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| Type | Label | Description |
|---|---|---|
| Statement | ||
| Theorem | ccatcl 14601 | 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 14602 | 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 14603 | 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 14604 | 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 14605 | 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 14606 | 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 14607 | 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 14608 | 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 14609 | 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 14610 | 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 14611 | 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 14612 | 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 14613 | 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 14614 | 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 14615 | 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 14616 | 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 14617 | 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 14618 | Concatenation of the empty word by the empty word. (Contributed by AV, 26-Mar-2022.) |
| ⊢ (∅ ++ ∅) = ∅ | ||
| Theorem | lswccatn0lsw 14619 | 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 14620 | 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 14621 | 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 14622 | 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 14623 | Syntax for the singleton word constructor. |
| class 〈“𝐴”〉 | ||
| Definition | df-s1 14624 | 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 14632, and not, as maybe expected, the empty word, see also s1nz 14635. (Contributed by Stefan O'Rear, 15-Aug-2015.) (Revised by Mario Carneiro, 26-Feb-2016.) |
| ⊢ 〈“𝐴”〉 = {〈0, ( I ‘𝐴)〉} | ||
| Theorem | ids1 14625 | Identity function protection for a singleton word. (Contributed by Mario Carneiro, 26-Feb-2016.) |
| ⊢ 〈“𝐴”〉 = 〈“( I ‘𝐴)”〉 | ||
| Theorem | s1val 14626 | 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 14627 | The range of a singleton word. (Contributed by Mario Carneiro, 18-Jul-2016.) |
| ⊢ (𝐴 ∈ 𝑉 → ran 〈“𝐴”〉 = {𝐴}) | ||
| Theorem | s1eq 14628 | Equality theorem for a singleton word. (Contributed by Mario Carneiro, 26-Feb-2016.) |
| ⊢ (𝐴 = 𝐵 → 〈“𝐴”〉 = 〈“𝐵”〉) | ||
| Theorem | s1eqd 14629 | Equality theorem for a singleton word. (Contributed by Mario Carneiro, 26-Feb-2016.) |
| ⊢ (𝜑 → 𝐴 = 𝐵) ⇒ ⊢ (𝜑 → 〈“𝐴”〉 = 〈“𝐵”〉) | ||
| Theorem | s1cl 14630 | 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 14631 | A singleton word is a word. (Contributed by Mario Carneiro, 26-Feb-2016.) |
| ⊢ (𝜑 → 𝐴 ∈ 𝐵) ⇒ ⊢ (𝜑 → 〈“𝐴”〉 ∈ Word 𝐵) | ||
| Theorem | s1prc 14632 | Value of a singleton word if the symbol is a proper class. (Contributed by AV, 26-Mar-2022.) |
| ⊢ (¬ 𝐴 ∈ V → 〈“𝐴”〉 = 〈“∅”〉) | ||
| Theorem | s1cli 14633 | A singleton word is a word. (Contributed by Mario Carneiro, 26-Feb-2016.) |
| ⊢ 〈“𝐴”〉 ∈ Word V | ||
| Theorem | s1len 14634 | Length of a singleton word. (Contributed by Stefan O'Rear, 15-Aug-2015.) (Revised by Mario Carneiro, 26-Feb-2016.) |
| ⊢ (♯‘〈“𝐴”〉) = 1 | ||
| Theorem | s1nz 14635 | 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 14636 | The domain of a singleton word is a singleton. (Contributed by AV, 9-Jan-2020.) |
| ⊢ dom 〈“𝐴”〉 = {0} | ||
| Theorem | s1dmALT 14637 | Alternate version of s1dm 14636, 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 14638 | 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 14639 | The last symbol of a singleton word is its symbol. (Contributed by AV, 22-Oct-2018.) |
| ⊢ (𝐴 ∈ 𝑉 → (lastS‘〈“𝐴”〉) = 𝐴) | ||
| Theorem | eqs1 14640 | 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 14641* | 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 14642 | 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 14643 | 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 14644 | 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 14645 | 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 14646 | 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 14647 | 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 14648 | 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 14649 | 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 14650 | 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 14651 | 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 14652 | 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 14653 | 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 14654 | 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 14655 | 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 14656 | 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 14657 | 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 14658 | 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 14659 | 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 14660 | 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 14661 | 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 14662 | 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 14663 | 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 14664 | 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 14665 | 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 14666 | 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 14667 | 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 14668 | Syntax for the subword operator. |
| class substr | ||
| Definition | df-substr 14669* | 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 14670 | 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 6903). (Contributed by AV, 2-Dec-2022.) (New usage is discouraged.) |
| ⊢ (¬ (𝐹 ∈ ℤ ∧ 𝐿 ∈ ℤ) → (𝑆 substr 〈𝐹, 𝐿〉) = ∅) | ||
| Theorem | swrdval 14671* | 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 14672 | A zero length substring. (Contributed by Stefan O'Rear, 27-Aug-2015.) |
| ⊢ (𝑆 substr 〈𝑋, 𝑋〉) = ∅ | ||
| Theorem | swrdcl 14673 | 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 14674* | 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 14675 | 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 14676 | 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 14677 | 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 14678 | 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 14679 | 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 14680 | 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 14681 | 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 14682 | 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 14683 | 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 14684 | 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 14685 | 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 14686 | 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 14687 | 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 14688 | 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 14689 | 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 14690 | 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 14691 | 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 14692 | 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 14693* | 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 14694 | 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 14695 | 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 14696 | 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 14697 | 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 14698 | Syntax for the prefix operator. |
| class prefix | ||
| Definition | df-pfx 14699* | 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 14700 | 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 6903). (Contributed by AV, 3-Dec-2022.) (New usage is discouraged.) |
| ⊢ (¬ (𝑆 ∈ V ∧ 𝐿 ∈ ℕ0) → (𝑆 prefix 𝐿) = ∅) | ||
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