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Theorem List for Metamath Proof Explorer - 13901-14000   *Has distinct variable group(s)
TypeLabelDescription
Statement

Theoremeqwrd 13901* 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 13902* 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))

Theoremelovmptnn0wrd 13903* Implications for the value of an operation defined by the maps-to notation with a function of nonnegative integers into 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 and 𝑛. (Contributed by AV, 16-Jul-2018.) (Revised by AV, 16-May-2019.)
𝑂 = (𝑣 ∈ V, 𝑦 ∈ V ↦ (𝑛 ∈ ℕ0 ↦ {𝑧 ∈ Word 𝑣𝜑}))       (𝑍 ∈ ((𝑉𝑂𝑌)‘𝑁) → ((𝑉 ∈ V ∧ 𝑌 ∈ V) ∧ (𝑁 ∈ ℕ0𝑍 ∈ Word 𝑉)))

Theoremwrdred1 13904 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 13905 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))

5.7.2  Last symbol of a word

Syntaxclsw 13906 Extend class notation with the Last Symbol of a word.
class lastS

Definitiondf-lsw 13907 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)))

Theoremlsw 13908 Extract the last symbol of a word. May be not meaningful for other sets which are not words. (Contributed by Alexander van der Vekens, 18-Mar-2018.)
(𝑊𝑋 → (lastS‘𝑊) = (𝑊‘((♯‘𝑊) − 1)))

Theoremlsw0 13909 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 13910 The last symbol of an empty word does not exist. (Contributed by Alexander van der Vekens, 11-Nov-2018.)
(lastS‘∅) = ∅

Theoremlsw1 13911 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 13912 Closure of the last symbol: the last symbol of a not empty 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)

Theoremlswlgt0cl 13913 The last symbol of a nonempty word is 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)

5.7.3  Concatenations of words

Syntaxcconcat 13914 Syntax for the concatenation operator.
class ++

Definitiondf-concat 13915* 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))))))

Theoremccatfn 13916 The concatenation operator is a two-argument function. (Contributed by Mario Carneiro, 27-Sep-2015.) (Proof shortened by AV, 29-Apr-2020.)
++ Fn (V × V)

Theoremccatfval 13917* Value of the concatenation operator. (Contributed by Stefan O'Rear, 15-Aug-2015.)
((𝑆𝑉𝑇𝑊) → (𝑆 ++ 𝑇) = (𝑥 ∈ (0..^((♯‘𝑆) + (♯‘𝑇))) ↦ if(𝑥 ∈ (0..^(♯‘𝑆)), (𝑆𝑥), (𝑇‘(𝑥 − (♯‘𝑆))))))

Theoremccatcl 13918 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)

Theoremccatlen 13919 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)))

TheoremccatlenOLD 13920 Obsolete version of ccatlen 13919 as of 1-Jan-2024. The length of a concatenated word. (Contributed by Stefan O'Rear, 15-Aug-2015.) (Proof modification is discouraged.) (New usage is discouraged.)
((𝑆 ∈ Word 𝐵𝑇 ∈ Word 𝐵) → (♯‘(𝑆 ++ 𝑇)) = ((♯‘𝑆) + (♯‘𝑇)))

Theoremccat0 13921 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 13922 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))

Theoremccatval1OLD 13923 Obsolete version of ccatval1 13922 as of 18-Jan-2024. 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.) (Proof modification is discouraged.) (New usage is discouraged.)
((𝑆 ∈ Word 𝐵𝑇 ∈ Word 𝐵𝐼 ∈ (0..^(♯‘𝑆))) → ((𝑆 ++ 𝑇)‘𝐼) = (𝑆𝐼))

Theoremccatval2 13924 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 13925 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 13926 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 13927 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 13928 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 13929 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 13930 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 13931 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 13932 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 13933 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 13934 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 13935 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 13936 Concatenation of the empty word by the empty word. (Contributed by AV, 26-Mar-2022.)
(∅ ++ ∅) = ∅

Theoremlswccatn0lsw 13937 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 13938 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 13939 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 13940 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)

