P. 147 of Theorem List - Metamath Proof Explorer

Theorem List for Metamath Proof Explorer - 14601-14700   *Has distinct variable group(s)

Type	Label	Description
Statement
Theorem	elovmpowrd 14601*	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 𝑣 ∣ 𝜑})    ⇒   ⊢ (𝑍 ∈ (𝑉𝑂𝑌) → (𝑉 ∈ V ∧ 𝑌 ∈ V ∧ 𝑍 ∈ Word 𝑉))
Theorem	elovmptnn0wrd 14602*	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 𝑣 ∣ 𝜑}))    ⇒   ⊢ (𝑍 ∈ ((𝑉𝑂𝑌)‘𝑁) → ((𝑉 ∈ V ∧ 𝑌 ∈ V) ∧ (𝑁 ∈ ℕ0 ∧ 𝑍 ∈ Word 𝑉)))
Theorem	wrdred1 14603	A word truncated by a symbol is a word. (Contributed by AV, 29-Jan-2021.)
⊢ (𝐹 ∈ Word 𝑆 → (𝐹 ↾ (0..^((♯‘𝐹) − 1))) ∈ Word 𝑆)
Theorem	wrdred1hash 14604	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
Syntax	clsw 14605	Extend class notation with the Last Symbol of a word.
class lastS
Definition	df-lsw 14606	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)))
Theorem	lsw 14607	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.)
⊢ (𝑊 ∈ 𝑋 → (lastS‘𝑊) = (𝑊‘((♯‘𝑊) − 1)))
Theorem	lsw0 14608	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) = ∅)
Theorem	lsw0g 14609	The last symbol of an empty word does not exist. (Contributed by Alexander van der Vekens, 11-Nov-2018.)
⊢ (lastS‘∅) = ∅
Theorem	lsw1 14610	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))
Theorem	lswcl 14611	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 𝑉 ∧ 𝑊 ≠ ∅) → (lastS‘𝑊) ∈ 𝑉)
Theorem	lswlgt0cl 14612	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)
5.7.3  Concatenations of words
Syntax	cconcat 14613	Syntax for the concatenation operator.
class ++
Definition	df-concat 14614*	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)), (𝑠‘𝑥), (𝑡‘(𝑥 − (♯‘𝑠))))))
Theorem	ccatfn 14615	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)
Theorem	ccatfval 14616*	Value of the concatenation operator. (Contributed by Stefan O'Rear, 15-Aug-2015.)
⊢ ((𝑆 ∈ 𝑉 ∧ 𝑇 ∈ 𝑊) → (𝑆 ++ 𝑇) = (𝑥 ∈ (0..^((♯‘𝑆) + (♯‘𝑇))) ↦ if(𝑥 ∈ (0..^(♯‘𝑆)), (𝑆‘𝑥), (𝑇‘(𝑥 − (♯‘𝑆))))))
Theorem	ccatcl 14617	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 𝐵 ∧ 𝑇 ∈ Word 𝐵) → (𝑆 ++ 𝑇) ∈ Word 𝐵)
Theorem	ccatlen 14618	The length of a concatenated word. (Contributed by Stefan O'Rear, 15-Aug-2015.) (Revised by JJ, 1-Jan-2024.)
