P. 113 of Theorem List - Intuitionistic Logic Explorer

Theorem List for Intuitionistic Logic Explorer - 11201-11300   *Has distinct variable group(s)

Type	Label	Description
Statement
Theorem	s111 11201	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 11202	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 11203	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 11204	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 11205	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	wrdlenccats1lenm1g 11206	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 ∧ 𝑆 ∈ 𝐵) → ((♯‘(𝑊 ++ ⟨“𝑆”⟩)) − 1) = (♯‘𝑊))
Theorem	ccatw2s1cl 11207	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	ccatw2s1leng 11208	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 ∧ 𝑋 ∈ 𝑉 ∧ 𝑌 ∈ 𝑉) → (♯‘((𝑊 ++ ⟨“𝑋”⟩) ++ ⟨“𝑌”⟩)) = ((♯‘𝑊) + 2))
Theorem	ccats1val1g 11209	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 11210	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 11211	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 11212	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 11213	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 11214	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	ccatw2s1p1g 11215	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) = 𝑁) ∧ (𝑋 ∈ 𝑉 ∧ 𝑌 ∈ 𝑉)) → (((𝑊 ++ ⟨“𝑋”⟩) ++ ⟨“𝑌”⟩)‘𝑁) = 𝑋)
Theorem	ccatw2s1p2 11216	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)
Theorem	ccat2s1fvwd 11217	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.)
⊢ (𝜑 → 𝑊 ∈ Word 𝑉)    &   ⊢ (𝜑 → 𝐼 ∈ ℕ0)    &   ⊢ (𝜑 → 𝐼 < (♯‘𝑊))    &   ⊢ (𝜑 → 𝑋 ∈ 𝐴)    &   ⊢ (𝜑 → 𝑌 ∈ 𝐵)    ⇒   ⊢ (𝜑 → (((𝑊 ++ ⟨“𝑋”⟩) ++ ⟨“𝑌”⟩)‘𝐼) = (𝑊‘𝐼))
Theorem	ccat2s1fstg 11218	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)) ∧ (𝑋 ∈ 𝐴 ∧ 𝑌 ∈ 𝐵)) → (((𝑊 ++ ⟨“𝑋”⟩) ++ ⟨“𝑌”⟩)‘0) = (𝑊‘0))
4.7.6  Subwords/substrings
Syntax	csubstr 11219	Syntax for the subword operator.
class substr
Definition	df-substr 11220*	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 11221	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 11222*	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 11223	A zero length substring. (Contributed by Stefan O'Rear, 27-Aug-2015.)
⊢ ((𝑆 ∈ 𝑉 ∧ 𝑋 ∈ ℤ) → (𝑆 substr ⟨𝑋, 𝑋⟩) = ∅)
Theorem	swrdclg 11224	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 11225*	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 11226	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 11227	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 11228	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 11229	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 11230	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 11231	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 11232	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 11233	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 11234	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 11235	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 11236	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 11237	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 11238	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 11239	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 11240	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 11241*	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 11242	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 11243	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 11244	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 11245	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 11246	Syntax for the prefix operator.
class prefix
Definition	df-pfx 11247*	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 11248	Value of a prefix operation. (Contributed by AV, 2-May-2020.)
⊢ ((𝑆 ∈ 𝑉 ∧ 𝐿 ∈ ℕ0) → (𝑆 prefix 𝐿) = (𝑆 substr ⟨0, 𝐿⟩))
Theorem	pfx00g 11249	The zero length prefix is the empty set. (Contributed by AV, 2-May-2020.)
⊢ (𝑆 ∈ 𝑉 → (𝑆 prefix 0) = ∅)
Theorem	pfx0g 11250	A prefix of an empty set is always the empty set. (Contributed by AV, 3-May-2020.)
⊢ (𝐿 ∈ ℕ0 → (∅ prefix 𝐿) = ∅)
Theorem	fnpfx 11251	The domain of the prefix extractor. (Contributed by Jim Kingdon, 8-Jan-2026.)
