Theorem List for Intuitionistic Logic Explorer - 11101-11200 *Has distinct variable
group(s)
| Type | Label | Description |
| Statement |
| |
| Theorem | elfzelfzccat 11101 |
An element of a finite set of sequential integers up to the length of a
word is an element of an extended finite set of sequential integers up to
the length of a concatenation of this word with another word.
(Contributed by Alexander van der Vekens, 28-Mar-2018.)
|
| ⊢ ((𝐴 ∈ Word 𝑉 ∧ 𝐵 ∈ Word 𝑉) → (𝑁 ∈ (0...(♯‘𝐴)) → 𝑁 ∈ (0...(♯‘(𝐴 ++ 𝐵))))) |
| |
| Theorem | ccatvalfn 11102 |
The concatenation of two words is a function over the half-open integer
range having the sum of the lengths of the word as length. (Contributed
by Alexander van der Vekens, 30-Mar-2018.)
|
| ⊢ ((𝐴 ∈ Word 𝑉 ∧ 𝐵 ∈ Word 𝑉) → (𝐴 ++ 𝐵) Fn (0..^((♯‘𝐴) + (♯‘𝐵)))) |
| |
| Theorem | ccatsymb 11103 |
The symbol at a given position in a concatenated word. (Contributed by
AV, 26-May-2018.) (Proof shortened by AV, 24-Nov-2018.)
|
| ⊢ ((𝐴 ∈ Word 𝑉 ∧ 𝐵 ∈ Word 𝑉 ∧ 𝐼 ∈ ℤ) → ((𝐴 ++ 𝐵)‘𝐼) = if(𝐼 < (♯‘𝐴), (𝐴‘𝐼), (𝐵‘(𝐼 − (♯‘𝐴))))) |
| |
| Theorem | ccatfv0 11104 |
The first symbol of a concatenation of two words is the first symbol of
the first word if the first word is not empty. (Contributed by Alexander
van der Vekens, 22-Sep-2018.)
|
| ⊢ ((𝐴 ∈ Word 𝑉 ∧ 𝐵 ∈ Word 𝑉 ∧ 0 < (♯‘𝐴)) → ((𝐴 ++ 𝐵)‘0) = (𝐴‘0)) |
| |
| Theorem | ccatval1lsw 11105 |
The last symbol of the left (nonempty) half of a concatenated word.
(Contributed by Alexander van der Vekens, 3-Oct-2018.) (Proof shortened
by AV, 1-May-2020.)
|
| ⊢ ((𝐴 ∈ Word 𝑉 ∧ 𝐵 ∈ Word 𝑉 ∧ 𝐴 ≠ ∅) → ((𝐴 ++ 𝐵)‘((♯‘𝐴) − 1)) = (lastS‘𝐴)) |
| |
| Theorem | ccatval21sw 11106 |
The first symbol of the right (nonempty) half of a concatenated word.
(Contributed by AV, 23-Apr-2022.)
|
| ⊢ ((𝐴 ∈ Word 𝑉 ∧ 𝐵 ∈ Word 𝑉 ∧ 𝐵 ≠ ∅) → ((𝐴 ++ 𝐵)‘(♯‘𝐴)) = (𝐵‘0)) |
| |
| Theorem | ccatlid 11107 |
Concatenation of a word by the empty word on the left. (Contributed by
Stefan O'Rear, 15-Aug-2015.) (Proof shortened by AV, 1-May-2020.)
|
| ⊢ (𝑆 ∈ Word 𝐵 → (∅ ++ 𝑆) = 𝑆) |
| |
| Theorem | ccatrid 11108 |
Concatenation of a word by the empty word on the right. (Contributed by
Stefan O'Rear, 15-Aug-2015.) (Proof shortened by AV, 1-May-2020.)
|
| ⊢ (𝑆 ∈ Word 𝐵 → (𝑆 ++ ∅) = 𝑆) |
| |
| Theorem | ccatass 11109 |
Associative law for concatenation of words. (Contributed by Stefan
O'Rear, 15-Aug-2015.)
|
| ⊢ ((𝑆 ∈ Word 𝐵 ∧ 𝑇 ∈ Word 𝐵 ∧ 𝑈 ∈ Word 𝐵) → ((𝑆 ++ 𝑇) ++ 𝑈) = (𝑆 ++ (𝑇 ++ 𝑈))) |
| |
| Theorem | ccatrn 11110 |
The range of a concatenated word. (Contributed by Stefan O'Rear,
15-Aug-2015.)
|
| ⊢ ((𝑆 ∈ Word 𝐵 ∧ 𝑇 ∈ Word 𝐵) → ran (𝑆 ++ 𝑇) = (ran 𝑆 ∪ ran 𝑇)) |
| |
| Theorem | ccatidid 11111 |
Concatenation of the empty word by the empty word. (Contributed by AV,
26-Mar-2022.)
|
| ⊢ (∅ ++ ∅) =
∅ |
| |
| Theorem | lswccatn0lsw 11112 |
The last symbol of a word concatenated with a nonempty word is the last
symbol of the nonempty word. (Contributed by AV, 22-Oct-2018.) (Proof
shortened by AV, 1-May-2020.)
|
| ⊢ ((𝐴 ∈ Word 𝑉 ∧ 𝐵 ∈ Word 𝑉 ∧ 𝐵 ≠ ∅) → (lastS‘(𝐴 ++ 𝐵)) = (lastS‘𝐵)) |
| |
| Theorem | lswccat0lsw 11113 |
The last symbol of a word concatenated with the empty word is the last
symbol of the word. (Contributed by AV, 22-Oct-2018.) (Proof shortened
by AV, 1-May-2020.)
|
| ⊢ (𝑊 ∈ Word 𝑉 → (lastS‘(𝑊 ++ ∅)) = (lastS‘𝑊)) |
| |
| 4.7.4 Singleton words
|
| |
| Syntax | cs1 11114 |
Syntax for the singleton word constructor.
|
| class 〈“𝐴”〉 |
| |
| Definition | df-s1 11115 |
Define the canonical injection from symbols to words. Although not
required, 𝐴 should usually be a set. Otherwise,
the singleton word
〈“𝐴”〉 would be the singleton
word consisting of the empty set, see
s1prc 11122, and not, as maybe expected, the empty word.
