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 ( 𝜑𝐴C )
chscl.2 ( 𝜑𝐵C )
chscl.3 ( 𝜑𝐵 ⊆ ( ⊥ ‘ 𝐴 ) )
chscl.4 ( 𝜑𝐻 : ℕ ⟶ ( 𝐴 + 𝐵 ) )
chscl.5 ( 𝜑𝐻𝑣 𝑢 )
chscl.6 𝐹 = ( 𝑛 ∈ ℕ ↦ ( ( proj𝐴 ) ‘ ( 𝐻𝑛 ) ) )
Assertion chscllem1 ( 𝜑𝐹 : ℕ ⟶ 𝐴 )

Proof

Step Hyp Ref Expression
1 chscl.1 ( 𝜑𝐴C )
2 chscl.2 ( 𝜑𝐵C )
3 chscl.3 ( 𝜑𝐵 ⊆ ( ⊥ ‘ 𝐴 ) )
4 chscl.4 ( 𝜑𝐻 : ℕ ⟶ ( 𝐴 + 𝐵 ) )
5 chscl.5 ( 𝜑𝐻𝑣 𝑢 )
6 chscl.6 𝐹 = ( 𝑛 ∈ ℕ ↦ ( ( proj𝐴 ) ‘ ( 𝐻𝑛 ) ) )
7 eqid ( ( proj𝐴 ) ‘ ( 𝐻𝑛 ) ) = ( ( proj𝐴 ) ‘ ( 𝐻𝑛 ) )
8 1 adantr ( ( 𝜑𝑛 ∈ ℕ ) → 𝐴C )
9 4 ffvelrnda ( ( 𝜑𝑛 ∈ ℕ ) → ( 𝐻𝑛 ) ∈ ( 𝐴 + 𝐵 ) )
10 chsh ( 𝐵C𝐵S )
11 2 10 syl ( 𝜑𝐵S )
12 chsh ( 𝐴C𝐴S )
13 1 12 syl ( 𝜑𝐴S )
14 shocsh ( 𝐴S → ( ⊥ ‘ 𝐴 ) ∈ S )
15 13 14 syl ( 𝜑 → ( ⊥ ‘ 𝐴 ) ∈ S )
16 shless ( ( ( 𝐵S ∧ ( ⊥ ‘ 𝐴 ) ∈ S𝐴S ) ∧ 𝐵 ⊆ ( ⊥ ‘ 𝐴 ) ) → ( 𝐵 + 𝐴 ) ⊆ ( ( ⊥ ‘ 𝐴 ) + 𝐴 ) )
17 11 15 13 3 16 syl31anc ( 𝜑 → ( 𝐵 + 𝐴 ) ⊆ ( ( ⊥ ‘ 𝐴 ) + 𝐴 ) )
18 shscom ( ( 𝐴S𝐵S ) → ( 𝐴 + 𝐵 ) = ( 𝐵 + 𝐴 ) )
19 13 11 18 syl2anc ( 𝜑 → ( 𝐴 + 𝐵 ) = ( 𝐵 + 𝐴 ) )
20 shscom ( ( 𝐴S ∧ ( ⊥ ‘ 𝐴 ) ∈ S ) → ( 𝐴 + ( ⊥ ‘ 𝐴 ) ) = ( ( ⊥ ‘ 𝐴 ) + 𝐴 ) )
21 13 15 20 syl2anc ( 𝜑 → ( 𝐴 + ( ⊥ ‘ 𝐴 ) ) = ( ( ⊥ ‘ 𝐴 ) + 𝐴 ) )
22 17 19 21 3sstr4d ( 𝜑 → ( 𝐴 + 𝐵 ) ⊆ ( 𝐴 + ( ⊥ ‘ 𝐴 ) ) )
23 22 sselda ( ( 𝜑 ∧ ( 𝐻𝑛 ) ∈ ( 𝐴 + 𝐵 ) ) → ( 𝐻𝑛 ) ∈ ( 𝐴 + ( ⊥ ‘ 𝐴 ) ) )
24 9 23 syldan ( ( 𝜑𝑛 ∈ ℕ ) → ( 𝐻𝑛 ) ∈ ( 𝐴 + ( ⊥ ‘ 𝐴 ) ) )
25 pjpreeq ( ( 𝐴C ∧ ( 𝐻𝑛 ) ∈ ( 𝐴 + ( ⊥ ‘ 𝐴 ) ) ) → ( ( ( proj𝐴 ) ‘ ( 𝐻𝑛 ) ) = ( ( proj𝐴 ) ‘ ( 𝐻𝑛 ) ) ↔ ( ( ( proj𝐴 ) ‘ ( 𝐻𝑛 ) ) ∈ 𝐴 ∧ ∃ 𝑥 ∈ ( ⊥ ‘ 𝐴 ) ( 𝐻𝑛 ) = ( ( ( proj𝐴 ) ‘ ( 𝐻𝑛 ) ) + 𝑥 ) ) ) )
26 8 24 25 syl2anc ( ( 𝜑𝑛 ∈ ℕ ) → ( ( ( proj𝐴 ) ‘ ( 𝐻𝑛 ) ) = ( ( proj𝐴 ) ‘ ( 𝐻𝑛 ) ) ↔ ( ( ( proj𝐴 ) ‘ ( 𝐻𝑛 ) ) ∈ 𝐴 ∧ ∃ 𝑥 ∈ ( ⊥ ‘ 𝐴 ) ( 𝐻𝑛 ) = ( ( ( proj𝐴 ) ‘ ( 𝐻𝑛 ) ) + 𝑥 ) ) ) )
27 7 26 mpbii ( ( 𝜑𝑛 ∈ ℕ ) → ( ( ( proj𝐴 ) ‘ ( 𝐻𝑛 ) ) ∈ 𝐴 ∧ ∃ 𝑥 ∈ ( ⊥ ‘ 𝐴 ) ( 𝐻𝑛 ) = ( ( ( proj𝐴 ) ‘ ( 𝐻𝑛 ) ) + 𝑥 ) ) )
28 27 simpld ( ( 𝜑𝑛 ∈ ℕ ) → ( ( proj𝐴 ) ‘ ( 𝐻𝑛 ) ) ∈ 𝐴 )
29 28 6 fmptd ( 𝜑𝐹 : ℕ ⟶ 𝐴 )