Theorem sbrbif 1877

Description: Introduce right biconditional inside of a substitution. (Contributed by NM, 18-Aug-1993.)

Hypotheses

Ref	Expression
sbrbif.1	|- ( ch -> A. 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 1876	. 2 |- ( [ y / x ] ( ph <-> ch ) <-> ( ps <-> [ y / x ] ch ) )
3		sbrbif.1	. . . 4 |- ( ch -> A. x ch )
4	3	sbh 1699	. . 3 |- ( [ y / x ] ch <-> ch )
5	4	bibi2i 225	. 2 |- ( ( ps <-> [ y / x ] ch ) <-> ( ps <-> ch ) )
6	2, 5	bitri 182	1 |- ( [ y / x ] ( ph <-> ch ) <-> ( ps <-> ch ) )

Syntax hints:   -> wi 4   <-> wb 103   A.wal 1282   [wsb 1685

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-io 662  ax-5 1376  ax-7 1377  ax-gen 1378  ax-ie1 1422  ax-ie2 1423  ax-8 1435  ax-10 1436  ax-11 1437  ax-i12 1438  ax-4 1440  ax-17 1459  ax-i9 1463  ax-ial 1467  ax-i5r 1468

This theorem depends on definitions:  df-bi 115  df-nf 1390  df-sb 1686

This theorem is referenced by: (None)