(Contributed by
Stefan O'Rear, 15-Aug-2015.) (Revised by Mario Carneiro, 26-Feb-2016.)
|
| ⊢ 〈“𝐴”〉 = {〈0, ( I ‘𝐴)〉} |
| |
| Theorem | s1val 11116 |
Value of a singleton word. (Contributed by Stefan O'Rear, 15-Aug-2015.)
(Revised by Mario Carneiro, 26-Feb-2016.)
|
| ⊢ (𝐴 ∈ 𝑉 → 〈“𝐴”〉 = {〈0, 𝐴〉}) |
| |
| Theorem | s1rn 11117 |
The range of a singleton word. (Contributed by Mario Carneiro,
18-Jul-2016.)
|
| ⊢ (𝐴 ∈ 𝑉 → ran 〈“𝐴”〉 = {𝐴}) |
| |
| Theorem | s1eq 11118 |
Equality theorem for a singleton word. (Contributed by Mario Carneiro,
26-Feb-2016.)
|
| ⊢ (𝐴 = 𝐵 → 〈“𝐴”〉 = 〈“𝐵”〉) |
| |
| Theorem | s1eqd 11119 |
Equality theorem for a singleton word. (Contributed by Mario Carneiro,
26-Feb-2016.)
|
| ⊢ (𝜑 → 𝐴 = 𝐵) ⇒ ⊢ (𝜑 → 〈“𝐴”〉 = 〈“𝐵”〉) |
| |
| Theorem | s1cl 11120 |
A singleton word is a word. (Contributed by Stefan O'Rear, 15-Aug-2015.)
(Revised by Mario Carneiro, 26-Feb-2016.) (Proof shortened by AV,
23-Nov-2018.)
|
| ⊢ (𝐴 ∈ 𝐵 → 〈“𝐴”〉 ∈ Word 𝐵) |
| |
| Theorem | s1cld 11121 |
A singleton word is a word. (Contributed by Mario Carneiro,
26-Feb-2016.)
|
| ⊢ (𝜑 → 𝐴 ∈ 𝐵) ⇒ ⊢ (𝜑 → 〈“𝐴”〉 ∈ Word 𝐵) |
| |
| Theorem | s1prc 11122 |
Value of a singleton word if the symbol is a proper class. (Contributed
by AV, 26-Mar-2022.)
|
| ⊢ (¬ 𝐴 ∈ V → 〈“𝐴”〉 =
〈“∅”〉) |
| |
| Theorem | s1leng 11123 |
Length of a singleton word. (Contributed by Stefan O'Rear, 15-Aug-2015.)
(Revised by Mario Carneiro, 26-Feb-2016.)
|
| ⊢ (𝐴 ∈ 𝑉 → (♯‘〈“𝐴”〉) =
1) |
| |
| Theorem | s1dmg 11124 |
The domain of a singleton word is a singleton. (Contributed by AV,
9-Jan-2020.)
|
| ⊢ (𝐴 ∈ 𝑆 → dom 〈“𝐴”〉 = {0}) |
| |
| Theorem | s1fv 11125 |
Sole symbol of a singleton word. (Contributed by Stefan O'Rear,
15-Aug-2015.) (Revised by Mario Carneiro, 26-Feb-2016.)
|
| ⊢ (𝐴 ∈ 𝐵 → (〈“𝐴”〉‘0) = 𝐴) |
| |
| Theorem | lsws1 11126 |
The last symbol of a singleton word is its symbol. (Contributed by AV,
22-Oct-2018.)
|
| ⊢ (𝐴 ∈ 𝑉 → (lastS‘〈“𝐴”〉) = 𝐴) |
| |
| Theorem | eqs1 11127 |
A word of length 1 is a singleton word. (Contributed by Stefan O'Rear,
23-Aug-2015.) (Proof shortened by AV, 1-May-2020.)
|
| ⊢ ((𝑊 ∈ Word 𝐴 ∧ (♯‘𝑊) = 1) → 𝑊 = 〈“(𝑊‘0)”〉) |
| |
| Theorem | wrdl1exs1 11128* |
A word of length 1 is a singleton word. (Contributed by AV,
24-Jan-2021.)
|
| ⊢ ((𝑊 ∈ Word 𝑆 ∧ (♯‘𝑊) = 1) → ∃𝑠 ∈ 𝑆 𝑊 = 〈“𝑠”〉) |
| |
| Theorem | wrdl1s1 11129 |
A word of length 1 is a singleton word consisting of the first symbol of
the word. (Contributed by AV, 22-Jul-2018.) (Proof shortened by AV,
14-Oct-2018.)
|
| ⊢ (𝑆 ∈ 𝑉 → (𝑊 = 〈“𝑆”〉 ↔ (𝑊 ∈ Word 𝑉 ∧ (♯‘𝑊) = 1 ∧ (𝑊‘0) = 𝑆))) |
| |
| Theorem | s111 11130 |
The singleton word function is injective. (Contributed by Mario Carneiro,
1-Oct-2015.) (Revised by Mario Carneiro, 26-Feb-2016.)
