Metamath Proof Explorer


Theorem constrimcl

Description: Constructible numbers are closed under taking the imaginary part. (Contributed by Thierry Arnoux, 5-Nov-2025)

Ref Expression
Hypothesis constrcjcl.1 ⊢ φ → X ∈ Constr
Assertion constrimcl ⊢ φ → ℑ ⁡ X ∈ Constr

Proof

Step Hyp Ref Expression
1 constrcjcl.1 ⊢ φ → X ∈ Constr
2 0zd ⊢ φ → 0 ∈ ℤ
3 2 zconstr ⊢ φ → 0 ∈ Constr
4 1zzd ⊢ φ → 1 ∈ ℤ
5 4 zconstr ⊢ φ → 1 ∈ Constr
6 1 constrcn ⊢ φ → X ∈ ℂ
7 6 recld ⊢ φ → ℜ ⁡ X ∈ ℝ
8 7 recnd ⊢ φ → ℜ ⁡ X ∈ ℂ
9 ax-icn ⊢ i ∈ ℂ
10 9 a1i ⊢ φ → i ∈ ℂ
11 6 imcld ⊢ φ → ℑ ⁡ X ∈ ℝ
12 11 recnd ⊢ φ → ℑ ⁡ X ∈ ℂ
13 10 12 mulcld ⊢ φ → i ⁢ ℑ ⁡ X ∈ ℂ
14 6 replimd ⊢ φ → X = ℜ ⁡ X + i ⁢ ℑ ⁡ X
15 8 13 14 mvrladdd ⊢ φ → X − ℜ ⁡ X = i ⁢ ℑ ⁡ X
16 6 8 negsubd ⊢ φ → X + − ℜ ⁡ X = X − ℜ ⁡ X
17 1 constrrecl ⊢ φ → ℜ ⁡ X ∈ Constr
18 17 constrnegcl ⊢ φ → − ℜ ⁡ X ∈ Constr
19 1 18 constraddcl ⊢ φ → X + − ℜ ⁡ X ∈ Constr
20 16 19 eqeltrrd ⊢ φ → X − ℜ ⁡ X ∈ Constr
21 15 20 eqeltrrd ⊢ φ → i ⁢ ℑ ⁡ X ∈ Constr
22 1m0e1 ⊢ 1 − 0 = 1
23 1cnd ⊢ φ → 1 ∈ ℂ
24 22 23 eqeltrid ⊢ φ → 1 − 0 ∈ ℂ
25 12 24 mulcld ⊢ φ → ℑ ⁡ X ⁢ 1 − 0 ∈ ℂ
26 25 addlidd ⊢ φ → 0 + ℑ ⁡ X ⁢ 1 − 0 = ℑ ⁡ X ⁢ 1 − 0
27 22 a1i ⊢ φ → 1 − 0 = 1
28 27 oveq2d ⊢ φ → ℑ ⁡ X ⁢ 1 − 0 = ℑ ⁡ X ⋅ 1
29 12 mulridd ⊢ φ → ℑ ⁡ X ⋅ 1 = ℑ ⁡ X
30 26 28 29 3eqtrrd ⊢ φ → ℑ ⁡ X = 0 + ℑ ⁡ X ⁢ 1 − 0
31 10 12 absmuld ⊢ φ → i ⁢ ℑ ⁡ X = i ⁢ ℑ ⁡ X
32 absi ⊢ i = 1
33 32 a1i ⊢ φ → i = 1
34 33 oveq1d ⊢ φ → i ⁢ ℑ ⁡ X = 1 ⁢ ℑ ⁡ X
35 12 abscld ⊢ φ → ℑ ⁡ X ∈ ℝ
36 35 recnd ⊢ φ → ℑ ⁡ X ∈ ℂ
37 36 mullidd ⊢ φ → 1 ⁢ ℑ ⁡ X = ℑ ⁡ X
38 31 34 37 3eqtrd ⊢ φ → i ⁢ ℑ ⁡ X = ℑ ⁡ X
39 13 subid1d ⊢ φ → i ⁢ ℑ ⁡ X − 0 = i ⁢ ℑ ⁡ X
40 39 fveq2d ⊢ φ → i ⁢ ℑ ⁡ X − 0 = i ⁢ ℑ ⁡ X
41 12 subid1d ⊢ φ → ℑ ⁡ X − 0 = ℑ ⁡ X
42 41 fveq2d ⊢ φ → ℑ ⁡ X − 0 = ℑ ⁡ X
43 38 40 42 3eqtr4rd ⊢ φ → ℑ ⁡ X − 0 = i ⁢ ℑ ⁡ X − 0
44 3 5 3 21 3 11 12 30 43 constrlccl ⊢ φ → ℑ ⁡ X ∈ Constr