Metamath Proof Explorer


Theorem addsq2nreurex

Description: For each complex number C , there is no unique complex number a added to the square of another complex number b resulting in the given complex number C . (Contributed by AV, 2-Jul-2023)

Ref Expression
Assertion addsq2nreurex ⊢ C ∈ ℂ → ¬ ∃! a ∈ ℂ ∃ b ∈ ℂ a + b 2 = C

Proof

Step Hyp Ref Expression
1 peano2cnm ⊢ C ∈ ℂ → C − 1 ∈ ℂ
2 id ⊢ C ∈ ℂ → C ∈ ℂ
3 4cn ⊢ 4 ∈ ℂ
4 3 a1i ⊢ C ∈ ℂ → 4 ∈ ℂ
5 2 4 subcld ⊢ C ∈ ℂ → C − 4 ∈ ℂ
6 1cnd ⊢ C ∈ ℂ → 1 ∈ ℂ
7 1re ⊢ 1 ∈ ℝ
8 1lt4 ⊢ 1 < 4
9 7 8 ltneii ⊢ 1 ≠ 4
10 9 a1i ⊢ C ∈ ℂ → 1 ≠ 4
11 2 6 4 10 subneintrd ⊢ C ∈ ℂ → C − 1 ≠ C − 4
12 oveq1 ⊢ b = 1 → b 2 = 1 2
13 12 oveq2d ⊢ b = 1 → C - 1 + b 2 = C - 1 + 1 2
14 13 eqeq1d ⊢ b = 1 → C - 1 + b 2 = C ↔ C - 1 + 1 2 = C
15 14 adantl ⊢ C ∈ ℂ ∧ b = 1 → C - 1 + b 2 = C ↔ C - 1 + 1 2 = C
16 sq1 ⊢ 1 2 = 1
17 16 oveq2i ⊢ C - 1 + 1 2 = C - 1 + 1
18 npcan1 ⊢ C ∈ ℂ → C - 1 + 1 = C
19 17 18 eqtrid ⊢ C ∈ ℂ → C - 1 + 1 2 = C
20 6 15 19 rspcedvd ⊢ C ∈ ℂ → ∃ b ∈ ℂ C - 1 + b 2 = C
21 2cnd ⊢ C ∈ ℂ → 2 ∈ ℂ
22 oveq1 ⊢ b = 2 → b 2 = 2 2
23 22 oveq2d ⊢ b = 2 → C - 4 + b 2 = C - 4 + 2 2
24 23 eqeq1d ⊢ b = 2 → C - 4 + b 2 = C ↔ C - 4 + 2 2 = C
25 24 adantl ⊢ C ∈ ℂ ∧ b = 2 → C - 4 + b 2 = C ↔ C - 4 + 2 2 = C
26 2cn ⊢ 2 ∈ ℂ
27 26 sqcli ⊢ 2 2 ∈ ℂ
28 27 a1i ⊢ C ∈ ℂ → 2 2 ∈ ℂ
29 2 4 28 subadd23d ⊢ C ∈ ℂ → C - 4 + 2 2 = C + 2 2 - 4
30 sq2 ⊢ 2 2 = 4
31 30 a1i ⊢ C ∈ ℂ → 2 2 = 4
32 28 31 subeq0bd ⊢ C ∈ ℂ → 2 2 − 4 = 0
33 27 3 subcli ⊢ 2 2 − 4 ∈ ℂ
34 addid0 ⊢ C ∈ ℂ ∧ 2 2 − 4 ∈ ℂ → C + 2 2 - 4 = C ↔ 2 2 − 4 = 0
35 33 34 mpan2 ⊢ C ∈ ℂ → C + 2 2 - 4 = C ↔ 2 2 − 4 = 0
36 32 35 mpbird ⊢ C ∈ ℂ → C + 2 2 - 4 = C
37 29 36 eqtrd ⊢ C ∈ ℂ → C - 4 + 2 2 = C
38 21 25 37 rspcedvd ⊢ C ∈ ℂ → ∃ b ∈ ℂ C - 4 + b 2 = C
39 oveq1 ⊢ a = C − 1 → a + b 2 = C - 1 + b 2
40 39 eqeq1d ⊢ a = C − 1 → a + b 2 = C ↔ C - 1 + b 2 = C
41 40 rexbidv ⊢ a = C − 1 → ∃ b ∈ ℂ a + b 2 = C ↔ ∃ b ∈ ℂ C - 1 + b 2 = C
42 oveq1 ⊢ a = C − 4 → a + b 2 = C - 4 + b 2
43 42 eqeq1d ⊢ a = C − 4 → a + b 2 = C ↔ C - 4 + b 2 = C
44 43 rexbidv ⊢ a = C − 4 → ∃ b ∈ ℂ a + b 2 = C ↔ ∃ b ∈ ℂ C - 4 + b 2 = C
45 41 44 2nreu ⊢ C − 1 ∈ ℂ ∧ C − 4 ∈ ℂ ∧ C − 1 ≠ C − 4 → ∃ b ∈ ℂ C - 1 + b 2 = C ∧ ∃ b ∈ ℂ C - 4 + b 2 = C → ¬ ∃! a ∈ ℂ ∃ b ∈ ℂ a + b 2 = C
46 45 imp ⊢ C − 1 ∈ ℂ ∧ C − 4 ∈ ℂ ∧ C − 1 ≠ C − 4 ∧ ∃ b ∈ ℂ C - 1 + b 2 = C ∧ ∃ b ∈ ℂ C - 4 + b 2 = C → ¬ ∃! a ∈ ℂ ∃ b ∈ ℂ a + b 2 = C
47 1 5 11 20 38 46 syl32anc ⊢ C ∈ ℂ → ¬ ∃! a ∈ ℂ ∃ b ∈ ℂ a + b 2 = C