P. 135 of Theorem List - Metamath Proof Explorer

Theorem List for Metamath Proof Explorer - 13401-13500   *Has distinct variable group(s)

Type	Label	Description
Statement
Theorem	ccatw2s1cl 13401	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	ccatw2s1len 13402	The length of the concatenation of a word with two singleton words. (Contributed by Alexander van der Vekens, 22-Sep-2018.)
⊢ ((𝑊 ∈ Word 𝑉 ∧ 𝑋 ∈ 𝑉 ∧ 𝑌 ∈ 𝑉) → (#‘((𝑊 ++ ⟨“𝑋”⟩) ++ ⟨“𝑌”⟩)) = ((#‘𝑊) + 2))
Theorem	ccats1val1 13403	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.)
⊢ ((𝑊 ∈ Word 𝑉 ∧ 𝑆 ∈ 𝑉 ∧ 𝐼 ∈ (0..^(#‘𝑊))) → ((𝑊 ++ ⟨“𝑆”⟩)‘𝐼) = (𝑊‘𝐼))
Theorem	ccats1val2 13404	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	ccat2s1p1 13405	Extract the first of two concatenated singleton words. (Contributed by Alexander van der Vekens, 22-Sep-2018.)
⊢ ((𝑋 ∈ 𝑉 ∧ 𝑌 ∈ 𝑉) → ((⟨“𝑋”⟩ ++ ⟨“𝑌”⟩)‘0) = 𝑋)
Theorem	ccat2s1p2 13406	Extract the second of two concatenated singleton words. (Contributed by Alexander van der Vekens, 22-Sep-2018.)
⊢ ((𝑋 ∈ 𝑉 ∧ 𝑌 ∈ 𝑉) → ((⟨“𝑋”⟩ ++ ⟨“𝑌”⟩)‘1) = 𝑌)
Theorem	ccatw2s1ass 13407	Associative law for a concatenation of a word with two singleton words. (Contributed by Alexander van der Vekens, 22-Sep-2018.)
⊢ ((𝑊 ∈ Word 𝑉 ∧ 𝑋 ∈ 𝑉 ∧ 𝑌 ∈ 𝑉) → ((𝑊 ++ ⟨“𝑋”⟩) ++ ⟨“𝑌”⟩) = (𝑊 ++ (⟨“𝑋”⟩ ++ ⟨“𝑌”⟩)))
Theorem	ccatws1lenrevOLD 13408	Obsolete theorem as of 24-Jun-2022. Use wrdlenccats1lenm1 13399 instead. The length of a word concatenated with a singleton word. (Contributed by Alexander van der Vekens, 3-Oct-2018.) (New usage is discouraged.) (Proof modification is discouraged.)
⊢ ((𝑊 ∈ Word 𝑉 ∧ 𝑋 ∈ 𝑉) → ((#‘(𝑊 ++ ⟨“𝑋”⟩)) = 𝑁 → (#‘𝑊) = (𝑁 − 1)))
Theorem	ccatws1n0 13409	The concatenation of a word with a singleton word is not the empty set. (Contributed by Alexander van der Vekens, 29-Sep-2018.)
⊢ ((𝑊 ∈ Word 𝑉 ∧ 𝑋 ∈ 𝑉) → (𝑊 ++ ⟨“𝑋”⟩) ≠ ∅)
Theorem	ccatws1ls 13410	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 13411	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 𝑉 ∧ 𝑆 ∈ 𝑉) → ( lastS ‘(𝑊 ++ ⟨“𝑆”⟩)) = 𝑆)
Theorem	lswccats1fst 13412	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	ccatw2s1p1 13413	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.)
⊢ (((𝑊 ∈ Word 𝑉 ∧ (#‘𝑊) = 𝑁) ∧ (𝑋 ∈ 𝑉 ∧ 𝑌 ∈ 𝑉)) → (((𝑊 ++ ⟨“𝑋”⟩) ++ ⟨“𝑌”⟩)‘𝑁) = 𝑋)
Theorem	ccatw2s1p2 13414	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	ccat2s1fvw 13415	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.)
⊢ (((𝑊 ∈ Word 𝑉 ∧ 𝐼 ∈ ℕ0 ∧ 𝐼 < (#‘𝑊)) ∧ (𝑋 ∈ 𝑉 ∧ 𝑌 ∈ 𝑉)) → (((𝑊 ++ ⟨“𝑋”⟩) ++ ⟨“𝑌”⟩)‘𝐼) = (𝑊‘𝐼))
Theorem	ccat2s1fst 13416	The first symbol of the concatenation of a word with two single symbols. (Contributed by Alexander van der Vekens, 22-Sep-2018.)
