Theorem sbbi 1833

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 368	. . 3 |- ( ( ph <-> ps ) <-> ( ( ph -> ps ) /\ ( ps -> ph ) ) )
2	1	sbbii 1648	. 2 |- ( [ y / x ] ( ph <-> ps ) <-> [ y / x ] ( ( ph -> ps ) /\ ( ps -> ph ) ) )
3		sbim 1827	. . . 4 |- ( [ y / x ] ( ph -> ps ) <-> ( [ y / x ] ph -> [ y / x ] ps ) )
4		sbim 1827	. . . 4 |- ( [ y / x ] ( ps -> ph ) <-> ( [ y / x ] ps -> [ y / x ] ph ) )
5	3, 4	anbi12i 433	. . 3 |- ( ( [ y / x ] ( ph -> ps ) /\ [ y / x ] ( ps -> ph ) ) <-> ( ( [ y / x ] ph -> [ y / x ] ps ) /\ ( [ y / x ] ps -> [ y / x ] ph ) ) )
6		sban 1829	. . 3 |- ( [ y / x ] ( ( ph -> ps ) /\ ( ps -> ph ) ) <-> ( [ y / x ] ( ph -> ps ) /\ [ y / x ] ( ps -> ph ) ) )
7		dfbi2 368	. . 3 |- ( ( [ y / x ] ph <-> [ y / x ] ps ) <-> ( ( [ y / x ] ph -> [ y / x ] ps ) /\ ( [ y / x ] ps -> [ y / x ] ph ) ) )
8	5, 6, 7	3bitr4i 201	. 2 |- ( [ y / x ] ( ( ph -> ps ) /\ ( ps -> ph ) ) <-> ( [ y / x ] ph <-> [ y / x ] ps ) )
9	2, 8	bitri 173	1 |- ( [ y / x ] ( ph <-> ps ) <-> ( [ y / x ] ph <-> [ y / x ] ps ) )

Syntax hints:   -> wi 4   /\ wa 97   <-> wb 98   [wsb 1645

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-io 630  ax-5 1336  ax-7 1337  ax-gen 1338  ax-ie1 1382  ax-ie2 1383  ax-8 1395  ax-10 1396  ax-11 1397  ax-i12 1398  ax-4 1400  ax-17 1419  ax-i9 1423  ax-ial 1427  ax-i5r 1428

This theorem depends on definitions:  df-bi 110  df-nf 1350  df-sb 1646

This theorem is referenced by:  sblbis  1834  sbrbis  1835  sbco  1842  sbcocom  1844  elsb3  1852  elsb4  1853  sb8eu  1913  sb8euh  1923  pm13.183  2681  sbcbig  2809  sb8iota  4874