Theorem sbbi 1874

Description: Equivalence inside and outside of a substitution are equivalent. (Contributed by NM, 5-Aug-1993.)

Assertion

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

Proof of Theorem sbbi

Step	Hyp	Ref	Expression
1		dfbi2 380	. . 3 |- ( ( ph <-> ps ) <-> ( ( ph -> ps ) /\ ( ps -> ph ) ) )
2	1	sbbii 1688	. 2 |- ( [ y / x ] ( ph <-> ps ) <-> [ y / x ] ( ( ph -> ps ) /\ ( ps -> ph ) ) )
3		sbim 1868	. . . 4 |- ( [ y / x ] ( ph -> ps ) <-> ( [ y / x ] ph -> [ y / x ] ps ) )
4		sbim 1868	. . . 4 |- ( [ y / x ] ( ps -> ph ) <-> ( [ y / x ] ps -> [ y / x ] ph ) )
5	3, 4	anbi12i 447	. . 3 |- ( ( [ y / x ] ( ph -> ps ) /\ [ y / x ] ( ps -> ph ) ) <-> ( ( [ y / x ] ph -> [ y / x ] ps ) /\ ( [ y / x ] ps -> [ y / x ] ph ) ) )
6		sban 1870	. . 3 |- ( [ y / x ] ( ( ph -> ps ) /\ ( ps -> ph ) ) <-> ( [ y / x ] ( ph -> ps ) /\ [ y / x ] ( ps -> ph ) ) )
7		dfbi2 380	. . 3 |- ( ( [ y / x ] ph <-> [ y / x ] ps ) <-> ( ( [ y / x ] ph -> [ y / x ] ps ) /\ ( [ y / x ] ps -> [ y / x ] ph ) ) )
8	5, 6, 7	3bitr4i 210	. 2 |- ( [ y / x ] ( ( ph -> ps ) /\ ( ps -> ph ) ) <-> ( [ y / x ] ph <-> [ y / x ] ps ) )
9	2, 8	bitri 182	1 |- ( [ y / x ] ( ph <-> ps ) <-> ( [ y / x ] ph <-> [ y / x ] ps ) )

Syntax hints:   -> wi 4   /\ wa 102   <-> wb 103   [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-io 662  ax-5 1376  ax-7 1377  ax-gen 1378  ax-ie1 1422  ax-ie2 1423  ax-8 1435  ax-10 1436  ax-11 1437  ax-i12 1438  ax-4 1440  ax-17 1459  ax-i9 1463  ax-ial 1467  ax-i5r 1468

This theorem depends on definitions:  df-bi 115  df-nf 1390  df-sb 1686

This theorem is referenced by:  sblbis  1875  sbrbis  1876  sbco  1883  sbcocom  1885  elsb3  1893  elsb4  1894  sb8eu  1954  sb8euh  1964  pm13.183  2732  sbcbig  2860  sb8iota  4894