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Metamath Proof Explorer
Description: Deduction eliminating an antecedent. (Contributed by NM, 27-Apr-1994)
(Proof shortened by Wolf Lammen, 12-Sep-2013)
|
|
Ref |
Expression |
|
Hypotheses |
pm2.61d.1 |
⊢ ( 𝜑 → ( 𝜓 → 𝜒 ) ) |
|
|
pm2.61d.2 |
⊢ ( 𝜑 → ( ¬ 𝜓 → 𝜒 ) ) |
|
Assertion |
pm2.61d |
⊢ ( 𝜑 → 𝜒 ) |
Proof
| Step |
Hyp |
Ref |
Expression |
| 1 |
|
pm2.61d.1 |
⊢ ( 𝜑 → ( 𝜓 → 𝜒 ) ) |
| 2 |
|
pm2.61d.2 |
⊢ ( 𝜑 → ( ¬ 𝜓 → 𝜒 ) ) |
| 3 |
2
|
con1d |
⊢ ( 𝜑 → ( ¬ 𝜒 → 𝜓 ) ) |
| 4 |
3 1
|
syld |
⊢ ( 𝜑 → ( ¬ 𝜒 → 𝜒 ) ) |
| 5 |
4
|
pm2.18d |
⊢ ( 𝜑 → 𝜒 ) |