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Theorem List for Intuitionistic Logic Explorer - 11101-11200   *Has distinct variable group(s)
TypeLabelDescription
Statement
 
Theoremwrdexb 11101 The set of words over a set is a set, bidirectional version. (Contributed by Mario Carneiro, 26-Feb-2016.) (Proof shortened by AV, 23-Nov-2018.)
(𝑆 ∈ V ↔ Word 𝑆 ∈ V)
 
Theoremwrdexi 11102 The set of words over a set is a set, inference form. (Contributed by AV, 23-May-2021.)
𝑆 ∈ V       Word 𝑆 ∈ V
 
Theoremwrdsymbcl 11103 A symbol within a word over an alphabet belongs to the alphabet. (Contributed by Alexander van der Vekens, 28-Jun-2018.)
((𝑊 ∈ Word 𝑉𝐼 ∈ (0..^(♯‘𝑊))) → (𝑊𝐼) ∈ 𝑉)
 
Theoremwrdfn 11104 A word is a function with a zero-based sequence of integers as domain. (Contributed by Alexander van der Vekens, 13-Apr-2018.)
(𝑊 ∈ Word 𝑆𝑊 Fn (0..^(♯‘𝑊)))
 
Theoremwrdv 11105 A word over an alphabet is a word over the universal class. (Contributed by AV, 8-Feb-2021.) (Proof shortened by JJ, 18-Nov-2022.)
(𝑊 ∈ Word 𝑉𝑊 ∈ Word V)
 
Theoremwrdlndm 11106 The length of a word is not in the domain of the word (regarded as a function). (Contributed by AV, 3-Mar-2021.) (Proof shortened by JJ, 18-Nov-2022.)
(𝑊 ∈ Word 𝑉 → (♯‘𝑊) ∉ dom 𝑊)
 
Theoremiswrdsymb 11107* An arbitrary word is a word over an alphabet if all of its symbols belong to the alphabet. (Contributed by AV, 23-Jan-2021.)
((𝑊 ∈ Word V ∧ ∀𝑖 ∈ (0..^(♯‘𝑊))(𝑊𝑖) ∈ 𝑉) → 𝑊 ∈ Word 𝑉)
 
Theoremwrdfin 11108 A word is a finite set. (Contributed by Stefan O'Rear, 2-Nov-2015.) (Proof shortened by AV, 18-Nov-2018.)
(𝑊 ∈ Word 𝑆𝑊 ∈ Fin)
 
Theoremlennncl 11109 The length of a nonempty word is a positive integer. (Contributed by Mario Carneiro, 1-Oct-2015.)
((𝑊 ∈ Word 𝑆𝑊 ≠ ∅) → (♯‘𝑊) ∈ ℕ)
 
Theoremwrdffz 11110 A word is a function from a finite interval of integers. (Contributed by AV, 10-Feb-2021.)
(𝑊 ∈ Word 𝑆𝑊:(0...((♯‘𝑊) − 1))⟶𝑆)
 
Theoremwrdeq 11111 Equality theorem for the set of words. (Contributed by Mario Carneiro, 26-Feb-2016.)
(𝑆 = 𝑇 → Word 𝑆 = Word 𝑇)
 
Theoremwrdeqi 11112 Equality theorem for the set of words, inference form. (Contributed by AV, 23-May-2021.)
𝑆 = 𝑇       Word 𝑆 = Word 𝑇
 
Theoremiswrddm0 11113 A function with empty domain is a word. (Contributed by AV, 13-Oct-2018.)
(𝑊:∅⟶𝑆𝑊 ∈ Word 𝑆)
 
Theoremwrd0 11114 The empty set is a word (the empty word, frequently denoted ε in this context). This corresponds to the definition in Section 9.1 of [AhoHopUll] p. 318. (Contributed by Stefan O'Rear, 15-Aug-2015.) (Proof shortened by AV, 13-May-2020.)
∅ ∈ Word 𝑆
 
Theorem0wrd0 11115 The empty word is the only word over an empty alphabet. (Contributed by AV, 25-Oct-2018.)
(𝑊 ∈ Word ∅ ↔ 𝑊 = ∅)
 