⊢ ((𝑆 ∈ Word 𝐴 ∧ 𝑇 ∈ Word 𝐵) → (♯‘(𝑆 ++ 𝑇)) = ((♯‘𝑆) + (♯‘𝑇)))
Theorem	ccat0 14619	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 𝐴 ∧ 𝑇 ∈ Word 𝐵) → ((𝑆 ++ 𝑇) = ∅ ↔ (𝑆 = ∅ ∧ 𝑇 = ∅)))
Theorem	ccatval1 14620	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 𝐴 ∧ 𝑇 ∈ Word 𝐵 ∧ 𝐼 ∈ (0..^(♯‘𝑆))) → ((𝑆 ++ 𝑇)‘𝐼) = (𝑆‘𝐼))
Theorem	ccatval2 14621	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 𝐵 ∧ 𝑇 ∈ Word 𝐵 ∧ 𝐼 ∈ ((♯‘𝑆)..^((♯‘𝑆) + (♯‘𝑇)))) → ((𝑆 ++ 𝑇)‘𝐼) = (𝑇‘(𝐼 − (♯‘𝑆))))
Theorem	ccatval3 14622	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 𝐵 ∧ 𝑇 ∈ Word 𝐵 ∧ 𝐼 ∈ (0..^(♯‘𝑇))) → ((𝑆 ++ 𝑇)‘(𝐼 + (♯‘𝑆))) = (𝑇‘𝐼))
Theorem	elfzelfzccat 14623	An element of a finite set of sequential integers up to the length of a word is an element of an extended finite set of sequential integers up to the length of a concatenation of this word with another word. (Contributed by Alexander van der Vekens, 28-Mar-2018.)
⊢ ((𝐴 ∈ Word 𝑉 ∧ 𝐵 ∈ Word 𝑉) → (𝑁 ∈ (0...(♯‘𝐴)) → 𝑁 ∈ (0...(♯‘(𝐴 ++ 𝐵)))))
Theorem	ccatvalfn 14624	The concatenation of two words is a function over the half-open integer range having the sum of the lengths of the word as length. (Contributed by Alexander van der Vekens, 30-Mar-2018.)
⊢ ((𝐴 ∈ Word 𝑉 ∧ 𝐵 ∈ Word 𝑉) → (𝐴 ++ 𝐵) Fn (0..^((♯‘𝐴) + (♯‘𝐵))))
Theorem	ccatdmss 14625	The domain of a concatenated word is a superset of the domain of the first word. (Contributed by Thierry Arnoux, 19-Jun-2025.)
⊢ (𝜑 → 𝐴 ∈ Word 𝑆)    &   ⊢ (𝜑 → 𝐵 ∈ Word 𝑆)    ⇒   ⊢ (𝜑 → dom 𝐴 ⊆ dom (𝐴 ++ 𝐵))
Theorem	ccatsymb 14626	The symbol at a given position in a concatenated word. (Contributed by AV, 26-May-2018.) (Proof shortened by AV, 24-Nov-2018.)
⊢ ((𝐴 ∈ Word 𝑉 ∧ 𝐵 ∈ Word 𝑉 ∧ 𝐼 ∈ ℤ) → ((𝐴 ++ 𝐵)‘𝐼) = if(𝐼 < (♯‘𝐴), (𝐴‘𝐼), (𝐵‘(𝐼 − (♯‘𝐴)))))
Theorem	ccatfv0 14627	The first symbol of a concatenation of two words is the first symbol of the first word if the first word is not empty. (Contributed by Alexander van der Vekens, 22-Sep-2018.)
⊢ ((𝐴 ∈ Word 𝑉 ∧ 𝐵 ∈ Word 𝑉 ∧ 0 < (♯‘𝐴)) → ((𝐴 ++ 𝐵)‘0) = (𝐴‘0))
Theorem	ccatval1lsw 14628	The last symbol of the left (nonempty) half of a concatenated word. (Contributed by Alexander van der Vekens, 3-Oct-2018.) (Proof shortened by AV, 1-May-2020.)
⊢ ((𝐴 ∈ Word 𝑉 ∧ 𝐵 ∈ Word 𝑉 ∧ 𝐴 ≠ ∅) → ((𝐴 ++ 𝐵)‘((♯‘𝐴) − 1)) = (lastS‘𝐴))
Theorem	ccatval21sw 14629	The first symbol of the right (nonempty) half of a concatenated word. (Contributed by AV, 23-Apr-2022.)
⊢ ((𝐴 ∈ Word 𝑉 ∧ 𝐵 ∈ Word 𝑉 ∧ 𝐵 ≠ ∅) → ((𝐴 ++ 𝐵)‘(♯‘𝐴)) = (𝐵‘0))
Theorem	ccatlid 14630	Concatenation of a word by the empty word on the left. (Contributed by Stefan O'Rear, 15-Aug-2015.) (Proof shortened by AV, 1-May-2020.)
