Theorem jaob 328

Description: Disjunction of antecedents. Compare Theorem *4.77 of [WhiteheadRussell] p. 121.

Assertion

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

Proof of Theorem jaob

Step	Hyp	Ref	Expression
1		orc 225	. . . 4 |- (ph -> (ph \/ ch))
2	1	syl4 19	. . 3 |- (((ph \/ ch) -> ps) -> (ph -> ps))
3		olc 224	. . . 4 |- (ch -> (ph \/ ch))
4	3	syl4 19	. . 3 |- (((ph \/ ch) -> ps) -> (ch -> ps))
5	2, 4	jca 236	. 2 |- (((ph \/ ch) -> ps) -> ((ph -> ps) /\ (ch -> ps)))
6		jao 274	. . 3 |- ((ph -> ps) -> ((ch -> ps) -> ((ph \/ ch) -> ps)))
7	6	imp 277	. 2 |- (((ph -> ps) /\ (ch -> ps)) -> ((ph \/ ch) -> ps))
8	5, 7	impbi 139	1 |- (((ph \/ ch) -> ps) <-> ((ph -> ps) /\ (ch -> ps)))

Syntax hints:   -> wi 2   <-> wb 127   \/ wo 195   /\ wa 196

This theorem is referenced by:  unss 1632  prsspw 1858  intun 1989  intpr 1990  ordsseleq 2227

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

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