Theoremffz0iswrdnn0 11116 A sequence with zero-based indices is a word. (Contributed by AV, 31-Jan-2018.) (Proof shortened by AV, 13-Oct-2018.) (Proof shortened by JJ, 18-Nov-2022.)
((𝑊:(0...𝐿)⟶𝑆𝐿 ∈ ℕ0) → 𝑊 ∈ Word 𝑆)
 
Theoremwrdsymb 11117 A word is a word over the symbols it consists of. (Contributed by AV, 1-Dec-2022.)
(𝑆 ∈ Word 𝐴𝑆 ∈ Word (𝑆 “ (0..^(♯‘𝑆))))
 
Theoremnfwrd 11118 Hypothesis builder for Word 𝑆. (Contributed by Mario Carneiro, 26-Feb-2016.)
𝑥𝑆       𝑥Word 𝑆
 
Theoremcsbwrdg 11119* Class substitution for the symbols of a word. (Contributed by Alexander van der Vekens, 15-Jul-2018.)
(𝑆𝑉𝑆 / 𝑥Word 𝑥 = Word 𝑆)
 
Theoremwrdnval 11120* Words of a fixed length are mappings from a fixed half-open integer interval. (Contributed by Alexander van der Vekens, 25-Mar-2018.) (Proof shortened by AV, 13-May-2020.)
((𝑉𝑋𝑁 ∈ ℕ0) → {𝑤 ∈ Word 𝑉 ∣ (♯‘𝑤) = 𝑁} = (𝑉𝑚 (0..^𝑁)))
 
Theoremwrdmap 11121 Words as a mapping. (Contributed by Thierry Arnoux, 4-Mar-2020.)
((𝑉𝑋𝑁 ∈ ℕ0) → ((𝑊 ∈ Word 𝑉 ∧ (♯‘𝑊) = 𝑁) ↔ 𝑊 ∈ (𝑉𝑚 (0..^𝑁))))
 
Theoremwrdsymb0 11122 A symbol at a position "outside" of a word. (Contributed by Alexander van der Vekens, 26-May-2018.) (Proof shortened by AV, 2-May-2020.)
((𝑊 ∈ Word 𝑉𝐼 ∈ ℤ) → ((𝐼 < 0 ∨ (♯‘𝑊) ≤ 𝐼) → (𝑊𝐼) = ∅))
 
Theoremwrdlenge1n0 11123 A word with length at least 1 is not empty. (Contributed by AV, 14-Oct-2018.)
(𝑊 ∈ Word 𝑉 → (𝑊 ≠ ∅ ↔ 1 ≤ (♯‘𝑊)))
 
Theoremlen0nnbi 11124 The length of a word is a positive integer iff the word is not empty. (Contributed by AV, 22-Mar-2022.)
(𝑊 ∈ Word 𝑆 → (𝑊 ≠ ∅ ↔ (♯‘𝑊) ∈ ℕ))
 
Theoremwrdlenge2n0 11125 A word with length at least 2 is not empty. (Contributed by AV, 18-Jun-2018.) (Proof shortened by AV, 14-Oct-2018.)
((𝑊 ∈ Word 𝑉 ∧ 2 ≤ (♯‘𝑊)) → 𝑊 ≠ ∅)
 
Theoremwrdsymb1 11126 The first symbol of a nonempty word over an alphabet belongs to the alphabet. (Contributed by Alexander van der Vekens, 28-Jun-2018.)
((𝑊 ∈ Word 𝑉 ∧ 1 ≤ (♯‘𝑊)) → (𝑊‘0) ∈ 𝑉)
 
Theoremwrdlen1 11127* A word of length 1 starts with a symbol. (Contributed by AV, 20-Jul-2018.) (Proof shortened by AV, 19-Oct-2018.)
((𝑊 ∈ Word 𝑉 ∧ (♯‘𝑊) = 1) → ∃𝑣𝑉 (𝑊‘0) = 𝑣)
 
Theoremfstwrdne 11128 The first symbol of a nonempty word is element of the alphabet for the word. (Contributed by AV, 28-Sep-2018.) (Proof shortened by AV, 14-Oct-2018.)
((𝑊 ∈ Word 𝑉𝑊 ≠ ∅) → (𝑊‘0) ∈ 𝑉)
 