⊢ (𝑆 ∈ Word 𝐵 → (∅ ++ 𝑆) = 𝑆)
Theorem	ccatrid 14631	Concatenation of a word by the empty word on the right. (Contributed by Stefan O'Rear, 15-Aug-2015.) (Proof shortened by AV, 1-May-2020.)
⊢ (𝑆 ∈ Word 𝐵 → (𝑆 ++ ∅) = 𝑆)
Theorem	ccatass 14632	Associative law for concatenation of words. (Contributed by Stefan O'Rear, 15-Aug-2015.)
⊢ ((𝑆 ∈ Word 𝐵 ∧ 𝑇 ∈ Word 𝐵 ∧ 𝑈 ∈ Word 𝐵) → ((𝑆 ++ 𝑇) ++ 𝑈) = (𝑆 ++ (𝑇 ++ 𝑈)))
Theorem	ccatrn 14633	The range of a concatenated word. (Contributed by Stefan O'Rear, 15-Aug-2015.)
⊢ ((𝑆 ∈ Word 𝐵 ∧ 𝑇 ∈ Word 𝐵) → ran (𝑆 ++ 𝑇) = (ran 𝑆 ∪ ran 𝑇))
Theorem	ccatidid 14634	Concatenation of the empty word by the empty word. (Contributed by AV, 26-Mar-2022.)
⊢ (∅ ++ ∅) = ∅
Theorem	lswccatn0lsw 14635	The last symbol of a word concatenated with a nonempty word is the last symbol of the nonempty word. (Contributed by AV, 22-Oct-2018.) (Proof shortened by AV, 1-May-2020.)
⊢ ((𝐴 ∈ Word 𝑉 ∧ 𝐵 ∈ Word 𝑉 ∧ 𝐵 ≠ ∅) → (lastS‘(𝐴 ++ 𝐵)) = (lastS‘𝐵))
Theorem	lswccat0lsw 14636	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))
Theorem	ccatalpha 14637	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 𝑆 ∧ 𝐵 ∈ Word 𝑆)))
Theorem	ccatrcl1 14638	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 𝑋 ∧ 𝐵 ∈ Word 𝑌 ∧ (𝑊 = (𝐴 ++ 𝐵) ∧ 𝑊 ∈ Word 𝑆)) → 𝐴 ∈ Word 𝑆)
5.7.4  Singleton words
Syntax	cs1 14639	Syntax for the singleton word constructor.
class ⟨“𝐴”⟩
Definition	df-s1 14640	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 14648, and not, as maybe expected, the empty word, see also s1nz 14651. (Contributed by Stefan O'Rear, 15-Aug-2015.) (Revised by Mario Carneiro, 26-Feb-2016.)
⊢ ⟨“𝐴”⟩ = {⟨0, ( I ‘𝐴)⟩}
Theorem	ids1 14641	Identity function protection for a singleton word. (Contributed by Mario Carneiro, 26-Feb-2016.)
⊢ ⟨“𝐴”⟩ = ⟨“( I ‘𝐴)”⟩
Theorem	s1val 14642	Value of a singleton word. (Contributed by Stefan O'Rear, 15-Aug-2015.) (Revised by Mario Carneiro, 26-Feb-2016.)
⊢ (𝐴 ∈ 𝑉 → ⟨“𝐴”⟩ = {⟨0, 𝐴⟩})
Theorem	s1rn 14643	The range of a singleton word. (Contributed by Mario Carneiro, 18-Jul-2016.)
⊢ (𝐴 ∈ 𝑉 → ran ⟨“𝐴”⟩ = {𝐴})
Theorem	s1eq 14644	Equality theorem for a singleton word. (Contributed by Mario Carneiro, 26-Feb-2016.)
⊢ (𝐴 = 𝐵 → ⟨“𝐴”⟩ = ⟨“𝐵”⟩)
Theorem	s1eqd 14645	Equality theorem for a singleton word. (Contributed by Mario Carneiro, 26-Feb-2016.)
