Metamath Proof Explorer


Theorem 3ecase

Description: Inference for elimination by cases. (Contributed by NM, 13-Jul-2005)

Ref Expression
Hypotheses 3ecase.1
|- ( -. ph -> th )
3ecase.2
|- ( -. ps -> th )
3ecase.3
|- ( -. ch -> th )
3ecase.4
|- ( ( ph /\ ps /\ ch ) -> th )
Assertion 3ecase
|- th

Proof

Step Hyp Ref Expression
1 3ecase.1
 |-  ( -. ph -> th )
2 3ecase.2
 |-  ( -. ps -> th )
3 3ecase.3
 |-  ( -. ch -> th )
4 3ecase.4
 |-  ( ( ph /\ ps /\ ch ) -> th )
5 4 3exp
 |-  ( ph -> ( ps -> ( ch -> th ) ) )
6 1 2a1d
 |-  ( -. ph -> ( ps -> ( ch -> th ) ) )
7 5 6 pm2.61i
 |-  ( ps -> ( ch -> th ) )
8 7 2 3 pm2.61nii
 |-  th