Metamath Proof Explorer


Theorem sub1cncfd

Description: Subtracting a constant is a continuous function. (Contributed by Glauco Siliprandi, 8-Apr-2021)

Ref Expression
Hypotheses sub1cncfd.1 ( 𝜑𝐴 ∈ ℂ )
sub1cncfd.2 𝐹 = ( 𝑥 ∈ ℂ ↦ ( 𝑥𝐴 ) )
Assertion sub1cncfd ( 𝜑𝐹 ∈ ( ℂ –cn→ ℂ ) )

Proof

Step Hyp Ref Expression
1 sub1cncfd.1 ( 𝜑𝐴 ∈ ℂ )
2 sub1cncfd.2 𝐹 = ( 𝑥 ∈ ℂ ↦ ( 𝑥𝐴 ) )
3 ssid ℂ ⊆ ℂ
4 cncfmptid ( ( ℂ ⊆ ℂ ∧ ℂ ⊆ ℂ ) → ( 𝑥 ∈ ℂ ↦ 𝑥 ) ∈ ( ℂ –cn→ ℂ ) )
5 3 3 4 mp2an ( 𝑥 ∈ ℂ ↦ 𝑥 ) ∈ ( ℂ –cn→ ℂ )
6 5 a1i ( 𝜑 → ( 𝑥 ∈ ℂ ↦ 𝑥 ) ∈ ( ℂ –cn→ ℂ ) )
7 3 a1i ( 𝜑 → ℂ ⊆ ℂ )
8 cncfmptc ( ( 𝐴 ∈ ℂ ∧ ℂ ⊆ ℂ ∧ ℂ ⊆ ℂ ) → ( 𝑥 ∈ ℂ ↦ 𝐴 ) ∈ ( ℂ –cn→ ℂ ) )
9 1 7 7 8 syl3anc ( 𝜑 → ( 𝑥 ∈ ℂ ↦ 𝐴 ) ∈ ( ℂ –cn→ ℂ ) )
10 6 9 subcncf ( 𝜑 → ( 𝑥 ∈ ℂ ↦ ( 𝑥𝐴 ) ) ∈ ( ℂ –cn→ ℂ ) )
11 2 10 eqeltrid ( 𝜑𝐹 ∈ ( ℂ –cn→ ℂ ) )