Theorem sbc5 3009

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 2994	. 2 |- ( [. A / x ]. ph -> A e. _V )
2		exsimpl 1628	. . 3 |- ( E. x ( x = A /\ ph ) -> E. x x = A )
3		isset 2766	. . 3 |- ( A e. _V <-> E. x x = A )
4	2, 3	sylibr 134	. 2 |- ( E. x ( x = A /\ ph ) -> A e. _V )
5		dfsbcq2 2988	. . 3 |- ( y = A -> ( [ y / x ] ph <-> [. A / x ]. ph ) )
6		eqeq2 2203	. . . . 5 |- ( y = A -> ( x = y <-> x = A ) )
7	6	anbi1d 465	. . . 4 |- ( y = A -> ( ( x = y /\ ph ) <-> ( x = A /\ ph ) ) )
8	7	exbidv 1836	. . 3 |- ( y = A -> ( E. x ( x = y /\ ph ) <-> E. x ( x = A /\ ph ) ) )
9		sb5 1899	. . 3 |- ( [ y / x ] ph <-> E. x ( x = y /\ ph ) )
10	5, 8, 9	vtoclbg 2821	. 2 |- ( A e. _V -> ( [. A / x ]. ph <-> E. x ( x = A /\ ph ) ) )
11	1, 4, 10	pm5.21nii 705	1 |- ( [. A / x ]. ph <-> E. x ( x = A /\ ph ) )

Syntax hints:   /\ wa 104   <-> wb 105   = wceq 1364   E.wex 1503   [wsb 1773   e. wcel 2164   _Vcvv 2760   [.wsbc 2985

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 710  ax-5 1458  ax-7 1459  ax-gen 1460  ax-ie1 1504  ax-ie2 1505  ax-8 1515  ax-10 1516  ax-11 1517  ax-i12 1518  ax-bndl 1520  ax-4 1521  ax-17 1537  ax-i9 1541  ax-ial 1545  ax-i5r 1546  ax-ext 2175

This theorem depends on definitions:  df-bi 117  df-tru 1367  df-nf 1472  df-sb 1774  df-clab 2180  df-cleq 2186  df-clel 2189  df-nfc 2325  df-v 2762  df-sbc 2986

This theorem is referenced by:  sbc6g  3010  sbc7  3012  sbciegft  3016  sbccomlem  3060  csb2  3082  rexsns  3657