Metamath Proof Explorer


Theorem bj-ccinftydisj

Description: The circle at infinity is disjoint from the set of complex numbers. (Contributed by BJ, 22-Jun-2019)

Ref Expression
Assertion bj-ccinftydisj ⊢ ℂ ∩ ℂ ∞ = ∅

Proof

Step Hyp Ref Expression
1 bj-inftyexpidisj ⊢ ¬ inftyexpi ⁡ y ∈ ℂ
2 1 nex ⊢ ¬ ∃ y inftyexpi ⁡ y ∈ ℂ
3 elin ⊢ x ∈ ℂ ∩ ℂ ∞ ↔ x ∈ ℂ ∧ x ∈ ℂ ∞
4 df-bj-inftyexpi ⊢ inftyexpi = z ∈ − π π ⟼ z ℂ
5 4 funmpt2 ⊢ Fun ⁡ inftyexpi
6 elrnrexdm ⊢ Fun ⁡ inftyexpi → x ∈ ran ⁡ inftyexpi → ∃ y ∈ dom ⁡ inftyexpi x = inftyexpi ⁡ y
7 5 6 ax-mp ⊢ x ∈ ran ⁡ inftyexpi → ∃ y ∈ dom ⁡ inftyexpi x = inftyexpi ⁡ y
8 rexex ⊢ ∃ y ∈ dom ⁡ inftyexpi x = inftyexpi ⁡ y → ∃ y x = inftyexpi ⁡ y
9 7 8 syl ⊢ x ∈ ran ⁡ inftyexpi → ∃ y x = inftyexpi ⁡ y
10 df-bj-ccinfty ⊢ ℂ ∞ = ran ⁡ inftyexpi
11 9 10 eleq2s ⊢ x ∈ ℂ ∞ → ∃ y x = inftyexpi ⁡ y
12 11 anim2i ⊢ x ∈ ℂ ∧ x ∈ ℂ ∞ → x ∈ ℂ ∧ ∃ y x = inftyexpi ⁡ y
13 3 12 sylbi ⊢ x ∈ ℂ ∩ ℂ ∞ → x ∈ ℂ ∧ ∃ y x = inftyexpi ⁡ y
14 ancom ⊢ x ∈ ℂ ∧ ∃ y x = inftyexpi ⁡ y ↔ ∃ y x = inftyexpi ⁡ y ∧ x ∈ ℂ
15 exancom ⊢ ∃ y x ∈ ℂ ∧ x = inftyexpi ⁡ y ↔ ∃ y x = inftyexpi ⁡ y ∧ x ∈ ℂ
16 19.41v ⊢ ∃ y x = inftyexpi ⁡ y ∧ x ∈ ℂ ↔ ∃ y x = inftyexpi ⁡ y ∧ x ∈ ℂ
17 15 16 bitri ⊢ ∃ y x ∈ ℂ ∧ x = inftyexpi ⁡ y ↔ ∃ y x = inftyexpi ⁡ y ∧ x ∈ ℂ
18 14 17 sylbb2 ⊢ x ∈ ℂ ∧ ∃ y x = inftyexpi ⁡ y → ∃ y x ∈ ℂ ∧ x = inftyexpi ⁡ y
19 13 18 syl ⊢ x ∈ ℂ ∩ ℂ ∞ → ∃ y x ∈ ℂ ∧ x = inftyexpi ⁡ y
20 eleq1 ⊢ x = inftyexpi ⁡ y → x ∈ ℂ ↔ inftyexpi ⁡ y ∈ ℂ
21 20 biimpac ⊢ x ∈ ℂ ∧ x = inftyexpi ⁡ y → inftyexpi ⁡ y ∈ ℂ
22 21 eximi ⊢ ∃ y x ∈ ℂ ∧ x = inftyexpi ⁡ y → ∃ y inftyexpi ⁡ y ∈ ℂ
23 19 22 syl ⊢ x ∈ ℂ ∩ ℂ ∞ → ∃ y inftyexpi ⁡ y ∈ ℂ
24 2 23 mto ⊢ ¬ x ∈ ℂ ∩ ℂ ∞
25 24 nel0 ⊢ ℂ ∩ ℂ ∞ = ∅