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
|
syl5bi |
⊢ ( 𝜑 → ( ¬ 𝐴 ≠ 𝐵 → 𝜓 ) ) |
4 |
3
|
con1d |
⊢ ( 𝜑 → ( ¬ 𝜓 → 𝐴 ≠ 𝐵 ) ) |