Theorem sb3an 1875

Description: Conjunction inside and outside of a substitution are equivalent. (Contributed by NM, 14-Dec-2006.)

Assertion

Ref	Expression
sb3an	|- ( [ y / x ] ( ph /\ ps /\ ch ) <-> ( [ y / x ] ph /\ [ y / x ] ps /\ [ y / x ] ch ) )

Proof of Theorem sb3an

Step	Hyp	Ref	Expression
1		df-3an 922	. . 3 |- ( ( ph /\ ps /\ ch ) <-> ( ( ph /\ ps ) /\ ch ) )
2	1	sbbii 1690	. 2 |- ( [ y / x ] ( ph /\ ps /\ ch ) <-> [ y / x ] ( ( ph /\ ps ) /\ ch ) )
3		sban 1872	. 2 |- ( [ y / x ] ( ( ph /\ ps ) /\ ch ) <-> ( [ y / x ] ( ph /\ ps ) /\ [ y / x ] ch ) )
4		sban 1872	. . . 4 |- ( [ y / x ] ( ph /\ ps ) <-> ( [ y / x ] ph /\ [ y / x ] ps ) )
5	4	anbi1i 446	. . 3 |- ( ( [ y / x ] ( ph /\ ps ) /\ [ y / x ] ch ) <-> ( ( [ y / x ] ph /\ [ y / x ] ps ) /\ [ y / x ] ch ) )
6		df-3an 922	. . 3 |- ( ( [ y / x ] ph /\ [ y / x ] ps /\ [ y / x ] ch ) <-> ( ( [ y / x ] ph /\ [ y / x ] ps ) /\ [ y / x ] ch ) )
7	5, 6	bitr4i 185	. 2 |- ( ( [ y / x ] ( ph /\ ps ) /\ [ y / x ] ch ) <-> ( [ y / x ] ph /\ [ y / x ] ps /\ [ y / x ] ch ) )
8	2, 3, 7	3bitri 204	1 |- ( [ y / x ] ( ph /\ ps /\ ch ) <-> ( [ y / x ] ph /\ [ y / x ] ps /\ [ y / x ] ch ) )

Syntax hints:   /\ wa 102   <-> wb 103   /\ w3a 920   [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-3an 922  df-nf 1391  df-sb 1688

This theorem is referenced by: (None)