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 ) |