|
| ⊢ ((𝑆 ∈ 𝐴 ∧ 𝑇 ∈ 𝐴) → (〈“𝑆”〉 = 〈“𝑇”〉 ↔ 𝑆 = 𝑇)) |
| |
| 4.7.5 Concatenations with singleton
words
|
| |
| Theorem | ccatws1cl 11131 |
The concatenation of a word with a singleton word is a word. (Contributed
by Alexander van der Vekens, 22-Sep-2018.)
|
| ⊢ ((𝑊 ∈ Word 𝑉 ∧ 𝑋 ∈ 𝑉) → (𝑊 ++ 〈“𝑋”〉) ∈ Word 𝑉) |
| |
| Theorem | ccat2s1cl 11132 |
The concatenation of two singleton words is a word. (Contributed by
Alexander van der Vekens, 22-Sep-2018.)
|
| ⊢ ((𝑋 ∈ 𝑉 ∧ 𝑌 ∈ 𝑉) → (〈“𝑋”〉 ++ 〈“𝑌”〉) ∈ Word
𝑉) |
| |
| Theorem | ccatws1leng 11133 |
The length of the concatenation of a word with a singleton word.
(Contributed by Alexander van der Vekens, 22-Sep-2018.) (Revised by AV,
4-Mar-2022.)
|
| ⊢ ((𝑊 ∈ Word 𝑉 ∧ 𝑋 ∈ 𝑌) → (♯‘(𝑊 ++ 〈“𝑋”〉)) = ((♯‘𝑊) + 1)) |
| |
| Theorem | ccatws1lenp1bg 11134 |
The length of a word is 𝑁 iff the length of the concatenation
of the
word with a singleton word is 𝑁 + 1. (Contributed by AV,
4-Mar-2022.)
|
| ⊢ ((𝑊 ∈ Word 𝑉 ∧ 𝑋 ∈ 𝑌 ∧ 𝑁 ∈ ℕ0) →
((♯‘(𝑊 ++
〈“𝑋”〉)) = (𝑁 + 1) ↔ (♯‘𝑊) = 𝑁)) |
| |
| Theorem | ccatw2s1cl 11135 |
The concatenation of a word with two singleton words is a word.
(Contributed by Alexander van der Vekens, 22-Sep-2018.)
|
| ⊢ ((𝑊 ∈ Word 𝑉 ∧ 𝑋 ∈ 𝑉 ∧ 𝑌 ∈ 𝑉) → ((𝑊 ++ 〈“𝑋”〉) ++ 〈“𝑌”〉) ∈ Word
𝑉) |
| |
| Theorem | ccats1val1g 11136 |
Value of a symbol in the left half of a word concatenated with a single
symbol. (Contributed by Alexander van der Vekens, 5-Aug-2018.) (Revised
by JJ, 20-Jan-2024.)
|
| ⊢ ((𝑊 ∈ Word 𝑉 ∧ 𝑆 ∈ 𝑌 ∧ 𝐼 ∈ (0..^(♯‘𝑊))) → ((𝑊 ++ 〈“𝑆”〉)‘𝐼) = (𝑊‘𝐼)) |
| |
| Theorem | ccats1val2 11137 |
Value of the symbol concatenated with a word. (Contributed by Alexander
van der Vekens, 5-Aug-2018.) (Proof shortened by Alexander van der
Vekens, 14-Oct-2018.)
|
| ⊢ ((𝑊 ∈ Word 𝑉 ∧ 𝑆 ∈ 𝑉 ∧ 𝐼 = (♯‘𝑊)) → ((𝑊 ++ 〈“𝑆”〉)‘𝐼) = 𝑆) |
| |
| Theorem | ccat1st1st 11138 |
The first symbol of a word concatenated with its first symbol is the first
symbol of the word. This theorem holds even if 𝑊 is the empty word.
(Contributed by AV, 26-Mar-2022.)
|
| ⊢ (𝑊 ∈ Word 𝑉 → ((𝑊 ++ 〈“(𝑊‘0)”〉)‘0) = (𝑊‘0)) |
| |
| Theorem | ccatws1ls 11139 |
The last symbol of the concatenation of a word with a singleton word is
the symbol of the singleton word. (Contributed by AV, 29-Sep-2018.)
(Proof shortened by AV, 14-Oct-2018.)
|
| ⊢ ((𝑊 ∈ Word 𝑉 ∧ 𝑋 ∈ 𝑉) → ((𝑊 ++ 〈“𝑋”〉)‘(♯‘𝑊)) = 𝑋) |
| |
| Theorem | lswccats1 11140 |
The last symbol of a word concatenated with a singleton word is the symbol
of the singleton word. (Contributed by AV, 6-Aug-2018.)
|
| ⊢ ((𝑊 ∈ Word 𝑉 ∧ 𝑆 ∈ 𝑉) → (lastS‘(𝑊 ++ 〈“𝑆”〉)) = 𝑆) |
| |
| Theorem | lswccats1fst 11141 |
The last symbol of a nonempty word concatenated with its first symbol is
the first symbol. (Contributed by AV, 28-Jun-2018.) (Proof shortened by
AV, 1-May-2020.)
|
| ⊢ ((𝑃 ∈ Word 𝑉 ∧ 1 ≤ (♯‘𝑃)) → (lastS‘(𝑃 ++ 〈“(𝑃‘0)”〉)) =
((𝑃 ++ 〈“(𝑃‘0)”〉)‘0)) |
| |
| Theorem | ccatw2s1p2 11142 |
Extract the second of two single symbols concatenated with a word.
