Metamath Proof Explorer
Description: Deduction eliminating disjunct. (Contributed by Thierry Arnoux, 19-Dec-2016)
|
|
Ref |
Expression |
|
Hypothesis |
3o1cs.1 |
⊢ ( ( 𝜑 ∨ 𝜓 ∨ 𝜒 ) → 𝜃 ) |
|
Assertion |
3o1cs |
⊢ ( 𝜑 → 𝜃 ) |
Proof
| Step |
Hyp |
Ref |
Expression |
| 1 |
|
3o1cs.1 |
⊢ ( ( 𝜑 ∨ 𝜓 ∨ 𝜒 ) → 𝜃 ) |
| 2 |
|
df-3or |
⊢ ( ( 𝜑 ∨ 𝜓 ∨ 𝜒 ) ↔ ( ( 𝜑 ∨ 𝜓 ) ∨ 𝜒 ) ) |
| 3 |
2 1
|
sylbir |
⊢ ( ( ( 𝜑 ∨ 𝜓 ) ∨ 𝜒 ) → 𝜃 ) |
| 4 |
3
|
orcs |
⊢ ( ( 𝜑 ∨ 𝜓 ) → 𝜃 ) |
| 5 |
4
|
orcs |
⊢ ( 𝜑 → 𝜃 ) |