Theorem sbrbis 1890

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 1888	. 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 1699

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-io 668  ax-5 1388  ax-7 1389  ax-gen 1390  ax-ie1 1434  ax-ie2 1435  ax-8 1447  ax-10 1448  ax-11 1449  ax-i12 1450  ax-4 1452  ax-17 1471  ax-i9 1475  ax-ial 1479  ax-i5r 1480

This theorem depends on definitions:  df-bi 116  df-nf 1402  df-sb 1700

This theorem is referenced by:  sbrbif  1891  sbabel  2261