Theorem bj-sbel1 32900

Description: Version of sbcel1g 3987 when substituting a set. (Note: one could have a corresponding version of sbcel12 3983 when substituting a set, but the point here is that the antecedent of sbcel1g 3987 is not needed when substituting a set.) (Contributed by BJ, 6-Oct-2018.)

Assertion

Ref	Expression
bj-sbel1	|- ( [ y / x ] A e. B <-> [_ y / x ]_ A e. B )

Distinct variable group:   x, B

Allowed substitution hints:   A( x, y)   B( y)

Proof of Theorem bj-sbel1

Step	Hyp	Ref	Expression
1		sbsbc 3439	. 2 |- ( [ y / x ] A e. B <-> [. y / x ]. A e. B )
2		vex 3203	. . 3 |- y e. _V
3		sbcel1g 3987	. . 3 |- ( y e. _V -> ( [. y / x ]. A e. B <-> [_ y / x ]_ A e. B ) )
4	2, 3	ax-mp 5	. 2 |- ( [. y / x ]. A e. B <-> [_ y / x ]_ A e. B )
5	1, 4	bitri 264	1 |- ( [ y / x ] A e. B <-> [_ y / x ]_ A e. B )

Syntax hints:   <-> wb 196   [wsb 1880   e. wcel 1990   _Vcvv 3200   [.wsbc 3435   [_csb 3533

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1722  ax-4 1737  ax-5 1839  ax-6 1888  ax-7 1935  ax-9 1999  ax-10 2019  ax-11 2034  ax-12 2047  ax-13 2246  ax-ext 2602

This theorem depends on definitions:  df-bi 197  df-or 385  df-an 386  df-tru 1486  df-fal 1489  df-ex 1705  df-nf 1710  df-sb 1881  df-clab 2609  df-cleq 2615  df-clel 2618  df-nfc 2753  df-v 3202  df-sbc 3436  df-csb 3534  df-dif 3577  df-nul 3916

This theorem is referenced by:  bj-snsetex  32951