii 693	1 |- ( [. A / x ]. ph <-> E. x ( x = A /\ ph ) )

Syntax hints:   /\ wa 103   <-> wb 104   = wceq 1331   E.wex 1468   e. wcel 1480   [wsb 1735   _Vcvv 2686   [.wsbc 2909

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-io 698  ax-5 1423  ax-7 1424  ax-gen 1425  ax-ie1 1469  ax-ie2 1470  ax-8 1482  ax-10 1483  ax-11 1484  ax-i12 1485  ax-bndl 1486  ax-4 1487  ax-17 1506  ax-i9 1510  ax-ial 1514  ax-i5r 1515  ax-ext 2121

This theorem depends on definitions:  df-bi 116  df-tru 1334  df-nf 1437  df-sb 1736  df-clab 2126  df-cleq 2132  df-clel 2135  df-nfc 2270  df-v 2688  df-sbc 2910

This theorem is referenced by:  sbc6g  2933  sbc7  2935  sbciegft  2939  sbccomlem  2983  csb2  3005  rexsns  3563