Metamath Proof Explorer


Theorem pythagreim

Description: A simplified version of the Pythagorean theorem, where the points A and B respectively lie on the imaginary and real axes, and the right angle is at the origin. (Contributed by Thierry Arnoux, 2-Nov-2025)

Ref Expression
Hypotheses pythagreim.1 ⊢ φ → A ∈ ℝ
pythagreim.2 ⊢ φ → B ∈ ℝ
Assertion pythagreim ⊢ φ → B − i ⁢ A 2 = A 2 + B 2

Proof

Step Hyp Ref Expression
1 pythagreim.1 ⊢ φ → A ∈ ℝ
2 pythagreim.2 ⊢ φ → B ∈ ℝ
3 cjreim2 ⊢ B ∈ ℝ ∧ A ∈ ℝ → B − i ⁢ A ‾ = B + i ⁢ A
4 2 1 3 syl2anc ⊢ φ → B − i ⁢ A ‾ = B + i ⁢ A
5 4 oveq2d ⊢ φ → B − i ⁢ A ⁢ B − i ⁢ A ‾ = B − i ⁢ A ⁢ B + i ⁢ A
6 2 recnd ⊢ φ → B ∈ ℂ
7 ax-icn ⊢ i ∈ ℂ
8 7 a1i ⊢ φ → i ∈ ℂ
9 1 recnd ⊢ φ → A ∈ ℂ
10 8 9 mulcld ⊢ φ → i ⁢ A ∈ ℂ
11 6 10 subcld ⊢ φ → B − i ⁢ A ∈ ℂ
12 6 10 addcld ⊢ φ → B + i ⁢ A ∈ ℂ
13 11 12 mulcomd ⊢ φ → B − i ⁢ A ⁢ B + i ⁢ A = B + i ⁢ A ⁢ B − i ⁢ A
14 5 13 eqtrd ⊢ φ → B − i ⁢ A ⁢ B − i ⁢ A ‾ = B + i ⁢ A ⁢ B − i ⁢ A
15 11 absvalsqd ⊢ φ → B − i ⁢ A 2 = B − i ⁢ A ⁢ B − i ⁢ A ‾
16 8 9 sqmuld ⊢ φ → i ⁢ A 2 = i 2 ⁢ A 2
17 i2 ⊢ i 2 = − 1
18 17 oveq1i ⊢ i 2 ⁢ A 2 = -1 ⁢ A 2
19 16 18 eqtrdi ⊢ φ → i ⁢ A 2 = -1 ⁢ A 2
20 9 sqcld ⊢ φ → A 2 ∈ ℂ
21 20 mulm1d ⊢ φ → -1 ⁢ A 2 = − A 2
22 19 21 eqtrd ⊢ φ → i ⁢ A 2 = − A 2
23 22 oveq2d ⊢ φ → B 2 − i ⁢ A 2 = B 2 − − A 2
24 6 sqcld ⊢ φ → B 2 ∈ ℂ
25 24 20 subnegd ⊢ φ → B 2 − − A 2 = B 2 + A 2
26 24 20 addcomd ⊢ φ → B 2 + A 2 = A 2 + B 2
27 23 25 26 3eqtrd ⊢ φ → B 2 − i ⁢ A 2 = A 2 + B 2
28 subsq ⊢ B ∈ ℂ ∧ i ⁢ A ∈ ℂ → B 2 − i ⁢ A 2 = B + i ⁢ A ⁢ B − i ⁢ A
29 6 10 28 syl2anc ⊢ φ → B 2 − i ⁢ A 2 = B + i ⁢ A ⁢ B − i ⁢ A
30 27 29 eqtr3d ⊢ φ → A 2 + B 2 = B + i ⁢ A ⁢ B − i ⁢ A
31 14 15 30 3eqtr4d ⊢ φ → B − i ⁢ A 2 = A 2 + B 2