(Contributed by Alexander van der Vekens, 22-Sep-2018.) (Proof shortened
by AV, 1-May-2020.)
|
| ⊢ (((𝑊 ∈ Word 𝑉 ∧ (♯‘𝑊) = 𝑁) ∧ (𝑋 ∈ 𝑉 ∧ 𝑌 ∈ 𝑉)) → (((𝑊 ++ 〈“𝑋”〉) ++ 〈“𝑌”〉)‘(𝑁 + 1)) = 𝑌) |
| |
| 4.7.6 Subwords/substrings
|
| |
| Syntax | csubstr 11143 |
Syntax for the subword operator.
|
| class substr |
| |
| Definition | df-substr 11144* |
Define an operation which extracts portions (called subwords or
substrings) of words. Definition in Section 9.1 of [AhoHopUll]
p. 318. (Contributed by Stefan O'Rear, 15-Aug-2015.)
|
| ⊢ substr = (𝑠 ∈ V, 𝑏 ∈ (ℤ × ℤ) ↦
if(((1st ‘𝑏)..^(2nd ‘𝑏)) ⊆ dom 𝑠, (𝑥 ∈ (0..^((2nd ‘𝑏) − (1st
‘𝑏))) ↦ (𝑠‘(𝑥 + (1st ‘𝑏)))), ∅)) |
| |
| Theorem | fzowrddc 11145 |
Decidability of whether a range of integers is a subset of a word's
domain. (Contributed by Jim Kingdon, 23-Dec-2025.)
|
| ⊢ ((𝑆 ∈ Word 𝐴 ∧ 𝐹 ∈ ℤ ∧ 𝐿 ∈ ℤ) →
DECID (𝐹..^𝐿) ⊆ dom 𝑆) |
| |
| Theorem | swrdval 11146* |
Value of a subword. (Contributed by Stefan O'Rear, 15-Aug-2015.)
|
| ⊢ ((𝑆 ∈ 𝑉 ∧ 𝐹 ∈ ℤ ∧ 𝐿 ∈ ℤ) → (𝑆 substr 〈𝐹, 𝐿〉) = if((𝐹..^𝐿) ⊆ dom 𝑆, (𝑥 ∈ (0..^(𝐿 − 𝐹)) ↦ (𝑆‘(𝑥 + 𝐹))), ∅)) |
| |
| Theorem | swrd00g 11147 |
A zero length substring. (Contributed by Stefan O'Rear,
27-Aug-2015.)
|
| ⊢ ((𝑆 ∈ 𝑉 ∧ 𝑋 ∈ ℤ) → (𝑆 substr 〈𝑋, 𝑋〉) = ∅) |
| |
| Theorem | swrdclg 11148 |
Closure of the subword extractor. (Contributed by Stefan O'Rear,
16-Aug-2015.) (Revised by Mario Carneiro, 26-Feb-2016.)
|
| ⊢ ((𝑆 ∈ Word 𝐴 ∧ 𝐹 ∈ ℤ ∧ 𝐿 ∈ ℤ) → (𝑆 substr 〈𝐹, 𝐿〉) ∈ Word 𝐴) |
| |
| Theorem | swrdval2 11149* |
Value of the subword extractor in its intended domain. (Contributed by
Stefan O'Rear, 15-Aug-2015.) (Proof shortened by AV, 2-May-2020.)
|
| ⊢ ((𝑆 ∈ Word 𝐴 ∧ 𝐹 ∈ (0...𝐿) ∧ 𝐿 ∈ (0...(♯‘𝑆))) → (𝑆 substr 〈𝐹, 𝐿〉) = (𝑥 ∈ (0..^(𝐿 − 𝐹)) ↦ (𝑆‘(𝑥 + 𝐹)))) |
| |
| Theorem | swrdlen 11150 |
Length of an extracted subword. (Contributed by Stefan O'Rear,
16-Aug-2015.)
|
| ⊢ ((𝑆 ∈ Word 𝐴 ∧ 𝐹 ∈ (0...𝐿) ∧ 𝐿 ∈ (0...(♯‘𝑆))) →
(♯‘(𝑆 substr
〈𝐹, 𝐿〉)) = (𝐿 − 𝐹)) |
| |
| Theorem | swrdfv 11151 |
A symbol in an extracted subword, indexed using the subword's indices.
(Contributed by Stefan O'Rear, 16-Aug-2015.)
|
| ⊢ (((𝑆 ∈ Word 𝐴 ∧ 𝐹 ∈ (0...𝐿) ∧ 𝐿 ∈ (0...(♯‘𝑆))) ∧ 𝑋 ∈ (0..^(𝐿 − 𝐹))) → ((𝑆 substr 〈𝐹, 𝐿〉)‘𝑋) = (𝑆‘(𝑋 + 𝐹))) |
| |
| Theorem | swrdfv0 11152 |
The first symbol in an extracted subword. (Contributed by AV,
27-Apr-2022.)
|
| ⊢ ((𝑆 ∈ Word 𝐴 ∧ 𝐹 ∈ (0..^𝐿) ∧ 𝐿 ∈ (0...(♯‘𝑆))) → ((𝑆 substr 〈𝐹, 𝐿〉)‘0) = (𝑆‘𝐹)) |
| |
| Theorem | swrdf 11153 |
A subword of a word is a function from a half-open range of nonnegative
integers of the same length as the subword to the set of symbols for the
original word. (Contributed by AV, 13-Nov-2018.)
|
| ⊢ ((𝑊 ∈ Word 𝑉 ∧ 𝑀 ∈ (0...𝑁) ∧ 𝑁 ∈ (0...(♯‘𝑊))) → (𝑊 substr 〈𝑀, 𝑁〉):(0..^(𝑁 − 𝑀))⟶𝑉) |
| |
| Theorem | swrdvalfn 11154 |
Value of the subword extractor as function with domain. (Contributed by
Alexander van der Vekens, 28-Mar-2018.) (Proof shortened by AV,
2-May-2020.)