Theoremfstwrdne0 11129 The first symbol of a nonempty word is element of the alphabet for the word. (Contributed by AV, 29-Sep-2018.) (Proof shortened by AV, 14-Oct-2018.)
((𝑁 ∈ ℕ ∧ (𝑊 ∈ Word 𝑉 ∧ (♯‘𝑊) = 𝑁)) → (𝑊‘0) ∈ 𝑉)
 
Theoremeqwrd 11130* Two words are equal iff they have the same length and the same symbol at each position. (Contributed by AV, 13-Apr-2018.) (Revised by JJ, 30-Dec-2023.)
((𝑈 ∈ Word 𝑆𝑊 ∈ Word 𝑇) → (𝑈 = 𝑊 ↔ ((♯‘𝑈) = (♯‘𝑊) ∧ ∀𝑖 ∈ (0..^(♯‘𝑈))(𝑈𝑖) = (𝑊𝑖))))
 
Theoremelovmpowrd 11131* Implications for the value of an operation defined by the maps-to notation with a class abstraction of words as a result having an element. Note that 𝜑 may depend on 𝑧 as well as on 𝑣 and 𝑦. (Contributed by Alexander van der Vekens, 15-Jul-2018.)
𝑂 = (𝑣 ∈ V, 𝑦 ∈ V ↦ {𝑧 ∈ Word 𝑣𝜑})       (𝑍 ∈ (𝑉𝑂𝑌) → (𝑉 ∈ V ∧ 𝑌 ∈ V ∧ 𝑍 ∈ Word 𝑉))
 
Theoremwrdred1 11132 A word truncated by a symbol is a word. (Contributed by AV, 29-Jan-2021.)
(𝐹 ∈ Word 𝑆 → (𝐹 ↾ (0..^((♯‘𝐹) − 1))) ∈ Word 𝑆)
 
Theoremwrdred1hash 11133 The length of a word truncated by a symbol. (Contributed by Alexander van der Vekens, 1-Nov-2017.) (Revised by AV, 29-Jan-2021.)
((𝐹 ∈ Word 𝑆 ∧ 1 ≤ (♯‘𝐹)) → (♯‘(𝐹 ↾ (0..^((♯‘𝐹) − 1)))) = ((♯‘𝐹) − 1))
 
4.7.2  Last symbol of a word
 
Syntaxclsw 11134 Extend class notation with the Last Symbol of a word.
class lastS
 
Definitiondf-lsw 11135 Extract the last symbol of a word. May be not meaningful for other sets which are not words. The name lastS (as abbreviation of "lastSymbol") is a compromise between usually used names for corresponding functions in computer programs (as last() or lastChar()), the terminology used for words in set.mm ("symbol" instead of "character") and brevity ("lastS" is shorter than "lastChar" and "lastSymbol"). Labels of theorems about last symbols of a word will contain the abbreviation "lsw" (Last Symbol of a Word). (Contributed by Alexander van der Vekens, 18-Mar-2018.)
lastS = (𝑤 ∈ V ↦ (𝑤‘((♯‘𝑤) − 1)))
 
Theoremlswwrd 11136 Extract the last symbol of a word. (Contributed by Alexander van der Vekens, 18-Mar-2018.) (Revised by Jim Kingdon, 18-Dec-2025.)
(𝑊 ∈ Word 𝑉 → (lastS‘𝑊) = (𝑊‘((♯‘𝑊) − 1)))
 
Theoremlsw0 11137 The last symbol of an empty word does not exist. (Contributed by Alexander van der Vekens, 19-Mar-2018.) (Proof shortened by AV, 2-May-2020.)
((𝑊 ∈ Word 𝑉 ∧ (♯‘𝑊) = 0) → (lastS‘𝑊) = ∅)
 
Theoremlsw0g 11138 The last symbol of an empty word does not exist. (Contributed by Alexander van der Vekens, 11-Nov-2018.)
(lastS‘∅) = ∅
 
Theoremlsw1 11139 The last symbol of a word of length 1 is the first symbol of this word. (Contributed by Alexander van der Vekens, 19-Mar-2018.)
((𝑊 ∈ Word 𝑉 ∧ (♯‘𝑊) = 1) → (lastS‘𝑊) = (𝑊‘0))
 
