Step |
Hyp |
Ref |
Expression |
1 |
|
simp1 |
⊢ ( ( 𝑆 ∈ Word 𝐴 ∧ 𝐹 ∈ ( 0 ... 𝐿 ) ∧ 𝐿 ∈ ( 0 ... ( ♯ ‘ 𝑆 ) ) ) → 𝑆 ∈ Word 𝐴 ) |
2 |
|
elfzelz |
⊢ ( 𝐹 ∈ ( 0 ... 𝐿 ) → 𝐹 ∈ ℤ ) |
3 |
2
|
3ad2ant2 |
⊢ ( ( 𝑆 ∈ Word 𝐴 ∧ 𝐹 ∈ ( 0 ... 𝐿 ) ∧ 𝐿 ∈ ( 0 ... ( ♯ ‘ 𝑆 ) ) ) → 𝐹 ∈ ℤ ) |
4 |
|
elfzelz |
⊢ ( 𝐿 ∈ ( 0 ... ( ♯ ‘ 𝑆 ) ) → 𝐿 ∈ ℤ ) |
5 |
4
|
3ad2ant3 |
⊢ ( ( 𝑆 ∈ Word 𝐴 ∧ 𝐹 ∈ ( 0 ... 𝐿 ) ∧ 𝐿 ∈ ( 0 ... ( ♯ ‘ 𝑆 ) ) ) → 𝐿 ∈ ℤ ) |
6 |
|
swrdval |
⊢ ( ( 𝑆 ∈ Word 𝐴 ∧ 𝐹 ∈ ℤ ∧ 𝐿 ∈ ℤ ) → ( 𝑆 substr 〈 𝐹 , 𝐿 〉 ) = if ( ( 𝐹 ..^ 𝐿 ) ⊆ dom 𝑆 , ( 𝑥 ∈ ( 0 ..^ ( 𝐿 − 𝐹 ) ) ↦ ( 𝑆 ‘ ( 𝑥 + 𝐹 ) ) ) , ∅ ) ) |
7 |
1 3 5 6
|
syl3anc |
⊢ ( ( 𝑆 ∈ Word 𝐴 ∧ 𝐹 ∈ ( 0 ... 𝐿 ) ∧ 𝐿 ∈ ( 0 ... ( ♯ ‘ 𝑆 ) ) ) → ( 𝑆 substr 〈 𝐹 , 𝐿 〉 ) = if ( ( 𝐹 ..^ 𝐿 ) ⊆ dom 𝑆 , ( 𝑥 ∈ ( 0 ..^ ( 𝐿 − 𝐹 ) ) ↦ ( 𝑆 ‘ ( 𝑥 + 𝐹 ) ) ) , ∅ ) ) |
8 |
|
elfzuz |
⊢ ( 𝐹 ∈ ( 0 ... 𝐿 ) → 𝐹 ∈ ( ℤ≥ ‘ 0 ) ) |
9 |
8
|
3ad2ant2 |
⊢ ( ( 𝑆 ∈ Word 𝐴 ∧ 𝐹 ∈ ( 0 ... 𝐿 ) ∧ 𝐿 ∈ ( 0 ... ( ♯ ‘ 𝑆 ) ) ) → 𝐹 ∈ ( ℤ≥ ‘ 0 ) ) |
10 |
|
fzoss1 |
⊢ ( 𝐹 ∈ ( ℤ≥ ‘ 0 ) → ( 𝐹 ..^ 𝐿 ) ⊆ ( 0 ..^ 𝐿 ) ) |
11 |
9 10
|
syl |
⊢ ( ( 𝑆 ∈ Word 𝐴 ∧ 𝐹 ∈ ( 0 ... 𝐿 ) ∧ 𝐿 ∈ ( 0 ... ( ♯ ‘ 𝑆 ) ) ) → ( 𝐹 ..^ 𝐿 ) ⊆ ( 0 ..^ 𝐿 ) ) |
12 |
|
elfzuz3 |