⊢ (((𝑊 ∈ Word 𝑉 ∧ 0 < (#‘𝑊)) ∧ (𝑋 ∈ 𝑉 ∧ 𝑌 ∈ 𝑉)) → (((𝑊 ++ ⟨“𝑋”⟩) ++ ⟨“𝑌”⟩)‘0) = (𝑊‘0))
5.7.6  Subwords
Theorem	swrdval 13417*	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	swrd00 13418	A zero length substring. (Contributed by Stefan O'Rear, 27-Aug-2015.)
⊢ (𝑆 substr ⟨𝑋, 𝑋⟩) = ∅
Theorem	swrdcl 13419	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)
Theorem	swrdval2 13420*	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	swrd0val 13421	Value of the subword extractor for left-anchored subwords. (Contributed by Stefan O'Rear, 24-Aug-2015.)
⊢ ((𝑆 ∈ Word 𝐴 ∧ 𝐿 ∈ (0...(#‘𝑆))) → (𝑆 substr ⟨0, 𝐿⟩) = (𝑆 ↾ (0..^𝐿)))
Theorem	swrd0len 13422	Length of a left-anchored subword. (Contributed by Stefan O'Rear, 24-Aug-2015.)
⊢ ((𝑆 ∈ Word 𝐴 ∧ 𝐿 ∈ (0...(#‘𝑆))) → (#‘(𝑆 substr ⟨0, 𝐿⟩)) = 𝐿)
Theorem	swrdlen 13423	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 13424	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	swrdf 13425	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 13426	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	swrd0f 13427	A left-anchored 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, 2-May-2020.)
⊢ ((𝑊 ∈ Word 𝑉 ∧ 𝑁 ∈ (0...(#‘𝑊))) → (𝑊 substr ⟨0, 𝑁⟩):(0..^𝑁)⟶𝑉)
Theorem	swrdid 13428	A word is a subword of itself. (Contributed by Stefan O'Rear, 16-Aug-2015.) (Proof shortened by AV, 2-May-2020.)
⊢ (𝑆 ∈ Word 𝐴 → (𝑆 substr ⟨0, (#‘𝑆)⟩) = 𝑆)
Theorem	swrdrn 13429	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	swrdn0 13430	A prefixing subword consisting of at least one symbol is not empty. (Contributed by Alexander van der Vekens, 4-Aug-2018.) (Proof shortened by AV, 2-May-2020.)
⊢ ((𝑊 ∈ Word 𝑉 ∧ 𝑁 ∈ ℕ ∧ 𝑁 ≤ (#‘𝑊)) → (𝑊 substr ⟨0, 𝑁⟩) ≠ ∅)
Theorem	swrdlend 13431	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 13432	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	swrdnd2 13433	Value of the subword extractor outside its intended domain. (Contributed by Alexander van der Vekens, 24-May-2018.)
⊢ ((𝑊 ∈ Word 𝑉 ∧ 𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ) → ((𝐵 ≤ 𝐴 ∨ (#‘𝑊) ≤ 𝐴 ∨ 𝐵 ≤ 0) → (𝑊 substr ⟨𝐴, 𝐵⟩) = ∅))
Theorem	swrd0 13434	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.)
⊢ (∅ substr ⟨𝐹, 𝐿⟩) = ∅
Theorem	swrdrlen 13435	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	swrd0len0 13436	Length of a prefix of a word reduced by a single symbol, analogous to swrd0len 13422. (Contributed by AV, 4-Aug-2018.) (Proof shortened by AV, 14-Oct-2018.)
⊢ ((𝑊 ∈ Word 𝑉 ∧ 𝑁 ∈ ℕ0 ∧ (#‘𝑊) = (𝑁 + 1)) → (#‘(𝑊 substr ⟨0, 𝑁⟩)) = 𝑁)
Theorem	addlenrevswrd 13437	The sum of the lengths of two reversed parts of a word is the length of the word. (Contributed by Alexander van der Vekens, 1-Apr-2018.)
⊢ ((𝑊 ∈ Word 𝑉 ∧ 𝑀 ∈ (0...(#‘𝑊))) → ((#‘(𝑊 substr ⟨𝑀, (#‘𝑊)⟩)) + (#‘(𝑊 substr ⟨0, 𝑀⟩))) = (#‘𝑊))
Theorem	addlenswrd 13438	The sum of the lengths of two parts of a word is the length of the word. (Contributed by AV, 21-Oct-2018.)
