P. 144 of Theorem List - Metamath Proof Explorer

Theorem List for Metamath Proof Explorer - 14301-14400   *Has distinct variable group(s)

Type	Label	Description
Statement
Theorem	ccatrid 14301	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 14302	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 14303	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 14304	Concatenation of the empty word by the empty word. (Contributed by AV, 26-Mar-2022.)
⊢ (∅ ++ ∅) = ∅
Theorem	lswccatn0lsw 14305	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 14306	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 14307	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 14308	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 14309	Syntax for the singleton word constructor.
class ⟨“𝐴”⟩
Definition	df-s1 14310	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 14318, and not, as maybe expected, the empty word, see also s1nz 14321. (Contributed by Stefan O'Rear, 15-Aug-2015.) (Revised by Mario Carneiro, 26-Feb-2016.)
⊢ ⟨“𝐴”⟩ = {⟨0, ( I ‘𝐴)⟩}
Theorem	ids1 14311	Identity function protection for a singleton word. (Contributed by Mario Carneiro, 26-Feb-2016.)
⊢ ⟨“𝐴”⟩ = ⟨“( I ‘𝐴)”⟩
Theorem	s1val 14312	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 14313	The range of a singleton word. (Contributed by Mario Carneiro, 18-Jul-2016.)
⊢ (𝐴 ∈ 𝑉 → ran ⟨“𝐴”⟩ = {𝐴})
Theorem	s1eq 14314	Equality theorem for a singleton word. (Contributed by Mario Carneiro, 26-Feb-2016.)
⊢ (𝐴 = 𝐵 → ⟨“𝐴”⟩ = ⟨“𝐵”⟩)
Theorem	s1eqd 14315	Equality theorem for a singleton word. (Contributed by Mario Carneiro, 26-Feb-2016.)
⊢ (𝜑 → 𝐴 = 𝐵)    ⇒   ⊢ (𝜑 → ⟨“𝐴”⟩ = ⟨“𝐵”⟩)
Theorem	s1cl 14316	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 14317	A singleton word is a word. (Contributed by Mario Carneiro, 26-Feb-2016.)
⊢ (𝜑 → 𝐴 ∈ 𝐵)    ⇒   ⊢ (𝜑 → ⟨“𝐴”⟩ ∈ Word 𝐵)
Theorem	s1prc 14318	Value of a singleton word if the symbol is a proper class. (Contributed by AV, 26-Mar-2022.)
⊢ (¬ 𝐴 ∈ V → ⟨“𝐴”⟩ = ⟨“∅”⟩)
Theorem	s1cli 14319	A singleton word is a word. (Contributed by Mario Carneiro, 26-Feb-2016.)
⊢ ⟨“𝐴”⟩ ∈ Word V
Theorem	s1len 14320	Length of a singleton word. (Contributed by Stefan O'Rear, 15-Aug-2015.) (Revised by Mario Carneiro, 26-Feb-2016.)
⊢ (♯‘⟨“𝐴”⟩) = 1
Theorem	s1nz 14321	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 14322	The domain of a singleton word is a singleton. (Contributed by AV, 9-Jan-2020.)
⊢ dom ⟨“𝐴”⟩ = {0}
Theorem	s1dmALT 14323	Alternate version of s1dm 14322, 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 14324	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 14325	The last symbol of a singleton word is its symbol. (Contributed by AV, 22-Oct-2018.)
⊢ (𝐴 ∈ 𝑉 → (lastS‘⟨“𝐴”⟩) = 𝐴)
Theorem	eqs1 14326	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 14327*	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 14328	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 14329	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 14330	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 14331	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 14332	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 14333	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 14334	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 14335	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 14336	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 14337	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	ccat2s1lenOLD 14338	Obsolete version of ccat2s1len 14337 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.)
⊢ ((𝑋 ∈ 𝑉 ∧ 𝑌 ∈ 𝑉) → (♯‘(⟨“𝑋”⟩ ++ ⟨“𝑌”⟩)) = 2)
Theorem	ccatw2s1cl 14339	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 14340	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 14341	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	ccats1val1OLD 14342	Obsolete version of ccats1val1 14341 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 𝑉 ∧ 𝑆 ∈ 𝑉 ∧ 𝐼 ∈ (0..^(♯‘𝑊))) → ((𝑊 ++ ⟨“𝑆”⟩)‘𝐼) = (𝑊‘𝐼))
Theorem	ccats1val2 14343	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 14344	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 14345	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 14346	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	ccat2s1p1OLD 14347	Obsolete version of ccat2s1p1 14345 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.)
⊢ ((𝑋 ∈ 𝑉 ∧ 𝑌 ∈ 𝑉) → ((⟨“𝑋”⟩ ++ ⟨“𝑌”⟩)‘0) = 𝑋)
Theorem	ccat2s1p2OLD 14348	Obsolete version of ccat2s1p2 14346 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.)
