Metamath Proof Explorer


Theorem swrdnnn0nd

Description: The value of a subword operation for arguments not being nonnegative integers is the empty set. (Contributed by AV, 2-Dec-2022)

Ref Expression
Assertion swrdnnn0nd ⊢ S ∈ Word V ∧ ¬ F ∈ ℕ 0 ∧ L ∈ ℕ 0 → S substr F L = ∅

Proof

Step Hyp Ref Expression
1 ianor ⊢ ¬ F ∈ ℕ 0 ∧ L ∈ ℕ 0 ↔ ¬ F ∈ ℕ 0 ∨ ¬ L ∈ ℕ 0
2 ianor ⊢ ¬ F ∈ ℤ ∧ 0 ≤ F ↔ ¬ F ∈ ℤ ∨ ¬ 0 ≤ F
3 elnn0z ⊢ F ∈ ℕ 0 ↔ F ∈ ℤ ∧ 0 ≤ F
4 2 3 xchnxbir ⊢ ¬ F ∈ ℕ 0 ↔ ¬ F ∈ ℤ ∨ ¬ 0 ≤ F
5 ianor ⊢ ¬ L ∈ ℤ ∧ 0 ≤ L ↔ ¬ L ∈ ℤ ∨ ¬ 0 ≤ L
6 elnn0z ⊢ L ∈ ℕ 0 ↔ L ∈ ℤ ∧ 0 ≤ L
7 5 6 xchnxbir ⊢ ¬ L ∈ ℕ 0 ↔ ¬ L ∈ ℤ ∨ ¬ 0 ≤ L
8 4 7 orbi12i ⊢ ¬ F ∈ ℕ 0 ∨ ¬ L ∈ ℕ 0 ↔ ¬ F ∈ ℤ ∨ ¬ 0 ≤ F ∨ ¬ L ∈ ℤ ∨ ¬ 0 ≤ L
9 or4 ⊢ ¬ F ∈ ℤ ∨ ¬ 0 ≤ F ∨ ¬ L ∈ ℤ ∨ ¬ 0 ≤ L ↔ ¬ F ∈ ℤ ∨ ¬ L ∈ ℤ ∨ ¬ 0 ≤ F ∨ ¬ 0 ≤ L
10 ianor ⊢ ¬ F ∈ ℤ ∧ L ∈ ℤ ↔ ¬ F ∈ ℤ ∨ ¬ L ∈ ℤ
11 10 bicomi ⊢ ¬ F ∈ ℤ ∨ ¬ L ∈ ℤ ↔ ¬ F ∈ ℤ ∧ L ∈ ℤ
12 11 orbi1i ⊢ ¬ F ∈ ℤ ∨ ¬ L ∈ ℤ ∨ ¬ 0 ≤ F ∨ ¬ 0 ≤ L ↔ ¬ F ∈ ℤ ∧ L ∈ ℤ ∨ ¬ 0 ≤ F ∨ ¬ 0 ≤ L
13 9 12 bitri ⊢ ¬ F ∈ ℤ ∨ ¬ 0 ≤ F ∨ ¬ L ∈ ℤ ∨ ¬ 0 ≤ L ↔ ¬ F ∈ ℤ ∧ L ∈ ℤ ∨ ¬ 0 ≤ F ∨ ¬ 0 ≤ L
14 8 13 bitri ⊢ ¬ F ∈ ℕ 0 ∨ ¬ L ∈ ℕ 0 ↔ ¬ F ∈ ℤ ∧ L ∈ ℤ ∨ ¬ 0 ≤ F ∨ ¬ 0 ≤ L
15 1 14 bitri ⊢ ¬ F ∈ ℕ 0 ∧ L ∈ ℕ 0 ↔ ¬ F ∈ ℤ ∧ L ∈ ℤ ∨ ¬ 0 ≤ F ∨ ¬ 0 ≤ L
16 swrdnznd ⊢ ¬ F ∈ ℤ ∧ L ∈ ℤ → S substr F L = ∅
17 16 a1d ⊢ ¬ F ∈ ℤ ∧ L ∈ ℤ → S ∈ Word V → S substr F L = ∅
18 notnotb ⊢ F ∈ ℤ ∧ L ∈ ℤ ↔ ¬ ¬ F ∈ ℤ ∧ L ∈ ℤ
19 zre ⊢ F ∈ ℤ → F ∈ ℝ
