Theorem sbbid 1860

Description: Deduction substituting both sides of a biconditional. (Contributed by NM, 30-Jun-1993.)

Hypotheses

Ref	Expression
sbbid.1	|- F/ x ph
sbbid.2	|- ( ph -> ( ps <-> ch ) )

Assertion

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

Proof of Theorem sbbid

Step	Hyp	Ref	Expression
1		sbbid.1	. . 3 |- F/ x ph
2		sbbid.2	. . 3 |- ( ph -> ( ps <-> ch ) )
3	1, 2	alrimi 1536	. 2 |- ( ph -> A. x ( ps <-> ch ) )
4		spsbbi 1858	. 2 |- ( A. x ( ps <-> ch ) -> ( [ y / x ] ps <-> [ y / x ] ch ) )
5	3, 4	syl 14	1 |- ( ph -> ( [ y / x ] ps <-> [ y / x ] ch ) )

Syntax hints:   -> wi 4   <-> wb 105   A.wal 1362   F/wnf 1474   [wsb 1776

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-5 1461  ax-gen 1463  ax-ie1 1507  ax-ie2 1508  ax-4 1524  ax-ial 1548

This theorem depends on definitions:  df-bi 117  df-nf 1475  df-sb 1777

This theorem is referenced by:  bezoutlemmain  12141