Theoremlswcl 11140 Closure of the last symbol: the last symbol of a nonempty word belongs to the alphabet for the word. (Contributed by AV, 2-Aug-2018.) (Proof shortened by AV, 29-Apr-2020.)
((𝑊 ∈ Word 𝑉𝑊 ≠ ∅) → (lastS‘𝑊) ∈ 𝑉)
 
Theoremlswex 11141 Existence of the last symbol. The last symbol of a word is a set. See lsw0g 11138 or lswcl 11140 if you want more specific results for empty or nonempty words, respectively. (Contributed by Jim Kingdon, 27-Dec-2025.)
(𝑊 ∈ Word 𝑉 → (lastS‘𝑊) ∈ V)
 
Theoremlswlgt0cl 11142 The last symbol of a nonempty word is an element of the alphabet for the word. (Contributed by Alexander van der Vekens, 1-Oct-2018.) (Proof shortened by AV, 29-Apr-2020.)
((𝑁 ∈ ℕ ∧ (𝑊 ∈ Word 𝑉 ∧ (♯‘𝑊) = 𝑁)) → (lastS‘𝑊) ∈ 𝑉)
 
4.7.3  Concatenations of words
 
Syntaxcconcat 11143 Syntax for the concatenation operator.
class ++
 
Definitiondf-concat 11144* Define the concatenation operator which combines two words. Definition in Section 9.1 of [AhoHopUll] p. 318. (Contributed by FL, 14-Jan-2014.) (Revised by Stefan O'Rear, 15-Aug-2015.)
++ = (𝑠 ∈ V, 𝑡 ∈ V ↦ (𝑥 ∈ (0..^((♯‘𝑠) + (♯‘𝑡))) ↦ if(𝑥 ∈ (0..^(♯‘𝑠)), (𝑠𝑥), (𝑡‘(𝑥 − (♯‘𝑠))))))
 
Theoremccatfvalfi 11145* Value of the concatenation operator. (Contributed by Stefan O'Rear, 15-Aug-2015.)
((𝑆 ∈ Fin ∧ 𝑇 ∈ Fin) → (𝑆 ++ 𝑇) = (𝑥 ∈ (0..^((♯‘𝑆) + (♯‘𝑇))) ↦ if(𝑥 ∈ (0..^(♯‘𝑆)), (𝑆𝑥), (𝑇‘(𝑥 − (♯‘𝑆))))))
 
Theoremccatcl 11146 The concatenation of two words is a word. (Contributed by FL, 2-Feb-2014.) (Proof shortened by Stefan O'Rear, 15-Aug-2015.) (Proof shortened by AV, 29-Apr-2020.)
((𝑆 ∈ Word 𝐵𝑇 ∈ Word 𝐵) → (𝑆 ++ 𝑇) ∈ Word 𝐵)
 
Theoremccatclab 11147 The concatenation of words over two sets is a word over the union of those sets. (Contributed by Jim Kingdon, 19-Dec-2025.)
((𝑆 ∈ Word 𝐴𝑇 ∈ Word 𝐵) → (𝑆 ++ 𝑇) ∈ Word (𝐴𝐵))
 
Theoremccatlen 11148 The length of a concatenated word. (Contributed by Stefan O'Rear, 15-Aug-2015.) (Revised by JJ, 1-Jan-2024.)
((𝑆 ∈ Word 𝐴𝑇 ∈ Word 𝐵) → (♯‘(𝑆 ++ 𝑇)) = ((♯‘𝑆) + (♯‘𝑇)))
 
Theoremccat0 11149 The concatenation of two words is empty iff the two words are empty. (Contributed by AV, 4-Mar-2022.) (Revised by JJ, 18-Jan-2024.)
((𝑆 ∈ Word 𝐴𝑇 ∈ Word 𝐵) → ((𝑆 ++ 𝑇) = ∅ ↔ (𝑆 = ∅ ∧ 𝑇 = ∅)))
 
Theoremccatval1 11150 Value of a symbol in the left half of a concatenated word. (Contributed by Stefan O'Rear, 15-Aug-2015.) (Revised by Mario Carneiro, 22-Sep-2015.) (Proof shortened by AV, 30-Apr-2020.) (Revised by JJ, 18-Jan-2024.)
((𝑆 ∈ Word 𝐴𝑇 ∈ Word 𝐵𝐼 ∈ (0..^(♯‘𝑆))) → ((𝑆 ++ 𝑇)‘𝐼) = (𝑆𝐼))
 