⊢ ((𝑊 ∈ Word 𝑉 ∧ 𝑀 ∈ (0...(#‘𝑊))) → ((#‘(𝑊 substr ⟨0, 𝑀⟩)) + (#‘(𝑊 substr ⟨𝑀, (#‘𝑊)⟩))) = (#‘𝑊))
Theorem	swrd0fv 13439	A symbol in an left-anchored subword, indexed using the subword's indices. (Contributed by Alexander van der Vekens, 16-Jun-2018.)
⊢ ((𝑊 ∈ Word 𝑉 ∧ 𝐿 ∈ (0...(#‘𝑊)) ∧ 𝐼 ∈ (0..^𝐿)) → ((𝑊 substr ⟨0, 𝐿⟩)‘𝐼) = (𝑊‘𝐼))
Theorem	swrd0fv0 13440	The first symbol in a left-anchored subword. (Contributed by Alexander van der Vekens, 16-Jun-2018.)
⊢ ((𝑊 ∈ Word 𝑉 ∧ 𝐼 ∈ (1...(#‘𝑊))) → ((𝑊 substr ⟨0, 𝐼⟩)‘0) = (𝑊‘0))
Theorem	swrdtrcfv 13441	A symbol in a word truncated by one symbol. (Contributed by Alexander van der Vekens, 16-Jun-2018.)
⊢ ((𝑊 ∈ Word 𝑉 ∧ 𝑊 ≠ ∅ ∧ 𝐼 ∈ (0..^((#‘𝑊) − 1))) → ((𝑊 substr ⟨0, ((#‘𝑊) − 1)⟩)‘𝐼) = (𝑊‘𝐼))
Theorem	swrdtrcfv0 13442	The first symbol in a word truncated by one symbol. (Contributed by Alexander van der Vekens, 16-Jun-2018.)
⊢ ((𝑊 ∈ Word 𝑉 ∧ 2 ≤ (#‘𝑊)) → ((𝑊 substr ⟨0, ((#‘𝑊) − 1)⟩)‘0) = (𝑊‘0))
Theorem	swrd0fvlsw 13443	The last symbol in a left-anchored subword. (Contributed by Alexander van der Vekens, 24-Jun-2018.)
⊢ ((𝑊 ∈ Word 𝑉 ∧ 𝐿 ∈ (1...(#‘𝑊))) → ( lastS ‘(𝑊 substr ⟨0, 𝐿⟩)) = (𝑊‘(𝐿 − 1)))
Theorem	swrdeq 13444*	Two subwords of 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.)
⊢ (((𝑊 ∈ Word 𝑉 ∧ 𝑈 ∈ Word 𝑉) ∧ (𝑀 ∈ ℕ0 ∧ 𝑁 ∈ ℕ0) ∧ (𝑀 ≤ (#‘𝑊) ∧ 𝑁 ≤ (#‘𝑈))) → ((𝑊 substr ⟨0, 𝑀⟩) = (𝑈 substr ⟨0, 𝑁⟩) ↔ (𝑀 = 𝑁 ∧ ∀𝑖 ∈ (0..^𝑀)(𝑊‘𝑖) = (𝑈‘𝑖))))
Theorem	swrdlen2 13445	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 13446	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	swrdsb0eq 13447	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 13448	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 13449*	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	swrdtrcfvl 13450	The last symbol in a word truncated by one symbol. (Contributed by AV, 16-Jun-2018.) (Proof shortened by Mario Carneiro/AV, 14-Oct-2018.)
⊢ ((𝑊 ∈ Word 𝑉 ∧ 2 ≤ (#‘𝑊)) → ( lastS ‘(𝑊 substr ⟨0, ((#‘𝑊) − 1)⟩)) = (𝑊‘((#‘𝑊) − 2)))
Theorem	swrds1 13451	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 13452	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	2swrdeqwrdeq 13453	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.)
⊢ ((𝑊 ∈ Word 𝑉 ∧ 𝑆 ∈ Word 𝑉 ∧ 𝐼 ∈ (0..^(#‘𝑊))) → (𝑊 = 𝑆 ↔ ((#‘𝑊) = (#‘𝑆) ∧ ((𝑊 substr ⟨0, 𝐼⟩) = (𝑆 substr ⟨0, 𝐼⟩) ∧ (𝑊 substr ⟨𝐼, (#‘𝑊)⟩) = (𝑆 substr ⟨𝐼, (#‘𝑊)⟩)))))
Theorem	2swrd1eqwrdeq 13454	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 Mario Carneiro/AV, 23-Oct-2018.)