5.7.4  Singleton words

Syntaxcs1 13941 Syntax for the singleton word constructor.
class ⟨“𝐴”⟩

Definitiondf-s1 13942 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 13950, and not, as maybe expected, the empty word, see also s1nz 13953. (Contributed by Stefan O'Rear, 15-Aug-2015.) (Revised by Mario Carneiro, 26-Feb-2016.)
⟨“𝐴”⟩ = {⟨0, ( I ‘𝐴)⟩}

Theoremids1 13943 Identity function protection for a singleton word. (Contributed by Mario Carneiro, 26-Feb-2016.)
⟨“𝐴”⟩ = ⟨“( I ‘𝐴)”⟩

Theorems1val 13944 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 13945 The range of a singleton word. (Contributed by Mario Carneiro, 18-Jul-2016.)
(𝐴𝑉 → ran ⟨“𝐴”⟩ = {𝐴})

Theorems1eq 13946 Equality theorem for a singleton word. (Contributed by Mario Carneiro, 26-Feb-2016.)
(𝐴 = 𝐵 → ⟨“𝐴”⟩ = ⟨“𝐵”⟩)

Theorems1eqd 13947 Equality theorem for a singleton word. (Contributed by Mario Carneiro, 26-Feb-2016.)
(𝜑𝐴 = 𝐵)       (𝜑 → ⟨“𝐴”⟩ = ⟨“𝐵”⟩)

Theorems1cl 13948 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 13949 A singleton word is a word. (Contributed by Mario Carneiro, 26-Feb-2016.)
(𝜑𝐴𝐵)       (𝜑 → ⟨“𝐴”⟩ ∈ Word 𝐵)

Theorems1prc 13950 Value of a singleton word if the symbol is a proper class. (Contributed by AV, 26-Mar-2022.)
𝐴 ∈ V → ⟨“𝐴”⟩ = ⟨“∅”⟩)

Theorems1cli 13951 A singleton word is a word. (Contributed by Mario Carneiro, 26-Feb-2016.)
⟨“𝐴”⟩ ∈ Word V

Theorems1len 13952 Length of a singleton word. (Contributed by Stefan O'Rear, 15-Aug-2015.) (Revised by Mario Carneiro, 26-Feb-2016.)
(♯‘⟨“𝐴”⟩) = 1

Theorems1nz 13953 A singleton word is not the empty string. (Contributed by Mario Carneiro, 27-Feb-2016.) (Proof shortened by Kyle Wyonch, 18-Jul-2021.)
⟨“𝐴”⟩ ≠ ∅

Theorems1dm 13954 The domain of a singleton word is a singleton. (Contributed by AV, 9-Jan-2020.)
dom ⟨“𝐴”⟩ = {0}

Theorems1dmALT 13955 Alternate version of s1dm 13954, having a shorter proof, but requiring that 𝐴 is a set. (Contributed by AV, 9-Jan-2020.) (Proof modification is discouraged.) (New usage is discouraged.)
(𝐴𝑆 → dom ⟨“𝐴”⟩ = {0})

Theorems1fv 13956 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 13957 The last symbol of a singleton word is its symbol. (Contributed by AV, 22-Oct-2018.)
(𝐴𝑉 → (lastS‘⟨“𝐴”⟩) = 𝐴)

Theoremeqs1 13958 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 13959* A word of length 1 is a singleton word. (Contributed by AV, 24-Jan-2021.)
((𝑊 ∈ Word 𝑆 ∧ (♯‘𝑊) = 1) → ∃𝑠𝑆 𝑊 = ⟨“𝑠”⟩)

Theoremwrdl1s1 13960 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 13961 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))

5.7.5  Concatenations with singleton words

Theoremccatws1cl 13962 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)

Theoremccatws1clv 13963 The concatenation of a word with a singleton word (which can be over a different alphabet) is a word. (Contributed by AV, 5-Mar-2022.)
(𝑊 ∈ Word 𝑉 → (𝑊 ++ ⟨“𝑋”⟩) ∈ Word V)

