Metamath Proof Explorer


Theorem reusq0

Description: A complex number is the square of exactly one complex number iff the given complex number is zero. (Contributed by AV, 21-Jun-2023)

Ref Expression
Assertion reusq0 ⊢ X ∈ ℂ → ∃! x ∈ ℂ x 2 = X ↔ X = 0

Proof

Step Hyp Ref Expression
1 2a1 ⊢ X = 0 → X ∈ ℂ → ∃! x ∈ ℂ x 2 = X → X = 0
2 sqrtcl ⊢ X ∈ ℂ → X ∈ ℂ
3 2 adantr ⊢ X ∈ ℂ ∧ ¬ X = 0 → X ∈ ℂ
4 2 negcld ⊢ X ∈ ℂ → − X ∈ ℂ
5 4 adantr ⊢ X ∈ ℂ ∧ ¬ X = 0 → − X ∈ ℂ
6 2 eqnegd ⊢ X ∈ ℂ → X = − X ↔ X = 0
7 simpl ⊢ X ∈ ℂ ∧ X = 0 → X ∈ ℂ
8 simpr ⊢ X ∈ ℂ ∧ X = 0 → X = 0
9 7 8 sqr00d ⊢ X ∈ ℂ ∧ X = 0 → X = 0
10 9 ex ⊢ X ∈ ℂ → X = 0 → X = 0
11 6 10 sylbid ⊢ X ∈ ℂ → X = − X → X = 0
12 11 necon3bd ⊢ X ∈ ℂ → ¬ X = 0 → X ≠ − X
13 12 imp ⊢ X ∈ ℂ ∧ ¬ X = 0 → X ≠ − X
14 3 5 13 3jca ⊢ X ∈ ℂ ∧ ¬ X = 0 → X ∈ ℂ ∧ − X ∈ ℂ ∧ X ≠ − X
15 sqrtth ⊢ X ∈ ℂ → X 2 = X
16 sqneg ⊢ X ∈ ℂ → − X 2 = X 2
17 2 16 syl ⊢ X ∈ ℂ → − X 2 = X 2
18 17 15 eqtrd ⊢ X ∈ ℂ → − X 2 = X
19 15 18 jca ⊢ X ∈ ℂ → X 2 = X ∧ − X 2 = X
20 19 adantr ⊢ X ∈ ℂ ∧ ¬ X = 0 → X 2 = X ∧ − X 2 = X
21 oveq1 ⊢ x = X → x 2 = X 2
22 21 eqeq1d ⊢ x = X → x 2 = X ↔ X 2 = X
23 oveq1 ⊢ x = − X → x 2 = − X 2
24 23 eqeq1d ⊢ x = − X → x 2 = X ↔ − X 2 = X
25 22 24 2nreu ⊢ X ∈ ℂ ∧ − X ∈ ℂ ∧ X ≠ − X → X 2 = X ∧ − X 2 = X → ¬ ∃! x ∈ ℂ x 2 = X
26 14 20 25 sylc ⊢ X ∈ ℂ ∧ ¬ X = 0 → ¬ ∃! x ∈ ℂ x 2 = X
27 26 pm2.21d ⊢ X ∈ ℂ ∧ ¬ X = 0 → ∃! x ∈ ℂ x 2 = X → X = 0
28 27 expcom ⊢ ¬ X = 0 → X ∈ ℂ → ∃! x ∈ ℂ x 2 = X → X = 0
29 1 28 pm2.61i ⊢ X ∈ ℂ → ∃! x ∈ ℂ x 2 = X → X = 0
30 2nn ⊢ 2 ∈ ℕ
31 0cnd ⊢ 2 ∈ ℕ → 0 ∈ ℂ
32 oveq1 ⊢ x = 0 → x 2 = 0 2
33 32 eqeq1d ⊢ x = 0 → x 2 = 0 ↔ 0 2 = 0
