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Metamath Proof Explorer

Theorem necom

Description: Commutation of inequality. (Contributed by NM, 14-May-1999)

Ref Expression
Assertion necom 
|- ( A =/= B <-> B =/= A )

Proof

Step Hyp Ref Expression
1 eqcom 
 |-  ( A = B <-> B = A )
2 1 necon3bii 
 |-  ( A =/= B <-> B =/= A )