Theoremccatval2 11151 Value of a symbol in the right half of a concatenated word. (Contributed by Stefan O'Rear, 15-Aug-2015.) (Revised by Mario Carneiro, 22-Sep-2015.)
((𝑆 ∈ Word 𝐵𝑇 ∈ Word 𝐵𝐼 ∈ ((♯‘𝑆)..^((♯‘𝑆) + (♯‘𝑇)))) → ((𝑆 ++ 𝑇)‘𝐼) = (𝑇‘(𝐼 − (♯‘𝑆))))
 
Theoremccatval3 11152 Value of a symbol in the right half of a concatenated word, using an index relative to the subword. (Contributed by Stefan O'Rear, 16-Aug-2015.) (Proof shortened by AV, 30-Apr-2020.)
((𝑆 ∈ Word 𝐵𝑇 ∈ Word 𝐵𝐼 ∈ (0..^(♯‘𝑇))) → ((𝑆 ++ 𝑇)‘(𝐼 + (♯‘𝑆))) = (𝑇𝐼))
 
Theoremelfzelfzccat 11153 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 𝑉𝐵 ∈ Word 𝑉) → (𝑁 ∈ (0...(♯‘𝐴)) → 𝑁 ∈ (0...(♯‘(𝐴 ++ 𝐵)))))
 
Theoremccatvalfn 11154 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 𝑉𝐵 ∈ Word 𝑉) → (𝐴 ++ 𝐵) Fn (0..^((♯‘𝐴) + (♯‘𝐵))))
 
Theoremccatsymb 11155 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 𝑉𝐵 ∈ Word 𝑉𝐼 ∈ ℤ) → ((𝐴 ++ 𝐵)‘𝐼) = if(𝐼 < (♯‘𝐴), (𝐴𝐼), (𝐵‘(𝐼 − (♯‘𝐴)))))
 
Theoremccatfv0 11156 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 𝑉𝐵 ∈ Word 𝑉 ∧ 0 < (♯‘𝐴)) → ((𝐴 ++ 𝐵)‘0) = (𝐴‘0))
 
Theoremccatval1lsw 11157 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 𝑉𝐵 ∈ Word 𝑉𝐴 ≠ ∅) → ((𝐴 ++ 𝐵)‘((♯‘𝐴) − 1)) = (lastS‘𝐴))
 
Theoremccatval21sw 11158 The first symbol of the right (nonempty) half of a concatenated word. (Contributed by AV, 23-Apr-2022.)
((𝐴 ∈ Word 𝑉𝐵 ∈ Word 𝑉𝐵 ≠ ∅) → ((𝐴 ++ 𝐵)‘(♯‘𝐴)) = (𝐵‘0))
 
Theoremccatlid 11159 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)
 
Theoremccatrid 11160 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)
 
Theoremccatass 11161 Associative law for concatenation of words. (Contributed by Stefan O'Rear, 15-Aug-2015.)
((𝑆 ∈ Word 𝐵𝑇 ∈ Word 𝐵𝑈 ∈ Word 𝐵) → ((𝑆 ++ 𝑇) ++ 𝑈) = (𝑆 ++ (𝑇 ++ 𝑈)))
 
Theoremccatrn 11162 The range of a concatenated word. (Contributed by Stefan O'Rear, 15-Aug-2015.)
((𝑆 ∈ Word 𝐵𝑇 ∈ Word 𝐵) → ran (𝑆 ++ 𝑇) = (ran 𝑆 ∪ ran 𝑇))
 
Theoremccatidid 11163 Concatenation of the empty word by the empty word. (Contributed by AV, 26-Mar-2022.)
(∅ ++ ∅) = ∅
 
Theoremlswccatn0lsw 11164 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 𝑉𝐵 ∈ Word 𝑉𝐵 ≠ ∅) → (lastS‘(𝐴 ++ 𝐵)) = (lastS‘𝐵))
 
Theoremlswccat0lsw 11165 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))
 
