Metamath Proof Explorer


Theorem sub1cncf

Description: Subtracting a constant is a continuous function. (Contributed by Jeff Madsen, 2-Sep-2009) (Proof shortened by Mario Carneiro, 12-Sep-2015)

Ref Expression
Hypothesis sub1cncf.1 F = x x A
Assertion sub1cncf A F : cn

Proof

Step Hyp Ref Expression
1 sub1cncf.1 F = x x A
2 eqid TopOpen fld = TopOpen fld
3 2 subcn TopOpen fld × t TopOpen fld Cn TopOpen fld
4 3 a1i A TopOpen fld × t TopOpen fld Cn TopOpen fld
5 eqid x x = x x
6 5 idcncf x x : cn
7 6 a1i A x x : cn
8 ssid
9 cncfmptc A x A : cn
10 8 8 9 mp3an23 A x A : cn
11 2 4 7 10 cncfmpt2f A x x A : cn
12 1 11 eqeltrid A F : cn