Theorem 3jaob 1302

Description: Disjunction of 3 antecedents. (Contributed by NM, 13-Sep-2011.)

Assertion

Ref	Expression
3jaob	|- ( ( ( ph \/ ch \/ th ) -> ps ) <-> ( ( ph -> ps ) /\ ( ch -> ps ) /\ ( th -> ps ) ) )

Proof of Theorem 3jaob

Step	Hyp	Ref	Expression
1		3mix1 1166	. . . 4 |- ( ph -> ( ph \/ ch \/ th ) )
2	1	imim1i 60	. . 3 |- ( ( ( ph \/ ch \/ th ) -> ps ) -> ( ph -> ps ) )
3		3mix2 1167	. . . 4 |- ( ch -> ( ph \/ ch \/ th ) )
4	3	imim1i 60	. . 3 |- ( ( ( ph \/ ch \/ th ) -> ps ) -> ( ch -> ps ) )
5		3mix3 1168	. . . 4 |- ( th -> ( ph \/ ch \/ th ) )
6	5	imim1i 60	. . 3 |- ( ( ( ph \/ ch \/ th ) -> ps ) -> ( th -> ps ) )
7	2, 4, 6	3jca 1177	. 2 |- ( ( ( ph \/ ch \/ th ) -> ps ) -> ( ( ph -> ps ) /\ ( ch -> ps ) /\ ( th -> ps ) ) )
8		3jao 1301	. 2 |- ( ( ( ph -> ps ) /\ ( ch -> ps ) /\ ( th -> ps ) ) -> ( ( ph \/ ch \/ th ) -> ps ) )
9	7, 8	impbii 126	1 |- ( ( ( ph \/ ch \/ th ) -> ps ) <-> ( ( ph -> ps ) /\ ( ch -> ps ) /\ ( th -> ps ) ) )

Syntax hints:   -> wi 4   <-> wb 105   \/ w3o 977   /\ w3a 978

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 709

This theorem depends on definitions:  df-bi 117  df-3or 979  df-3an 980

This theorem is referenced by: (None)