Metamath Proof Explorer


Theorem cphdivcl

Description: The scalar field of a subcomplex pre-Hilbert space is closed under reciprocal. (Contributed by Mario Carneiro, 11-Oct-2015)

Ref Expression
Hypotheses cphsca.f ⊢ F = Scalar ⁡ W
cphsca.k ⊢ K = Base F
Assertion cphdivcl ⊢ W ∈ CPreHil ∧ A ∈ K ∧ B ∈ K ∧ B ≠ 0 → A B ∈ K

Proof

Step Hyp Ref Expression
1 cphsca.f ⊢ F = Scalar ⁡ W
2 cphsca.k ⊢ K = Base F
3 1 2 cphsubrg ⊢ W ∈ CPreHil → K ∈ SubRing ⁡ ℂ fld
4 3 adantr ⊢ W ∈ CPreHil ∧ A ∈ K ∧ B ∈ K ∧ B ≠ 0 → K ∈ SubRing ⁡ ℂ fld
5 cnfldbas ⊢ ℂ = Base ℂ fld
6 5 subrgss ⊢ K ∈ SubRing ⁡ ℂ fld → K ⊆ ℂ
7 4 6 syl ⊢ W ∈ CPreHil ∧ A ∈ K ∧ B ∈ K ∧ B ≠ 0 → K ⊆ ℂ
8 simpr1 ⊢ W ∈ CPreHil ∧ A ∈ K ∧ B ∈ K ∧ B ≠ 0 → A ∈ K
9 7 8 sseldd ⊢ W ∈ CPreHil ∧ A ∈ K ∧ B ∈ K ∧ B ≠ 0 → A ∈ ℂ
10 simpr2 ⊢ W ∈ CPreHil ∧ A ∈ K ∧ B ∈ K ∧ B ≠ 0 → B ∈ K
11 7 10 sseldd ⊢ W ∈ CPreHil ∧ A ∈ K ∧ B ∈ K ∧ B ≠ 0 → B ∈ ℂ
12 simpr3 ⊢ W ∈ CPreHil ∧ A ∈ K ∧ B ∈ K ∧ B ≠ 0 → B ≠ 0
13 9 11 12 divrecd ⊢ W ∈ CPreHil ∧ A ∈ K ∧ B ∈ K ∧ B ≠ 0 → A B = A ⁢ 1 B
14 1 2 cphreccl ⊢ W ∈ CPreHil ∧ B ∈ K ∧ B ≠ 0 → 1 B ∈ K
15 14 3adant3r1 ⊢ W ∈ CPreHil ∧ A ∈ K ∧ B ∈ K ∧ B ≠ 0 → 1 B ∈ K
16 cnfldmul ⊢ × = ⋅ ℂ fld
17 16 subrgmcl ⊢ K ∈ SubRing ⁡ ℂ fld ∧ A ∈ K ∧ 1 B ∈ K → A ⁢ 1 B ∈ K
18 4 8 15 17 syl3anc ⊢ W ∈ CPreHil ∧ A ∈ K ∧ B ∈ K ∧ B ≠ 0 → A ⁢ 1 B ∈ K
19 13 18 eqeltrd ⊢ W ∈ CPreHil ∧ A ∈ K ∧ B ∈ K ∧ B ≠ 0 → A B ∈ K