Theoremccat2s1cl 13964 The concatenation of two singleton words is a word. (Contributed by Alexander van der Vekens, 22-Sep-2018.)
((𝑋𝑉𝑌𝑉) → (⟨“𝑋”⟩ ++ ⟨“𝑌”⟩) ∈ Word 𝑉)

Theoremccats1alpha 13965 A concatenation of a word with a singleton word is a word over an alphabet 𝑆 iff the symbols of both words belong to the alphabet 𝑆. (Contributed by AV, 27-Mar-2022.)
((𝐴 ∈ Word 𝑉𝑋𝑈) → ((𝐴 ++ ⟨“𝑋”⟩) ∈ Word 𝑆 ↔ (𝐴 ∈ Word 𝑆𝑋𝑆)))

Theoremccatws1len 13966 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 → (♯‘(𝑊 ++ ⟨“𝑋”⟩)) = ((♯‘𝑊) + 1))

Theoremccatws1lenp1b 13967 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 𝑉𝑁 ∈ ℕ0) → ((♯‘(𝑊 ++ ⟨“𝑋”⟩)) = (𝑁 + 1) ↔ (♯‘𝑊) = 𝑁))

Theoremwrdlenccats1lenm1 13968 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 𝑉 → ((♯‘(𝑊 ++ ⟨“𝑆”⟩)) − 1) = (♯‘𝑊))

Theoremccat2s1len 13969 The length of the concatenation of two singleton words. (Contributed by Alexander van der Vekens, 22-Sep-2018.) (Revised by JJ, 14-Jan-2024.)
(♯‘(⟨“𝑋”⟩ ++ ⟨“𝑌”⟩)) = 2

Theoremccat2s1lenOLD 13970 Obsolete version of ccat2s1len 13969 as of 14-Jan-2024. The length of the concatenation of two singleton words. (Contributed by Alexander van der Vekens, 22-Sep-2018.) (Proof modification is discouraged.) (New usage is discouraged.)
((𝑋𝑉𝑌𝑉) → (♯‘(⟨“𝑋”⟩ ++ ⟨“𝑌”⟩)) = 2)

Theoremccatw2s1cl 13971 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)

Theoremccatw2s1len 13972 The length of the concatenation of a word with two singleton words. (Contributed by Alexander van der Vekens, 22-Sep-2018.) (Revised by AV, 5-Mar-2022.)
(𝑊 ∈ Word 𝑉 → (♯‘((𝑊 ++ ⟨“𝑋”⟩) ++ ⟨“𝑌”⟩)) = ((♯‘𝑊) + 2))

Theoremccats1val1 13973 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 𝑉𝐼 ∈ (0..^(♯‘𝑊))) → ((𝑊 ++ ⟨“𝑆”⟩)‘𝐼) = (𝑊𝐼))

Theoremccats1val1OLD 13974 Obsolete version of ccats1val1 13973 as of 20-Jan-2024. 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.) (Proof modification is discouraged.) (New usage is discouraged.)
((𝑊 ∈ Word 𝑉𝑆𝑉𝐼 ∈ (0..^(♯‘𝑊))) → ((𝑊 ++ ⟨“𝑆”⟩)‘𝐼) = (𝑊𝐼))

Theoremccats1val2 13975 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 13976 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))

Theoremccat2s1p1 13977 Extract the first of two concatenated singleton words. (Contributed by Alexander van der Vekens, 22-Sep-2018.) (Revised by JJ, 20-Jan-2024.)
(𝑋𝑉 → ((⟨“𝑋”⟩ ++ ⟨“𝑌”⟩)‘0) = 𝑋)

Theoremccat2s1p2 13978 Extract the second of two concatenated singleton words. (Contributed by Alexander van der Vekens, 22-Sep-2018.) (Revised by JJ, 20-Jan-2024.)
(𝑌𝑉 → ((⟨“𝑋”⟩ ++ ⟨“𝑌”⟩)‘1) = 𝑌)

Theoremccat2s1p1OLD 13979 Obsolete version of ccat2s1p1 13977 as of 20-Jan-2024. Extract the first of two concatenated singleton words. (Contributed by Alexander van der Vekens, 22-Sep-2018.) (Proof modification is discouraged.) (New usage is discouraged.)
((𝑋𝑉𝑌𝑉) → ((⟨“𝑋”⟩ ++ ⟨“𝑌”⟩)‘0) = 𝑋)

