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Metamath Proof Explorer


Theorem necon3abii

Description: Deduction from equality to inequality. (Contributed by NM, 9-Nov-2007)

Ref Expression
Hypothesis necon3abii.1
|- ( A = B <-> ph )
Assertion necon3abii
|- ( A =/= B <-> -. ph )

Proof

Step Hyp Ref Expression
1 necon3abii.1
 |-  ( A = B <-> ph )
2 df-ne
 |-  ( A =/= B <-> -. A = B )
3 2 1 xchbinx
 |-  ( A =/= B <-> -. ph )