Metamath Proof Explorer


Theorem subcncf

Description: The addition of two continuous complex functions is continuous. (Contributed by Glauco Siliprandi, 11-Dec-2019)

Ref Expression
Hypotheses subcncf.a
|- ( ph -> ( x e. X |-> A ) e. ( X -cn-> CC ) )
subcncf.b
|- ( ph -> ( x e. X |-> B ) e. ( X -cn-> CC ) )
Assertion subcncf
|- ( ph -> ( x e. X |-> ( A - B ) ) e. ( X -cn-> CC ) )

Proof

Step Hyp Ref Expression
1 subcncf.a
 |-  ( ph -> ( x e. X |-> A ) e. ( X -cn-> CC ) )
2 subcncf.b
 |-  ( ph -> ( x e. X |-> B ) e. ( X -cn-> CC ) )
3 eqid
 |-  ( TopOpen ` CCfld ) = ( TopOpen ` CCfld )
4 3 subcn
 |-  - e. ( ( ( TopOpen ` CCfld ) tX ( TopOpen ` CCfld ) ) Cn ( TopOpen ` CCfld ) )
5 4 a1i
 |-  ( ph -> - e. ( ( ( TopOpen ` CCfld ) tX ( TopOpen ` CCfld ) ) Cn ( TopOpen ` CCfld ) ) )
6 3 5 1 2 cncfmpt2f
 |-  ( ph -> ( x e. X |-> ( A - B ) ) e. ( X -cn-> CC ) )