Metamath Proof Explorer


Theorem shsel3

Description: Membership in the subspace sum of two Hilbert subspaces, using vector subtraction. (Contributed by NM, 20-Jan-2007) (New usage is discouraged.)

Ref Expression
Assertion shsel3 ⊢ A ∈ S ℋ ∧ B ∈ S ℋ → C ∈ A + ℋ B ↔ ∃ x ∈ A ∃ y ∈ B C = x - ℎ y

Proof

Step Hyp Ref Expression
1 shsel ⊢ A ∈ S ℋ ∧ B ∈ S ℋ → C ∈ A + ℋ B ↔ ∃ x ∈ A ∃ z ∈ B C = x + ℎ z
2 id ⊢ C = x + ℎ z → C = x + ℎ z
3 shel ⊢ A ∈ S ℋ ∧ x ∈ A → x ∈ ℋ
4 shel ⊢ B ∈ S ℋ ∧ z ∈ B → z ∈ ℋ
5 hvaddsubval ⊢ x ∈ ℋ ∧ z ∈ ℋ → x + ℎ z = x - ℎ -1 ⋅ ℎ z
6 3 4 5 syl2an ⊢ A ∈ S ℋ ∧ x ∈ A ∧ B ∈ S ℋ ∧ z ∈ B → x + ℎ z = x - ℎ -1 ⋅ ℎ z
7 6 an4s ⊢ A ∈ S ℋ ∧ B ∈ S ℋ ∧ x ∈ A ∧ z ∈ B → x + ℎ z = x - ℎ -1 ⋅ ℎ z
8 7 anassrs ⊢ A ∈ S ℋ ∧ B ∈ S ℋ ∧ x ∈ A ∧ z ∈ B → x + ℎ z = x - ℎ -1 ⋅ ℎ z
9 2 8 sylan9eqr ⊢ A ∈ S ℋ ∧ B ∈ S ℋ ∧ x ∈ A ∧ z ∈ B ∧ C = x + ℎ z → C = x - ℎ -1 ⋅ ℎ z
10 neg1cn ⊢ − 1 ∈ ℂ
11 shmulcl ⊢ B ∈ S ℋ ∧ − 1 ∈ ℂ ∧ z ∈ B → -1 ⋅ ℎ z ∈ B
12 10 11 mp3an2 ⊢ B ∈ S ℋ ∧ z ∈ B → -1 ⋅ ℎ z ∈ B
13 12 adantll ⊢ A ∈ S ℋ ∧ B ∈ S ℋ ∧ z ∈ B → -1 ⋅ ℎ z ∈ B
14 13 adantlr ⊢ A ∈ S ℋ ∧ B ∈ S ℋ ∧ x ∈ A ∧ z ∈ B → -1 ⋅ ℎ z ∈ B
15 oveq2 ⊢ y = -1 ⋅ ℎ z → x - ℎ y = x - ℎ -1 ⋅ ℎ z
16 15 rspceeqv ⊢ -1 ⋅ ℎ z ∈ B ∧ C = x - ℎ -1 ⋅ ℎ z → ∃ y ∈ B C = x - ℎ y
17 14 16 sylan ⊢ A ∈ S ℋ ∧ B ∈ S ℋ ∧ x ∈ A ∧ z ∈ B ∧ C = x - ℎ -1 ⋅ ℎ z → ∃ y ∈ B C = x - ℎ y
18 9 17 syldan ⊢ A ∈ S ℋ ∧ B ∈ S ℋ ∧ x ∈ A ∧ z ∈ B ∧ C = x + ℎ z → ∃ y ∈ B C = x - ℎ y
19 18 rexlimdva2 ⊢ A ∈ S ℋ ∧ B ∈ S ℋ ∧ x ∈ A → ∃ z ∈ B C = x + ℎ z → ∃ y ∈ B C = x - ℎ y
20 id ⊢ C = x - ℎ y → C = x - ℎ y
21 shel ⊢ B ∈ S ℋ ∧ y ∈ B → y ∈ ℋ
22 hvsubval ⊢ x ∈ ℋ ∧ y ∈ ℋ → x - ℎ y = x + ℎ -1 ⋅ ℎ y
23 3 21 22 syl2an ⊢ A ∈ S ℋ ∧ x ∈ A ∧ B ∈ S ℋ ∧ y ∈ B → x - ℎ y = x + ℎ -1 ⋅ ℎ y
24 23 an4s ⊢ A ∈ S ℋ ∧ B ∈ S ℋ ∧ x ∈ A ∧ y ∈ B → x - ℎ y = x + ℎ -1 ⋅ ℎ y
25 24 anassrs ⊢ A ∈ S ℋ ∧ B ∈ S ℋ ∧ x ∈ A ∧ y ∈ B → x - ℎ y = x + ℎ -1 ⋅ ℎ y
26 20 25 sylan9eqr ⊢ A ∈ S ℋ ∧ B ∈ S ℋ ∧ x ∈ A ∧ y ∈ B ∧ C = x - ℎ y → C = x + ℎ -1 ⋅ ℎ y
27 shmulcl ⊢ B ∈ S ℋ ∧ − 1 ∈ ℂ ∧ y ∈ B → -1 ⋅ ℎ y ∈ B
28 10 27 mp3an2 ⊢ B ∈ S ℋ ∧ y ∈ B → -1 ⋅ ℎ y ∈ B
29 28 adantll ⊢ A ∈ S ℋ ∧ B ∈ S ℋ ∧ y ∈ B → -1 ⋅ ℎ y ∈ B
30 29 adantlr ⊢ A ∈ S ℋ ∧ B ∈ S ℋ ∧ x ∈ A ∧ y ∈ B → -1 ⋅ ℎ y ∈ B
31 oveq2 ⊢ z = -1 ⋅ ℎ y → x + ℎ z = x + ℎ -1 ⋅ ℎ y
32 31 rspceeqv ⊢ -1 ⋅ ℎ y ∈ B ∧ C = x + ℎ -1 ⋅ ℎ y → ∃ z ∈ B C = x + ℎ z
33 30 32 sylan ⊢ A ∈ S ℋ ∧ B ∈ S ℋ ∧ x ∈ A ∧ y ∈ B ∧ C = x + ℎ -1 ⋅ ℎ y → ∃ z ∈ B C = x + ℎ z
34 26 33 syldan ⊢ A ∈ S ℋ ∧ B ∈ S ℋ ∧ x ∈ A ∧ y ∈ B ∧ C = x - ℎ y → ∃ z ∈ B C = x + ℎ z
35 34 rexlimdva2 ⊢ A ∈ S ℋ ∧ B ∈ S ℋ ∧ x ∈ A → ∃ y ∈ B C = x - ℎ y → ∃ z ∈ B C = x + ℎ z
36 19 35 impbid ⊢ A ∈ S ℋ ∧ B ∈ S ℋ ∧ x ∈ A → ∃ z ∈ B C = x + ℎ z ↔ ∃ y ∈ B C = x - ℎ y
37 36 rexbidva ⊢ A ∈ S ℋ ∧ B ∈ S ℋ → ∃ x ∈ A ∃ z ∈ B C = x + ℎ z ↔ ∃ x ∈ A ∃ y ∈ B C = x - ℎ y
38 1 37 bitrd ⊢ A ∈ S ℋ ∧ B ∈ S ℋ → C ∈ A + ℋ B ↔ ∃ x ∈ A ∃ y ∈ B C = x - ℎ y