⊢ (𝜑 → 𝐴 = 𝐵)    ⇒   ⊢ (𝜑 → ⟨“𝐴”⟩ = ⟨“𝐵”⟩)
Theorem	s1cl 14646	A singleton word is a word. (Contributed by Stefan O'Rear, 15-Aug-2015.) (Revised by Mario Carneiro, 26-Feb-2016.) (Proof shortened by AV, 23-Nov-2018.)
⊢ (𝐴 ∈ 𝐵 → ⟨“𝐴”⟩ ∈ Word 𝐵)
Theorem	s1cld 14647	A singleton word is a word. (Contributed by Mario Carneiro, 26-Feb-2016.)
⊢ (𝜑 → 𝐴 ∈ 𝐵)    ⇒   ⊢ (𝜑 → ⟨“𝐴”⟩ ∈ Word 𝐵)
Theorem	s1prc 14648	Value of a singleton word if the symbol is a proper class. (Contributed by AV, 26-Mar-2022.)
⊢ (¬ 𝐴 ∈ V → ⟨“𝐴”⟩ = ⟨“∅”⟩)
Theorem	s1cli 14649	A singleton word is a word. (Contributed by Mario Carneiro, 26-Feb-2016.)
⊢ ⟨“𝐴”⟩ ∈ Word V
Theorem	s1len 14650	Length of a singleton word. (Contributed by Stefan O'Rear, 15-Aug-2015.) (Revised by Mario Carneiro, 26-Feb-2016.)
⊢ (♯‘⟨“𝐴”⟩) = 1
Theorem	s1nz 14651	A singleton word is not the empty string. (Contributed by Mario Carneiro, 27-Feb-2016.) (Proof shortened by Kyle Wyonch, 18-Jul-2021.)
⊢ ⟨“𝐴”⟩ ≠ ∅
Theorem	s1dm 14652	The domain of a singleton word is a singleton. (Contributed by AV, 9-Jan-2020.)
⊢ dom ⟨“𝐴”⟩ = {0}
Theorem	s1dmALT 14653	Alternate version of s1dm 14652, 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.)
⊢ (𝐴 ∈ 𝑆 → dom ⟨“𝐴”⟩ = {0})
Theorem	s1fv 14654	Sole symbol of a singleton word. (Contributed by Stefan O'Rear, 15-Aug-2015.) (Revised by Mario Carneiro, 26-Feb-2016.)
⊢ (𝐴 ∈ 𝐵 → (⟨“𝐴”⟩‘0) = 𝐴)
Theorem	lsws1 14655	The last symbol of a singleton word is its symbol. (Contributed by AV, 22-Oct-2018.)
⊢ (𝐴 ∈ 𝑉 → (lastS‘⟨“𝐴”⟩) = 𝐴)
Theorem	eqs1 14656	A word of length 1 is a singleton word. (Contributed by Stefan O'Rear, 23-Aug-2015.) (Proof shortened by AV, 1-May-2020.)
⊢ ((𝑊 ∈ Word 𝐴 ∧ (♯‘𝑊) = 1) → 𝑊 = ⟨“(𝑊‘0)”⟩)
Theorem	wrdl1exs1 14657*	A word of length 1 is a singleton word. (Contributed by AV, 24-Jan-2021.)
⊢ ((𝑊 ∈ Word 𝑆 ∧ (♯‘𝑊) = 1) → ∃𝑠 ∈ 𝑆 𝑊 = ⟨“𝑠”⟩)
Theorem	wrdl1s1 14658	A word of length 1 is a singleton word consisting of the first symbol of the word. (Contributed by AV, 22-Jul-2018.) (Proof shortened by AV, 14-Oct-2018.)