⊢ ((𝑋 ∈ 𝑉 ∧ 𝑌 ∈ 𝑉) → ((⟨“𝑋”⟩ ++ ⟨“𝑌”⟩)‘1) = 𝑌)
Theorem	ccatw2s1ass 14349	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	ccatw2s1assOLD 14350	Obsolete version of ccatw2s1ass 14349 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 𝑉 ∧ 𝑋 ∈ 𝑉 ∧ 𝑌 ∈ 𝑉) → ((𝑊 ++ ⟨“𝑋”⟩) ++ ⟨“𝑌”⟩) = (𝑊 ++ (⟨“𝑋”⟩ ++ ⟨“𝑌”⟩)))
Theorem	ccatws1n0 14351	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 14352	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 14353	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 14354	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 14355	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	ccatw2s1p1OLD 14356	Obsolete version of ccatw2s1p1 14355 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) ∧ (𝑋 ∈ 𝑉 ∧ 𝑌 ∈ 𝑉)) → (((𝑊 ++ ⟨“𝑋”⟩) ++ ⟨“𝑌”⟩)‘𝑁) = 𝑋)
Theorem	ccatw2s1p2 14357	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 14358	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	ccat2s1fvwOLD 14359	Obsolete version of ccat2s1fvw 14358 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 𝑉 ∧ 𝐼 ∈ ℕ0 ∧ 𝐼 < (♯‘𝑊)) ∧ (𝑋 ∈ 𝑉 ∧ 𝑌 ∈ 𝑉)) → (((𝑊 ++ ⟨“𝑋”⟩) ++ ⟨“𝑌”⟩)‘𝐼) = (𝑊‘𝐼))
Theorem	ccat2s1fst 14360	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))
Theorem	ccat2s1fstOLD 14361	Obsolete version of ccat2s1fst 14360 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)) ∧ (𝑋 ∈ 𝑉 ∧ 𝑌 ∈ 𝑉)) → (((𝑊 ++ ⟨“𝑋”⟩) ++ ⟨“𝑌”⟩)‘0) = (𝑊‘0))
5.7.6  Subwords/substrings
Syntax	csubstr 14362	Syntax for the subword operator.
class substr
Definition	df-substr 14363*	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 14364	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 6813). (Contributed by AV, 2-Dec-2022.) (New usage is discouraged.)
⊢ (¬ (𝐹 ∈ ℤ ∧ 𝐿 ∈ ℤ) → (𝑆 substr ⟨𝐹, 𝐿⟩) = ∅)
Theorem	swrdval 14365*	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 14366	A zero length substring. (Contributed by Stefan O'Rear, 27-Aug-2015.)
⊢ (𝑆 substr ⟨𝑋, 𝑋⟩) = ∅
Theorem	swrdcl 14367	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 14368*	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 14369	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 14370	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 14371	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 14372	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 14373	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 14374	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 14375	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 14376	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 14377	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 14378	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⟩) = ∅)
Theorem	swrdnd0 14379	The value of a subword operation for inproper arguments is the empty set. (Contributed by AV, 2-Dec-2022.)
⊢ (𝑆 ∈ Word 𝑉 → (¬ (𝐹 ∈ (0...𝐿) ∧ 𝐿 ∈ (0...(♯‘𝑆))) → (𝑆 substr ⟨𝐹, 𝐿⟩) = ∅))
Theorem	swrd0 14380	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 14381	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 14382	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 14383	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	swrdwrdsymb 14384	A subword is a word over the symbols it consists of. (Contributed by AV, 2-Dec-2022.)
⊢ (𝑆 ∈ Word 𝐴 → (𝑆 substr ⟨𝑀, 𝑁⟩) ∈ Word (𝑆 “ (𝑀..^𝑁)))
Theorem	swrdsb0eq 14385	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 14386	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 14387*	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 14388	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 14389	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 14390	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 14391	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  Prefixes of a word
Syntax	cpfx 14392	Syntax for the prefix operator.
class prefix
Definition	df-pfx 14393*	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	pfxnndmnd 14394	The value of a prefix operation for out-of-domain arguments. (This is due to our definition of function values for out-of-domain arguments, see ndmfv 6813). (Contributed by AV, 3-Dec-2022.) (New usage is discouraged.)
⊢ (¬ (𝑆 ∈ V ∧ 𝐿 ∈ ℕ0) → (𝑆 prefix 𝐿) = ∅)
Theorem	pfxval 14395	Value of a prefix operation. (Contributed by AV, 2-May-2020.)
⊢ ((𝑆 ∈ 𝑉 ∧ 𝐿 ∈ ℕ0) → (𝑆 prefix 𝐿) = (𝑆 substr ⟨0, 𝐿⟩))
Theorem	pfx00 14396	The zero length prefix is the empty set. (Contributed by AV, 2-May-2020.)
⊢ (𝑆 prefix 0) = ∅
Theorem	pfx0 14397	A prefix of an empty set is always the empty set. (Contributed by AV, 3-May-2020.)
⊢ (∅ prefix 𝐿) = ∅
Theorem	pfxval0 14398	Value of a prefix operation. This theorem should only be used in proofs if 𝐿 ∈ ℕ0 is not available. Otherwise (and usually), pfxval 14395 should be used. (Contributed by AV, 3-Dec-2022.) (New usage is discouraged.)
⊢ (𝑆 ∈ Word 𝐴 → (𝑆 prefix 𝐿) = (𝑆 substr ⟨0, 𝐿⟩))
Theorem	pfxcl 14399	Closure of the prefix extractor. (Contributed by AV, 2-May-2020.)
⊢ (𝑆 ∈ Word 𝐴 → (𝑆 prefix 𝐿) ∈ Word 𝐴)
Theorem	pfxmpt 14400*	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)))