Description: Deduction from equality to inequality. (Contributed by NM, 23-Feb-2005) (Proof shortened by Andrew Salmon, 25-May-2011)
Ref | Expression | ||
---|---|---|---|
Hypothesis | necon3bid.1 | |- ( ph -> ( A = B <-> C = D ) ) |
|
Assertion | necon3bid | |- ( ph -> ( A =/= B <-> C =/= D ) ) |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | necon3bid.1 | |- ( ph -> ( A = B <-> C = D ) ) |
|
2 | df-ne | |- ( A =/= B <-> -. A = B ) |
|
3 | 1 | necon3bbid | |- ( ph -> ( -. A = B <-> C =/= D ) ) |
4 | 2 3 | syl5bb | |- ( ph -> ( A =/= B <-> C =/= D ) ) |