Theoremccat2s1p2OLD 13980 Obsolete version of ccat2s1p2 13978 as of 20-Jan-2024. Extract the second of two concatenated singleton words. (Contributed by Alexander van der Vekens, 22-Sep-2018.) (Proof modification is discouraged.) (New usage is discouraged.)
((𝑋𝑉𝑌𝑉) → ((⟨“𝑋”⟩ ++ ⟨“𝑌”⟩)‘1) = 𝑌)

Theoremccatw2s1ass 13981 Associative law for a concatenation of a word with two singleton words. (Contributed by Alexander van der Vekens, 22-Sep-2018.)
(𝑊 ∈ Word 𝑉 → ((𝑊 ++ ⟨“𝑋”⟩) ++ ⟨“𝑌”⟩) = (𝑊 ++ (⟨“𝑋”⟩ ++ ⟨“𝑌”⟩)))

Theoremccatw2s1assOLD 13982 Obsolete version of ccatw2s1ass 13981 as of 29-Jan-2024. Associative law for a concatenation of a word with two singleton words. (Contributed by Alexander van der Vekens, 22-Sep-2018.) (Proof modification is discouraged.) (New usage is discouraged.)
((𝑊 ∈ Word 𝑉𝑋𝑉𝑌𝑉) → ((𝑊 ++ ⟨“𝑋”⟩) ++ ⟨“𝑌”⟩) = (𝑊 ++ (⟨“𝑋”⟩ ++ ⟨“𝑌”⟩)))

Theoremccatws1n0 13983 The concatenation of a word with a singleton word is not the empty set. (Contributed by Alexander van der Vekens, 29-Sep-2018.) (Revised by AV, 5-Mar-2022.)
(𝑊 ∈ Word 𝑉 → (𝑊 ++ ⟨“𝑋”⟩) ≠ ∅)

Theoremccatws1ls 13984 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 13985 The last symbol of a word concatenated with a singleton word is the symbol of the singleton word. (Contributed by AV, 6-Aug-2018.) (Proof shortened by AV, 22-Oct-2018.)
((𝑊 ∈ Word 𝑉𝑆𝑉) → (lastS‘(𝑊 ++ ⟨“𝑆”⟩)) = 𝑆)

Theoremlswccats1fst 13986 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))

Theoremccatw2s1p1 13987 Extract the symbol of the first singleton word of a word concatenated with this singleton word and another singleton word. (Contributed by Alexander van der Vekens, 22-Sep-2018.) (Proof shortened by AV, 1-May-2020.) (Revised by AV, 1-May-2020.) (Revised by AV, 29-Jan-2024.)
((𝑊 ∈ Word 𝑉 ∧ (♯‘𝑊) = 𝑁𝑋𝑉) → (((𝑊 ++ ⟨“𝑋”⟩) ++ ⟨“𝑌”⟩)‘𝑁) = 𝑋)

Theoremccatw2s1p1OLD 13988 Obsolete version of ccatw2s1p1 13987 as of 29-Jan-2024. Extract the symbol of the first singleton word of a word concatenated with this singleton word and another singleton word. (Contributed by Alexander van der Vekens, 22-Sep-2018.) (Proof shortened by AV, 1-May-2020.) (Proof modification is discouraged.) (New usage is discouraged.)
(((𝑊 ∈ Word 𝑉 ∧ (♯‘𝑊) = 𝑁) ∧ (𝑋𝑉𝑌𝑉)) → (((𝑊 ++ ⟨“𝑋”⟩) ++ ⟨“𝑌”⟩)‘𝑁) = 𝑋)

Theoremccatw2s1p2 13989 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)

