Metamath Proof Explorer


Theorem negsubd

Description: Relationship between subtraction and negative. Theorem I.3 of Apostol p. 18. (Contributed by Mario Carneiro, 27-May-2016)

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

Proof

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