Metamath Proof Explorer
Description: Contrapositive law deduction for inequality. (Contributed by NM, 2-Apr-2007) (Proof shortened by Andrew Salmon, 25-May-2011)
|
|
Ref |
Expression |
|
Hypothesis |
necon3bd.1 |
⊢ ( 𝜑 → ( 𝐴 = 𝐵 → 𝜓 ) ) |
|
Assertion |
necon3bd |
⊢ ( 𝜑 → ( ¬ 𝜓 → 𝐴 ≠ 𝐵 ) ) |
Proof
| Step |
Hyp |
Ref |
Expression |
| 1 |
|
necon3bd.1 |
⊢ ( 𝜑 → ( 𝐴 = 𝐵 → 𝜓 ) ) |
| 2 |
|
nne |
⊢ ( ¬ 𝐴 ≠ 𝐵 ↔ 𝐴 = 𝐵 ) |
| 3 |
2 1
|
biimtrid |
⊢ ( 𝜑 → ( ¬ 𝐴 ≠ 𝐵 → 𝜓 ) ) |
| 4 |
3
|
con1d |
⊢ ( 𝜑 → ( ¬ 𝜓 → 𝐴 ≠ 𝐵 ) ) |