Theorem sbrbis 2021

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 2019	. 2 |- ( [ y / x ] ( ph <-> ch ) <-> ( [ y / x ] ph <-> [ y / x ] ch ) )
2		sbrbis.1	. . 3 |- ( [ y / x ] ph <-> ps )
3	2	bibi1i 228	. 2 |- ( ( [ y / x ] ph <-> [ y / x ] ch ) <-> ( ps <-> [ y / x ] ch ) )
4	1, 3	bitri 184	1 |- ( [ y / x ] ( ph <-> ch ) <-> ( ps <-> [ y / x ] ch ) )

Syntax hints:   <-> wb 105   [wsb 1815

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 721  ax-5 1500  ax-7 1501  ax-gen 1502  ax-ie1 1546  ax-ie2 1547  ax-8 1557  ax-10 1558  ax-11 1559  ax-i12 1560  ax-4 1563  ax-17 1579  ax-i9 1583  ax-ial 1587  ax-i5r 1588

This theorem depends on definitions:  df-bi 117  df-nf 1514  df-sb 1816

This theorem is referenced by:  sbrbif  2022  sbabel  2419