Theorem 3jao 1389

Description: Disjunction of three antecedents. (Contributed by NM, 8-Apr-1994.)

Assertion

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

Proof of Theorem 3jao

Step	Hyp	Ref	Expression
1		df-3or 1038	. 2 |- ( ( ph \/ ch \/ th ) <-> ( ( ph \/ ch ) \/ th ) )
2		jao 534	. . . 4 |- ( ( ph -> ps ) -> ( ( ch -> ps ) -> ( ( ph \/ ch ) -> ps ) ) )
3		jao 534	. . . 4 |- ( ( ( ph \/ ch ) -> ps ) -> ( ( th -> ps ) -> ( ( ( ph \/ ch ) \/ th ) -> ps ) ) )
4	2, 3	syl6 35	. . 3 |- ( ( ph -> ps ) -> ( ( ch -> ps ) -> ( ( th -> ps ) -> ( ( ( ph \/ ch ) \/ th ) -> ps ) ) ) )
5	4	3imp 1256	. 2 |- ( ( ( ph -> ps ) /\ ( ch -> ps ) /\ ( th -> ps ) ) -> ( ( ( ph \/ ch ) \/ th ) -> ps ) )
6	1, 5	syl5bi 232	1 |- ( ( ( ph -> ps ) /\ ( ch -> ps ) /\ ( th -> ps ) ) -> ( ( ph \/ ch \/ th ) -> ps ) )

Syntax hints:   -> wi 4   \/ wo 383   \/ w3o 1036   /\ w3a 1037

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8

This theorem depends on definitions:  df-bi 197  df-or 385  df-an 386  df-3or 1038  df-3an 1039

This theorem is referenced by:  3jaob  1390  3jaoi  1391  3jaod  1392