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| Type | Label | Description |
|---|---|---|
| Statement | ||
| Theorem | ccat0 14501 | 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 14502 | 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 14503 | 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 14504 | 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 14505 | 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 14506 | 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 | ccatsymb 14507 | 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 14508 | 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 14509 | 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 14510 | 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 14511 | 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 14512 | 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 14513 | 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 14514 | 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 14515 | Concatenation of the empty word by the empty word. (Contributed by AV, 26-Mar-2022.) |
| ⊢ (∅ ++ ∅) = ∅ | ||
| Theorem | lswccatn0lsw 14516 | 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 14517 | 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 14518 | 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 14519 | 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) | ||
| Syntax | cs1 14520 | Syntax for the singleton word constructor. |
| class 〈“𝐴”〉 | ||
| Definition | df-s1 14521 | 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 14529, and not, as maybe expected, the empty word, see also s1nz 14532. (Contributed by Stefan O'Rear, 15-Aug-2015.) (Revised by Mario Carneiro, 26-Feb-2016.) |
| ⊢ 〈“𝐴”〉 = {〈0, ( I ‘𝐴)〉} | ||
| Theorem | ids1 14522 | Identity function protection for a singleton word. (Contributed by Mario Carneiro, 26-Feb-2016.) |
| ⊢ 〈“𝐴”〉 = 〈“( I ‘𝐴)”〉 | ||
| Theorem | s1val 14523 | 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 14524 | The range of a singleton word. (Contributed by Mario Carneiro, 18-Jul-2016.) |
| ⊢ (𝐴 ∈ 𝑉 → ran 〈“𝐴”〉 = {𝐴}) | ||
| Theorem | s1eq 14525 | Equality theorem for a singleton word. (Contributed by Mario Carneiro, 26-Feb-2016.) |
| ⊢ (𝐴 = 𝐵 → 〈“𝐴”〉 = 〈“𝐵”〉) | ||
| Theorem | s1eqd 14526 | Equality theorem for a singleton word. (Contributed by Mario Carneiro, 26-Feb-2016.) |
| ⊢ (𝜑 → 𝐴 = 𝐵) ⇒ ⊢ (𝜑 → 〈“𝐴”〉 = 〈“𝐵”〉) | ||
| Theorem | s1cl 14527 | 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 14528 | A singleton word is a word. (Contributed by Mario Carneiro, 26-Feb-2016.) |
| ⊢ (𝜑 → 𝐴 ∈ 𝐵) ⇒ ⊢ (𝜑 → 〈“𝐴”〉 ∈ Word 𝐵) | ||
| Theorem | s1prc 14529 | Value of a singleton word if the symbol is a proper class. (Contributed by AV, 26-Mar-2022.) |
| ⊢ (¬ 𝐴 ∈ V → 〈“𝐴”〉 = 〈“∅”〉) | ||
| Theorem | s1cli 14530 | A singleton word is a word. (Contributed by Mario Carneiro, 26-Feb-2016.) |
| ⊢ 〈“𝐴”〉 ∈ Word V | ||
| Theorem | s1len 14531 | Length of a singleton word. (Contributed by Stefan O'Rear, 15-Aug-2015.) (Revised by Mario Carneiro, 26-Feb-2016.) |
| ⊢ (♯‘〈“𝐴”〉) = 1 | ||
| Theorem | s1nz 14532 | 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 14533 | The domain of a singleton word is a singleton. (Contributed by AV, 9-Jan-2020.) |
| ⊢ dom 〈“𝐴”〉 = {0} | ||
| Theorem | s1dmALT 14534 | Alternate version of s1dm 14533, 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 14535 | 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 14536 | The last symbol of a singleton word is its symbol. (Contributed by AV, 22-Oct-2018.) |
| ⊢ (𝐴 ∈ 𝑉 → (lastS‘〈“𝐴”〉) = 𝐴) | ||
| Theorem | eqs1 14537 | 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 14538* | 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 14539 | 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 14540 | 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)) | ||
| Theorem | ccatws1cl 14541 | 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 14542 | 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 14543 | 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 14544 | 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 14545 | 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 14546 | 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 14547 | 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 14548 | 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 14549 | 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 14550 | 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 14551 | 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 14552 | 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 14553 | 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 14554 | 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 14555 | 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 14556 | 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 14557 | 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 14558 | 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 14559 | 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 14560 | 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 14561 | 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 14562 | 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 14563 | 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 14564 | 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)) | ||
| Syntax | csubstr 14565 | Syntax for the subword operator. |
| class substr | ||
| Definition | df-substr 14566* | 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 14567 | 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 6859). (Contributed by AV, 2-Dec-2022.) (New usage is discouraged.) |
| ⊢ (¬ (𝐹 ∈ ℤ ∧ 𝐿 ∈ ℤ) → (𝑆 substr 〈𝐹, 𝐿〉) = ∅) | ||
| Theorem | swrdval 14568* | 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 14569 | A zero length substring. (Contributed by Stefan O'Rear, 27-Aug-2015.) |
| ⊢ (𝑆 substr 〈𝑋, 𝑋〉) = ∅ | ||
| Theorem | swrdcl 14570 | 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 14571* | 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 14572 | 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 14573 | 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 14574 | 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 14575 | 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 14576 | 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 14577 | 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 14578 | 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 14579 | 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 14580 | 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 14581 | 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 14582 | 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 14583 | 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 14584 | 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 14585 | 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 14586 | 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 14587 | 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 14588 | 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 14589 | 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 14590* | 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 14591 | 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 14592 | 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 14593 | 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 14594 | 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) | ||
| Syntax | cpfx 14595 | Syntax for the prefix operator. |
| class prefix | ||
| Definition | df-pfx 14596* | 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 14597 | 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 6859). (Contributed by AV, 3-Dec-2022.) (New usage is discouraged.) |
| ⊢ (¬ (𝑆 ∈ V ∧ 𝐿 ∈ ℕ0) → (𝑆 prefix 𝐿) = ∅) | ||
| Theorem | pfxval 14598 | Value of a prefix operation. (Contributed by AV, 2-May-2020.) |
| ⊢ ((𝑆 ∈ 𝑉 ∧ 𝐿 ∈ ℕ0) → (𝑆 prefix 𝐿) = (𝑆 substr 〈0, 𝐿〉)) | ||
| Theorem | pfx00 14599 | The zero length prefix is the empty set. (Contributed by AV, 2-May-2020.) |
| ⊢ (𝑆 prefix 0) = ∅ | ||
| Theorem | pfx0 14600 | A prefix of an empty set is always the empty set. (Contributed by AV, 3-May-2020.) |
| ⊢ (∅ prefix 𝐿) = ∅ | ||
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