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 |
|
nne |
⊢ ( ¬ 𝐴 ≠ 𝐵 ↔ 𝐴 = 𝐵 ) |
4 |
3 1
|
syl5bi |
⊢ ( 𝜑 → ( ¬ 𝐴 ≠ 𝐵 → 𝜓 ) ) |
5 |
2 4
|
pm2.61d |
⊢ ( 𝜑 → 𝜓 ) |