Metamath Proof Explorer
Description: Deduction for elimination by cases. (Contributed by NM, 24-May-2013)
|
|
Ref |
Expression |
|
Hypotheses |
ecase3ad.1 |
⊢ ( 𝜑 → ( 𝜓 → 𝜃 ) ) |
|
|
ecase3ad.2 |
⊢ ( 𝜑 → ( 𝜒 → 𝜃 ) ) |
|
|
ecase3ad.3 |
⊢ ( 𝜑 → ( ( ¬ 𝜓 ∧ ¬ 𝜒 ) → 𝜃 ) ) |
|
Assertion |
ecase3ad |
⊢ ( 𝜑 → 𝜃 ) |
Proof
Step |
Hyp |
Ref |
Expression |
1 |
|
ecase3ad.1 |
⊢ ( 𝜑 → ( 𝜓 → 𝜃 ) ) |
2 |
|
ecase3ad.2 |
⊢ ( 𝜑 → ( 𝜒 → 𝜃 ) ) |
3 |
|
ecase3ad.3 |
⊢ ( 𝜑 → ( ( ¬ 𝜓 ∧ ¬ 𝜒 ) → 𝜃 ) ) |
4 |
|
notnotr |
⊢ ( ¬ ¬ 𝜓 → 𝜓 ) |
5 |
4 1
|
syl5 |
⊢ ( 𝜑 → ( ¬ ¬ 𝜓 → 𝜃 ) ) |
6 |
|
notnotr |
⊢ ( ¬ ¬ 𝜒 → 𝜒 ) |
7 |
6 2
|
syl5 |
⊢ ( 𝜑 → ( ¬ ¬ 𝜒 → 𝜃 ) ) |
8 |
5 7 3
|
ecased |
⊢ ( 𝜑 → 𝜃 ) |