Metamath Proof Explorer


Theorem chscllem1

Description: Lemma for chscl . (Contributed by Mario Carneiro, 19-May-2014) (New usage is discouraged.)

Ref Expression
Hypotheses chscl.1 ⊢ φ → A ∈ C ℋ
chscl.2 ⊢ φ → B ∈ C ℋ
chscl.3 ⊢ φ → B ⊆ ⊥ ⁡ A
chscl.4 ⊢ φ → H : ℕ ⟶ A + ℋ B
chscl.5 ⊢ φ → H ⇝v u
chscl.6 ⊢ F = n ∈ ℕ ⟼ proj ℎ ⁡ A ⁡ H ⁡ n
Assertion chscllem1 ⊢ φ → F : ℕ ⟶ A

Proof

Step Hyp Ref Expression
1 chscl.1 ⊢ φ → A ∈ C ℋ
2 chscl.2 ⊢ φ → B ∈ C ℋ
3 chscl.3 ⊢ φ → B ⊆ ⊥ ⁡ A
4 chscl.4 ⊢ φ → H : ℕ ⟶ A + ℋ B
5 chscl.5 ⊢ φ → H ⇝v u
6 chscl.6 ⊢ F = n ∈ ℕ ⟼ proj ℎ ⁡ A ⁡ H ⁡ n
7 eqid ⊢ proj ℎ ⁡ A ⁡ H ⁡ n = proj ℎ ⁡ A ⁡ H ⁡ n
8 1 adantr ⊢ φ ∧ n ∈ ℕ → A ∈ C ℋ
9 4 ffvelcdmda ⊢ φ ∧ n ∈ ℕ → H ⁡ n ∈ A + ℋ B
10 chsh ⊢ B ∈ C ℋ → B ∈ S ℋ
11 2 10 syl ⊢ φ → B ∈ S ℋ
12 chsh ⊢ A ∈ C ℋ → A ∈ S ℋ
13 1 12 syl ⊢ φ → A ∈ S ℋ
14 shocsh ⊢ A ∈ S ℋ → ⊥ ⁡ A ∈ S ℋ
15 13 14 syl ⊢ φ → ⊥ ⁡ A ∈ S ℋ
16 shless ⊢ B ∈ S ℋ ∧ ⊥ ⁡ A ∈ S ℋ ∧ A ∈ S ℋ ∧ B ⊆ ⊥ ⁡ A → B + ℋ A ⊆ ⊥ ⁡ A + ℋ A
17 11 15 13 3 16 syl31anc ⊢ φ → B + ℋ A ⊆ ⊥ ⁡ A + ℋ A
18 shscom ⊢ A ∈ S ℋ ∧ B ∈ S ℋ → A + ℋ B = B + ℋ A
19 13 11 18 syl2anc ⊢ φ → A + ℋ B = B + ℋ A
20 shscom ⊢ A ∈ S ℋ ∧ ⊥ ⁡ A ∈ S ℋ → A + ℋ ⊥ ⁡ A = ⊥ ⁡ A + ℋ A
21 13 15 20 syl2anc ⊢ φ → A + ℋ ⊥ ⁡ A = ⊥ ⁡ A + ℋ A
22 17 19 21 3sstr4d ⊢ φ → A + ℋ B ⊆ A + ℋ ⊥ ⁡ A
23 22 sselda ⊢ φ ∧ H ⁡ n ∈ A + ℋ B → H ⁡ n ∈ A + ℋ ⊥ ⁡ A
24 9 23 syldan ⊢ φ ∧ n ∈ ℕ → H ⁡ n ∈ A + ℋ ⊥ ⁡ A
25 pjpreeq ⊢ A ∈ C ℋ ∧ H ⁡ n ∈ A + ℋ ⊥ ⁡ A → proj ℎ ⁡ A ⁡ H ⁡ n = proj ℎ ⁡ A ⁡ H ⁡ n ↔ proj ℎ ⁡ A ⁡ H ⁡ n ∈ A ∧ ∃ x ∈ ⊥ ⁡ A H ⁡ n = proj ℎ ⁡ A ⁡ H ⁡ n + ℎ x
26 8 24 25 syl2anc ⊢ φ ∧ n ∈ ℕ → proj ℎ ⁡ A ⁡ H ⁡ n = proj ℎ ⁡ A ⁡ H ⁡ n ↔ proj ℎ ⁡ A ⁡ H ⁡ n ∈ A ∧ ∃ x ∈ ⊥ ⁡ A H ⁡ n = proj ℎ ⁡ A ⁡ H ⁡ n + ℎ x
27 7 26 mpbii ⊢ φ ∧ n ∈ ℕ → proj ℎ ⁡ A ⁡ H ⁡ n ∈ A ∧ ∃ x ∈ ⊥ ⁡ A H ⁡ n = proj ℎ ⁡ A ⁡ H ⁡ n + ℎ x
28 27 simpld ⊢ φ ∧ n ∈ ℕ → proj ℎ ⁡ A ⁡ H ⁡ n ∈ A
29 28 6 fmptd ⊢ φ → F : ℕ ⟶ A