Theorem sbc5 3052

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 3037	. 2 |- ( [. A / x ]. ph -> A e. _V )
2		exsimpl 1663	. . 3 |- ( E. x ( x = A /\ ph ) -> E. x x = A )
3		isset 2806	. . 3 |- ( A e. _V <-> E. x x = A )
4	2, 3	sylibr 134	. 2 |- ( E. x ( x = A /\ ph ) -> A e. _V )
5		dfsbcq2 3031	. . 3 |- ( y = A -> ( [ y / x ] ph <-> [. A / x ]. ph ) )
6		eqeq2 2239	. . . . 5 |- ( y = A -> ( x = y <-> x = A ) )
7	6	anbi1d 465	. . . 4 |- ( y = A -> ( ( x = y /\ ph ) <-> ( x = A /\ ph ) ) )
8	7	exbidv 1871	. . 3 |- ( y = A -> ( E. x ( x = y /\ ph ) <-> E. x ( x = A /\ ph ) ) )
9		sb5 1934	. . 3 |- ( [ y / x ] ph <-> E. x ( x = y /\ ph ) )
10	5, 8, 9	vtoclbg 2862	. 2 |- ( A e. _V -> ( [. A / x ]. ph <-> E. x ( x = A /\ ph ) ) )
11	1, 4, 10	pm5.21nii 709	1 |- ( [. A / x ]. ph <-> E. x ( x = A /\ ph ) )

Syntax hints:   /\ wa 104   <-> wb 105   = wceq 1395   E.wex 1538   [wsb 1808   e. wcel 2200   _Vcvv 2799   [.wsbc 3028

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 714  ax-5 1493  ax-7 1494  ax-gen 1495  ax-ie1 1539  ax-ie2 1540  ax-8 1550  ax-10 1551  ax-11 1552  ax-i12 1553  ax-bndl 1555  ax-4 1556  ax-17 1572  ax-i9 1576  ax-ial 1580  ax-i5r 1581  ax-ext 2211

This theorem depends on definitions:  df-bi 117  df-tru 1398  df-nf 1507  df-sb 1809  df-clab 2216  df-cleq 2222  df-clel 2225  df-nfc 2361  df-v 2801  df-sbc 3029

This theorem is referenced by:  sbc6g  3053  sbc7  3055  sbciegft  3059  sbccomlem  3103  csb2  3126  rexsns  3705