Theorem 3jao 883

Description: Disjunction of 3 antecedents.

Assertion

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

Proof of Theorem 3jao

Step	Hyp	Ref	Expression
1		jao 340	. . . 4 |- ((ph -> ps) -> ((ch -> ps) -> ((ph \/ ch) -> ps)))
2		jao 340	. . . 4 |- (((ph \/ ch) -> ps) -> ((th -> ps) -> (((ph \/ ch) \/ th) -> ps)))
3	1, 2	syl6 22	. . 3 |- ((ph -> ps) -> ((ch -> ps) -> ((th -> ps) -> (((ph \/ ch) \/ th) -> ps))))
4	3	3imp 825	. 2 |- (((ph -> ps) /\ (ch -> ps) /\ (th -> ps)) -> (((ph \/ ch) \/ th) -> ps))
5		df-3or 774	. 2 |- ((ph \/ ch \/ th) <-> ((ph \/ ch) \/ th))
6	4, 5	syl5ib 206	1 |- (((ph -> ps) /\ (ch -> ps) /\ (th -> ps)) -> ((ph \/ ch \/ th) -> ps))

Syntax hints:   -> wi 3   \/ wo 222   \/ w3o 772   /\ w3a 773

This theorem is referenced by:  3jaoi 884  3jaod 885  tpss 2467  fr3nr 2916

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7

This theorem depends on definitions:  df-bi 147  df-or 224  df-an 225  df-3or 774  df-3an 775

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