Theorem sbbi 1888

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 381	. . 3 |- ( ( ph <-> ps ) <-> ( ( ph -> ps ) /\ ( ps -> ph ) ) )
2	1	sbbii 1702	. 2 |- ( [ y / x ] ( ph <-> ps ) <-> [ y / x ] ( ( ph -> ps ) /\ ( ps -> ph ) ) )
3		sbim 1882	. . . 4 |- ( [ y / x ] ( ph -> ps ) <-> ( [ y / x ] ph -> [ y / x ] ps ) )
4		sbim 1882	. . . 4 |- ( [ y / x ] ( ps -> ph ) <-> ( [ y / x ] ps -> [ y / x ] ph ) )
5	3, 4	anbi12i 449	. . 3 |- ( ( [ y / x ] ( ph -> ps ) /\ [ y / x ] ( ps -> ph ) ) <-> ( ( [ y / x ] ph -> [ y / x ] ps ) /\ ( [ y / x ] ps -> [ y / x ] ph ) ) )
6		sban 1884	. . 3 |- ( [ y / x ] ( ( ph -> ps ) /\ ( ps -> ph ) ) <-> ( [ y / x ] ( ph -> ps ) /\ [ y / x ] ( ps -> ph ) ) )
7		dfbi2 381	. . 3 |- ( ( [ y / x ] ph <-> [ y / x ] ps ) <-> ( ( [ y / x ] ph -> [ y / x ] ps ) /\ ( [ y / x ] ps -> [ y / x ] ph ) ) )
8	5, 6, 7	3bitr4i 211	. 2 |- ( [ y / x ] ( ( ph -> ps ) /\ ( ps -> ph ) ) <-> ( [ y / x ] ph <-> [ y / x ] ps ) )
9	2, 8	bitri 183	1 |- ( [ y / x ] ( ph <-> ps ) <-> ( [ y / x ] ph <-> [ y / x ] ps ) )

Syntax hints:   -> wi 4   /\ wa 103   <-> wb 104   [wsb 1699

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-io 668  ax-5 1388  ax-7 1389  ax-gen 1390  ax-ie1 1434  ax-ie2 1435  ax-8 1447  ax-10 1448  ax-11 1449  ax-i12 1450  ax-4 1452  ax-17 1471  ax-i9 1475  ax-ial 1479  ax-i5r 1480

This theorem depends on definitions:  df-bi 116  df-nf 1402  df-sb 1700

This theorem is referenced by:  sblbis  1889  sbrbis  1890  sbco  1897  sbcocom  1899  elsb3  1907  elsb4  1908  sb8eu  1968  sb8euh  1978  pm13.183  2768  sbcbig  2899  sb8iota  5021