Metamath Proof Explorer


Theorem subcld

Description: Closure law for subtraction. (Contributed by Mario Carneiro, 27-May-2016)

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

Proof

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