Metamath Proof Explorer


Theorem pjssge0ii

Description: Theorem 4.5(iv)->(v) of Beran p. 112. (Contributed by NM, 13-Aug-2000) (New usage is discouraged.)

Ref Expression
Hypotheses pjidm.1 ⊢ H ∈ C ℋ
pjidm.2 ⊢ A ∈ ℋ
pjsslem.1 ⊢ G ∈ C ℋ
Assertion pjssge0ii ⊢ proj ℎ ⁡ G ⁡ A - ℎ proj ℎ ⁡ H ⁡ A = proj ℎ ⁡ G ∩ ⊥ ⁡ H ⁡ A → 0 ≤ proj ℎ ⁡ G ⁡ A - ℎ proj ℎ ⁡ H ⁡ A ⋅ ih A

Proof

Step Hyp Ref Expression
1 pjidm.1 ⊢ H ∈ C ℋ
2 pjidm.2 ⊢ A ∈ ℋ
3 pjsslem.1 ⊢ G ∈ C ℋ
4 1 choccli ⊢ ⊥ ⁡ H ∈ C ℋ
5 3 4 chincli ⊢ G ∩ ⊥ ⁡ H ∈ C ℋ
6 5 2 pjhclii ⊢ proj ℎ ⁡ G ∩ ⊥ ⁡ H ⁡ A ∈ ℋ
7 6 normcli ⊢ norm ℎ ⁡ proj ℎ ⁡ G ∩ ⊥ ⁡ H ⁡ A ∈ ℝ
8 7 sqge0i ⊢ 0 ≤ norm ℎ ⁡ proj ℎ ⁡ G ∩ ⊥ ⁡ H ⁡ A 2
9 oveq1 ⊢ proj ℎ ⁡ G ⁡ A - ℎ proj ℎ ⁡ H ⁡ A = proj ℎ ⁡ G ∩ ⊥ ⁡ H ⁡ A → proj ℎ ⁡ G ⁡ A - ℎ proj ℎ ⁡ H ⁡ A ⋅ ih A = proj ℎ ⁡ G ∩ ⊥ ⁡ H ⁡ A ⋅ ih A
10 5 2 pjinormii ⊢ proj ℎ ⁡ G ∩ ⊥ ⁡ H ⁡ A ⋅ ih A = norm ℎ ⁡ proj ℎ ⁡ G ∩ ⊥ ⁡ H ⁡ A 2
11 9 10 eqtrdi ⊢ proj ℎ ⁡ G ⁡ A - ℎ proj ℎ ⁡ H ⁡ A = proj ℎ ⁡ G ∩ ⊥ ⁡ H ⁡ A → proj ℎ ⁡ G ⁡ A - ℎ proj ℎ ⁡ H ⁡ A ⋅ ih A = norm ℎ ⁡ proj ℎ ⁡ G ∩ ⊥ ⁡ H ⁡ A 2
12 8 11 breqtrrid ⊢ proj ℎ ⁡ G ⁡ A - ℎ proj ℎ ⁡ H ⁡ A = proj ℎ ⁡ G ∩ ⊥ ⁡ H ⁡ A → 0 ≤ proj ℎ ⁡ G ⁡ A - ℎ proj ℎ ⁡ H ⁡ A ⋅ ih A