⊢ prefix Fn (V × ℕ0)
Theorem	pfxclg 11252	Closure of the prefix extractor. (Contributed by AV, 2-May-2020.)
⊢ ((𝑆 ∈ Word 𝐴 ∧ 𝐿 ∈ ℕ0) → (𝑆 prefix 𝐿) ∈ Word 𝐴)
Theorem	pfxclz 11253	Closure of the prefix extractor. This extends pfxclg 11252 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 11254*	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 11255	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 11256	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 11257	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 11258	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 11259	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 11260	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 11261	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 11262	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 11263	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 5666). (Contributed by AV, 3-May-2020.)
⊢ ((𝑊 ∈ Word 𝑉 ∧ 𝐿 ∈ ℕ0 ∧ (♯‘𝑊) < 𝐿) → (𝑊 prefix 𝐿) = ∅)
Theorem	pfxwrdsymbg 11264	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 11265	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 11266	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 11267	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 11268	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 11269	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 11270*	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 11271	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 11272	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 11273	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 11274*	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 11275	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 11276	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)
Theorem	pfx1 11277	The prefix of length one of a nonempty word expressed as a singleton word. (Contributed by AV, 15-May-2020.)
⊢ ((𝑊 ∈ Word 𝑉 ∧ 𝑊 ≠ ∅) → (𝑊 prefix 1) = ⟨“(𝑊‘0)”⟩)
4.7.8  Subwords of subwords
Theorem	swrdswrdlem 11278	Lemma for swrdswrd 11279. (Contributed by Alexander van der Vekens, 4-Apr-2018.)
⊢ (((𝑊 ∈ Word 𝑉 ∧ 𝑁 ∈ (0...(♯‘𝑊)) ∧ 𝑀 ∈ (0...𝑁)) ∧ (𝐾 ∈ (0...(𝑁 − 𝑀)) ∧ 𝐿 ∈ (𝐾...(𝑁 − 𝑀)))) → (𝑊 ∈ Word 𝑉 ∧ (𝑀 + 𝐾) ∈ (0...(𝑀 + 𝐿)) ∧ (𝑀 + 𝐿) ∈ (0...(♯‘𝑊))))
Theorem	swrdswrd 11279	A subword of a subword is a subword. (Contributed by Alexander van der Vekens, 4-Apr-2018.)
⊢ ((𝑊 ∈ Word 𝑉 ∧ 𝑁 ∈ (0...(♯‘𝑊)) ∧ 𝑀 ∈ (0...𝑁)) → ((𝐾 ∈ (0...(𝑁 − 𝑀)) ∧ 𝐿 ∈ (𝐾...(𝑁 − 𝑀))) → ((𝑊 substr ⟨𝑀, 𝑁⟩) substr ⟨𝐾, 𝐿⟩) = (𝑊 substr ⟨(𝑀 + 𝐾), (𝑀 + 𝐿)⟩)))
Theorem	pfxswrd 11280	A prefix of a subword is a subword. (Contributed by AV, 2-Apr-2018.) (Revised by AV, 8-May-2020.)
⊢ ((𝑊 ∈ Word 𝑉 ∧ 𝑁 ∈ (0...(♯‘𝑊)) ∧ 𝑀 ∈ (0...𝑁)) → (𝐿 ∈ (0...(𝑁 − 𝑀)) → ((𝑊 substr ⟨𝑀, 𝑁⟩) prefix 𝐿) = (𝑊 substr ⟨𝑀, (𝑀 + 𝐿)⟩)))
Theorem	swrdpfx 11281	A subword of a prefix is a subword. (Contributed by Alexander van der Vekens, 6-Apr-2018.) (Revised by AV, 8-May-2020.)