20 0red ⊢ F ∈ ℤ → 0 ∈ ℝ
21 19 20 jca ⊢ F ∈ ℤ → F ∈ ℝ ∧ 0 ∈ ℝ
22 21 3ad2ant2 ⊢ S ∈ Word V ∧ F ∈ ℤ ∧ L ∈ ℤ → F ∈ ℝ ∧ 0 ∈ ℝ
23 ltnle ⊢ F ∈ ℝ ∧ 0 ∈ ℝ → F < 0 ↔ ¬ 0 ≤ F
24 22 23 syl ⊢ S ∈ Word V ∧ F ∈ ℤ ∧ L ∈ ℤ → F < 0 ↔ ¬ 0 ≤ F
25 orc ⊢ F < 0 → F < 0 ∨ L ≤ F
26 24 25 biimtrrdi ⊢ S ∈ Word V ∧ F ∈ ℤ ∧ L ∈ ℤ → ¬ 0 ≤ F → F < 0 ∨ L ≤ F
27 26 com12 ⊢ ¬ 0 ≤ F → S ∈ Word V ∧ F ∈ ℤ ∧ L ∈ ℤ → F < 0 ∨ L ≤ F
28 notnotb ⊢ 0 ≤ F ↔ ¬ ¬ 0 ≤ F
29 28 a1i ⊢ S ∈ Word V ∧ F ∈ ℤ ∧ L ∈ ℤ → 0 ≤ F ↔ ¬ ¬ 0 ≤ F
30 zre ⊢ L ∈ ℤ → L ∈ ℝ
31 0red ⊢ L ∈ ℤ → 0 ∈ ℝ
32 30 31 jca ⊢ L ∈ ℤ → L ∈ ℝ ∧ 0 ∈ ℝ
33 32 3ad2ant3 ⊢ S ∈ Word V ∧ F ∈ ℤ ∧ L ∈ ℤ → L ∈ ℝ ∧ 0 ∈ ℝ
34 ltnle ⊢ L ∈ ℝ ∧ 0 ∈ ℝ → L < 0 ↔ ¬ 0 ≤ L
35 33 34 syl ⊢ S ∈ Word V ∧ F ∈ ℤ ∧ L ∈ ℤ → L < 0 ↔ ¬ 0 ≤ L
36 29 35 anbi12d ⊢ S ∈ Word V ∧ F ∈ ℤ ∧ L ∈ ℤ → 0 ≤ F ∧ L < 0 ↔ ¬ ¬ 0 ≤ F ∧ ¬ 0 ≤ L
37 30 3ad2ant3 ⊢ S ∈ Word V ∧ F ∈ ℤ ∧ L ∈ ℤ → L ∈ ℝ
38 0red ⊢ S ∈ Word V ∧ F ∈ ℤ ∧ L ∈ ℤ → 0 ∈ ℝ
39 19 3ad2ant2 ⊢ S ∈ Word V ∧ F ∈ ℤ ∧ L ∈ ℤ → F ∈ ℝ
40 ltleletr ⊢ L ∈ ℝ ∧ 0 ∈ ℝ ∧ F ∈ ℝ → L < 0 ∧ 0 ≤ F → L ≤ F
41 37 38 39 40 syl3anc ⊢ S ∈ Word V ∧ F ∈ ℤ ∧ L ∈ ℤ → L < 0 ∧ 0 ≤ F → L ≤ F
42 olc ⊢ L ≤ F → F < 0 ∨ L ≤ F
43 41 42 syl6 ⊢ S ∈ Word V ∧ F ∈ ℤ ∧ L ∈ ℤ → L < 0 ∧ 0 ≤ F → F < 0 ∨ L ≤ F
44 43 ancomsd ⊢ S ∈ Word V ∧ F ∈ ℤ ∧ L ∈ ℤ → 0 ≤ F ∧ L < 0 → F < 0 ∨ L ≤ F
45 36 44 sylbird ⊢ S ∈ Word V ∧ F ∈ ℤ ∧ L ∈ ℤ → ¬ ¬ 0 ≤ F ∧ ¬ 0 ≤ L → F < 0 ∨ L ≤ F
