Theorem sbbidh 1868

Description: Deduction substituting both sides of a biconditional. New proofs should use sbbid 1869 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 1492	. 2 |- ( ph -> A. x ( ps <-> ch ) )
4		spsbbi 1867	. 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 1371   [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-5 1470  ax-gen 1472  ax-ie1 1516  ax-ie2 1517  ax-4 1533  ax-ial 1557

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

This theorem is referenced by:  sbcomxyyz  2000