⊢ ((𝑊 ∈ Word 𝑉 ∧ 𝑁 ∈ (0...(♯‘𝑊))) → ((𝐾 ∈ (0...𝑁) ∧ 𝐿 ∈ (𝐾...𝑁)) → ((𝑊 prefix 𝑁) substr ⟨𝐾, 𝐿⟩) = (𝑊 substr ⟨𝐾, 𝐿⟩)))
Theorem	pfxpfx 11282	A prefix of a prefix is a prefix. (Contributed by Alexander van der Vekens, 7-Apr-2018.) (Revised by AV, 8-May-2020.)
⊢ ((𝑊 ∈ Word 𝑉 ∧ 𝑁 ∈ (0...(♯‘𝑊)) ∧ 𝐿 ∈ (0...𝑁)) → ((𝑊 prefix 𝑁) prefix 𝐿) = (𝑊 prefix 𝐿))
Theorem	pfxpfxid 11283	A prefix of a prefix with the same length is the original prefix. In other words, the operation "prefix of length 𝑁 " is idempotent. (Contributed by AV, 5-Apr-2018.) (Revised by AV, 8-May-2020.)
⊢ ((𝑊 ∈ Word 𝑉 ∧ 𝑁 ∈ (0...(♯‘𝑊))) → ((𝑊 prefix 𝑁) prefix 𝑁) = (𝑊 prefix 𝑁))
4.7.9  Subwords and concatenations
Theorem	pfxcctswrd 11284	The concatenation of the prefix of a word and the rest of the word yields the word itself. (Contributed by AV, 21-Oct-2018.) (Revised by AV, 9-May-2020.)
⊢ ((𝑊 ∈ Word 𝑉 ∧ 𝑀 ∈ (0...(♯‘𝑊))) → ((𝑊 prefix 𝑀) ++ (𝑊 substr ⟨𝑀, (♯‘𝑊)⟩)) = 𝑊)
Theorem	lenpfxcctswrd 11285	The length of the concatenation of the prefix of a word and the rest of the word is the length of the word. (Contributed by AV, 21-Oct-2018.) (Revised by AV, 9-May-2020.)
⊢ ((𝑊 ∈ Word 𝑉 ∧ 𝑀 ∈ (0...(♯‘𝑊))) → (♯‘((𝑊 prefix 𝑀) ++ (𝑊 substr ⟨𝑀, (♯‘𝑊)⟩))) = (♯‘𝑊))
Theorem	lenrevpfxcctswrd 11286	The length of the concatenation of the rest of a word and the prefix of the word is the length of the word. (Contributed by Alexander van der Vekens, 1-Apr-2018.) (Revised by AV, 9-May-2020.)
⊢ ((𝑊 ∈ Word 𝑉 ∧ 𝑀 ∈ (0...(♯‘𝑊))) → (♯‘((𝑊 substr ⟨𝑀, (♯‘𝑊)⟩) ++ (𝑊 prefix 𝑀))) = (♯‘𝑊))
Theorem	pfxlswccat 11287	Reconstruct a nonempty word from its prefix and last symbol. (Contributed by Alexander van der Vekens, 5-Aug-2018.) (Revised by AV, 9-May-2020.)
⊢ ((𝑊 ∈ Word 𝑉 ∧ 𝑊 ≠ ∅) → ((𝑊 prefix ((♯‘𝑊) − 1)) ++ ⟨“(lastS‘𝑊)”⟩) = 𝑊)
Theorem	ccats1pfxeq 11288	The last symbol of a word concatenated with the word with the last symbol removed results in the word itself. (Contributed by Alexander van der Vekens, 24-Oct-2018.) (Revised by AV, 9-May-2020.)
⊢ ((𝑊 ∈ Word 𝑉 ∧ 𝑈 ∈ Word 𝑉 ∧ (♯‘𝑈) = ((♯‘𝑊) + 1)) → (𝑊 = (𝑈 prefix (♯‘𝑊)) → 𝑈 = (𝑊 ++ ⟨“(lastS‘𝑈)”⟩)))
Theorem	ccats1pfxeqrex 11289*	There exists a symbol such that its concatenation after the prefix obtained by deleting the last symbol of a nonempty word results in the word itself. (Contributed by AV, 5-Oct-2018.) (Revised by AV, 9-May-2020.)