⊢ ( 𝐿 ∈ ( 0 ... ( ♯ ‘ 𝑆 ) ) → ( ♯ ‘ 𝑆 ) ∈ ( ℤ≥ ‘ 𝐿 ) ) |
13 |
12
|
3ad2ant3 |
⊢ ( ( 𝑆 ∈ Word 𝐴 ∧ 𝐹 ∈ ( 0 ... 𝐿 ) ∧ 𝐿 ∈ ( 0 ... ( ♯ ‘ 𝑆 ) ) ) → ( ♯ ‘ 𝑆 ) ∈ ( ℤ≥ ‘ 𝐿 ) ) |
14 |
|
fzoss2 |
⊢ ( ( ♯ ‘ 𝑆 ) ∈ ( ℤ≥ ‘ 𝐿 ) → ( 0 ..^ 𝐿 ) ⊆ ( 0 ..^ ( ♯ ‘ 𝑆 ) ) ) |
15 |
13 14
|
syl |
⊢ ( ( 𝑆 ∈ Word 𝐴 ∧ 𝐹 ∈ ( 0 ... 𝐿 ) ∧ 𝐿 ∈ ( 0 ... ( ♯ ‘ 𝑆 ) ) ) → ( 0 ..^ 𝐿 ) ⊆ ( 0 ..^ ( ♯ ‘ 𝑆 ) ) ) |
16 |
11 15
|
sstrd |
⊢ ( ( 𝑆 ∈ Word 𝐴 ∧ 𝐹 ∈ ( 0 ... 𝐿 ) ∧ 𝐿 ∈ ( 0 ... ( ♯ ‘ 𝑆 ) ) ) → ( 𝐹 ..^ 𝐿 ) ⊆ ( 0 ..^ ( ♯ ‘ 𝑆 ) ) ) |
17 |
|
wrddm |
⊢ ( 𝑆 ∈ Word 𝐴 → dom 𝑆 = ( 0 ..^ ( ♯ ‘ 𝑆 ) ) ) |
18 |
17
|
3ad2ant1 |
⊢ ( ( 𝑆 ∈ Word 𝐴 ∧ 𝐹 ∈ ( 0 ... 𝐿 ) ∧ 𝐿 ∈ ( 0 ... ( ♯ ‘ 𝑆 ) ) ) → dom 𝑆 = ( 0 ..^ ( ♯ ‘ 𝑆 ) ) ) |
19 |
16 18
|
sseqtrrd |
⊢ ( ( 𝑆 ∈ Word 𝐴 ∧ 𝐹 ∈ ( 0 ... 𝐿 ) ∧ 𝐿 ∈ ( 0 ... ( ♯ ‘ 𝑆 ) ) ) → ( 𝐹 ..^ 𝐿 ) ⊆ dom 𝑆 ) |
20 |
19
|
iftrued |
⊢ ( ( 𝑆 ∈ Word 𝐴 ∧ 𝐹 ∈ ( 0 ... 𝐿 ) ∧ 𝐿 ∈ ( 0 ... ( ♯ ‘ 𝑆 ) ) ) → if ( ( 𝐹 ..^ 𝐿 ) ⊆ dom 𝑆 , ( 𝑥 ∈ ( 0 ..^ ( 𝐿 − 𝐹 ) ) ↦ ( 𝑆 ‘ ( 𝑥 + 𝐹 ) ) ) , ∅ ) = ( 𝑥 ∈ ( 0 ..^ ( 𝐿 − 𝐹 ) ) ↦ ( 𝑆 ‘ ( 𝑥 + 𝐹 ) ) ) ) |
21 |
7 20
|
eqtrd |
⊢ ( ( 𝑆 ∈ Word 𝐴 ∧ 𝐹 ∈ ( 0 ... 𝐿 ) ∧ 𝐿 ∈ ( 0 ... ( ♯ ‘ 𝑆 ) ) ) → ( 𝑆 substr 〈 𝐹 , 𝐿 〉 ) = ( 𝑥 ∈ ( 0 ..^ ( 𝐿 − 𝐹 ) ) ↦ ( 𝑆 ‘ ( 𝑥 + 𝐹 ) ) ) ) |