Theorem csbprg 4244

Description: Distribute proper substitution through a pair of classes. (Contributed by Alexander van der Vekens, 4-Sep-2018.)

Assertion

Ref	Expression
csbprg	|- ( C e. V -> [_ C / x ]_ { A , B } = { [_ C / x ]_ A , [_ C / x ]_ B } )

Proof of Theorem csbprg

Step	Hyp	Ref	Expression
1		csbun 4009	. . 3 |- [_ C / x ]_ ( { A } u. { B } ) = ( [_ C / x ]_ { A } u. [_ C / x ]_ { B } )
2		csbsng 4243	. . . 4 |- ( C e. V -> [_ C / x ]_ { A } = { [_ C / x ]_ A } )
3		csbsng 4243	. . . 4 |- ( C e. V -> [_ C / x ]_ { B } = { [_ C / x ]_ B } )
4	2, 3	uneq12d 3768	. . 3 |- ( C e. V -> ( [_ C / x ]_ { A } u. [_ C / x ]_ { B } ) = ( { [_ C / x ]_ A } u. { [_ C / x ]_ B } ) )
5	1, 4	syl5eq 2668	. 2 |- ( C e. V -> [_ C / x ]_ ( { A } u. { B } ) = ( { [_ C / x ]_ A } u. { [_ C / x ]_ B } ) )
6		df-pr 4180	. . 3 |- { A , B } = ( { A } u. { B } )
7	6	csbeq2i 3993	. 2 |- [_ C / x ]_ { A , B } = [_ C / x ]_ ( { A } u. { B } )
8		df-pr 4180	. 2 |- { [_ C / x ]_ A , [_ C / x ]_ B } = ( { [_ C / x ]_ A } u. { [_ C / x ]_ B } )
9	5, 7, 8	3eqtr4g 2681	1 |- ( C e. V -> [_ C / x ]_ { A , B } = { [_ C / x ]_ A , [_ C / x ]_ B } )

Syntax hints:   -> wi 4   = wceq 1483   e. wcel 1990   [_csb 3533   u. cun 3572   {csn 4177   {cpr 4179

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-3an 1039  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-un 3579  df-nul 3916  df-sn 4178  df-pr 4180

This theorem is referenced by:  csbopg  4420