⊢ ((𝑊 ∈ Word 𝑉 ∧ 𝑈 ∈ Word 𝑉 ∧ (♯‘𝑈) = ((♯‘𝑊) + 1)) → (𝑊 = (𝑈 prefix (♯‘𝑊)) → ∃𝑠 ∈ 𝑉 𝑈 = (𝑊 ++ ⟨“𝑠”⟩)))
Theorem	ccatopth 11290	An opth 4327-like theorem for recovering the two halves of a concatenated word. (Contributed by Mario Carneiro, 1-Oct-2015.) (Proof shortened by AV, 12-Oct-2022.)
⊢ (((𝐴 ∈ Word 𝑋 ∧ 𝐵 ∈ Word 𝑋) ∧ (𝐶 ∈ Word 𝑋 ∧ 𝐷 ∈ Word 𝑋) ∧ (♯‘𝐴) = (♯‘𝐶)) → ((𝐴 ++ 𝐵) = (𝐶 ++ 𝐷) ↔ (𝐴 = 𝐶 ∧ 𝐵 = 𝐷)))
Theorem	ccatopth2 11291	An opth 4327-like theorem for recovering the two halves of a concatenated word. (Contributed by Mario Carneiro, 1-Oct-2015.)
⊢ (((𝐴 ∈ Word 𝑋 ∧ 𝐵 ∈ Word 𝑋) ∧ (𝐶 ∈ Word 𝑋 ∧ 𝐷 ∈ Word 𝑋) ∧ (♯‘𝐵) = (♯‘𝐷)) → ((𝐴 ++ 𝐵) = (𝐶 ++ 𝐷) ↔ (𝐴 = 𝐶 ∧ 𝐵 = 𝐷)))
Theorem	ccatlcan 11292	Concatenation of words is left-cancellative. (Contributed by Mario Carneiro, 2-Oct-2015.)
⊢ ((𝐴 ∈ Word 𝑋 ∧ 𝐵 ∈ Word 𝑋 ∧ 𝐶 ∈ Word 𝑋) → ((𝐶 ++ 𝐴) = (𝐶 ++ 𝐵) ↔ 𝐴 = 𝐵))
Theorem	ccatrcan 11293	Concatenation of words is right-cancellative. (Contributed by Mario Carneiro, 2-Oct-2015.)
⊢ ((𝐴 ∈ Word 𝑋 ∧ 𝐵 ∈ Word 𝑋 ∧ 𝐶 ∈ Word 𝑋) → ((𝐴 ++ 𝐶) = (𝐵 ++ 𝐶) ↔ 𝐴 = 𝐵))
Theorem	wrdeqs1cat 11294	Decompose a nonempty word by separating off the first symbol. (Contributed by Stefan O'Rear, 25-Aug-2015.) (Revised by Mario Carneiro, 1-Oct-2015.) (Proof shortened by AV, 12-Oct-2022.)
⊢ ((𝑊 ∈ Word 𝐴 ∧ 𝑊 ≠ ∅) → 𝑊 = (⟨“(𝑊‘0)”⟩ ++ (𝑊 substr ⟨1, (♯‘𝑊)⟩)))
Theorem	cats1un 11295	Express a word with an extra symbol as the union of the word and the new value. (Contributed by Mario Carneiro, 28-Feb-2016.)
⊢ ((𝐴 ∈ Word 𝑋 ∧ 𝐵 ∈ 𝑋) → (𝐴 ++ ⟨“𝐵”⟩) = (𝐴 ∪ {⟨(♯‘𝐴), 𝐵⟩}))
Theorem	wrdind 11296*	Perform induction over the structure of a word. (Contributed by Mario Carneiro, 27-Sep-2015.) (Revised by Mario Carneiro, 26-Feb-2016.) (Proof shortened by AV, 12-Oct-2022.)
