Metamath Proof Explorer
Description: Contrapositive law deduction for inequality. (Contributed by NM, 28-Dec-2008) (Proof shortened by Andrew Salmon, 25-May-2011)
|
|
Ref |
Expression |
|
Hypothesis |
necon1d.1 |
⊢ ( 𝜑 → ( 𝐴 ≠ 𝐵 → 𝐶 = 𝐷 ) ) |
|
Assertion |
necon1d |
⊢ ( 𝜑 → ( 𝐶 ≠ 𝐷 → 𝐴 = 𝐵 ) ) |
Proof
Step |
Hyp |
Ref |
Expression |
1 |
|
necon1d.1 |
⊢ ( 𝜑 → ( 𝐴 ≠ 𝐵 → 𝐶 = 𝐷 ) ) |
2 |
|
nne |
⊢ ( ¬ 𝐶 ≠ 𝐷 ↔ 𝐶 = 𝐷 ) |
3 |
1 2
|
syl6ibr |
⊢ ( 𝜑 → ( 𝐴 ≠ 𝐵 → ¬ 𝐶 ≠ 𝐷 ) ) |
4 |
3
|
necon4ad |
⊢ ( 𝜑 → ( 𝐶 ≠ 𝐷 → 𝐴 = 𝐵 ) ) |