Metamath Proof Explorer
Description: Deduction for elimination by cases. (Contributed by NM, 21-Apr-1994)
(Proof shortened by Wolf Lammen, 19-Sep-2024)
|
|
Ref |
Expression |
|
Hypotheses |
ecase2d.1 |
⊢ ( 𝜑 → 𝜓 ) |
|
|
ecase2d.2 |
⊢ ( 𝜑 → ¬ ( 𝜓 ∧ 𝜒 ) ) |
|
|
ecase2d.3 |
⊢ ( 𝜑 → ¬ ( 𝜓 ∧ 𝜃 ) ) |
|
|
ecase2d.4 |
⊢ ( 𝜑 → ( 𝜏 ∨ ( 𝜒 ∨ 𝜃 ) ) ) |
|
Assertion |
ecase2d |
⊢ ( 𝜑 → 𝜏 ) |
Proof
Step |
Hyp |
Ref |
Expression |
1 |
|
ecase2d.1 |
⊢ ( 𝜑 → 𝜓 ) |
2 |
|
ecase2d.2 |
⊢ ( 𝜑 → ¬ ( 𝜓 ∧ 𝜒 ) ) |
3 |
|
ecase2d.3 |
⊢ ( 𝜑 → ¬ ( 𝜓 ∧ 𝜃 ) ) |
4 |
|
ecase2d.4 |
⊢ ( 𝜑 → ( 𝜏 ∨ ( 𝜒 ∨ 𝜃 ) ) ) |
5 |
1 2
|
mpnanrd |
⊢ ( 𝜑 → ¬ 𝜒 ) |
6 |
1 3
|
mpnanrd |
⊢ ( 𝜑 → ¬ 𝜃 ) |
7 |
4
|
ord |
⊢ ( 𝜑 → ( ¬ 𝜏 → ( 𝜒 ∨ 𝜃 ) ) ) |
8 |
5 6 7
|
mtord |
⊢ ( 𝜑 → ¬ ¬ 𝜏 ) |
9 |
8
|
notnotrd |
⊢ ( 𝜑 → 𝜏 ) |