Metamath Proof Explorer
Description: Deduction for elimination by cases. (Contributed by NM, 2-May-1996)
(Proof shortened by Andrew Salmon, 7-May-2011)
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Ref |
Expression |
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Hypotheses |
ecase3d.1 |
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|
ecase3d.2 |
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|
ecase3d.3 |
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Assertion |
ecase3d |
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Proof
Step |
Hyp |
Ref |
Expression |
1 |
|
ecase3d.1 |
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2 |
|
ecase3d.2 |
|
3 |
|
ecase3d.3 |
|
4 |
1 2
|
jaod |
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5 |
4 3
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pm2.61d |
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