Theorem sbbi 1876

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 1690	. 2 |- ( [ y / x ] ( ph <-> ps ) <-> [ y / x ] ( ( ph -> ps ) /\ ( ps -> ph ) ) )
3		sbim 1870	. . . 4 |- ( [ y / x ] ( ph -> ps ) <-> ( [ y / x ] ph -> [ y / x ] ps ) )
4		sbim 1870	. . . 4 |- ( [ y / x ] ( ps -> ph ) <-> ( [ y / x ] ps -> [ y / x ] ph ) )
5	3, 4	anbi12i 448	. . 3 |- ( ( [ y / x ] ( ph -> ps ) /\ [ y / x ] ( ps -> ph ) ) <-> ( ( [ y / x ] ph -> [ y / x ] ps ) /\ ( [ y / x ] ps -> [ y / x ] ph ) ) )
6		sban 1872	. . 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 1687

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 663  ax-5 1377  ax-7 1378  ax-gen 1379  ax-ie1 1423  ax-ie2 1424  ax-8 1436  ax-10 1437  ax-11 1438  ax-i12 1439  ax-4 1441  ax-17 1460  ax-i9 1464  ax-ial 1468  ax-i5r 1469

This theorem depends on definitions:  df-bi 115  df-nf 1391  df-sb 1688

This theorem is referenced by:  sblbis  1877  sbrbis  1878  sbco  1885  sbcocom  1887  elsb3  1895  elsb4  1896  sb8eu  1956  sb8euh  1966  pm13.183  2741  sbcbig  2870  sb8iota  4925