Theorem sbrbif

Description: Introduce right biconditional inside of a substitution. (Contributed by NM, 18-Aug-1993) (Revised by Mario Carneiro, 4-Oct-2016)

	Ref	Expression
Hypotheses	sbrbif.1	|- F/ x ch
	sbrbif.2	|- ( [ y / x ] ph <-> ps )
Assertion	sbrbif	|- ( [ y / x ] ( ph <-> ch ) <-> ( ps <-> ch ) )

Proof

Step	Hyp	Ref	Expression
1		sbrbif.1	|- F/ x ch
2		sbrbif.2	|- ( [ y / x ] ph <-> ps )
3	2	sbrbis	|- ( [ y / x ] ( ph <-> ch ) <-> ( ps <-> [ y / x ] ch ) )
4	1	sbf	|- ( [ y / x ] ch <-> ch )
5	4	bibi2i	|- ( ( ps <-> [ y / x ] ch ) <-> ( ps <-> ch ) )
6	3 5	bitri	|- ( [ y / x ] ( ph <-> ch ) <-> ( ps <-> ch ) )