⊢ ((𝑊 ∈ Word 𝑉 ∧ 𝑈 ∈ Word 𝑉 ∧ 0 < (#‘𝑊)) → (𝑊 = 𝑈 ↔ ((#‘𝑊) = (#‘𝑈) ∧ ((𝑊 substr ⟨0, ((#‘𝑊) − 1)⟩) = (𝑈 substr ⟨0, ((#‘𝑊) − 1)⟩) ∧ ( lastS ‘𝑊) = ( lastS ‘𝑈)))))
Theorem	disjxwrd 13455*	Sets of words are disjoint if each set contains extensions of distinct words of a fixed length. (Contributed by AV, 29-Jul-2018.) (Proof shortened by AV, 7-May-2020.)
⊢ Disj 𝑦 ∈ 𝑊 {𝑥 ∈ Word 𝑉 ∣ (𝑥 substr ⟨0, 𝑁⟩) = 𝑦}
Theorem	ccatswrd 13456	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	swrdccat1 13457	Recover the left half of a concatenated word. (Contributed by Mario Carneiro, 27-Sep-2015.)
⊢ ((𝑆 ∈ Word 𝐵 ∧ 𝑇 ∈ Word 𝐵) → ((𝑆 ++ 𝑇) substr ⟨0, (#‘𝑆)⟩) = 𝑆)
Theorem	swrdccat2 13458	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)
5.7.7  Subwords of subwords
Theorem	swrdswrdlem 13459	Lemma for swrdswrd 13460. (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 13460	A subword of 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	swrd0swrd 13461	A prefix of a subword. (Contributed by AV, 2-Apr-2018.) (Proof shortened by AV, 21-Oct-2018.)
⊢ ((𝑊 ∈ Word 𝑉 ∧ 𝑁 ∈ (0...(#‘𝑊)) ∧ 𝑀 ∈ (0...𝑁)) → (𝐿 ∈ (0...(𝑁 − 𝑀)) → ((𝑊 substr ⟨𝑀, 𝑁⟩) substr ⟨0, 𝐿⟩) = (𝑊 substr ⟨𝑀, (𝑀 + 𝐿)⟩)))
Theorem	swrdswrd0 13462	A subword of a prefix. (Contributed by Alexander van der Vekens, 6-Apr-2018.)
⊢ ((𝑊 ∈ Word 𝑉 ∧ 𝑁 ∈ (0...(#‘𝑊))) → ((𝐾 ∈ (0...𝑁) ∧ 𝐿 ∈ (𝐾...𝑁)) → ((𝑊 substr ⟨0, 𝑁⟩) substr ⟨𝐾, 𝐿⟩) = (𝑊 substr ⟨𝐾, 𝐿⟩)))
Theorem	swrd0swrd0 13463	A prefix of a prefix. (Contributed by Alexander van der Vekens, 7-Apr-2018.)
⊢ ((𝑊 ∈ Word 𝑉 ∧ 𝑁 ∈ (0...(#‘𝑊)) ∧ 𝐿 ∈ (0...𝑁)) → ((𝑊 substr ⟨0, 𝑁⟩) substr ⟨0, 𝐿⟩) = (𝑊 substr ⟨0, 𝐿⟩))
Theorem	swrd0swrdid 13464	A prefix of a prefix with the same length is the prefix. (Contributed by AV, 5-Apr-2018.) (Proof shortened by AV, 14-Oct-2018.)
⊢ ((𝑊 ∈ Word 𝑉 ∧ 𝑁 ∈ (0...(#‘𝑊))) → ((𝑊 substr ⟨0, 𝑁⟩) substr ⟨0, 𝑁⟩) = (𝑊 substr ⟨0, 𝑁⟩))
5.7.8  Subwords and concatenations
Theorem	wrdcctswrd 13465	The concatenation of two parts of a word yields the word itself. (Contributed by AV, 21-Oct-2018.)
⊢ ((𝑊 ∈ Word 𝑉 ∧ 𝑀 ∈ (0...(#‘𝑊))) → ((𝑊 substr ⟨0, 𝑀⟩) ++ (𝑊 substr ⟨𝑀, (#‘𝑊)⟩)) = 𝑊)
Theorem	lencctswrd 13466	The length of two concatenated parts of a word is the length of the word. (Contributed by AV, 21-Oct-2018.)
⊢ ((𝑊 ∈ Word 𝑉 ∧ 𝑀 ∈ (0...(#‘𝑊))) → (#‘((𝑊 substr ⟨0, 𝑀⟩) ++ (𝑊 substr ⟨𝑀, (#‘𝑊)⟩))) = (#‘𝑊))
Theorem	lenrevcctswrd 13467	The length of two reversely concatenated parts of a word is the length of the word. (Contributed by Alexander van der Vekens, 1-Apr-2018.)
