Metamath Proof Explorer


Theorem cnrecnv

Description: The inverse to the canonical bijection from ( RR X. RR ) to CC from cnref1o . (Contributed by Mario Carneiro, 25-Aug-2014)

Ref Expression
Hypothesis cnrecnv.1 ⊢ F = x ∈ ℝ , y ∈ ℝ ⟼ x + i ⁢ y
Assertion cnrecnv ⊢ F -1 = z ∈ ℂ ⟼ ℜ ⁡ z ℑ ⁡ z

Proof

Step Hyp Ref Expression
1 cnrecnv.1 ⊢ F = x ∈ ℝ , y ∈ ℝ ⟼ x + i ⁢ y
2 1 cnref1o ⊢ F : ℝ 2 ⟶ 1-1 onto ℂ
3 f1ocnv ⊢ F : ℝ 2 ⟶ 1-1 onto ℂ → F -1 : ℂ ⟶ 1-1 onto ℝ 2
4 f1of ⊢ F -1 : ℂ ⟶ 1-1 onto ℝ 2 → F -1 : ℂ ⟶ ℝ 2
5 2 3 4 mp2b ⊢ F -1 : ℂ ⟶ ℝ 2
6 5 a1i ⊢ ⊤ → F -1 : ℂ ⟶ ℝ 2
7 6 feqmptd ⊢ ⊤ → F -1 = z ∈ ℂ ⟼ F -1 ⁡ z
8 7 mptru ⊢ F -1 = z ∈ ℂ ⟼ F -1 ⁡ z
9 df-ov ⊢ ℜ ⁡ z F ℑ ⁡ z = F ⁡ ℜ ⁡ z ℑ ⁡ z
10 recl ⊢ z ∈ ℂ → ℜ ⁡ z ∈ ℝ
11 imcl ⊢ z ∈ ℂ → ℑ ⁡ z ∈ ℝ
12 oveq1 ⊢ x = ℜ ⁡ z → x + i ⁢ y = ℜ ⁡ z + i ⁢ y
13 oveq2 ⊢ y = ℑ ⁡ z → i ⁢ y = i ⁢ ℑ ⁡ z
14 13 oveq2d ⊢ y = ℑ ⁡ z → ℜ ⁡ z + i ⁢ y = ℜ ⁡ z + i ⁢ ℑ ⁡ z
15 ovex ⊢ ℜ ⁡ z + i ⁢ ℑ ⁡ z ∈ V
16 12 14 1 15 ovmpo ⊢ ℜ ⁡ z ∈ ℝ ∧ ℑ ⁡ z ∈ ℝ → ℜ ⁡ z F ℑ ⁡ z = ℜ ⁡ z + i ⁢ ℑ ⁡ z
17 10 11 16 syl2anc ⊢ z ∈ ℂ → ℜ ⁡ z F ℑ ⁡ z = ℜ ⁡ z + i ⁢ ℑ ⁡ z
18 9 17 eqtr3id ⊢ z ∈ ℂ → F ⁡ ℜ ⁡ z ℑ ⁡ z = ℜ ⁡ z + i ⁢ ℑ ⁡ z
19 replim ⊢ z ∈ ℂ → z = ℜ ⁡ z + i ⁢ ℑ ⁡ z
20 18 19 eqtr4d ⊢ z ∈ ℂ → F ⁡ ℜ ⁡ z ℑ ⁡ z = z
21 20 fveq2d ⊢ z ∈ ℂ → F -1 ⁡ F ⁡ ℜ ⁡ z ℑ ⁡ z = F -1 ⁡ z
22 10 11 opelxpd ⊢ z ∈ ℂ → ℜ ⁡ z ℑ ⁡ z ∈ ℝ 2
23 f1ocnvfv1 ⊢ F : ℝ 2 ⟶ 1-1 onto ℂ ∧ ℜ ⁡ z ℑ ⁡ z ∈ ℝ 2 → F -1 ⁡ F ⁡ ℜ ⁡ z ℑ ⁡ z = ℜ ⁡ z ℑ ⁡ z
24 2 22 23 sylancr ⊢ z ∈ ℂ → F -1 ⁡ F ⁡ ℜ ⁡ z ℑ ⁡ z = ℜ ⁡ z ℑ ⁡ z
25 21 24 eqtr3d ⊢ z ∈ ℂ → F -1 ⁡ z = ℜ ⁡ z ℑ ⁡ z
26 25 mpteq2ia ⊢ z ∈ ℂ ⟼ F -1 ⁡ z = z ∈ ℂ ⟼ ℜ ⁡ z ℑ ⁡ z
27 8 26 eqtri ⊢ F -1 = z ∈ ℂ ⟼ ℜ ⁡ z ℑ ⁡ z