Metamath Proof Explorer


Theorem iscmet3lem3

Description: Lemma for iscmet3 . (Contributed by Mario Carneiro, 15-Oct-2015)

Ref Expression
Hypothesis iscmet3.1 ⊢ Z = ℤ ≥ M
Assertion iscmet3lem3 ⊢ M ∈ ℤ ∧ R ∈ ℝ + → ∃ j ∈ Z ∀ k ∈ ℤ ≥ j 1 2 k < R

Proof

Step Hyp Ref Expression
1 iscmet3.1 ⊢ Z = ℤ ≥ M
2 simpl ⊢ M ∈ ℤ ∧ R ∈ ℝ + → M ∈ ℤ
3 simpr ⊢ M ∈ ℤ ∧ R ∈ ℝ + → R ∈ ℝ +
4 eluzelz ⊢ k ∈ ℤ ≥ M → k ∈ ℤ
5 4 1 eleq2s ⊢ k ∈ Z → k ∈ ℤ
6 5 adantl ⊢ M ∈ ℤ ∧ R ∈ ℝ + ∧ k ∈ Z → k ∈ ℤ
7 oveq2 ⊢ n = k → 1 2 n = 1 2 k
8 eqid ⊢ n ∈ ℤ ⟼ 1 2 n = n ∈ ℤ ⟼ 1 2 n
9 ovex ⊢ 1 2 k ∈ V
10 7 8 9 fvmpt ⊢ k ∈ ℤ → n ∈ ℤ ⟼ 1 2 n ⁡ k = 1 2 k
11 6 10 syl ⊢ M ∈ ℤ ∧ R ∈ ℝ + ∧ k ∈ Z → n ∈ ℤ ⟼ 1 2 n ⁡ k = 1 2 k
12 nn0uz ⊢ ℕ 0 = ℤ ≥ 0
13 12 reseq2i ⊢ n ∈ ℤ ⟼ 1 2 n ↾ ℕ 0 = n ∈ ℤ ⟼ 1 2 n ↾ ℤ ≥ 0
14 nn0ssz ⊢ ℕ 0 ⊆ ℤ
15 resmpt ⊢ ℕ 0 ⊆ ℤ → n ∈ ℤ ⟼ 1 2 n ↾ ℕ 0 = n ∈ ℕ 0 ⟼ 1 2 n
16 14 15 ax-mp ⊢ n ∈ ℤ ⟼ 1 2 n ↾ ℕ 0 = n ∈ ℕ 0 ⟼ 1 2 n
17 13 16 eqtr3i ⊢ n ∈ ℤ ⟼ 1 2 n ↾ ℤ ≥ 0 = n ∈ ℕ 0 ⟼ 1 2 n
18 halfcn ⊢ 1 2 ∈ ℂ
19 18 a1i ⊢ M ∈ ℤ ∧ R ∈ ℝ + → 1 2 ∈ ℂ
20 halfre ⊢ 1 2 ∈ ℝ
21 halfge0 ⊢ 0 ≤ 1 2
22 absid ⊢ 1 2 ∈ ℝ ∧ 0 ≤ 1 2 → 1 2 = 1 2
23 20 21 22 mp2an ⊢ 1 2 = 1 2
24 halflt1 ⊢ 1 2 < 1
25 23 24 eqbrtri ⊢ 1 2 < 1
26 25 a1i ⊢ M ∈ ℤ ∧ R ∈ ℝ + → 1 2 < 1
27 19 26 expcnv ⊢ M ∈ ℤ ∧ R ∈ ℝ + → n ∈ ℕ 0 ⟼ 1 2 n ⇝ 0
28 17 27 eqbrtrid ⊢ M ∈ ℤ ∧ R ∈ ℝ + → n ∈ ℤ ⟼ 1 2 n ↾ ℤ ≥ 0 ⇝ 0
29 0z ⊢ 0 ∈ ℤ
30 zex ⊢ ℤ ∈ V
31 30 mptex ⊢ n ∈ ℤ ⟼ 1 2 n ∈ V
32 31 a1i ⊢ M ∈ ℤ ∧ R ∈ ℝ + → n ∈ ℤ ⟼ 1 2 n ∈ V
33 climres ⊢ 0 ∈ ℤ ∧ n ∈ ℤ ⟼ 1 2 n ∈ V → n ∈ ℤ ⟼ 1 2 n ↾ ℤ ≥ 0 ⇝ 0 ↔ n ∈ ℤ ⟼ 1 2 n ⇝ 0
34 29 32 33 sylancr ⊢ M ∈ ℤ ∧ R ∈ ℝ + → n ∈ ℤ ⟼ 1 2 n ↾ ℤ ≥ 0 ⇝ 0 ↔ n ∈ ℤ ⟼ 1 2 n ⇝ 0
35 28 34 mpbid ⊢ M ∈ ℤ ∧ R ∈ ℝ + → n ∈ ℤ ⟼ 1 2 n ⇝ 0
36 1 2 3 11 35 climi0 ⊢ M ∈ ℤ ∧ R ∈ ℝ + → ∃ j ∈ Z ∀ k ∈ ℤ ≥ j 1 2 k < R
37 1 uztrn2 ⊢ j ∈ Z ∧ k ∈ ℤ ≥ j → k ∈ Z
38 1rp ⊢ 1 ∈ ℝ +
39 rphalfcl ⊢ 1 ∈ ℝ + → 1 2 ∈ ℝ +
40 38 39 ax-mp ⊢ 1 2 ∈ ℝ +
41 rpexpcl ⊢ 1 2 ∈ ℝ + ∧ k ∈ ℤ → 1 2 k ∈ ℝ +
42 40 6 41 sylancr ⊢ M ∈ ℤ ∧ R ∈ ℝ + ∧ k ∈ Z → 1 2 k ∈ ℝ +
43 rpre ⊢ 1 2 k ∈ ℝ + → 1 2 k ∈ ℝ
44 rpge0 ⊢ 1 2 k ∈ ℝ + → 0 ≤ 1 2 k
45 43 44 absidd ⊢ 1 2 k ∈ ℝ + → 1 2 k = 1 2 k
46 42 45 syl ⊢ M ∈ ℤ ∧ R ∈ ℝ + ∧ k ∈ Z → 1 2 k = 1 2 k
47 46 breq1d ⊢ M ∈ ℤ ∧ R ∈ ℝ + ∧ k ∈ Z → 1 2 k < R ↔ 1 2 k < R
48 37 47 sylan2 ⊢ M ∈ ℤ ∧ R ∈ ℝ + ∧ j ∈ Z ∧ k ∈ ℤ ≥ j → 1 2 k < R ↔ 1 2 k < R
49 48 anassrs ⊢ M ∈ ℤ ∧ R ∈ ℝ + ∧ j ∈ Z ∧ k ∈ ℤ ≥ j → 1 2 k < R ↔ 1 2 k < R
50 49 ralbidva ⊢ M ∈ ℤ ∧ R ∈ ℝ + ∧ j ∈ Z → ∀ k ∈ ℤ ≥ j 1 2 k < R ↔ ∀ k ∈ ℤ ≥ j 1 2 k < R
51 50 rexbidva ⊢ M ∈ ℤ ∧ R ∈ ℝ + → ∃ j ∈ Z ∀ k ∈ ℤ ≥ j 1 2 k < R ↔ ∃ j ∈ Z ∀ k ∈ ℤ ≥ j 1 2 k < R
52 36 51 mpbid ⊢ M ∈ ℤ ∧ R ∈ ℝ + → ∃ j ∈ Z ∀ k ∈ ℤ ≥ j 1 2 k < R