Metamath Proof Explorer


Theorem normpythi

Description: Analogy to Pythagorean theorem for orthogonal vectors. Remark 3.4(C) of Beran p. 98. (Contributed by NM, 17-Oct-1999) (New usage is discouraged.)

Ref Expression
Hypotheses normsub.1 ⊢ A ∈ ℋ
normsub.2 ⊢ B ∈ ℋ
Assertion normpythi ⊢ A ⋅ ih B = 0 → norm ℎ ⁡ A + ℎ B 2 = norm ℎ ⁡ A 2 + norm ℎ ⁡ B 2

Proof

Step Hyp Ref Expression
1 normsub.1 ⊢ A ∈ ℋ
2 normsub.2 ⊢ B ∈ ℋ
3 1 2 1 2 normlem8 ⊢ A + ℎ B ⋅ ih A + ℎ B = A ⋅ ih A + B ⋅ ih B + A ⋅ ih B + B ⋅ ih A
4 id ⊢ A ⋅ ih B = 0 → A ⋅ ih B = 0
5 orthcom ⊢ A ∈ ℋ ∧ B ∈ ℋ → A ⋅ ih B = 0 ↔ B ⋅ ih A = 0
6 1 2 5 mp2an ⊢ A ⋅ ih B = 0 ↔ B ⋅ ih A = 0
7 6 biimpi ⊢ A ⋅ ih B = 0 → B ⋅ ih A = 0
8 4 7 oveq12d ⊢ A ⋅ ih B = 0 → A ⋅ ih B + B ⋅ ih A = 0 + 0
9 00id ⊢ 0 + 0 = 0
10 8 9 eqtrdi ⊢ A ⋅ ih B = 0 → A ⋅ ih B + B ⋅ ih A = 0
11 10 oveq2d ⊢ A ⋅ ih B = 0 → A ⋅ ih A + B ⋅ ih B + A ⋅ ih B + B ⋅ ih A = A ⋅ ih A + B ⋅ ih B + 0
12 1 1 hicli ⊢ A ⋅ ih A ∈ ℂ
13 2 2 hicli ⊢ B ⋅ ih B ∈ ℂ
14 12 13 addcli ⊢ A ⋅ ih A + B ⋅ ih B ∈ ℂ
15 14 addridi ⊢ A ⋅ ih A + B ⋅ ih B + 0 = A ⋅ ih A + B ⋅ ih B
16 11 15 eqtrdi ⊢ A ⋅ ih B = 0 → A ⋅ ih A + B ⋅ ih B + A ⋅ ih B + B ⋅ ih A = A ⋅ ih A + B ⋅ ih B
17 3 16 eqtrid ⊢ A ⋅ ih B = 0 → A + ℎ B ⋅ ih A + ℎ B = A ⋅ ih A + B ⋅ ih B
18 1 2 hvaddcli ⊢ A + ℎ B ∈ ℋ
19 18 normsqi ⊢ norm ℎ ⁡ A + ℎ B 2 = A + ℎ B ⋅ ih A + ℎ B
20 1 normsqi ⊢ norm ℎ ⁡ A 2 = A ⋅ ih A
21 2 normsqi ⊢ norm ℎ ⁡ B 2 = B ⋅ ih B
22 20 21 oveq12i ⊢ norm ℎ ⁡ A 2 + norm ℎ ⁡ B 2 = A ⋅ ih A + B ⋅ ih B
23 17 19 22 3eqtr4g ⊢ A ⋅ ih B = 0 → norm ℎ ⁡ A + ℎ B 2 = norm ℎ ⁡ A 2 + norm ℎ ⁡ B 2