⊢ (𝑆 ∈ 𝑉 → (𝑊 = ⟨“𝑆”⟩ ↔ (𝑊 ∈ Word 𝑉 ∧ (♯‘𝑊) = 1 ∧ (𝑊‘0) = 𝑆)))
Theorem	s111 14659	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))
5.7.5  Concatenations with singleton words
Theorem	ccatws1cl 14660	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	ccatws1clv 14661	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)
Theorem	ccat2s1cl 14662	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	ccats1alpha 14663	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 𝑉 ∧ 𝑋 ∈ 𝑈) → ((𝐴 ++ ⟨“𝑋”⟩) ∈ Word 𝑆 ↔ (𝐴 ∈ Word 𝑆 ∧ 𝑋 ∈ 𝑆)))
Theorem	ccatws1len 14664	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))
Theorem	ccatws1lenp1b 14665	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 ∧ 𝑁 ∈ ℕ0) → ((♯‘(𝑊 ++ ⟨“𝑋”⟩)) = (𝑁 + 1) ↔ (♯‘𝑊) = 𝑁))
Theorem	wrdlenccats1lenm1 14666	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))
Theorem	ccat2s1len 14667	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
Theorem	ccatw2s1cl 14668	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 14669	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))
Theorem	ccats1val1 14670	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 ∧ 𝐼 ∈ (0..^(♯‘𝑊))) → ((𝑊 ++ ⟨“𝑆”⟩)‘𝐼) = (𝑊‘𝐼))
Theorem	ccats1val2 14671	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 14672	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	ccat2s1p1 14673	Extract the first of two concatenated singleton words. (Contributed by Alexander van der Vekens, 22-Sep-2018.) (Revised by JJ, 20-Jan-2024.)
⊢ (𝑋 ∈ 𝑉 → ((⟨“𝑋”⟩ ++ ⟨“𝑌”⟩)‘0) = 𝑋)
Theorem	ccat2s1p2 14674	Extract the second of two concatenated singleton words. (Contributed by Alexander van der Vekens, 22-Sep-2018.) (Revised by JJ, 20-Jan-2024.)
⊢ (𝑌 ∈ 𝑉 → ((⟨“𝑋”⟩ ++ ⟨“𝑌”⟩)‘1) = 𝑌)
Theorem	ccatw2s1ass 14675	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”⟩)))
Theorem	ccatws1n0 14676	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”⟩) ≠ ∅)
Theorem	ccatws1ls 14677	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 14678	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 14679	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 14680	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) → (((𝑊 ++ ⟨“𝑋”⟩) ++ ⟨“𝑌”⟩)‘𝑁) = 𝑋)
Theorem	ccatw2s1p2 14681	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 14682	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 𝑉 ∧ 𝐼 ∈ ℕ0 ∧ 𝐼 < (♯‘𝑊)) → (((𝑊 ++ ⟨“𝑋”⟩) ++ ⟨“𝑌”⟩)‘𝐼) = (𝑊‘𝐼))
Theorem	ccat2s1fst 14683	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))
5.7.6  Subwords/substrings
Syntax	csubstr 14684	Syntax for the subword operator.
class substr
Definition	df-substr 14685*	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	swrdnznd 14686	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 6914). (Contributed by AV, 2-Dec-2022.) (New usage is discouraged.)
⊢ (¬ (𝐹 ∈ ℤ ∧ 𝐿 ∈ ℤ) → (𝑆 substr ⟨𝐹, 𝐿⟩) = ∅)
Theorem	swrdval 14687*	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 14688	A zero length substring. (Contributed by Stefan O'Rear, 27-Aug-2015.)
⊢ (𝑆 substr ⟨𝑋, 𝑋⟩) = ∅
Theorem	swrdcl 14689	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 14690*	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 14691	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 14692	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 14693	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 14694	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 14695	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 14696	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 14697	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 14698	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 14699	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	swrdnnn0nd 14700	The value of a subword operation for arguments not being nonnegative integers is the empty set. (Contributed by AV, 2-Dec-2022.)
⊢ ((𝑆 ∈ Word 𝑉 ∧ ¬ (𝐹 ∈ ℕ0 ∧ 𝐿 ∈ ℕ0)) → (𝑆 substr ⟨𝐹, 𝐿⟩) = ∅)