Metamath Proof Explorer


Theorem negsubdi2d

Description: Distribution of negative over subtraction. (Contributed by Mario Carneiro, 27-May-2016)

Ref Expression
Hypotheses negidd.1
|- ( ph -> A e. CC )
pncand.2
|- ( ph -> B e. CC )
Assertion negsubdi2d
|- ( ph -> -u ( A - B ) = ( B - A ) )

Proof

Step Hyp Ref Expression
1 negidd.1
 |-  ( ph -> A e. CC )
2 pncand.2
 |-  ( ph -> B e. CC )
3 negsubdi2
 |-  ( ( A e. CC /\ B e. CC ) -> -u ( A - B ) = ( B - A ) )
4 1 2 3 syl2anc
 |-  ( ph -> -u ( A - B ) = ( B - A ) )