Theorem sbbi 2012

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 388	. . 3 |- ( ( ph <-> ps ) <-> ( ( ph -> ps ) /\ ( ps -> ph ) ) )
2	1	sbbii 1813	. 2 |- ( [ y / x ] ( ph <-> ps ) <-> [ y / x ] ( ( ph -> ps ) /\ ( ps -> ph ) ) )
3		sbim 2006	. . . 4 |- ( [ y / x ] ( ph -> ps ) <-> ( [ y / x ] ph -> [ y / x ] ps ) )
4		sbim 2006	. . . 4 |- ( [ y / x ] ( ps -> ph ) <-> ( [ y / x ] ps -> [ y / x ] ph ) )
5	3, 4	anbi12i 460	. . 3 |- ( ( [ y / x ] ( ph -> ps ) /\ [ y / x ] ( ps -> ph ) ) <-> ( ( [ y / x ] ph -> [ y / x ] ps ) /\ ( [ y / x ] ps -> [ y / x ] ph ) ) )
6		sban 2008	. . 3 |- ( [ y / x ] ( ( ph -> ps ) /\ ( ps -> ph ) ) <-> ( [ y / x ] ( ph -> ps ) /\ [ y / x ] ( ps -> ph ) ) )
7		dfbi2 388	. . 3 |- ( ( [ y / x ] ph <-> [ y / x ] ps ) <-> ( ( [ y / x ] ph -> [ y / x ] ps ) /\ ( [ y / x ] ps -> [ y / x ] ph ) ) )
8	5, 6, 7	3bitr4i 212	. 2 |- ( [ y / x ] ( ( ph -> ps ) /\ ( ps -> ph ) ) <-> ( [ y / x ] ph <-> [ y / x ] ps ) )
9	2, 8	bitri 184	1 |- ( [ y / x ] ( ph <-> ps ) <-> ( [ y / x ] ph <-> [ y / x ] ps ) )

Syntax hints:   -> wi 4   /\ wa 104   <-> wb 105   [wsb 1810

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-io 716  ax-5 1495  ax-7 1496  ax-gen 1497  ax-ie1 1541  ax-ie2 1542  ax-8 1552  ax-10 1553  ax-11 1554  ax-i12 1555  ax-4 1558  ax-17 1574  ax-i9 1578  ax-ial 1582  ax-i5r 1583

This theorem depends on definitions:  df-bi 117  df-nf 1509  df-sb 1811

This theorem is referenced by:  sblbis  2013  sbrbis  2014  sbco  2021  sbcocom  2023  sb8eu  2092  sb8euh  2102  elsb1  2209  elsb2  2210  pm13.183  2944  sbcbig  3078  sb8iota  5294