Metamath Proof Explorer


Theorem ovolshftlem2

Description: Lemma for ovolshft . (Contributed by Mario Carneiro, 22-Mar-2014)

Ref Expression
Hypotheses ovolshft.1 ⊢ φ → A ⊆ ℝ
ovolshft.2 ⊢ φ → C ∈ ℝ
ovolshft.3 ⊢ φ → B = x ∈ ℝ | x − C ∈ A
ovolshft.4 ⊢ M = y ∈ ℝ * | ∃ f ∈ ≤ ∩ ℝ 2 ℕ B ⊆ ⋃ ran ⁡ . ∘ f ∧ y = sup ran ⁡ seq 1 + abs ∘ − ∘ f ℝ * <
Assertion ovolshftlem2 ⊢ φ → z ∈ ℝ * | ∃ g ∈ ≤ ∩ ℝ 2 ℕ A ⊆ ⋃ ran ⁡ . ∘ g ∧ z = sup ran ⁡ seq 1 + abs ∘ − ∘ g ℝ * < ⊆ M

Proof

Step Hyp Ref Expression
1 ovolshft.1 ⊢ φ → A ⊆ ℝ
2 ovolshft.2 ⊢ φ → C ∈ ℝ
3 ovolshft.3 ⊢ φ → B = x ∈ ℝ | x − C ∈ A
4 ovolshft.4 ⊢ M = y ∈ ℝ * | ∃ f ∈ ≤ ∩ ℝ 2 ℕ B ⊆ ⋃ ran ⁡ . ∘ f ∧ y = sup ran ⁡ seq 1 + abs ∘ − ∘ f ℝ * <
5 1 ad3antrrr ⊢ φ ∧ z ∈ ℝ * ∧ g ∈ ≤ ∩ ℝ 2 ℕ ∧ A ⊆ ⋃ ran ⁡ . ∘ g → A ⊆ ℝ
6 2 ad3antrrr ⊢ φ ∧ z ∈ ℝ * ∧ g ∈ ≤ ∩ ℝ 2 ℕ ∧ A ⊆ ⋃ ran ⁡ . ∘ g → C ∈ ℝ
7 3 ad3antrrr ⊢ φ ∧ z ∈ ℝ * ∧ g ∈ ≤ ∩ ℝ 2 ℕ ∧ A ⊆ ⋃ ran ⁡ . ∘ g → B = x ∈ ℝ | x − C ∈ A
8 eqid ⊢ seq 1 + abs ∘ − ∘ g = seq 1 + abs ∘ − ∘ g
9 2fveq3 ⊢ m = n → 1 st ⁡ g ⁡ m = 1 st ⁡ g ⁡ n
10 9 oveq1d ⊢ m = n → 1 st ⁡ g ⁡ m + C = 1 st ⁡ g ⁡ n + C
11 2fveq3 ⊢ m = n → 2 nd ⁡ g ⁡ m = 2 nd ⁡ g ⁡ n
12 11 oveq1d ⊢ m = n → 2 nd ⁡ g ⁡ m + C = 2 nd ⁡ g ⁡ n + C
13 10 12 opeq12d ⊢ m = n → 1 st ⁡ g ⁡ m + C 2 nd ⁡ g ⁡ m + C = 1 st ⁡ g ⁡ n + C 2 nd ⁡ g ⁡ n + C
14 13 cbvmptv ⊢ m ∈ ℕ ⟼ 1 st ⁡ g ⁡ m + C 2 nd ⁡ g ⁡ m + C = n ∈ ℕ ⟼ 1 st ⁡ g ⁡ n + C 2 nd ⁡ g ⁡ n + C
15 simplr ⊢ φ ∧ z ∈ ℝ * ∧ g ∈ ≤ ∩ ℝ 2 ℕ ∧ A ⊆ ⋃ ran ⁡ . ∘ g → g ∈ ≤ ∩ ℝ 2 ℕ
16 elovolmlem ⊢ g ∈ ≤ ∩ ℝ 2 ℕ ↔ g : ℕ ⟶ ≤ ∩ ℝ 2
17 15 16 sylib ⊢ φ ∧ z ∈ ℝ * ∧ g ∈ ≤ ∩ ℝ 2 ℕ ∧ A ⊆ ⋃ ran ⁡ . ∘ g → g : ℕ ⟶ ≤ ∩ ℝ 2
18 simpr ⊢ φ ∧ z ∈ ℝ * ∧ g ∈ ≤ ∩ ℝ 2 ℕ ∧ A ⊆ ⋃ ran ⁡ . ∘ g → A ⊆ ⋃ ran ⁡ . ∘ g
19 5 6 7 4 8 14 17 18 ovolshftlem1 ⊢ φ ∧ z ∈ ℝ * ∧ g ∈ ≤ ∩ ℝ 2 ℕ ∧ A ⊆ ⋃ ran ⁡ . ∘ g → sup ran ⁡ seq 1 + abs ∘ − ∘ g ℝ * < ∈ M
20 eleq1a ⊢ sup ran ⁡ seq 1 + abs ∘ − ∘ g ℝ * < ∈ M → z = sup ran ⁡ seq 1 + abs ∘ − ∘ g ℝ * < → z ∈ M
21 19 20 syl ⊢ φ ∧ z ∈ ℝ * ∧ g ∈ ≤ ∩ ℝ 2 ℕ ∧ A ⊆ ⋃ ran ⁡ . ∘ g → z = sup ran ⁡ seq 1 + abs ∘ − ∘ g ℝ * < → z ∈ M
22 21 expimpd ⊢ φ ∧ z ∈ ℝ * ∧ g ∈ ≤ ∩ ℝ 2 ℕ → A ⊆ ⋃ ran ⁡ . ∘ g ∧ z = sup ran ⁡ seq 1 + abs ∘ − ∘ g ℝ * < → z ∈ M
23 22 rexlimdva ⊢ φ ∧ z ∈ ℝ * → ∃ g ∈ ≤ ∩ ℝ 2 ℕ A ⊆ ⋃ ran ⁡ . ∘ g ∧ z = sup ran ⁡ seq 1 + abs ∘ − ∘ g ℝ * < → z ∈ M
24 23 ralrimiva ⊢ φ → ∀ z ∈ ℝ * ∃ g ∈ ≤ ∩ ℝ 2 ℕ A ⊆ ⋃ ran ⁡ . ∘ g ∧ z = sup ran ⁡ seq 1 + abs ∘ − ∘ g ℝ * < → z ∈ M
25 rabss ⊢ z ∈ ℝ * | ∃ g ∈ ≤ ∩ ℝ 2 ℕ A ⊆ ⋃ ran ⁡ . ∘ g ∧ z = sup ran ⁡ seq 1 + abs ∘ − ∘ g ℝ * < ⊆ M ↔ ∀ z ∈ ℝ * ∃ g ∈ ≤ ∩ ℝ 2 ℕ A ⊆ ⋃ ran ⁡ . ∘ g ∧ z = sup ran ⁡ seq 1 + abs ∘ − ∘ g ℝ * < → z ∈ M
26 24 25 sylibr ⊢ φ → z ∈ ℝ * | ∃ g ∈ ≤ ∩ ℝ 2 ℕ A ⊆ ⋃ ran ⁡ . ∘ g ∧ z = sup ran ⁡ seq 1 + abs ∘ − ∘ g ℝ * < ⊆ M