|
| ⊢ ((𝑆 ∈ Word 𝑉 ∧ 𝐹 ∈ (0...𝐿) ∧ 𝐿 ∈ (0...(♯‘𝑆))) → (𝑆 substr 〈𝐹, 𝐿〉) Fn (0..^(𝐿 − 𝐹))) |
| |
| Theorem | swrdrn 11155 |
The range of a subword of a word is a subset of the set of symbols for the
word. (Contributed by AV, 13-Nov-2018.)
|
| ⊢ ((𝑊 ∈ Word 𝑉 ∧ 𝑀 ∈ (0...𝑁) ∧ 𝑁 ∈ (0...(♯‘𝑊))) → ran (𝑊 substr 〈𝑀, 𝑁〉) ⊆ 𝑉) |
| |
| Theorem | swrdlend 11156 |
The value of the subword extractor is the empty set (undefined) if the
range is not valid. (Contributed by Alexander van der Vekens,
16-Mar-2018.) (Proof shortened by AV, 2-May-2020.)
|
| ⊢ ((𝑊 ∈ Word 𝑉 ∧ 𝐹 ∈ ℤ ∧ 𝐿 ∈ ℤ) → (𝐿 ≤ 𝐹 → (𝑊 substr 〈𝐹, 𝐿〉) = ∅)) |
| |
| Theorem | swrdnd 11157 |
The value of the subword extractor is the empty set (undefined) if the
range is not valid. (Contributed by Alexander van der Vekens,
16-Mar-2018.) (Proof shortened by AV, 2-May-2020.)
|
| ⊢ ((𝑊 ∈ Word 𝑉 ∧ 𝐹 ∈ ℤ ∧ 𝐿 ∈ ℤ) → ((𝐹 < 0 ∨ 𝐿 ≤ 𝐹 ∨ (♯‘𝑊) < 𝐿) → (𝑊 substr 〈𝐹, 𝐿〉) = ∅)) |
| |
| Theorem | swrd0g 11158 |
A subword of an empty set is always the empty set. (Contributed by AV,
31-Mar-2018.) (Revised by AV, 20-Oct-2018.) (Proof shortened by AV,
2-May-2020.)
|
| ⊢ ((𝐹 ∈ ℤ ∧ 𝐿 ∈ ℤ) → (∅ substr
〈𝐹, 𝐿〉) = ∅) |
| |
| Theorem | swrdrlen 11159 |
Length of a right-anchored subword. (Contributed by Alexander van der
Vekens, 5-Apr-2018.)
|
| ⊢ ((𝑊 ∈ Word 𝑉 ∧ 𝐼 ∈ (0...(♯‘𝑊))) →
(♯‘(𝑊 substr
〈𝐼,
(♯‘𝑊)〉))
= ((♯‘𝑊)
− 𝐼)) |
| |
| Theorem | swrdlen2 11160 |
Length of an extracted subword. (Contributed by AV, 5-May-2020.)
|
| ⊢ ((𝑆 ∈ Word 𝑉 ∧ (𝐹 ∈ ℕ0 ∧ 𝐿 ∈
(ℤ≥‘𝐹)) ∧ 𝐿 ≤ (♯‘𝑆)) → (♯‘(𝑆 substr 〈𝐹, 𝐿〉)) = (𝐿 − 𝐹)) |
| |
| Theorem | swrdfv2 11161 |
A symbol in an extracted subword, indexed using the word's indices.
(Contributed by AV, 5-May-2020.)
|
| ⊢ (((𝑆 ∈ Word 𝑉 ∧ (𝐹 ∈ ℕ0 ∧ 𝐿 ∈
(ℤ≥‘𝐹)) ∧ 𝐿 ≤ (♯‘𝑆)) ∧ 𝑋 ∈ (𝐹..^𝐿)) → ((𝑆 substr 〈𝐹, 𝐿〉)‘(𝑋 − 𝐹)) = (𝑆‘𝑋)) |
| |
| Theorem | swrdwrdsymbg 11162 |
A subword is a word over the symbols it consists of. (Contributed by
AV, 2-Dec-2022.)
|
| ⊢ ((𝑆 ∈ Word 𝐴 ∧ 𝑀 ∈ (0...𝑁) ∧ 𝑁 ∈ (0...(♯‘𝑆))) → (𝑆 substr 〈𝑀, 𝑁〉) ∈ Word (𝑆 “ (𝑀..^𝑁))) |
| |
| Theorem | swrdsb0eq 11163 |
Two subwords with the same bounds are equal if the range is not valid.
(Contributed by AV, 4-May-2020.)
|
| ⊢ (((𝑊 ∈ Word 𝑉 ∧ 𝑈 ∈ Word 𝑉) ∧ (𝑀 ∈ ℕ0 ∧ 𝑁 ∈ ℕ0)
∧ 𝑁 ≤ 𝑀) → (𝑊 substr 〈𝑀, 𝑁〉) = (𝑈 substr 〈𝑀, 𝑁〉)) |
| |
| Theorem | swrdsbslen 11164 |
Two subwords with the same bounds have the same length. (Contributed by
AV, 4-May-2020.)
|
| ⊢ (((𝑊 ∈ Word 𝑉 ∧ 𝑈 ∈ Word 𝑉) ∧ (𝑀 ∈ ℕ0 ∧ 𝑁 ∈ ℕ0)
∧ (𝑁 ≤
(♯‘𝑊) ∧
𝑁 ≤
(♯‘𝑈))) →
(♯‘(𝑊 substr
〈𝑀, 𝑁〉)) = (♯‘(𝑈 substr 〈𝑀, 𝑁〉))) |
| |
| Theorem | swrdspsleq 11165* |
Two words have a common subword (starting at the same position with the
same length) iff they have the same symbols at each position.
