Theorem 3jaodan 1306

Description: Disjunction of 3 antecedents (deduction). (Contributed by NM, 14-Oct-2005.)

Hypotheses

Ref	Expression
3jaodan.1	|- ( ( ph /\ ps ) -> ch )
3jaodan.2	|- ( ( ph /\ th ) -> ch )
3jaodan.3	|- ( ( ph /\ ta ) -> ch )

Assertion

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

Proof of Theorem 3jaodan

Step	Hyp	Ref	Expression
1		3jaodan.1	. . . 4 |- ( ( ph /\ ps ) -> ch )
2	1	ex 115	. . 3 |- ( ph -> ( ps -> ch ) )
3		3jaodan.2	. . . 4 |- ( ( ph /\ th ) -> ch )
4	3	ex 115	. . 3 |- ( ph -> ( th -> ch ) )
5		3jaodan.3	. . . 4 |- ( ( ph /\ ta ) -> ch )
6	5	ex 115	. . 3 |- ( ph -> ( ta -> ch ) )
7	2, 4, 6	3jaod 1304	. 2 |- ( ph -> ( ( ps \/ th \/ ta ) -> ch ) )
8	7	imp 124	1 |- ( ( ph /\ ( ps \/ th \/ ta ) ) -> ch )

Syntax hints:   -> wi 4   /\ wa 104   \/ w3o 977

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 709

This theorem depends on definitions:  df-bi 117  df-3or 979  df-3an 980

This theorem is referenced by:  zeo  9360  xrltnsym  9795  xrlttr  9797  xrltso  9798  xrlttri3  9799  xltnegi  9837  xaddcom  9863  xnegdi  9870  xsubge0  9883  qbtwnxr  10260  blssioo  14130