Theorem sbrbis 1935

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

Hypothesis

Ref	Expression
sbrbis.1	|- ( [ y / x ] ph <-> ps )

Assertion

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

Proof of Theorem sbrbis

Step	Hyp	Ref	Expression
1		sbbi 1933	. 2 |- ( [ y / x ] ( ph <-> ch ) <-> ( [ y / x ] ph <-> [ y / x ] ch ) )
2		sbrbis.1	. . 3 |- ( [ y / x ] ph <-> ps )
3	2	bibi1i 227	. 2 |- ( ( [ y / x ] ph <-> [ y / x ] ch ) <-> ( ps <-> [ y / x ] ch ) )
4	1, 3	bitri 183	1 |- ( [ y / x ] ( ph <-> ch ) <-> ( ps <-> [ y / x ] ch ) )

Syntax hints:   <-> wb 104   [wsb 1736

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-io 699  ax-5 1424  ax-7 1425  ax-gen 1426  ax-ie1 1470  ax-ie2 1471  ax-8 1483  ax-10 1484  ax-11 1485  ax-i12 1486  ax-4 1488  ax-17 1507  ax-i9 1511  ax-ial 1515  ax-i5r 1516

This theorem depends on definitions:  df-bi 116  df-nf 1438  df-sb 1737

This theorem is referenced by:  sbrbif  1936  sbabel  2308