(Contributed by Alexander van der Vekens, 7-Aug-2018.) (Proof shortened
by AV, 7-May-2020.)
|
| ⊢ (((𝑊 ∈ Word 𝑉 ∧ 𝑈 ∈ Word 𝑉) ∧ (𝑀 ∈ ℕ0 ∧ 𝑁 ∈ ℕ0)
∧ (𝑁 ≤
(♯‘𝑊) ∧
𝑁 ≤
(♯‘𝑈))) →
((𝑊 substr 〈𝑀, 𝑁〉) = (𝑈 substr 〈𝑀, 𝑁〉) ↔ ∀𝑖 ∈ (𝑀..^𝑁)(𝑊‘𝑖) = (𝑈‘𝑖))) |
| |
| Theorem | swrds1 11166 |
Extract a single symbol from a word. (Contributed by Stefan O'Rear,
23-Aug-2015.)
|
| ⊢ ((𝑊 ∈ Word 𝐴 ∧ 𝐼 ∈ (0..^(♯‘𝑊))) → (𝑊 substr 〈𝐼, (𝐼 + 1)〉) = 〈“(𝑊‘𝐼)”〉) |
| |
| Theorem | swrdlsw 11167 |
Extract the last single symbol from a word. (Contributed by Alexander van
der Vekens, 23-Sep-2018.)
|
| ⊢ ((𝑊 ∈ Word 𝑉 ∧ 𝑊 ≠ ∅) → (𝑊 substr 〈((♯‘𝑊) − 1),
(♯‘𝑊)〉) =
〈“(lastS‘𝑊)”〉) |
| |
| Theorem | ccatswrd 11168 |
Joining two adjacent subwords makes a longer subword. (Contributed by
Stefan O'Rear, 20-Aug-2015.)
|
| ⊢ ((𝑆 ∈ Word 𝐴 ∧ (𝑋 ∈ (0...𝑌) ∧ 𝑌 ∈ (0...𝑍) ∧ 𝑍 ∈ (0...(♯‘𝑆)))) → ((𝑆 substr 〈𝑋, 𝑌〉) ++ (𝑆 substr 〈𝑌, 𝑍〉)) = (𝑆 substr 〈𝑋, 𝑍〉)) |
| |
| Theorem | swrdccat2 11169 |
Recover the right half of a concatenated word. (Contributed by Mario
Carneiro, 27-Sep-2015.)
|
| ⊢ ((𝑆 ∈ Word 𝐵 ∧ 𝑇 ∈ Word 𝐵) → ((𝑆 ++ 𝑇) substr 〈(♯‘𝑆), ((♯‘𝑆) + (♯‘𝑇))〉) = 𝑇) |
| |
| 4.7.7 Prefixes of a word
|
| |
| Syntax | cpfx 11170 |
Syntax for the prefix operator.
|
| class prefix |
| |
| Definition | df-pfx 11171* |
Define an operation which extracts prefixes of words, i.e. subwords (or
substrings) starting at the beginning of a word (or string). In other
words, (𝑆 prefix 𝐿) is the prefix of the word 𝑆 of
length
𝐿. Definition in Section 9.1 of [AhoHopUll] p. 318. See also
Wikipedia "Substring" https://en.wikipedia.org/wiki/Substring#Prefix.
(Contributed by AV, 2-May-2020.)
|
| ⊢ prefix = (𝑠 ∈ V, 𝑙 ∈ ℕ0 ↦ (𝑠 substr 〈0, 𝑙〉)) |
| |
| Theorem | pfxval 11172 |
Value of a prefix operation. (Contributed by AV, 2-May-2020.)
|
| ⊢ ((𝑆 ∈ 𝑉 ∧ 𝐿 ∈ ℕ0) → (𝑆 prefix 𝐿) = (𝑆 substr 〈0, 𝐿〉)) |
| |
| Theorem | pfx00g 11173 |
The zero length prefix is the empty set. (Contributed by AV,
2-May-2020.)
|
| ⊢ (𝑆 ∈ 𝑉 → (𝑆 prefix 0) = ∅) |
| |
| Theorem | pfx0g 11174 |
A prefix of an empty set is always the empty set. (Contributed by AV,
3-May-2020.)
|
| ⊢ (𝐿 ∈ ℕ0 → (∅
prefix 𝐿) =
∅) |
| |
| Theorem | fnpfx 11175 |
The domain of the prefix extractor. (Contributed by Jim Kingdon,
8-Jan-2026.)
|
| ⊢ prefix Fn (V ×
ℕ0) |
| |
| Theorem | pfxclg 11176 |
Closure of the prefix extractor. (Contributed by AV, 2-May-2020.)
|
| ⊢ ((𝑆 ∈ Word 𝐴 ∧ 𝐿 ∈ ℕ0) → (𝑆 prefix 𝐿) ∈ Word 𝐴) |
| |
| Theorem | pfxclz 11177 |
Closure of the prefix extractor. This extends pfxclg 11176 from ℕ0 to
ℤ (negative lengths are trivial, resulting
in the empty word).
