Metamath Proof Explorer


Theorem negcon2

Description: Negative contraposition law. (Contributed by NM, 14-Nov-2004)

Ref Expression
Assertion negcon2
|- ( ( A e. CC /\ B e. CC ) -> ( A = -u B <-> B = -u A ) )

Proof

Step Hyp Ref Expression
1 eqcom
 |-  ( A = -u B <-> -u B = A )
2 negcon1
 |-  ( ( A e. CC /\ B e. CC ) -> ( -u A = B <-> -u B = A ) )
3 1 2 bitr4id
 |-  ( ( A e. CC /\ B e. CC ) -> ( A = -u B <-> -u A = B ) )
4 eqcom
 |-  ( -u A = B <-> B = -u A )
5 3 4 bitrdi
 |-  ( ( A e. CC /\ B e. CC ) -> ( A = -u B <-> B = -u A ) )