Theorem 3jaob 1159

Description: Disjunction of 3 antecedents.

Assertion

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

Proof of Theorem 3jaob

Step	Hyp	Ref	Expression
1		3mix1 1045	. . . 4 |- (ph -> (ph \/ ch \/ th))
2	1	imim1i 19	. . 3 |- (((ph \/ ch \/ th) -> ps) -> (ph -> ps))
3		3mix2 1046	. . . 4 |- (ch -> (ph \/ ch \/ th))
4	3	imim1i 19	. . 3 |- (((ph \/ ch \/ th) -> ps) -> (ch -> ps))
5		3mix3 1047	. . . 4 |- (th -> (ph \/ ch \/ th))
6	5	imim1i 19	. . 3 |- (((ph \/ ch \/ th) -> ps) -> (th -> ps))
7	2, 4, 6	3jca 1050	. 2 |- (((ph \/ ch \/ th) -> ps) -> ((ph -> ps) /\ (ch -> ps) /\ (th -> ps)))
8		3jao 1158	. 2 |- (((ph -> ps) /\ (ch -> ps) /\ (th -> ps)) -> ((ph \/ ch \/ th) -> ps))
9	7, 8	impbii 174	1 |- (((ph \/ ch \/ th) -> ps) <-> ((ph -> ps) /\ (ch -> ps) /\ (th -> ps)))

Syntax hints:   -> wi 3   <-> wb 163   \/ w3o 857   /\ w3a 858

This theorem is referenced by:  raltp 3083

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 164  df-or 241  df-an 242  df-3or 859  df-3an 860

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