(Contributed by Jim Kingdon, 8-Jan-2026.)
|
| ⊢ ((𝑆 ∈ Word 𝐴 ∧ 𝐿 ∈ ℤ) → (𝑆 prefix 𝐿) ∈ Word 𝐴) |
| |
| Theorem | pfxmpt 11178* |
Value of the prefix extractor as a mapping. (Contributed by AV,
2-May-2020.)
|
| ⊢ ((𝑆 ∈ Word 𝐴 ∧ 𝐿 ∈ (0...(♯‘𝑆))) → (𝑆 prefix 𝐿) = (𝑥 ∈ (0..^𝐿) ↦ (𝑆‘𝑥))) |
| |
| Theorem | pfxres 11179 |
Value of the prefix extractor as the restriction of a word.
(Contributed by Stefan O'Rear, 24-Aug-2015.) (Revised by AV,
2-May-2020.)
|
| ⊢ ((𝑆 ∈ Word 𝐴 ∧ 𝐿 ∈ (0...(♯‘𝑆))) → (𝑆 prefix 𝐿) = (𝑆 ↾ (0..^𝐿))) |
| |
| Theorem | pfxf 11180 |
A prefix of a word is a function from a half-open range of nonnegative
integers of the same length as the prefix to the set of symbols for the
original word. (Contributed by AV, 2-May-2020.)
|
| ⊢ ((𝑊 ∈ Word 𝑉 ∧ 𝐿 ∈ (0...(♯‘𝑊))) → (𝑊 prefix 𝐿):(0..^𝐿)⟶𝑉) |
| |
| Theorem | pfxfn 11181 |
Value of the prefix extractor as function with domain. (Contributed by
AV, 2-May-2020.)
|
| ⊢ ((𝑆 ∈ Word 𝑉 ∧ 𝐿 ∈ (0...(♯‘𝑆))) → (𝑆 prefix 𝐿) Fn (0..^𝐿)) |
| |
| Theorem | pfxfv 11182 |
A symbol in a prefix of a word, indexed using the prefix' indices.
(Contributed by Alexander van der Vekens, 16-Jun-2018.) (Revised by AV,
3-May-2020.)
|
| ⊢ ((𝑊 ∈ Word 𝑉 ∧ 𝐿 ∈ (0...(♯‘𝑊)) ∧ 𝐼 ∈ (0..^𝐿)) → ((𝑊 prefix 𝐿)‘𝐼) = (𝑊‘𝐼)) |
| |
| Theorem | pfxlen 11183 |
Length of a prefix. (Contributed by Stefan O'Rear, 24-Aug-2015.)
(Revised by AV, 2-May-2020.)
|
| ⊢ ((𝑆 ∈ Word 𝐴 ∧ 𝐿 ∈ (0...(♯‘𝑆))) →
(♯‘(𝑆 prefix
𝐿)) = 𝐿) |
| |
| Theorem | pfxid 11184 |
A word is a prefix of itself. (Contributed by Stefan O'Rear,
16-Aug-2015.) (Revised by AV, 2-May-2020.)
|
| ⊢ (𝑆 ∈ Word 𝐴 → (𝑆 prefix (♯‘𝑆)) = 𝑆) |
| |
| Theorem | pfxrn 11185 |
The range of a prefix of a word is a subset of the set of symbols for the
word. (Contributed by AV, 2-May-2020.)
|
| ⊢ ((𝑊 ∈ Word 𝑉 ∧ 𝐿 ∈ (0...(♯‘𝑊))) → ran (𝑊 prefix 𝐿) ⊆ 𝑉) |
| |
| Theorem | pfxn0 11186 |
A prefix consisting of at least one symbol is not empty. (Contributed by
Alexander van der Vekens, 4-Aug-2018.) (Revised by AV, 2-May-2020.)
|
| ⊢ ((𝑊 ∈ Word 𝑉 ∧ 𝐿 ∈ ℕ ∧ 𝐿 ≤ (♯‘𝑊)) → (𝑊 prefix 𝐿) ≠ ∅) |
| |
| Theorem | pfxnd 11187 |
The value of a prefix operation for a length argument larger than the word
length is the empty set. (This is due to our definition of function
values for out-of-domain arguments, see ndmfvg 5634). (Contributed by AV,
3-May-2020.)
|
| ⊢ ((𝑊 ∈ Word 𝑉 ∧ 𝐿 ∈ ℕ0 ∧
(♯‘𝑊) <
𝐿) → (𝑊 prefix 𝐿) = ∅) |
| |
| Theorem | pfxwrdsymbg 11188 |
A prefix of a word is a word over the symbols it consists of.
(Contributed by AV, 3-Dec-2022.)
|
| ⊢ ((𝑆 ∈ Word 𝐴 ∧ 𝐿 ∈ ℕ0) → (𝑆 prefix 𝐿) ∈ Word (𝑆 “ (0..^𝐿))) |
| |
| Theorem | addlenpfx 11189 |
The sum of the lengths of two parts of a word is the length of the word.
(Contributed by AV, 21-Oct-2018.) (Revised by AV, 3-May-2020.)
|
| ⊢ ((𝑊 ∈ Word 𝑉 ∧ 𝑀 ∈ (0...(♯‘𝑊))) →
((♯‘(𝑊 prefix
𝑀)) +
(♯‘(𝑊 substr
〈𝑀,
(♯‘𝑊)〉)))
= (♯‘𝑊)) |
| |
| Theorem | pfxfv0 11190 |
The first symbol of a prefix is the first symbol of the word.
