Theorem sb3an 2011

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 1007	. . 3 |- ( ( ph /\ ps /\ ch ) <-> ( ( ph /\ ps ) /\ ch ) )
2	1	sbbii 1813	. 2 |- ( [ y / x ] ( ph /\ ps /\ ch ) <-> [ y / x ] ( ( ph /\ ps ) /\ ch ) )
3		sban 2008	. 2 |- ( [ y / x ] ( ( ph /\ ps ) /\ ch ) <-> ( [ y / x ] ( ph /\ ps ) /\ [ y / x ] ch ) )
4		sban 2008	. . . 4 |- ( [ y / x ] ( ph /\ ps ) <-> ( [ y / x ] ph /\ [ y / x ] ps ) )
5	4	anbi1i 458	. . 3 |- ( ( [ y / x ] ( ph /\ ps ) /\ [ y / x ] ch ) <-> ( ( [ y / x ] ph /\ [ y / x ] ps ) /\ [ y / x ] ch ) )
6		df-3an 1007	. . 3 |- ( ( [ y / x ] ph /\ [ y / x ] ps /\ [ y / x ] ch ) <-> ( ( [ y / x ] ph /\ [ y / x ] ps ) /\ [ y / x ] ch ) )
7	5, 6	bitr4i 187	. 2 |- ( ( [ y / x ] ( ph /\ ps ) /\ [ y / x ] ch ) <-> ( [ y / x ] ph /\ [ y / x ] ps /\ [ y / x ] ch ) )
8	2, 3, 7	3bitri 206	1 |- ( [ y / x ] ( ph /\ ps /\ ch ) <-> ( [ y / x ] ph /\ [ y / x ] ps /\ [ y / x ] ch ) )

Syntax hints:   /\ wa 104   <-> wb 105   /\ w3a 1005   [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 717  ax-5 1496  ax-7 1497  ax-gen 1498  ax-ie1 1542  ax-ie2 1543  ax-8 1553  ax-10 1554  ax-11 1555  ax-i12 1556  ax-4 1559  ax-17 1575  ax-i9 1579  ax-ial 1583  ax-i5r 1584

This theorem depends on definitions:  df-bi 117  df-3an 1007  df-nf 1510  df-sb 1811

This theorem is referenced by: (None)