Theoremccatalpha 11166 A concatenation of two arbitrary words is a word over an alphabet iff the symbols of both words belong to the alphabet. (Contributed by AV, 28-Feb-2021.)
((𝐴 ∈ Word V ∧ 𝐵 ∈ Word V) → ((𝐴 ++ 𝐵) ∈ Word 𝑆 ↔ (𝐴 ∈ Word 𝑆𝐵 ∈ Word 𝑆)))
 
Theoremccatrcl1 11167 Reverse closure of a concatenation: If the concatenation of two arbitrary words is a word over an alphabet then the symbols of the first word belong to the alphabet. (Contributed by AV, 3-Mar-2021.)
((𝐴 ∈ Word 𝑋𝐵 ∈ Word 𝑌 ∧ (𝑊 = (𝐴 ++ 𝐵) ∧ 𝑊 ∈ Word 𝑆)) → 𝐴 ∈ Word 𝑆)
 
4.7.4  Singleton words
 
Syntaxcs1 11168 Syntax for the singleton word constructor.
class ⟨“𝐴”⟩
 
Definitiondf-s1 11169 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 11176, 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)⟩}
 
Theorems1val 11170 Value of a singleton word. (Contributed by Stefan O'Rear, 15-Aug-2015.) (Revised by Mario Carneiro, 26-Feb-2016.)
(𝐴𝑉 → ⟨“𝐴”⟩ = {⟨0, 𝐴⟩})
 
Theorems1rn 11171 The range of a singleton word. (Contributed by Mario Carneiro, 18-Jul-2016.)
(𝐴𝑉 → ran ⟨“𝐴”⟩ = {𝐴})
 
Theorems1eq 11172 Equality theorem for a singleton word. (Contributed by Mario Carneiro, 26-Feb-2016.)
(𝐴 = 𝐵 → ⟨“𝐴”⟩ = ⟨“𝐵”⟩)
 
Theorems1eqd 11173 Equality theorem for a singleton word. (Contributed by Mario Carneiro, 26-Feb-2016.)
(𝜑𝐴 = 𝐵)       (𝜑 → ⟨“𝐴”⟩ = ⟨“𝐵”⟩)
 
Theorems1cl 11174 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.)
(𝐴𝐵 → ⟨“𝐴”⟩ ∈ Word 𝐵)
 
Theorems1cld 11175 A singleton word is a word. (Contributed by Mario Carneiro, 26-Feb-2016.)
(𝜑𝐴𝐵)       (𝜑 → ⟨“𝐴”⟩ ∈ Word 𝐵)
 
Theorems1prc 11176 Value of a singleton word if the symbol is a proper class. (Contributed by AV, 26-Mar-2022.)
𝐴 ∈ V → ⟨“𝐴”⟩ = ⟨“∅”⟩)
 
Theorems1leng 11177 Length of a singleton word. (Contributed by Stefan O'Rear, 15-Aug-2015.) (Revised by Mario Carneiro, 26-Feb-2016.)
(𝐴𝑉 → (♯‘⟨“𝐴”⟩) = 1)
 
Theorems1dmg 11178 The domain of a singleton word is a singleton. (Contributed by AV, 9-Jan-2020.)
(𝐴𝑆 → dom ⟨“𝐴”⟩ = {0})
 
Theorems1fv 11179 Sole symbol of a singleton word. (Contributed by Stefan O'Rear, 15-Aug-2015.) (Revised by Mario Carneiro, 26-Feb-2016.)
(𝐴𝐵 → (⟨“𝐴”⟩‘0) = 𝐴)
 
Theoremlsws1 11180 The last symbol of a singleton word is its symbol. (Contributed by AV, 22-Oct-2018.)
(𝐴𝑉 → (lastS‘⟨“𝐴”⟩) = 𝐴)
 
Theoremeqs1 11181 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)”⟩)
 
Theoremwrdl1exs1 11182* A word of length 1 is a singleton word. (Contributed by AV, 24-Jan-2021.)
((𝑊 ∈ Word 𝑆 ∧ (♯‘𝑊) = 1) → ∃𝑠𝑆 𝑊 = ⟨“𝑠”⟩)
 
Theoremwrdl1s1 11183 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.)
(𝑆𝑉 → (𝑊 = ⟨“𝑆”⟩ ↔ (𝑊 ∈ Word 𝑉 ∧ (♯‘𝑊) = 1 ∧ (𝑊‘0) = 𝑆)))
 