(Contributed by Alexander van der Vekens, 16-Jun-2018.) (Revised by AV,
3-May-2020.)
|
| ⊢ ((𝑊 ∈ Word 𝑉 ∧ 𝐿 ∈ (1...(♯‘𝑊))) → ((𝑊 prefix 𝐿)‘0) = (𝑊‘0)) |
| |
| Theorem | pfxtrcfv 11191 |
A symbol in a word truncated by one symbol. (Contributed by Alexander van
der Vekens, 16-Jun-2018.) (Revised by AV, 3-May-2020.)
|
| ⊢ ((𝑊 ∈ Word 𝑉 ∧ 𝑊 ≠ ∅ ∧ 𝐼 ∈ (0..^((♯‘𝑊) − 1))) → ((𝑊 prefix ((♯‘𝑊) − 1))‘𝐼) = (𝑊‘𝐼)) |
| |
| Theorem | pfxtrcfv0 11192 |
The first symbol in a word truncated by one symbol. (Contributed by
Alexander van der Vekens, 16-Jun-2018.) (Revised by AV, 3-May-2020.)
|
| ⊢ ((𝑊 ∈ Word 𝑉 ∧ 2 ≤ (♯‘𝑊)) → ((𝑊 prefix ((♯‘𝑊) − 1))‘0) = (𝑊‘0)) |
| |
| Theorem | pfxfvlsw 11193 |
The last symbol in a nonempty prefix of a word. (Contributed by Alexander
van der Vekens, 24-Jun-2018.) (Revised by AV, 3-May-2020.)
|
| ⊢ ((𝑊 ∈ Word 𝑉 ∧ 𝐿 ∈ (1...(♯‘𝑊))) → (lastS‘(𝑊 prefix 𝐿)) = (𝑊‘(𝐿 − 1))) |
| |
| Theorem | pfxeq 11194* |
The prefixes of two words are equal iff they have the same length and
the same symbols at each position. (Contributed by Alexander van der
Vekens, 7-Aug-2018.) (Revised by AV, 4-May-2020.)
|
| ⊢ (((𝑊 ∈ Word 𝑉 ∧ 𝑈 ∈ Word 𝑉) ∧ (𝑀 ∈ ℕ0 ∧ 𝑁 ∈ ℕ0)
∧ (𝑀 ≤
(♯‘𝑊) ∧
𝑁 ≤
(♯‘𝑈))) →
((𝑊 prefix 𝑀) = (𝑈 prefix 𝑁) ↔ (𝑀 = 𝑁 ∧ ∀𝑖 ∈ (0..^𝑀)(𝑊‘𝑖) = (𝑈‘𝑖)))) |
| |
| Theorem | pfxtrcfvl 11195 |
The last symbol in a word truncated by one symbol. (Contributed by AV,
16-Jun-2018.) (Revised by AV, 5-May-2020.)
|
| ⊢ ((𝑊 ∈ Word 𝑉 ∧ 2 ≤ (♯‘𝑊)) → (lastS‘(𝑊 prefix ((♯‘𝑊) − 1))) = (𝑊‘((♯‘𝑊) − 2))) |
| |
| Theorem | pfxsuffeqwrdeq 11196 |
Two words are equal if and only if they have the same prefix and the
same suffix. (Contributed by Alexander van der Vekens, 23-Sep-2018.)
(Revised by AV, 5-May-2020.)
|
| ⊢ ((𝑊 ∈ Word 𝑉 ∧ 𝑆 ∈ Word 𝑉 ∧ 𝐼 ∈ (0..^(♯‘𝑊))) → (𝑊 = 𝑆 ↔ ((♯‘𝑊) = (♯‘𝑆) ∧ ((𝑊 prefix 𝐼) = (𝑆 prefix 𝐼) ∧ (𝑊 substr 〈𝐼, (♯‘𝑊)〉) = (𝑆 substr 〈𝐼, (♯‘𝑊)〉))))) |
| |
| Theorem | pfxsuff1eqwrdeq 11197 |
Two (nonempty) words are equal if and only if they have the same prefix
and the same single symbol suffix. (Contributed by Alexander van der
Vekens, 23-Sep-2018.) (Revised by AV, 6-May-2020.)
|
| ⊢ ((𝑊 ∈ Word 𝑉 ∧ 𝑈 ∈ Word 𝑉 ∧ 0 < (♯‘𝑊)) → (𝑊 = 𝑈 ↔ ((♯‘𝑊) = (♯‘𝑈) ∧ ((𝑊 prefix ((♯‘𝑊) − 1)) = (𝑈 prefix ((♯‘𝑊) − 1)) ∧ (lastS‘𝑊) = (lastS‘𝑈))))) |
| |
| Theorem | disjwrdpfx 11198* |
Sets of words are disjoint if each set contains exactly the extensions
of distinct words of a fixed length. Remark: A word 𝑊 is
called an
"extension" of a word 𝑃 if 𝑃 is a prefix of 𝑊.
(Contributed by AV, 29-Jul-2018.) (Revised by AV, 6-May-2020.)
|
| ⊢ Disj 𝑦 ∈ 𝑊 {𝑥 ∈ Word 𝑉 ∣ (𝑥 prefix 𝑁) = 𝑦} |
| |
| Theorem | ccatpfx 11199 |
Concatenating a prefix with an adjacent subword makes a longer prefix.
(Contributed by AV, 7-May-2020.)
|
| ⊢ ((𝑆 ∈ Word 𝐴 ∧ 𝑌 ∈ (0...𝑍) ∧ 𝑍 ∈ (0...(♯‘𝑆))) → ((𝑆 prefix 𝑌) ++ (𝑆 substr 〈𝑌, 𝑍〉)) = (𝑆 prefix 𝑍)) |
| |
| Theorem | pfxccat1 11200 |
Recover the left half of a concatenated word. (Contributed by Mario
Carneiro, 27-Sep-2015.) (Revised by AV, 6-May-2020.)
|
| ⊢ ((𝑆 ∈ Word 𝐵 ∧ 𝑇 ∈ Word 𝐵) → ((𝑆 ++ 𝑇) prefix (♯‘𝑆)) = 𝑆) |