Metamath Proof Explorer


Theorem divalglem10

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

Ref Expression
Hypotheses divalglem8.1 ⊢ N ∈ ℤ
divalglem8.2 ⊢ D ∈ ℤ
divalglem8.3 ⊢ D ≠ 0
divalglem8.4 ⊢ S = r ∈ ℕ 0 | D ∥ N − r
Assertion divalglem10 ⊢ ∃! r ∈ ℤ ∃ q ∈ ℤ 0 ≤ r ∧ r < D ∧ N = q ⁢ D + r

Proof

Step Hyp Ref Expression
1 divalglem8.1 ⊢ N ∈ ℤ
2 divalglem8.2 ⊢ D ∈ ℤ
3 divalglem8.3 ⊢ D ≠ 0
4 divalglem8.4 ⊢ S = r ∈ ℕ 0 | D ∥ N − r
5 eqid ⊢ inf S ℝ < = inf S ℝ <
6 1 2 3 4 5 divalglem9 ⊢ ∃! x ∈ S x < D
7 elnn0z ⊢ x ∈ ℕ 0 ↔ x ∈ ℤ ∧ 0 ≤ x
8 7 anbi2i ⊢ x < D ∧ x ∈ ℕ 0 ↔ x < D ∧ x ∈ ℤ ∧ 0 ≤ x
9 an12 ⊢ x < D ∧ x ∈ ℤ ∧ 0 ≤ x ↔ x ∈ ℤ ∧ x < D ∧ 0 ≤ x
10 ancom ⊢ x < D ∧ 0 ≤ x ↔ 0 ≤ x ∧ x < D
11 10 anbi2i ⊢ x ∈ ℤ ∧ x < D ∧ 0 ≤ x ↔ x ∈ ℤ ∧ 0 ≤ x ∧ x < D
12 9 11 bitri ⊢ x < D ∧ x ∈ ℤ ∧ 0 ≤ x ↔ x ∈ ℤ ∧ 0 ≤ x ∧ x < D
13 8 12 bitri ⊢ x < D ∧ x ∈ ℕ 0 ↔ x ∈ ℤ ∧ 0 ≤ x ∧ x < D
14 13 anbi1i ⊢ x < D ∧ x ∈ ℕ 0 ∧ ∃ q ∈ ℤ N = q ⁢ D + x ↔ x ∈ ℤ ∧ 0 ≤ x ∧ x < D ∧ ∃ q ∈ ℤ N = q ⁢ D + x
15 anass ⊢ x ∈ ℤ ∧ 0 ≤ x ∧ x < D ∧ ∃ q ∈ ℤ N = q ⁢ D + x ↔ x ∈ ℤ ∧ 0 ≤ x ∧ x < D ∧ ∃ q ∈ ℤ N = q ⁢ D + x
16 14 15 bitri ⊢ x < D ∧ x ∈ ℕ 0 ∧ ∃ q ∈ ℤ N = q ⁢ D + x ↔ x ∈ ℤ ∧ 0 ≤ x ∧ x < D ∧ ∃ q ∈ ℤ N = q ⁢ D + x
17 oveq2 ⊢ r = x → q ⁢ D + r = q ⁢ D + x
18 17 eqeq2d ⊢ r = x → N = q ⁢ D + r ↔ N = q ⁢ D + x
19 18 rexbidv ⊢ r = x → ∃ q ∈ ℤ N = q ⁢ D + r ↔ ∃ q ∈ ℤ N = q ⁢ D + x
20 1 2 3 4 divalglem4 ⊢ S = r ∈ ℕ 0 | ∃ q ∈ ℤ N = q ⁢ D + r
21 19 20 elrab2 ⊢ x ∈ S ↔ x ∈ ℕ 0 ∧ ∃ q ∈ ℤ N = q ⁢ D + x
22 21 anbi2i ⊢ x < D ∧ x ∈ S ↔ x < D ∧ x ∈ ℕ 0 ∧ ∃ q ∈ ℤ N = q ⁢ D + x
23 ancom ⊢ x ∈ S ∧ x < D ↔ x < D ∧ x ∈ S
