Theorem sbrbis 2405

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 2401	. 2 |- ( [ y / x ] ( ph <-> ch ) <-> ( [ y / x ] ph <-> [ y / x ] ch ) )
2		sbrbis.1	. . 3 |- ( [ y / x ] ph <-> ps )
3	2	bibi1i 328	. 2 |- ( ( [ y / x ] ph <-> [ y / x ] ch ) <-> ( ps <-> [ y / x ] ch ) )
4	1, 3	bitri 264	1 |- ( [ y / x ] ( ph <-> ch ) <-> ( ps <-> [ y / x ] ch ) )

Syntax hints:   <-> wb 196   [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:  sbrbif  2406  sbabel  2793