Metamath Proof Explorer


Theorem shs00i

Description: Two subspaces are zero iff their join is zero. (Contributed by NM, 7-Aug-2004) (New usage is discouraged.)

Ref Expression
Hypotheses shne0.1 ⊢ A ∈ S ℋ
shs00.2 ⊢ B ∈ S ℋ
Assertion shs00i ⊢ A = 0 ℋ ∧ B = 0 ℋ ↔ A + ℋ B = 0 ℋ

Proof

Step Hyp Ref Expression
1 shne0.1 ⊢ A ∈ S ℋ
2 shs00.2 ⊢ B ∈ S ℋ
3 oveq12 ⊢ A = 0 ℋ ∧ B = 0 ℋ → A + ℋ B = 0 ℋ + ℋ 0 ℋ
4 h0elsh ⊢ 0 ℋ ∈ S ℋ
5 4 shs0i ⊢ 0 ℋ + ℋ 0 ℋ = 0 ℋ
6 3 5 eqtrdi ⊢ A = 0 ℋ ∧ B = 0 ℋ → A + ℋ B = 0 ℋ
7 1 2 shsub1i ⊢ A ⊆ A + ℋ B
8 sseq2 ⊢ A + ℋ B = 0 ℋ → A ⊆ A + ℋ B ↔ A ⊆ 0 ℋ
9 7 8 mpbii ⊢ A + ℋ B = 0 ℋ → A ⊆ 0 ℋ
10 shle0 ⊢ A ∈ S ℋ → A ⊆ 0 ℋ ↔ A = 0 ℋ
11 1 10 ax-mp ⊢ A ⊆ 0 ℋ ↔ A = 0 ℋ
12 9 11 sylib ⊢ A + ℋ B = 0 ℋ → A = 0 ℋ
13 2 1 shsub2i ⊢ B ⊆ A + ℋ B
14 sseq2 ⊢ A + ℋ B = 0 ℋ → B ⊆ A + ℋ B ↔ B ⊆ 0 ℋ
15 13 14 mpbii ⊢ A + ℋ B = 0 ℋ → B ⊆ 0 ℋ
16 shle0 ⊢ B ∈ S ℋ → B ⊆ 0 ℋ ↔ B = 0 ℋ
17 2 16 ax-mp ⊢ B ⊆ 0 ℋ ↔ B = 0 ℋ
18 15 17 sylib ⊢ A + ℋ B = 0 ℋ → B = 0 ℋ
19 12 18 jca ⊢ A + ℋ B = 0 ℋ → A = 0 ℋ ∧ B = 0 ℋ
20 6 19 impbii ⊢ A = 0 ℋ ∧ B = 0 ℋ ↔ A + ℋ B = 0 ℋ