Metamath Proof Explorer


Theorem sq01

Description: If a complex number equals its square, it must be 0 or 1. (Contributed by NM, 6-Jun-2006)

Ref Expression
Assertion sq01 ⊢ A ∈ ℂ → A 2 = A ↔ A = 0 ∨ A = 1

Proof

Step Hyp Ref Expression
1 df-ne ⊢ A ≠ 0 ↔ ¬ A = 0
2 sqval ⊢ A ∈ ℂ → A 2 = A ⁢ A
3 mulrid ⊢ A ∈ ℂ → A ⋅ 1 = A
4 3 eqcomd ⊢ A ∈ ℂ → A = A ⋅ 1
5 2 4 eqeq12d ⊢ A ∈ ℂ → A 2 = A ↔ A ⁢ A = A ⋅ 1
6 5 adantr ⊢ A ∈ ℂ ∧ A ≠ 0 → A 2 = A ↔ A ⁢ A = A ⋅ 1
7 ax-1cn ⊢ 1 ∈ ℂ
8 mulcan ⊢ A ∈ ℂ ∧ 1 ∈ ℂ ∧ A ∈ ℂ ∧ A ≠ 0 → A ⁢ A = A ⋅ 1 ↔ A = 1
9 7 8 mp3an2 ⊢ A ∈ ℂ ∧ A ∈ ℂ ∧ A ≠ 0 → A ⁢ A = A ⋅ 1 ↔ A = 1
10 9 anabss5 ⊢ A ∈ ℂ ∧ A ≠ 0 → A ⁢ A = A ⋅ 1 ↔ A = 1
11 6 10 bitrd ⊢ A ∈ ℂ ∧ A ≠ 0 → A 2 = A ↔ A = 1
12 11 biimpd ⊢ A ∈ ℂ ∧ A ≠ 0 → A 2 = A → A = 1
13 12 impancom ⊢ A ∈ ℂ ∧ A 2 = A → A ≠ 0 → A = 1
14 1 13 biimtrrid ⊢ A ∈ ℂ ∧ A 2 = A → ¬ A = 0 → A = 1
15 14 orrd ⊢ A ∈ ℂ ∧ A 2 = A → A = 0 ∨ A = 1
16 15 ex ⊢ A ∈ ℂ → A 2 = A → A = 0 ∨ A = 1
17 sq0 ⊢ 0 2 = 0
18 oveq1 ⊢ A = 0 → A 2 = 0 2
19 id ⊢ A = 0 → A = 0
20 17 18 19 3eqtr4a ⊢ A = 0 → A 2 = A
21 sq1 ⊢ 1 2 = 1
22 oveq1 ⊢ A = 1 → A 2 = 1 2
23 id ⊢ A = 1 → A = 1
24 21 22 23 3eqtr4a ⊢ A = 1 → A 2 = A
25 20 24 jaoi ⊢ A = 0 ∨ A = 1 → A 2 = A
26 16 25 impbid1 ⊢ A ∈ ℂ → A 2 = A ↔ A = 0 ∨ A = 1