Metamath Proof Explorer


Theorem absabv

Description: The regular absolute value function on the complex numbers is in fact an absolute value under our definition. (Contributed by Mario Carneiro, 4-Dec-2014)

Ref Expression
Assertion absabv abs ∈ ( AbsVal ‘ ℂfld )

Proof

Step Hyp Ref Expression
1 eqidd ( ⊤ → ( AbsVal ‘ ℂfld ) = ( AbsVal ‘ ℂfld ) )
2 cnfldbas ℂ = ( Base ‘ ℂfld )
3 2 a1i ( ⊤ → ℂ = ( Base ‘ ℂfld ) )
4 cnfldadd + = ( +g ‘ ℂfld )
5 4 a1i ( ⊤ → + = ( +g ‘ ℂfld ) )
6 cnfldmul · = ( .r ‘ ℂfld )
7 6 a1i ( ⊤ → · = ( .r ‘ ℂfld ) )
8 cnfld0 0 = ( 0g ‘ ℂfld )
9 8 a1i ( ⊤ → 0 = ( 0g ‘ ℂfld ) )
10 cnring fld ∈ Ring
11 10 a1i ( ⊤ → ℂfld ∈ Ring )
12 absf abs : ℂ ⟶ ℝ
13 12 a1i ( ⊤ → abs : ℂ ⟶ ℝ )
14 abs0 ( abs ‘ 0 ) = 0
15 14 a1i ( ⊤ → ( abs ‘ 0 ) = 0 )
16 absgt0 ( 𝑥 ∈ ℂ → ( 𝑥 ≠ 0 ↔ 0 < ( abs ‘ 𝑥 ) ) )
17 16 biimpa ( ( 𝑥 ∈ ℂ ∧ 𝑥 ≠ 0 ) → 0 < ( abs ‘ 𝑥 ) )
18 17 3adant1 ( ( ⊤ ∧ 𝑥 ∈ ℂ ∧ 𝑥 ≠ 0 ) → 0 < ( abs ‘ 𝑥 ) )
19 absmul ( ( 𝑥 ∈ ℂ ∧ 𝑦 ∈ ℂ ) → ( abs ‘ ( 𝑥 · 𝑦 ) ) = ( ( abs ‘ 𝑥 ) · ( abs ‘ 𝑦 ) ) )
20 19 ad2ant2r ( ( ( 𝑥 ∈ ℂ ∧ 𝑥 ≠ 0 ) ∧ ( 𝑦 ∈ ℂ ∧ 𝑦 ≠ 0 ) ) → ( abs ‘ ( 𝑥 · 𝑦 ) ) = ( ( abs ‘ 𝑥 ) · ( abs ‘ 𝑦 ) ) )
21 20 3adant1 ( ( ⊤ ∧ ( 𝑥 ∈ ℂ ∧ 𝑥 ≠ 0 ) ∧ ( 𝑦 ∈ ℂ ∧ 𝑦 ≠ 0 ) ) → ( abs ‘ ( 𝑥 · 𝑦 ) ) = ( ( abs ‘ 𝑥 ) · ( abs ‘ 𝑦 ) ) )
22 abstri ( ( 𝑥 ∈ ℂ ∧ 𝑦 ∈ ℂ ) → ( abs ‘ ( 𝑥 + 𝑦 ) ) ≤ ( ( abs ‘ 𝑥 ) + ( abs ‘ 𝑦 ) ) )
23 22 ad2ant2r ( ( ( 𝑥 ∈ ℂ ∧ 𝑥 ≠ 0 ) ∧ ( 𝑦 ∈ ℂ ∧ 𝑦 ≠ 0 ) ) → ( abs ‘ ( 𝑥 + 𝑦 ) ) ≤ ( ( abs ‘ 𝑥 ) + ( abs ‘ 𝑦 ) ) )
24 23 3adant1 ( ( ⊤ ∧ ( 𝑥 ∈ ℂ ∧ 𝑥 ≠ 0 ) ∧ ( 𝑦 ∈ ℂ ∧ 𝑦 ≠ 0 ) ) → ( abs ‘ ( 𝑥 + 𝑦 ) ) ≤ ( ( abs ‘ 𝑥 ) + ( abs ‘ 𝑦 ) ) )
25 1 3 5 7 9 11 13 15 18 21 24 isabvd ( ⊤ → abs ∈ ( AbsVal ‘ ℂfld ) )
26 25 mptru abs ∈ ( AbsVal ‘ ℂfld )