Theorem sbcbr12g

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

	Ref	Expression
Assertion	sbcbr12g	|- ( A e. V -> ( [. A / x ]. B R C <-> [_ A / x ]_ B R [_ A / x ]_ C ) )

Proof

Step	Hyp	Ref	Expression
1		sbcbr123	|- ( [. A / x ]. B R C <-> [_ A / x ]_ B [_ A / x ]_ R [_ A / x ]_ C )
2		csbconstg	|- ( A e. V -> [_ A / x ]_ R = R )
3	2	breqd	|- ( A e. V -> ( [_ A / x ]_ B [_ A / x ]_ R [_ A / x ]_ C <-> [_ A / x ]_ B R [_ A / x ]_ C ) )
4	1 3	bitrid	|- ( A e. V -> ( [. A / x ]. B R C <-> [_ A / x ]_ B R [_ A / x ]_ C ) )