Theorem sbbidh 1766

Description: Deduction substituting both sides of a biconditional. New proofs should use sbbid 1767 instead. (Contributed by NM, 5-Aug-1993.) (New usage is discouraged.)

Hypotheses

Ref	Expression
sbbidh.1	|- ( ph -> A. x ph )
sbbidh.2	|- ( ph -> ( ps <-> ch ) )

Assertion

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

Proof of Theorem sbbidh

Step	Hyp	Ref	Expression
1		sbbidh.1	. . 3 |- ( ph -> A. x ph )
2		sbbidh.2	. . 3 |- ( ph -> ( ps <-> ch ) )
3	1, 2	alrimih 1398	. 2 |- ( ph -> A. x ( ps <-> ch ) )
4		spsbbi 1765	. 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 103   A.wal 1282   [wsb 1685

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-5 1376  ax-gen 1378  ax-ie1 1422  ax-ie2 1423  ax-4 1440  ax-ial 1467

This theorem depends on definitions:  df-bi 115  df-sb 1686

This theorem is referenced by:  sbcomxyyz  1887