⊢ ((𝑊 ∈ Word 𝑉 ∧ 𝑀 ∈ (0...(#‘𝑊))) → (#‘((𝑊 substr ⟨𝑀, (#‘𝑊)⟩) ++ (𝑊 substr ⟨0, 𝑀⟩))) = (#‘𝑊))
Theorem	swrdccatwrd 13468	Reconstruct a nonempty word from its prefix and last symbol. (Contributed by Alexander van der Vekens, 5-Aug-2018.)
⊢ ((𝑊 ∈ Word 𝑉 ∧ 𝑊 ≠ ∅) → ((𝑊 substr ⟨0, ((#‘𝑊) − 1)⟩) ++ ⟨“( lastS ‘𝑊)”⟩) = 𝑊)
Theorem	ccats1swrdeq 13469	The last symbol of a word concatenated with the subword of the word having length less by 1 than the word results in the word itself. (Contributed by Alexander van der Vekens, 24-Oct-2018.)
⊢ ((𝑊 ∈ Word 𝑉 ∧ 𝑈 ∈ Word 𝑉 ∧ (#‘𝑈) = ((#‘𝑊) + 1)) → (𝑊 = (𝑈 substr ⟨0, (#‘𝑊)⟩) → 𝑈 = (𝑊 ++ ⟨“( lastS ‘𝑈)”⟩)))
Theorem	ccatopth 13470	An opth 4945-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) ∧ (#‘𝐴) = (#‘𝐶)) → ((𝐴 ++ 𝐵) = (𝐶 ++ 𝐷) ↔ (𝐴 = 𝐶 ∧ 𝐵 = 𝐷)))
Theorem	ccatopth2 13471	An opth 4945-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 13472	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 13473	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 13474	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, 9-May-2020.)
⊢ ((𝑊 ∈ Word 𝐴 ∧ 𝑊 ≠ ∅) → 𝑊 = (⟨“(𝑊‘0)”⟩ ++ (𝑊 substr ⟨1, (#‘𝑊)⟩)))
Theorem	cats1un 13475	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 13476*	Perform induction over the structure of a word. (Contributed by Mario Carneiro, 27-Sep-2015.) (Revised by Mario Carneiro, 26-Feb-2016.)
⊢ (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 13477*	Perform induction over the structure of two words of the same length. (Contributed by AV, 23-Jan-2019.)
⊢ ((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)
Theorem	ccats1swrdeqrex 13478*	There exists a symbol such that its concatenation with the subword obtained by deleting the last symbol of a nonempty word results in the word itself. (Contributed by AV, 5-Oct-2018.) (Proof shortened by AV, 24-Oct-2018.)
⊢ ((𝑊 ∈ Word 𝑉 ∧ 𝑈 ∈ Word 𝑉 ∧ (#‘𝑈) = ((#‘𝑊) + 1)) → (𝑊 = (𝑈 substr ⟨0, (#‘𝑊)⟩) → ∃𝑠 ∈ 𝑉 𝑈 = (𝑊 ++ ⟨“𝑠”⟩)))
Theorem	reuccats1lem 13479*	Lemma for reuccats1 13480. (Contributed by Alexander van der Vekens, 5-Oct-2018.) (Proof shortened by AV, 15-Jan-2020.)
⊢ (((𝑊 ∈ Word 𝑉 ∧ 𝑈 ∈ 𝑋 ∧ (𝑊 ++ ⟨“𝑆”⟩) ∈ 𝑋) ∧ (∀𝑠 ∈ 𝑉 ((𝑊 ++ ⟨“𝑠”⟩) ∈ 𝑋 → 𝑆 = 𝑠) ∧ ∀𝑥 ∈ 𝑋 (𝑥 ∈ Word 𝑉 ∧ (#‘𝑥) = ((#‘𝑊) + 1)))) → (𝑊 = (𝑈 substr ⟨0, (#‘𝑊)⟩) → 𝑈 = (𝑊 ++ ⟨“𝑆”⟩)))
Theorem	reuccats1 13480*	A set of words having the length of a given word increased by 1 contains a unique word with the given word as prefix if there is a unique symbol which extends the given word to be a word of the set. (Contributed by Alexander van der Vekens, 6-Oct-2018.) (Revised by AV, 21-Jan-2022.)
⊢ Ⅎ𝑣𝑋    ⇒   ⊢ ((𝑊 ∈ Word 𝑉 ∧ ∀𝑥 ∈ 𝑋 (𝑥 ∈ Word 𝑉 ∧ (#‘𝑥) = ((#‘𝑊) + 1))) → (∃!𝑣 ∈ 𝑉 (𝑊 ++ ⟨“𝑣”⟩) ∈ 𝑋 → ∃!𝑥 ∈ 𝑋 𝑊 = (𝑥 substr ⟨0, (#‘𝑊)⟩)))
Theorem	reuccats1v 13481*	A set of words having the length of a given word increased by 1 contains a unique word with the given word as prefix if there is a unique symbol which extends the given word to be a word of the set. (Contributed by Alexander van der Vekens, 6-Oct-2018.) (Proof shortened by AV, 21-Jan-2022.)
