Theorem sbbidh 1856

Description: Deduction substituting both sides of a biconditional. New proofs should use sbbid 1857 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 1480	. 2 |- ( ph -> A. x ( ps <-> ch ) )
4		spsbbi 1855	. 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   [wsb 1773

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 1458  ax-gen 1460  ax-ie1 1504  ax-ie2 1505  ax-4 1521  ax-ial 1545

This theorem depends on definitions:  df-bi 117  df-sb 1774

This theorem is referenced by:  sbcomxyyz  1988