Metamath Proof Explorer


Theorem divalglem5

Description: Lemma for divalg . (Contributed by Paul Chapman, 21-Mar-2011) (Revised by AV, 2-Oct-2020)

Ref Expression
Hypotheses divalglem0.1 ⊢ N ∈ ℤ
divalglem0.2 ⊢ D ∈ ℤ
divalglem1.3 ⊢ D ≠ 0
divalglem2.4 ⊢ S = r ∈ ℕ 0 | D ∥ N − r
divalglem5.5 ⊢ R = inf S ℝ <
Assertion divalglem5 ⊢ 0 ≤ R ∧ R < D

Proof

Step Hyp Ref Expression
1 divalglem0.1 ⊢ N ∈ ℤ
2 divalglem0.2 ⊢ D ∈ ℤ
3 divalglem1.3 ⊢ D ≠ 0
4 divalglem2.4 ⊢ S = r ∈ ℕ 0 | D ∥ N − r
5 divalglem5.5 ⊢ R = inf S ℝ <
6 1 2 3 4 divalglem2 ⊢ inf S ℝ < ∈ S
7 5 6 eqeltri ⊢ R ∈ S
8 oveq2 ⊢ x = R → N − x = N − R
9 8 breq2d ⊢ x = R → D ∥ N − x ↔ D ∥ N − R
10 oveq2 ⊢ r = x → N − r = N − x
11 10 breq2d ⊢ r = x → D ∥ N − r ↔ D ∥ N − x
12 11 cbvrabv ⊢ r ∈ ℕ 0 | D ∥ N − r = x ∈ ℕ 0 | D ∥ N − x
13 4 12 eqtri ⊢ S = x ∈ ℕ 0 | D ∥ N − x
14 9 13 elrab2 ⊢ R ∈ S ↔ R ∈ ℕ 0 ∧ D ∥ N − R
15 7 14 mpbi ⊢ R ∈ ℕ 0 ∧ D ∥ N − R
16 15 simpli ⊢ R ∈ ℕ 0
17 16 nn0ge0i ⊢ 0 ≤ R
18 nnabscl ⊢ D ∈ ℤ ∧ D ≠ 0 → D ∈ ℕ
19 2 3 18 mp2an ⊢ D ∈ ℕ
20 19 nngt0i ⊢ 0 < D
21 0re ⊢ 0 ∈ ℝ
22 zcn ⊢ D ∈ ℤ → D ∈ ℂ
23 2 22 ax-mp ⊢ D ∈ ℂ
24 23 abscli ⊢ D ∈ ℝ
25 21 24 ltnlei ⊢ 0 < D ↔ ¬ D ≤ 0
26 20 25 mpbi ⊢ ¬ D ≤ 0
27 4 ssrab3 ⊢ S ⊆ ℕ 0
28 nn0uz ⊢ ℕ 0 = ℤ ≥ 0
29 27 28 sseqtri ⊢ S ⊆ ℤ ≥ 0
30 nn0abscl ⊢ D ∈ ℤ → D ∈ ℕ 0
31 2 30 ax-mp ⊢ D ∈ ℕ 0
32 nn0sub2 ⊢ D ∈ ℕ 0 ∧ R ∈ ℕ 0 ∧ D ≤ R → R − D ∈ ℕ 0
33 31 16 32 mp3an12 ⊢ D ≤ R → R − D ∈ ℕ 0
34 15 a1i ⊢ D ≤ R → R ∈ ℕ 0 ∧ D ∥ N − R
35 nn0z ⊢ R ∈ ℕ 0 → R ∈ ℤ
36 1z ⊢ 1 ∈ ℤ
37 1 2 divalglem0 ⊢ R ∈ ℤ ∧ 1 ∈ ℤ → D ∥ N − R → D ∥ N − R − 1 ⁢ D
38 36 37 mpan2 ⊢ R ∈ ℤ → D ∥ N − R → D ∥ N − R − 1 ⁢ D
39 24 recni ⊢ D ∈ ℂ
40 39 mullidi ⊢ 1 ⁢ D = D
41 40 oveq2i ⊢ R − 1 ⁢ D = R − D
42 41 oveq2i ⊢ N − R − 1 ⁢ D = N − R − D
43 42 breq2i ⊢ D ∥ N − R − 1 ⁢ D ↔ D ∥ N − R − D
44 38 43 imbitrdi ⊢ R ∈ ℤ → D ∥ N − R → D ∥ N − R − D
45 35 44 syl ⊢ R ∈ ℕ 0 → D ∥ N − R → D ∥ N − R − D
46 45 imp ⊢ R ∈ ℕ 0 ∧ D ∥ N − R → D ∥ N − R − D
47 34 46 syl ⊢ D ≤ R → D ∥ N − R − D
48 oveq2 ⊢ x = R − D → N − x = N − R − D
49 48 breq2d ⊢ x = R − D → D ∥ N − x ↔ D ∥ N − R − D
50 49 13 elrab2 ⊢ R − D ∈ S ↔ R − D ∈ ℕ 0 ∧ D ∥ N − R − D
51 33 47 50 sylanbrc ⊢ D ≤ R → R − D ∈ S
52 infssuzle ⊢ S ⊆ ℤ ≥ 0 ∧ R − D ∈ S → inf S ℝ < ≤ R − D
53 29 51 52 sylancr ⊢ D ≤ R → inf S ℝ < ≤ R − D
54 5 53 eqbrtrid ⊢ D ≤ R → R ≤ R − D
55 34 simpld ⊢ D ≤ R → R ∈ ℕ 0
56 55 nn0red ⊢ D ≤ R → R ∈ ℝ
57 lesub ⊢ R ∈ ℝ ∧ R ∈ ℝ ∧ D ∈ ℝ → R ≤ R − D ↔ D ≤ R − R
58 24 57 mp3an3 ⊢ R ∈ ℝ ∧ R ∈ ℝ → R ≤ R − D ↔ D ≤ R − R
59 56 56 58 syl2anc ⊢ D ≤ R → R ≤ R − D ↔ D ≤ R − R
60 56 recnd ⊢ D ≤ R → R ∈ ℂ
61 60 subidd ⊢ D ≤ R → R − R = 0
62 61 breq2d ⊢ D ≤ R → D ≤ R − R ↔ D ≤ 0
63 59 62 bitrd ⊢ D ≤ R → R ≤ R − D ↔ D ≤ 0
64 54 63 mpbid ⊢ D ≤ R → D ≤ 0
65 26 64 mto ⊢ ¬ D ≤ R
66 16 nn0rei ⊢ R ∈ ℝ
67 66 24 ltnlei ⊢ R < D ↔ ¬ D ≤ R
68 65 67 mpbir ⊢ R < D
69 17 68 pm3.2i ⊢ 0 ≤ R ∧ R < D