Theorem spsbbi 1765

Description: Specialization of biconditional. (Contributed by NM, 5-Aug-1993.) (Proof rewritten by Jim Kingdon, 21-Jan-2018.)

Assertion

Ref	Expression
spsbbi	|- ( A. x ( ph <-> ps ) -> ( [ y / x ] ph <-> [ y / x ] ps ) )

Proof of Theorem spsbbi

Step	Hyp	Ref	Expression
1		spsbim 1764	. . 3 |- ( A. x ( ph -> ps ) -> ( [ y / x ] ph -> [ y / x ] ps ) )
2		spsbim 1764	. . 3 |- ( A. x ( ps -> ph ) -> ( [ y / x ] ps -> [ y / x ] ph ) )
3	1, 2	anim12i 331	. 2 |- ( ( A. x ( ph -> ps ) /\ A. x ( ps -> ph ) ) -> ( ( [ y / x ] ph -> [ y / x ] ps ) /\ ( [ y / x ] ps -> [ y / x ] ph ) ) )
4		albiim 1416	. 2 |- ( A. x ( ph <-> ps ) <-> ( A. x ( ph -> ps ) /\ A. x ( ps -> ph ) ) )
5		dfbi2 380	. 2 |- ( ( [ y / x ] ph <-> [ y / x ] ps ) <-> ( ( [ y / x ] ph -> [ y / x ] ps ) /\ ( [ y / x ] ps -> [ y / x ] ph ) ) )
6	3, 4, 5	3imtr4i 199	1 |- ( A. x ( ph <-> ps ) -> ( [ y / x ] ph <-> [ y / x ] ps ) )

Syntax hints:   -> wi 4   /\ wa 102   <-> 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:  sbbidh  1766  sbbid  1767  relelfvdm  5226