24 anass ⊢ x < D ∧ x ∈ ℕ 0 ∧ ∃ q ∈ ℤ N = q ⁢ D + x ↔ x < D ∧ x ∈ ℕ 0 ∧ ∃ q ∈ ℤ N = q ⁢ D + x
25 22 23 24 3bitr4i ⊢ x ∈ S ∧ x < D ↔ x < D ∧ x ∈ ℕ 0 ∧ ∃ q ∈ ℤ N = q ⁢ D + x
26 df-3an ⊢ 0 ≤ x ∧ x < D ∧ N = q ⁢ D + x ↔ 0 ≤ x ∧ x < D ∧ N = q ⁢ D + x
27 26 rexbii ⊢ ∃ q ∈ ℤ 0 ≤ x ∧ x < D ∧ N = q ⁢ D + x ↔ ∃ q ∈ ℤ 0 ≤ x ∧ x < D ∧ N = q ⁢ D + x
28 r19.42v ⊢ ∃ q ∈ ℤ 0 ≤ x ∧ x < D ∧ N = q ⁢ D + x ↔ 0 ≤ x ∧ x < D ∧ ∃ q ∈ ℤ N = q ⁢ D + x
29 27 28 bitri ⊢ ∃ q ∈ ℤ 0 ≤ x ∧ x < D ∧ N = q ⁢ D + x ↔ 0 ≤ x ∧ x < D ∧ ∃ q ∈ ℤ N = q ⁢ D + x
30 29 anbi2i ⊢ x ∈ ℤ ∧ ∃ q ∈ ℤ 0 ≤ x ∧ x < D ∧ N = q ⁢ D + x ↔ x ∈ ℤ ∧ 0 ≤ x ∧ x < D ∧ ∃ q ∈ ℤ N = q ⁢ D + x
31 16 25 30 3bitr4i ⊢ x ∈ S ∧ x < D ↔ x ∈ ℤ ∧ ∃ q ∈ ℤ 0 ≤ x ∧ x < D ∧ N = q ⁢ D + x
32 31 eubii ⊢ ∃! x x ∈ S ∧ x < D ↔ ∃! x x ∈ ℤ ∧ ∃ q ∈ ℤ 0 ≤ x ∧ x < D ∧ N = q ⁢ D + x
33 df-reu ⊢ ∃! x ∈ S x < D ↔ ∃! x x ∈ S ∧ x < D
34 df-reu ⊢ ∃! x ∈ ℤ ∃ q ∈ ℤ 0 ≤ x ∧ x < D ∧ N = q ⁢ D + x ↔ ∃! x x ∈ ℤ ∧ ∃ q ∈ ℤ 0 ≤ x ∧ x < D ∧ N = q ⁢ D + x
35 32 33 34 3bitr4i ⊢ ∃! x ∈ S x < D ↔ ∃! x ∈ ℤ ∃ q ∈ ℤ 0 ≤ x ∧ x < D ∧ N = q ⁢ D + x
36 6 35 mpbi ⊢ ∃! x ∈ ℤ ∃ q ∈ ℤ 0 ≤ x ∧ x < D ∧ N = q ⁢ D + x
37 breq2 ⊢ x = r → 0 ≤ x ↔ 0 ≤ r
38 breq1 ⊢ x = r → x < D ↔ r < D
39 oveq2 ⊢ x = r → q ⁢ D + x = q ⁢ D + r
40 39 eqeq2d ⊢ x = r → N = q ⁢ D + x ↔ N = q ⁢ D + r
41 37 38 40 3anbi123d ⊢ x = r → 0 ≤ x ∧ x < D ∧ N = q ⁢ D + x ↔ 0 ≤ r ∧ r < D ∧ N = q ⁢ D + r
42 41 rexbidv ⊢ x = r → ∃ q ∈ ℤ 0 ≤ x ∧ x < D ∧ N = q ⁢ D + x ↔ ∃ q ∈ ℤ 0 ≤ r ∧ r < D ∧ N = q ⁢ D + r
43 42 cbvreuvw ⊢ ∃! x ∈ ℤ ∃ q ∈ ℤ 0 ≤ x ∧ x < D ∧ N = q ⁢ D + x ↔ ∃! r ∈ ℤ ∃ q ∈ ℤ 0 ≤ r ∧ r < D ∧ N = q ⁢ D + r
44 36 43 mpbi ⊢ ∃! r ∈ ℤ ∃ q ∈ ℤ 0 ≤ r ∧ r < D ∧ N = q ⁢ D + r