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Metamath Proof Explorer
Description: Deduction eliminating an inequality in an antecedent. (Contributed by NM, 1-Jun-2007) (Proof shortened by Andrew Salmon, 25-May-2011)
|
|
Ref |
Expression |
|
Hypotheses |
pm2.61dne.1 |
⊢ ( 𝜑 → ( 𝐴 = 𝐵 → 𝜓 ) ) |
|
|
pm2.61dne.2 |
⊢ ( 𝜑 → ( 𝐴 ≠ 𝐵 → 𝜓 ) ) |
|
Assertion |
pm2.61dne |
⊢ ( 𝜑 → 𝜓 ) |
Proof
| Step |
Hyp |
Ref |
Expression |
| 1 |
|
pm2.61dne.1 |
⊢ ( 𝜑 → ( 𝐴 = 𝐵 → 𝜓 ) ) |
| 2 |
|
pm2.61dne.2 |
⊢ ( 𝜑 → ( 𝐴 ≠ 𝐵 → 𝜓 ) ) |
| 3 |
1
|
com12 |
⊢ ( 𝐴 = 𝐵 → ( 𝜑 → 𝜓 ) ) |
| 4 |
2
|
com12 |
⊢ ( 𝐴 ≠ 𝐵 → ( 𝜑 → 𝜓 ) ) |
| 5 |
3 4
|
pm2.61ine |
⊢ ( 𝜑 → 𝜓 ) |