Theorem sbc5 3055

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 3040	. 2 |- ( [. A / x ]. ph -> A e. _V )
2		exsimpl 1665	. . 3 |- ( E. x ( x = A /\ ph ) -> E. x x = A )
3		isset 2809	. . 3 |- ( A e. _V <-> E. x x = A )
4	2, 3	sylibr 134	. 2 |- ( E. x ( x = A /\ ph ) -> A e. _V )
5		dfsbcq2 3034	. . 3 |- ( y = A -> ( [ y / x ] ph <-> [. A / x ]. ph ) )
6		eqeq2 2241	. . . . 5 |- ( y = A -> ( x = y <-> x = A ) )
7	6	anbi1d 465	. . . 4 |- ( y = A -> ( ( x = y /\ ph ) <-> ( x = A /\ ph ) ) )
8	7	exbidv 1873	. . 3 |- ( y = A -> ( E. x ( x = y /\ ph ) <-> E. x ( x = A /\ ph ) ) )
9		sb5 1936	. . 3 |- ( [ y / x ] ph <-> E. x ( x = y /\ ph ) )
10	5, 8, 9	vtoclbg 2865	. 2 |- ( A e. _V -> ( [. A / x ]. ph <-> E. x ( x = A /\ ph ) ) )
11	1, 4, 10	pm5.21nii 711	1 |- ( [. A / x ]. ph <-> E. x ( x = A /\ ph ) )

Syntax hints:   /\ wa 104   <-> wb 105   = wceq 1397   E.wex 1540   [wsb 1810   e. wcel 2202   _Vcvv 2802   [.wsbc 3031

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 716  ax-5 1495  ax-7 1496  ax-gen 1497  ax-ie1 1541  ax-ie2 1542  ax-8 1552  ax-10 1553  ax-11 1554  ax-i12 1555  ax-bndl 1557  ax-4 1558  ax-17 1574  ax-i9 1578  ax-ial 1582  ax-i5r 1583  ax-ext 2213

This theorem depends on definitions:  df-bi 117  df-tru 1400  df-nf 1509  df-sb 1811  df-clab 2218  df-cleq 2224  df-clel 2227  df-nfc 2363  df-v 2804  df-sbc 3032

This theorem is referenced by:  sbc6g  3056  sbc7  3058  sbciegft  3062  sbccomlem  3106  csb2  3129  rexsns  3708