⊢ ((𝑊 ∈ Word 𝑉 ∧ ∀𝑥 ∈ 𝑋 (𝑥 ∈ Word 𝑉 ∧ (#‘𝑥) = ((#‘𝑊) + 1))) → (∃!𝑣 ∈ 𝑉 (𝑊 ++ ⟨“𝑣”⟩) ∈ 𝑋 → ∃!𝑥 ∈ 𝑋 𝑊 = (𝑥 substr ⟨0, (#‘𝑊)⟩)))
5.7.9  Subwords of concatenations
Theorem	swrdccatfn 13482	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 13483	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	swrdccatin12lem1 13484	Lemma 1 for swrdccatin12 13491. (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)))))
Theorem	swrdccatin12lem2a 13485	Lemma 1 for swrdccatin12lem2 13489. (Contributed by AV, 30-Mar-2018.) (Revised by AV, 27-May-2018.)
⊢ ((𝑀 ∈ (0...𝐿) ∧ 𝑁 ∈ (𝐿...𝑋)) → ((𝐾 ∈ (0..^(𝑁 − 𝑀)) ∧ ¬ 𝐾 ∈ (0..^(𝐿 − 𝑀))) → (𝐾 + 𝑀) ∈ (𝐿..^𝑋)))
Theorem	swrdccatin12lem2b 13486	Lemma 2 for swrdccatin12lem2 13489. (Contributed by AV, 30-Mar-2018.) (Revised by AV, 27-May-2018.)
⊢ ((𝑀 ∈ (0...𝐿) ∧ 𝑁 ∈ (𝐿...𝑋)) → ((𝐾 ∈ (0..^(𝑁 − 𝑀)) ∧ ¬ 𝐾 ∈ (0..^(𝐿 − 𝑀))) → (𝐾 − (𝐿 − 𝑀)) ∈ (0..^((𝑁 − 𝐿) − 0))))
Theorem	swrdccatin2 13487	The subword of a concatenation of two words within the second of the concatenated words. (Contributed by Alexander van der Vekens, 28-Mar-2018.) (Revised by Alexander van der Vekens, 27-May-2018.)
⊢ 𝐿 = (#‘𝐴)    ⇒   ⊢ ((𝐴 ∈ Word 𝑉 ∧ 𝐵 ∈ Word 𝑉) → ((𝑀 ∈ (𝐿...𝑁) ∧ 𝑁 ∈ (𝐿...(𝐿 + (#‘𝐵)))) → ((𝐴 ++ 𝐵) substr ⟨𝑀, 𝑁⟩) = (𝐵 substr ⟨(𝑀 − 𝐿), (𝑁 − 𝐿)⟩)))
Theorem	swrdccatin12lem2c 13488	Lemma for swrdccatin12lem2 13489 and swrdccatin12lem3 13490. (Contributed by AV, 30-Mar-2018.) (Revised by AV, 27-May-2018.)
⊢ 𝐿 = (#‘𝐴)    ⇒   ⊢ (((𝐴 ∈ Word 𝑉 ∧ 𝐵 ∈ Word 𝑉) ∧ (𝑀 ∈ (0...𝐿) ∧ 𝑁 ∈ (𝐿...(𝐿 + (#‘𝐵))))) → ((𝐴 ++ 𝐵) ∈ Word 𝑉 ∧ 𝑀 ∈ (0...𝑁) ∧ 𝑁 ∈ (0...(#‘(𝐴 ++ 𝐵)))))
Theorem	swrdccatin12lem2 13489	Lemma 2 for swrdccatin12 13491. (Contributed by AV, 30-Mar-2018.) (Revised by AV, 27-May-2018.)
⊢ 𝐿 = (#‘𝐴)    ⇒   ⊢ (((𝐴 ∈ Word 𝑉 ∧ 𝐵 ∈ Word 𝑉) ∧ (𝑀 ∈ (0...𝐿) ∧ 𝑁 ∈ (𝐿...(𝐿 + (#‘𝐵))))) → ((𝐾 ∈ (0..^(𝑁 − 𝑀)) ∧ ¬ 𝐾 ∈ (0..^(𝐿 − 𝑀))) → (((𝐴 ++ 𝐵) substr ⟨𝑀, 𝑁⟩)‘𝐾) = ((𝐵 substr ⟨0, (𝑁 − 𝐿)⟩)‘(𝐾 − (#‘(𝐴 substr ⟨𝑀, 𝐿⟩))))))
Theorem	swrdccatin12lem3 13490	Lemma 3 for swrdccatin12 13491. (Contributed by AV, 30-Mar-2018.) (Revised by AV, 27-May-2018.)