⊢ (𝑥 = ∅ → (𝜑 ↔ 𝜓))    &   ⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜒))    &   ⊢ (𝑥 = (𝑦 ++ ⟨“𝑧”⟩) → (𝜑 ↔ 𝜃))    &   ⊢ (𝑥 = 𝐴 → (𝜑 ↔ 𝜏))    &   ⊢ 𝜓    &   ⊢ ((𝑦 ∈ Word 𝐵 ∧ 𝑧 ∈ 𝐵) → (𝜒 → 𝜃))    ⇒   ⊢ (𝐴 ∈ Word 𝐵 → 𝜏)
Theorem	wrd2ind 11297*	Perform induction over the structure of two words of the same length. (Contributed by AV, 23-Jan-2019.) (Proof shortened by AV, 12-Oct-2022.)
⊢ ((𝑥 = ∅ ∧ 𝑤 = ∅) → (𝜑 ↔ 𝜓))    &   ⊢ ((𝑥 = 𝑦 ∧ 𝑤 = 𝑢) → (𝜑 ↔ 𝜒))    &   ⊢ ((𝑥 = (𝑦 ++ ⟨“𝑧”⟩) ∧ 𝑤 = (𝑢 ++ ⟨“𝑠”⟩)) → (𝜑 ↔ 𝜃))    &   ⊢ (𝑥 = 𝐴 → (𝜌 ↔ 𝜏))    &   ⊢ (𝑤 = 𝐵 → (𝜑 ↔ 𝜌))    &   ⊢ 𝜓    &   ⊢ (((𝑦 ∈ Word 𝑋 ∧ 𝑧 ∈ 𝑋) ∧ (𝑢 ∈ Word 𝑌 ∧ 𝑠 ∈ 𝑌) ∧ (♯‘𝑦) = (♯‘𝑢)) → (𝜒 → 𝜃))    ⇒   ⊢ ((𝐴 ∈ Word 𝑋 ∧ 𝐵 ∈ Word 𝑌 ∧ (♯‘𝐴) = (♯‘𝐵)) → 𝜏)
4.7.10  Subwords of concatenations
Theorem	swrdccatfn 11298	The subword of a concatenation as function. (Contributed by Alexander van der Vekens, 27-May-2018.)
⊢ (((𝐴 ∈ Word 𝑉 ∧ 𝐵 ∈ Word 𝑉) ∧ (𝑀 ∈ (0...𝑁) ∧ 𝑁 ∈ (0...((♯‘𝐴) + (♯‘𝐵))))) → ((𝐴 ++ 𝐵) substr ⟨𝑀, 𝑁⟩) Fn (0..^(𝑁 − 𝑀)))
Theorem	swrdccatin1 11299	The subword of a concatenation of two words within the first of the concatenated words. (Contributed by Alexander van der Vekens, 28-Mar-2018.)
⊢ ((𝐴 ∈ Word 𝑉 ∧ 𝐵 ∈ Word 𝑉) → ((𝑀 ∈ (0...𝑁) ∧ 𝑁 ∈ (0...(♯‘𝐴))) → ((𝐴 ++ 𝐵) substr ⟨𝑀, 𝑁⟩) = (𝐴 substr ⟨𝑀, 𝑁⟩)))
Theorem	pfxccatin12lem4 11300	Lemma 4 for pfxccatin12 11307. (Contributed by Alexander van der Vekens, 30-Mar-2018.) (Revised by Alexander van der Vekens, 23-May-2018.)
⊢ ((𝐿 ∈ ℕ0 ∧ 𝑀 ∈ ℕ0 ∧ 𝑁 ∈ ℤ) → ((𝐾 ∈ (0..^(𝑁 − 𝑀)) ∧ ¬ 𝐾 ∈ (0..^(𝐿 − 𝑀))) → 𝐾 ∈ ((𝐿 − 𝑀)..^((𝐿 − 𝑀) + (𝑁 − 𝐿)))))