46 45 com12 ⊢ ¬ ¬ 0 ≤ F ∧ ¬ 0 ≤ L → S ∈ Word V ∧ F ∈ ℤ ∧ L ∈ ℤ → F < 0 ∨ L ≤ F
47 27 46 jaoi3 ⊢ ¬ 0 ≤ F ∨ ¬ 0 ≤ L → S ∈ Word V ∧ F ∈ ℤ ∧ L ∈ ℤ → F < 0 ∨ L ≤ F
48 47 impcom ⊢ S ∈ Word V ∧ F ∈ ℤ ∧ L ∈ ℤ ∧ ¬ 0 ≤ F ∨ ¬ 0 ≤ L → F < 0 ∨ L ≤ F
49 48 orcd ⊢ S ∈ Word V ∧ F ∈ ℤ ∧ L ∈ ℤ ∧ ¬ 0 ≤ F ∨ ¬ 0 ≤ L → F < 0 ∨ L ≤ F ∨ S < L
50 df-3or ⊢ F < 0 ∨ L ≤ F ∨ S < L ↔ F < 0 ∨ L ≤ F ∨ S < L
51 49 50 sylibr ⊢ S ∈ Word V ∧ F ∈ ℤ ∧ L ∈ ℤ ∧ ¬ 0 ≤ F ∨ ¬ 0 ≤ L → F < 0 ∨ L ≤ F ∨ S < L
52 swrdnd ⊢ S ∈ Word V ∧ F ∈ ℤ ∧ L ∈ ℤ → F < 0 ∨ L ≤ F ∨ S < L → S substr F L = ∅
53 52 imp ⊢ S ∈ Word V ∧ F ∈ ℤ ∧ L ∈ ℤ ∧ F < 0 ∨ L ≤ F ∨ S < L → S substr F L = ∅
54 51 53 syldan ⊢ S ∈ Word V ∧ F ∈ ℤ ∧ L ∈ ℤ ∧ ¬ 0 ≤ F ∨ ¬ 0 ≤ L → S substr F L = ∅
55 54 ex ⊢ S ∈ Word V ∧ F ∈ ℤ ∧ L ∈ ℤ → ¬ 0 ≤ F ∨ ¬ 0 ≤ L → S substr F L = ∅
56 55 3expb ⊢ S ∈ Word V ∧ F ∈ ℤ ∧ L ∈ ℤ → ¬ 0 ≤ F ∨ ¬ 0 ≤ L → S substr F L = ∅
57 56 expcom ⊢ F ∈ ℤ ∧ L ∈ ℤ → S ∈ Word V → ¬ 0 ≤ F ∨ ¬ 0 ≤ L → S substr F L = ∅
58 57 com23 ⊢ F ∈ ℤ ∧ L ∈ ℤ → ¬ 0 ≤ F ∨ ¬ 0 ≤ L → S ∈ Word V → S substr F L = ∅
59 18 58 sylbir ⊢ ¬ ¬ F ∈ ℤ ∧ L ∈ ℤ → ¬ 0 ≤ F ∨ ¬ 0 ≤ L → S ∈ Word V → S substr F L = ∅
60 59 imp ⊢ ¬ ¬ F ∈ ℤ ∧ L ∈ ℤ ∧ ¬ 0 ≤ F ∨ ¬ 0 ≤ L → S ∈ Word V → S substr F L = ∅
61 17 60 jaoi3 ⊢ ¬ F ∈ ℤ ∧ L ∈ ℤ ∨ ¬ 0 ≤ F ∨ ¬ 0 ≤ L → S ∈ Word V → S substr F L = ∅
62 15 61 sylbi ⊢ ¬ F ∈ ℕ 0 ∧ L ∈ ℕ 0 → S ∈ Word V → S substr F L = ∅
63 62 impcom ⊢ S ∈ Word V ∧ ¬ F ∈ ℕ 0 ∧ L ∈ ℕ 0 → S substr F L = ∅