⊢ 𝐿 = (#‘𝐴)    ⇒   ⊢ (((𝐴 ∈ Word 𝑉 ∧ 𝐵 ∈ Word 𝑉) ∧ (𝑀 ∈ (0...𝐿) ∧ 𝑁 ∈ (𝐿...(𝐿 + (#‘𝐵))))) → ((𝐾 ∈ (0..^(𝑁 − 𝑀)) ∧ 𝐾 ∈ (0..^(𝐿 − 𝑀))) → (((𝐴 ++ 𝐵) substr ⟨𝑀, 𝑁⟩)‘𝐾) = ((𝐴 substr ⟨𝑀, 𝐿⟩)‘𝐾)))
Theorem	swrdccatin12 13491	The subword of a concatenation of two words within both of the concatenated words. (Contributed by Alexander van der Vekens, 5-Apr-2018.) (Revised by Alexander van der Vekens, 27-May-2018.)
⊢ 𝐿 = (#‘𝐴)    ⇒   ⊢ ((𝐴 ∈ Word 𝑉 ∧ 𝐵 ∈ Word 𝑉) → ((𝑀 ∈ (0...𝐿) ∧ 𝑁 ∈ (𝐿...(𝐿 + (#‘𝐵)))) → ((𝐴 ++ 𝐵) substr ⟨𝑀, 𝑁⟩) = ((𝐴 substr ⟨𝑀, 𝐿⟩) ++ (𝐵 substr ⟨0, (𝑁 − 𝐿)⟩))))
Theorem	swrdccat3 13492	The subword of a concatenation is either a subword of the first concatenated word or a subword of the second concatenated word or a concatenation of a suffix of the first word with a prefix of the second word. (Contributed by Alexander van der Vekens, 30-Mar-2018.) (Revised by Alexander van der Vekens, 28-May-2018.)
⊢ 𝐿 = (#‘𝐴)    ⇒   ⊢ ((𝐴 ∈ Word 𝑉 ∧ 𝐵 ∈ Word 𝑉) → ((𝑀 ∈ (0...𝑁) ∧ 𝑁 ∈ (0...(𝐿 + (#‘𝐵)))) → ((𝐴 ++ 𝐵) substr ⟨𝑀, 𝑁⟩) = if(𝑁 ≤ 𝐿, (𝐴 substr ⟨𝑀, 𝑁⟩), if(𝐿 ≤ 𝑀, (𝐵 substr ⟨(𝑀 − 𝐿), (𝑁 − 𝐿)⟩), ((𝐴 substr ⟨𝑀, 𝐿⟩) ++ (𝐵 substr ⟨0, (𝑁 − 𝐿)⟩))))))
Theorem	swrdccat 13493	The subword of a concatenation of two words as concatenation of subwords of the two concatenated words. (Contributed by Alexander van der Vekens, 29-May-2018.)
⊢ 𝐿 = (#‘𝐴)    ⇒   ⊢ ((𝐴 ∈ Word 𝑉 ∧ 𝐵 ∈ Word 𝑉) → ((𝑀 ∈ (0...𝑁) ∧ 𝑁 ∈ (0...(𝐿 + (#‘𝐵)))) → ((𝐴 ++ 𝐵) substr ⟨𝑀, 𝑁⟩) = ((𝐴 substr ⟨𝑀, if(𝑁 ≤ 𝐿, 𝑁, 𝐿)⟩) ++ (𝐵 substr ⟨if(0 ≤ (𝑀 − 𝐿), (𝑀 − 𝐿), 0), (𝑁 − 𝐿)⟩))))
Theorem	swrdccat3a 13494	A prefix of a concatenation is either a prefix of the first concatenated word or a concatenation of the first word with a prefix of the second word. (Contributed by Alexander van der Vekens, 31-Mar-2018.) (Revised by Alexander van der Vekens, 29-May-2018.)
⊢ 𝐿 = (#‘𝐴)    ⇒   ⊢ ((𝐴 ∈ Word 𝑉 ∧ 𝐵 ∈ Word 𝑉) → (𝑁 ∈ (0...(𝐿 + (#‘𝐵))) → ((𝐴 ++ 𝐵) substr ⟨0, 𝑁⟩) = if(𝑁 ≤ 𝐿, (𝐴 substr ⟨0, 𝑁⟩), (𝐴 ++ (𝐵 substr ⟨0, (𝑁 − 𝐿)⟩)))))
Theorem	swrdccat3blem 13495	Lemma for swrdccat3b 13496. (Contributed by AV, 30-May-2018.)
