Metamath Proof Explorer
Description: Deduction from equality to inequality. (Contributed by NM, 13-Apr-2007)
|
|
Ref |
Expression |
|
Hypothesis |
necon3bbii.1 |
⊢ ( 𝜑 ↔ 𝐴 = 𝐵 ) |
|
Assertion |
necon3bbii |
⊢ ( ¬ 𝜑 ↔ 𝐴 ≠ 𝐵 ) |
Proof
| Step |
Hyp |
Ref |
Expression |
| 1 |
|
necon3bbii.1 |
⊢ ( 𝜑 ↔ 𝐴 = 𝐵 ) |
| 2 |
1
|
bicomi |
⊢ ( 𝐴 = 𝐵 ↔ 𝜑 ) |
| 3 |
2
|
necon3abii |
⊢ ( 𝐴 ≠ 𝐵 ↔ ¬ 𝜑 ) |
| 4 |
3
|
bicomi |
⊢ ( ¬ 𝜑 ↔ 𝐴 ≠ 𝐵 ) |