Theorem sbrbis 1989

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 1987	. 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 1785

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 711  ax-5 1470  ax-7 1471  ax-gen 1472  ax-ie1 1516  ax-ie2 1517  ax-8 1527  ax-10 1528  ax-11 1529  ax-i12 1530  ax-4 1533  ax-17 1549  ax-i9 1553  ax-ial 1557  ax-i5r 1558

This theorem depends on definitions:  df-bi 117  df-nf 1484  df-sb 1786

This theorem is referenced by:  sbrbif  1990  sbabel  2375