Theorem 3jao 632

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 274	. . . 4 |- ((ph -> ps) -> ((ch -> ps) -> ((ph \/ ch) -> ps)))
2		jao 274	. . . 4 |- (((ph \/ ch) -> ps) -> ((th -> ps) -> (((ph \/ ch) \/ th) -> ps)))
3	1, 2	syl6 23	. . 3 |- ((ph -> ps) -> ((ch -> ps) -> ((th -> ps) -> (((ph \/ ch) \/ th) -> ps))))
4	3	3imp 608	. 2 |- (((ph -> ps) /\ (ch -> ps) /\ (th -> ps)) -> (((ph \/ ch) \/ th) -> ps))
5		df-3or 582	. 2 |- ((ph \/ ch \/ th) <-> ((ph \/ ch) \/ th))
6	4, 5	syl5ib 181	1 |- (((ph -> ps) /\ (ch -> ps) /\ (th -> ps)) -> ((ph \/ ch \/ th) -> ps))

Syntax hints:   -> wi 2   \/ wo 195   \/ w3o 580   /\ w3a 581

This theorem is referenced by:  3jaoi 633  tpss 1855  fr3nr 2178

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

This theorem depends on definitions:  df-bi 128  df-or 197  df-an 198  df-3or 582  df-3an 583

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