Theoremccat2s1fvw 13990 Extract a symbol of a word from the concatenation of the word with two single symbols. (Contributed by AV, 22-Sep-2018.) (Revised by AV, 13-Jan-2020.) (Proof shortened by AV, 1-May-2020.) (Revised by AV, 28-Jan-2024.)
((𝑊 ∈ Word 𝑉𝐼 ∈ ℕ0𝐼 < (♯‘𝑊)) → (((𝑊 ++ ⟨“𝑋”⟩) ++ ⟨“𝑌”⟩)‘𝐼) = (𝑊𝐼))

Theoremccat2s1fvwOLD 13991 Obsolete version of ccat2s1fvw 13990 as of 28-Jan-2024. Extract a symbol of a word from the concatenation of the word with two single symbols. (Contributed by AV, 22-Sep-2018.) (Revised by AV, 13-Jan-2020.) (Proof shortened by AV, 1-May-2020.) (New usage is discouraged.) (Proof modification is discouraged.)
(((𝑊 ∈ Word 𝑉𝐼 ∈ ℕ0𝐼 < (♯‘𝑊)) ∧ (𝑋𝑉𝑌𝑉)) → (((𝑊 ++ ⟨“𝑋”⟩) ++ ⟨“𝑌”⟩)‘𝐼) = (𝑊𝐼))

Theoremccat2s1fst 13992 The first symbol of the concatenation of a word with two single symbols. (Contributed by Alexander van der Vekens, 22-Sep-2018.) (Revised by AV, 28-Jan-2024.)
((𝑊 ∈ Word 𝑉 ∧ 0 < (♯‘𝑊)) → (((𝑊 ++ ⟨“𝑋”⟩) ++ ⟨“𝑌”⟩)‘0) = (𝑊‘0))

Theoremccat2s1fstOLD 13993 Obsolete version of ccat2s1fst 13992 as of 28-Jan-2024. The first symbol of the concatenation of a word with two single symbols. (Contributed by Alexander van der Vekens, 22-Sep-2018.) (New usage is discouraged.) (Proof modification is discouraged.)
(((𝑊 ∈ Word 𝑉 ∧ 0 < (♯‘𝑊)) ∧ (𝑋𝑉𝑌𝑉)) → (((𝑊 ++ ⟨“𝑋”⟩) ++ ⟨“𝑌”⟩)‘0) = (𝑊‘0))

5.7.6  Subwords/substrings

Syntaxcsubstr 13994 Syntax for the subword operator.
class substr

Definitiondf-substr 13995* 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)))), ∅))

Theoremswrdnznd 13996 The value of a subword operation for noninteger arguments is the empty set. (This is due to our definition of function values for out-of-domain arguments, see ndmfv 6693). (Contributed by AV, 2-Dec-2022.) (New usage is discouraged.)
(¬ (𝐹 ∈ ℤ ∧ 𝐿 ∈ ℤ) → (𝑆 substr ⟨𝐹, 𝐿⟩) = ∅)

Theoremswrdval 13997* Value of a subword. (Contributed by Stefan O'Rear, 15-Aug-2015.)
((𝑆𝑉𝐹 ∈ ℤ ∧ 𝐿 ∈ ℤ) → (𝑆 substr ⟨𝐹, 𝐿⟩) = if((𝐹..^𝐿) ⊆ dom 𝑆, (𝑥 ∈ (0..^(𝐿𝐹)) ↦ (𝑆‘(𝑥 + 𝐹))), ∅))

Theoremswrd00 13998 A zero length substring. (Contributed by Stefan O'Rear, 27-Aug-2015.)
(𝑆 substr ⟨𝑋, 𝑋⟩) = ∅

Theoremswrdcl 13999 Closure of the subword extractor. (Contributed by Stefan O'Rear, 16-Aug-2015.) (Revised by Mario Carneiro, 26-Feb-2016.)
(𝑆 ∈ Word 𝐴 → (𝑆 substr ⟨𝐹, 𝐿⟩) ∈ Word 𝐴)

Theoremswrdval2 14000* 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 𝐴𝐹 ∈ (0...𝐿) ∧ 𝐿 ∈ (0...(♯‘𝑆))) → (𝑆 substr ⟨𝐹, 𝐿⟩) = (𝑥 ∈ (0..^(𝐿𝐹)) ↦ (𝑆‘(𝑥 + 𝐹))))

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