Theorem sb3an 1929

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 964	. . 3 |- ( ( ph /\ ps /\ ch ) <-> ( ( ph /\ ps ) /\ ch ) )
2	1	sbbii 1738	. 2 |- ( [ y / x ] ( ph /\ ps /\ ch ) <-> [ y / x ] ( ( ph /\ ps ) /\ ch ) )
3		sban 1926	. 2 |- ( [ y / x ] ( ( ph /\ ps ) /\ ch ) <-> ( [ y / x ] ( ph /\ ps ) /\ [ y / x ] ch ) )
4		sban 1926	. . . 4 |- ( [ y / x ] ( ph /\ ps ) <-> ( [ y / x ] ph /\ [ y / x ] ps ) )
5	4	anbi1i 453	. . 3 |- ( ( [ y / x ] ( ph /\ ps ) /\ [ y / x ] ch ) <-> ( ( [ y / x ] ph /\ [ y / x ] ps ) /\ [ y / x ] ch ) )
6		df-3an 964	. . 3 |- ( ( [ y / x ] ph /\ [ y / x ] ps /\ [ y / x ] ch ) <-> ( ( [ y / x ] ph /\ [ y / x ] ps ) /\ [ y / x ] ch ) )
7	5, 6	bitr4i 186	. 2 |- ( ( [ y / x ] ( ph /\ ps ) /\ [ y / x ] ch ) <-> ( [ y / x ] ph /\ [ y / x ] ps /\ [ y / x ] ch ) )
8	2, 3, 7	3bitri 205	1 |- ( [ y / x ] ( ph /\ ps /\ ch ) <-> ( [ y / x ] ph /\ [ y / x ] ps /\ [ y / x ] ch ) )

Syntax hints:   /\ wa 103   <-> wb 104   /\ w3a 962   [wsb 1735

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-io 698  ax-5 1423  ax-7 1424  ax-gen 1425  ax-ie1 1469  ax-ie2 1470  ax-8 1482  ax-10 1483  ax-11 1484  ax-i12 1485  ax-4 1487  ax-17 1506  ax-i9 1510  ax-ial 1514  ax-i5r 1515

This theorem depends on definitions:  df-bi 116  df-3an 964  df-nf 1437  df-sb 1736

This theorem is referenced by: (None)