Theorems111 11184 The singleton word function is injective. (Contributed by Mario Carneiro, 1-Oct-2015.) (Revised by Mario Carneiro, 26-Feb-2016.)
((𝑆𝐴𝑇𝐴) → (⟨“𝑆”⟩ = ⟨“𝑇”⟩ ↔ 𝑆 = 𝑇))
 
4.7.5  Concatenations with singleton words
 
Theoremccatws1cl 11185 The concatenation of a word with a singleton word is a word. (Contributed by Alexander van der Vekens, 22-Sep-2018.)
((𝑊 ∈ Word 𝑉𝑋𝑉) → (𝑊 ++ ⟨“𝑋”⟩) ∈ Word 𝑉)
 
Theoremccat2s1cl 11186 The concatenation of two singleton words is a word. (Contributed by Alexander van der Vekens, 22-Sep-2018.)
((𝑋𝑉𝑌𝑉) → (⟨“𝑋”⟩ ++ ⟨“𝑌”⟩) ∈ Word 𝑉)
 
Theoremccatws1leng 11187 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 𝑉𝑋𝑌) → (♯‘(𝑊 ++ ⟨“𝑋”⟩)) = ((♯‘𝑊) + 1))
 
Theoremccatws1lenp1bg 11188 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 𝑉𝑋𝑌𝑁 ∈ ℕ0) → ((♯‘(𝑊 ++ ⟨“𝑋”⟩)) = (𝑁 + 1) ↔ (♯‘𝑊) = 𝑁))
 
Theoremwrdlenccats1lenm1g 11189 The length of a word is the length of the word concatenated with a singleton word minus 1. (Contributed by AV, 28-Jun-2018.) (Revised by AV, 5-Mar-2022.)
((𝑊 ∈ Word 𝑉𝑆𝐵) → ((♯‘(𝑊 ++ ⟨“𝑆”⟩)) − 1) = (♯‘𝑊))
 
Theoremccatw2s1cl 11190 The concatenation of a word with two singleton words is a word. (Contributed by Alexander van der Vekens, 22-Sep-2018.)
((𝑊 ∈ Word 𝑉𝑋𝑉𝑌𝑉) → ((𝑊 ++ ⟨“𝑋”⟩) ++ ⟨“𝑌”⟩) ∈ Word 𝑉)
 
Theoremccats1val1g 11191 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 𝑉𝑆𝑌𝐼 ∈ (0..^(♯‘𝑊))) → ((𝑊 ++ ⟨“𝑆”⟩)‘𝐼) = (𝑊𝐼))
 
Theoremccats1val2 11192 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 𝑉𝑆𝑉𝐼 = (♯‘𝑊)) → ((𝑊 ++ ⟨“𝑆”⟩)‘𝐼) = 𝑆)
 
Theoremccat1st1st 11193 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))
 
Theoremccatws1ls 11194 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 𝑉𝑋𝑉) → ((𝑊 ++ ⟨“𝑋”⟩)‘(♯‘𝑊)) = 𝑋)
 
Theoremlswccats1 11195 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 𝑉𝑆𝑉) → (lastS‘(𝑊 ++ ⟨“𝑆”⟩)) = 𝑆)
 
Theoremlswccats1fst 11196 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))
 
Theoremccatw2s1p2 11197 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) ∧ (𝑋𝑉𝑌𝑉)) → (((𝑊 ++ ⟨“𝑋”⟩) ++ ⟨“𝑌”⟩)‘(𝑁 + 1)) = 𝑌)
 
4.7.6  Subwords/substrings
 
Syntaxcsubstr 11198 Syntax for the subword operator.
class substr
 
Definitiondf-substr 11199* 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(((1st𝑏)..^(2nd𝑏)) ⊆ dom 𝑠, (𝑥 ∈ (0..^((2nd𝑏) − (1st𝑏))) ↦ (𝑠‘(𝑥 + (1st𝑏)))), ∅))
 
Theoremfzowrddc 11200 Decidability of whether a range of integers is a subset of a word's domain. (Contributed by Jim Kingdon, 23-Dec-2025.)
((𝑆 ∈ Word 𝐴𝐹 ∈ ℤ ∧ 𝐿 ∈ ℤ) → DECID (𝐹..^𝐿) ⊆ dom 𝑆)
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