Theorem sbrbif 2406

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

Hypotheses

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

Assertion

Ref	Expression
sbrbif	|- ( [ y / x ] ( ph <-> ch ) <-> ( ps <-> ch ) )

Proof of Theorem sbrbif

Step	Hyp	Ref	Expression
1		sbrbif.2	. . 3 |- ( [ y / x ] ph <-> ps )
2	1	sbrbis 2405	. 2 |- ( [ y / x ] ( ph <-> ch ) <-> ( ps <-> [ y / x ] ch ) )
3		sbrbif.1	. . . 4 |- F/ x ch
4	3	sbf 2380	. . 3 |- ( [ y / x ] ch <-> ch )
5	4	bibi2i 327	. 2 |- ( ( ps <-> [ y / x ] ch ) <-> ( ps <-> ch ) )
6	2, 5	bitri 264	1 |- ( [ y / x ] ( ph <-> ch ) <-> ( ps <-> ch ) )

Syntax hints:   <-> wb 196   F/wnf 1708   [wsb 1880

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-10 2019  ax-12 2047  ax-13 2246

This theorem depends on definitions:  df-bi 197  df-or 385  df-an 386  df-ex 1705  df-nf 1710  df-sb 1881

This theorem is referenced by: (None)