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