⊢ 𝐿 = (#‘𝐴)    ⇒   ⊢ ((((𝐴 ∈ Word 𝑉 ∧ 𝐵 ∈ Word 𝑉) ∧ 𝑀 ∈ (0...(𝐿 + (#‘𝐵)))) ∧ (𝐿 + (#‘𝐵)) ≤ 𝐿) → if(𝐿 ≤ 𝑀, (𝐵 substr ⟨(𝑀 − 𝐿), (#‘𝐵)⟩), ((𝐴 substr ⟨𝑀, 𝐿⟩) ++ 𝐵)) = (𝐴 substr ⟨𝑀, (𝐿 + (#‘𝐵))⟩))
Theorem	swrdccat3b 13496	A suffix of a concatenation is either a suffix of the second concatenated word or a concatenation of a suffix of the first word with the second word. (Contributed by Alexander van der Vekens, 31-Mar-2018.) (Revised by Alexander van der Vekens, 30-May-2018.)
⊢ 𝐿 = (#‘𝐴)    ⇒   ⊢ ((𝐴 ∈ Word 𝑉 ∧ 𝐵 ∈ Word 𝑉) → (𝑀 ∈ (0...(𝐿 + (#‘𝐵))) → ((𝐴 ++ 𝐵) substr ⟨𝑀, (𝐿 + (#‘𝐵))⟩) = if(𝐿 ≤ 𝑀, (𝐵 substr ⟨(𝑀 − 𝐿), (#‘𝐵)⟩), ((𝐴 substr ⟨𝑀, 𝐿⟩) ++ 𝐵))))
Theorem	swrdccatid 13497	A prefix of a concatenation of length of the first concatenated word is the first word itself. (Contributed by Alexander van der Vekens, 20-Sep-2018.)
⊢ ((𝐴 ∈ Word 𝑉 ∧ 𝐵 ∈ Word 𝑉 ∧ 𝑁 = (#‘𝐴)) → ((𝐴 ++ 𝐵) substr ⟨0, 𝑁⟩) = 𝐴)
Theorem	ccats1swrdeqbi 13498	A word is a prefix of a word with length greater by 1 than the first word iff the second word is the first word concatenated with the last symbol of the second word. (Contributed by AV, 24-Oct-2018.)
⊢ ((𝑊 ∈ Word 𝑉 ∧ 𝑈 ∈ Word 𝑉 ∧ (#‘𝑈) = ((#‘𝑊) + 1)) → (𝑊 = (𝑈 substr ⟨0, (#‘𝑊)⟩) ↔ 𝑈 = (𝑊 ++ ⟨“( lastS ‘𝑈)”⟩)))
Theorem	swrdccatin1d 13499	The subword of a concatenation of two words within the first of the concatenated words. (Contributed by AV, 31-May-2018.) (Revised by Mario Carneiro/AV, 21-Oct-2018.)
⊢ (𝜑 → (#‘𝐴) = 𝐿)    &   ⊢ (𝜑 → (𝐴 ∈ Word 𝑉 ∧ 𝐵 ∈ Word 𝑉))    &   ⊢ (𝜑 → 𝑀 ∈ (0...𝑁))    &   ⊢ (𝜑 → 𝑁 ∈ (0...𝐿))    ⇒   ⊢ (𝜑 → ((𝐴 ++ 𝐵) substr ⟨𝑀, 𝑁⟩) = (𝐴 substr ⟨𝑀, 𝑁⟩))
Theorem	swrdccatin2d 13500	The subword of a concatenation of two words within the second of the concatenated words. (Contributed by AV, 31-May-2018.) (Revised by Mario Carneiro/AV, 21-Oct-2018.)
⊢ (𝜑 → (#‘𝐴) = 𝐿)    &   ⊢ (𝜑 → (𝐴 ∈ Word 𝑉 ∧ 𝐵 ∈ Word 𝑉))    &   ⊢ (𝜑 → 𝑀 ∈ (𝐿...𝑁))    &   ⊢ (𝜑 → 𝑁 ∈ (𝐿...(𝐿 + (#‘𝐵))))    ⇒   ⊢ (𝜑 → ((𝐴 ++ 𝐵) substr ⟨𝑀, 𝑁⟩) = (𝐵 substr ⟨(𝑀 − 𝐿), (𝑁 − 𝐿)⟩))