Metamath Proof Explorer
Description: Inference from equality to inequality. (Contributed by NM, 23-Feb-2005)
|
|
Ref |
Expression |
|
Hypothesis |
necon3bii.1 |
⊢ ( 𝐴 = 𝐵 ↔ 𝐶 = 𝐷 ) |
|
Assertion |
necon3bii |
⊢ ( 𝐴 ≠ 𝐵 ↔ 𝐶 ≠ 𝐷 ) |
Proof
| Step |
Hyp |
Ref |
Expression |
| 1 |
|
necon3bii.1 |
⊢ ( 𝐴 = 𝐵 ↔ 𝐶 = 𝐷 ) |
| 2 |
1
|
necon3abii |
⊢ ( 𝐴 ≠ 𝐵 ↔ ¬ 𝐶 = 𝐷 ) |
| 3 |
|
df-ne |
⊢ ( 𝐶 ≠ 𝐷 ↔ ¬ 𝐶 = 𝐷 ) |
| 4 |
2 3
|
bitr4i |
⊢ ( 𝐴 ≠ 𝐵 ↔ 𝐶 ≠ 𝐷 ) |