Theorem sbcbr2g 4188

Description: Move substitution in and out of a binary relation. (Contributed by NM, 13-Dec-2005.)

Assertion

Ref	Expression
sbcbr2g	|- ( A e. D -> ( [. A / x ]. B R C <-> B R [_ A / x ]_ C ) )

Distinct variable groups:   x, B   x, R

Allowed substitution hints:   A( x)   C( x)   D( x)

Proof of Theorem sbcbr2g

Step	Hyp	Ref	Expression
1		sbcbr12g 4186	. 2 |- ( A e. D -> ( [. A / x ]. B R C <-> [_ A / x ]_ B R [_ A / x ]_ C ) )
2		csbconstg 3161	. . 3 |- ( A e. D -> [_ A / x ]_ B = B )
3	2	breq1d 4140	. 2 |- ( A e. D -> ( [_ A / x ]_ B R [_ A / x ]_ C <-> B R [_ A / x ]_ C ) )
4	1, 3	bitrd 188	1 |- ( A e. D -> ( [. A / x ]. B R C <-> B R [_ A / x ]_ C ) )

This proof depends on syntax axioms:   -> wi 4   <-> wb 105   e. wcel 2209   [.wsbc 3051   [_csb 3147   class class class wbr 4130

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 721  ax-5 1500  ax-7 1501  ax-gen 1502  ax-ie1 1546  ax-ie2 1547  ax-8 1557  ax-10 1558  ax-11 1559  ax-i12 1560  ax-bndl 1562  ax-4 1563  ax-17 1579  ax-i9 1583  ax-ial 1587  ax-i5r 1588  ax-ext 2220

This proof depends on definitions:  df-bi 117  df-3an 1011  df-tru 1405  df-nf 1514  df-sb 1816  df-clab 2225  df-cleq 2231  df-clel 2234  df-nfc 2381  df-v 2823  df-sbc 3052  df-csb 3148  df-un 3224  df-sn 3715  df-pr 3716  df-op 3718  df-br 4131

This theorem is used by: (None)