34 eqeq1 ⊢ x = 0 → x = y ↔ 0 = y
35 34 imbi2d ⊢ x = 0 → y 2 = 0 → x = y ↔ y 2 = 0 → 0 = y
36 35 ralbidv ⊢ x = 0 → ∀ y ∈ ℂ y 2 = 0 → x = y ↔ ∀ y ∈ ℂ y 2 = 0 → 0 = y
37 33 36 anbi12d ⊢ x = 0 → x 2 = 0 ∧ ∀ y ∈ ℂ y 2 = 0 → x = y ↔ 0 2 = 0 ∧ ∀ y ∈ ℂ y 2 = 0 → 0 = y
38 37 adantl ⊢ 2 ∈ ℕ ∧ x = 0 → x 2 = 0 ∧ ∀ y ∈ ℂ y 2 = 0 → x = y ↔ 0 2 = 0 ∧ ∀ y ∈ ℂ y 2 = 0 → 0 = y
39 0exp ⊢ 2 ∈ ℕ → 0 2 = 0
40 sqeq0 ⊢ y ∈ ℂ → y 2 = 0 ↔ y = 0
41 40 biimpd ⊢ y ∈ ℂ → y 2 = 0 → y = 0
42 eqcom ⊢ 0 = y ↔ y = 0
43 41 42 imbitrrdi ⊢ y ∈ ℂ → y 2 = 0 → 0 = y
44 43 adantl ⊢ 2 ∈ ℕ ∧ y ∈ ℂ → y 2 = 0 → 0 = y
45 44 ralrimiva ⊢ 2 ∈ ℕ → ∀ y ∈ ℂ y 2 = 0 → 0 = y
46 39 45 jca ⊢ 2 ∈ ℕ → 0 2 = 0 ∧ ∀ y ∈ ℂ y 2 = 0 → 0 = y
47 31 38 46 rspcedvd ⊢ 2 ∈ ℕ → ∃ x ∈ ℂ x 2 = 0 ∧ ∀ y ∈ ℂ y 2 = 0 → x = y
48 30 47 mp1i ⊢ X = 0 → ∃ x ∈ ℂ x 2 = 0 ∧ ∀ y ∈ ℂ y 2 = 0 → x = y
49 eqeq2 ⊢ X = 0 → x 2 = X ↔ x 2 = 0
50 eqeq2 ⊢ X = 0 → y 2 = X ↔ y 2 = 0
51 50 imbi1d ⊢ X = 0 → y 2 = X → x = y ↔ y 2 = 0 → x = y
52 51 ralbidv ⊢ X = 0 → ∀ y ∈ ℂ y 2 = X → x = y ↔ ∀ y ∈ ℂ y 2 = 0 → x = y
53 49 52 anbi12d ⊢ X = 0 → x 2 = X ∧ ∀ y ∈ ℂ y 2 = X → x = y ↔ x 2 = 0 ∧ ∀ y ∈ ℂ y 2 = 0 → x = y
54 53 rexbidv ⊢ X = 0 → ∃ x ∈ ℂ x 2 = X ∧ ∀ y ∈ ℂ y 2 = X → x = y ↔ ∃ x ∈ ℂ x 2 = 0 ∧ ∀ y ∈ ℂ y 2 = 0 → x = y
55 48 54 mpbird ⊢ X = 0 → ∃ x ∈ ℂ x 2 = X ∧ ∀ y ∈ ℂ y 2 = X → x = y
56 55 a1i ⊢ X ∈ ℂ → X = 0 → ∃ x ∈ ℂ x 2 = X ∧ ∀ y ∈ ℂ y 2 = X → x = y
57 oveq1 ⊢ x = y → x 2 = y 2
58 57 eqeq1d ⊢ x = y → x 2 = X ↔ y 2 = X
59 58 reu8 ⊢ ∃! x ∈ ℂ x 2 = X ↔ ∃ x ∈ ℂ x 2 = X ∧ ∀ y ∈ ℂ y 2 = X → x = y
60 56 59 imbitrrdi ⊢ X ∈ ℂ → X = 0 → ∃! x ∈ ℂ x 2 = X
61 29 60 impbid ⊢ X ∈ ℂ → ∃! x ∈ ℂ x 2 = X ↔ X = 0