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Metamath Proof Explorer
Description: Deduction from equality to inequality. (Contributed by NM, 2-Jun-2007)
|
|
Ref |
Expression |
|
Hypothesis |
necon3bbid.1 |
⊢ ( 𝜑 → ( 𝜓 ↔ 𝐴 = 𝐵 ) ) |
|
Assertion |
necon3bbid |
⊢ ( 𝜑 → ( ¬ 𝜓 ↔ 𝐴 ≠ 𝐵 ) ) |
Proof
| Step |
Hyp |
Ref |
Expression |
| 1 |
|
necon3bbid.1 |
⊢ ( 𝜑 → ( 𝜓 ↔ 𝐴 = 𝐵 ) ) |
| 2 |
1
|
bicomd |
⊢ ( 𝜑 → ( 𝐴 = 𝐵 ↔ 𝜓 ) ) |
| 3 |
2
|
necon3abid |
⊢ ( 𝜑 → ( 𝐴 ≠ 𝐵 ↔ ¬ 𝜓 ) ) |
| 4 |
3
|
bicomd |
⊢ ( 𝜑 → ( ¬ 𝜓 ↔ 𝐴 ≠ 𝐵 ) ) |