Metamath Proof Explorer


Theorem nmop0

Description: The norm of the zero operator is zero. (Contributed by NM, 8-Feb-2006) (New usage is discouraged.)

Ref Expression
Assertion nmop0 ⊢ norm op ⁡ 0 hop = 0

Proof

Step Hyp Ref Expression
1 ho0f ⊢ 0 hop : ℋ ⟶ ℋ
2 nmopval ⊢ 0 hop : ℋ ⟶ ℋ → norm op ⁡ 0 hop = sup x | ∃ y ∈ ℋ norm ℎ ⁡ y ≤ 1 ∧ x = norm ℎ ⁡ 0 hop ⁡ y ℝ * <
3 1 2 ax-mp ⊢ norm op ⁡ 0 hop = sup x | ∃ y ∈ ℋ norm ℎ ⁡ y ≤ 1 ∧ x = norm ℎ ⁡ 0 hop ⁡ y ℝ * <
4 ho0val ⊢ y ∈ ℋ → 0 hop ⁡ y = 0 ℎ
5 4 fveq2d ⊢ y ∈ ℋ → norm ℎ ⁡ 0 hop ⁡ y = norm ℎ ⁡ 0 ℎ
6 norm0 ⊢ norm ℎ ⁡ 0 ℎ = 0
7 5 6 eqtrdi ⊢ y ∈ ℋ → norm ℎ ⁡ 0 hop ⁡ y = 0
8 7 eqeq2d ⊢ y ∈ ℋ → x = norm ℎ ⁡ 0 hop ⁡ y ↔ x = 0
9 8 anbi2d ⊢ y ∈ ℋ → norm ℎ ⁡ y ≤ 1 ∧ x = norm ℎ ⁡ 0 hop ⁡ y ↔ norm ℎ ⁡ y ≤ 1 ∧ x = 0
10 9 rexbiia ⊢ ∃ y ∈ ℋ norm ℎ ⁡ y ≤ 1 ∧ x = norm ℎ ⁡ 0 hop ⁡ y ↔ ∃ y ∈ ℋ norm ℎ ⁡ y ≤ 1 ∧ x = 0
11 ax-hv0cl ⊢ 0 ℎ ∈ ℋ
12 0le1 ⊢ 0 ≤ 1
13 fveq2 ⊢ y = 0 ℎ → norm ℎ ⁡ y = norm ℎ ⁡ 0 ℎ
14 13 6 eqtrdi ⊢ y = 0 ℎ → norm ℎ ⁡ y = 0
15 14 breq1d ⊢ y = 0 ℎ → norm ℎ ⁡ y ≤ 1 ↔ 0 ≤ 1
16 15 rspcev ⊢ 0 ℎ ∈ ℋ ∧ 0 ≤ 1 → ∃ y ∈ ℋ norm ℎ ⁡ y ≤ 1
17 11 12 16 mp2an ⊢ ∃ y ∈ ℋ norm ℎ ⁡ y ≤ 1
18 r19.41v ⊢ ∃ y ∈ ℋ norm ℎ ⁡ y ≤ 1 ∧ x = 0 ↔ ∃ y ∈ ℋ norm ℎ ⁡ y ≤ 1 ∧ x = 0
19 17 18 mpbiran ⊢ ∃ y ∈ ℋ norm ℎ ⁡ y ≤ 1 ∧ x = 0 ↔ x = 0
20 10 19 bitri ⊢ ∃ y ∈ ℋ norm ℎ ⁡ y ≤ 1 ∧ x = norm ℎ ⁡ 0 hop ⁡ y ↔ x = 0
21 20 abbii ⊢ x | ∃ y ∈ ℋ norm ℎ ⁡ y ≤ 1 ∧ x = norm ℎ ⁡ 0 hop ⁡ y = x | x = 0
22 df-sn ⊢ 0 = x | x = 0
23 21 22 eqtr4i ⊢ x | ∃ y ∈ ℋ norm ℎ ⁡ y ≤ 1 ∧ x = norm ℎ ⁡ 0 hop ⁡ y = 0
24 23 supeq1i ⊢ sup x | ∃ y ∈ ℋ norm ℎ ⁡ y ≤ 1 ∧ x = norm ℎ ⁡ 0 hop ⁡ y ℝ * < = sup 0 ℝ * <
25 xrltso ⊢ < Or ℝ *
26 0xr ⊢ 0 ∈ ℝ *
27 supsn ⊢ < Or ℝ * ∧ 0 ∈ ℝ * → sup 0 ℝ * < = 0
28 25 26 27 mp2an ⊢ sup 0 ℝ * < = 0
29 3 24 28 3eqtri ⊢ norm op ⁡ 0 hop = 0