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 e. CC |-> ( x - A ) )
Assertion	sub1cncf	|- ( A e. CC -> F e. ( CC -cn-> CC ) )

Proof

Step	Hyp	Ref	Expression
1		sub1cncf.1	|- F = ( x e. CC |-> ( x - A ) )
2		eqid	|- ( TopOpen ` CCfld ) = ( TopOpen ` CCfld )
3	2	subcn	|- - e. ( ( ( TopOpen ` CCfld ) tX ( TopOpen ` CCfld ) ) Cn ( TopOpen ` CCfld ) )
4	3	a1i	|- ( A e. CC -> - e. ( ( ( TopOpen ` CCfld ) tX ( TopOpen ` CCfld ) ) Cn ( TopOpen ` CCfld ) ) )
5		eqid	|- ( x e. CC |-> x ) = ( x e. CC |-> x )
6	5	idcncf	|- ( x e. CC |-> x ) e. ( CC -cn-> CC )
7	6	a1i	|- ( A e. CC -> ( x e. CC |-> x ) e. ( CC -cn-> CC ) )
8		ssid	|- CC C_ CC
9		cncfmptc	|- ( ( A e. CC /\ CC C_ CC /\ CC C_ CC ) -> ( x e. CC |-> A ) e. ( CC -cn-> CC ) )
10	8 8 9	mp3an23	|- ( A e. CC -> ( x e. CC |-> A ) e. ( CC -cn-> CC ) )
11	2 4 7 10	cncfmpt2f	|- ( A e. CC -> ( x e. CC |-> ( x - A ) ) e. ( CC -cn-> CC ) )
12	1 11	eqeltrid	|- ( A e. CC -> F e. ( CC -cn-> CC ) )