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Type | Label | Description |
---|---|---|
Statement | ||
Theorem | ccatalpha 14601 | 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 14602 | 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 14603 | Syntax for the singleton word constructor. |
class 〈“𝐴”〉 | ||
Definition | df-s1 14604 | 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 14612, and not, as maybe expected, the empty word, see also s1nz 14615. (Contributed by Stefan O'Rear, 15-Aug-2015.) (Revised by Mario Carneiro, 26-Feb-2016.) |
⊢ 〈“𝐴”〉 = {〈0, ( I ‘𝐴)〉} | ||
Theorem | ids1 14605 | Identity function protection for a singleton word. (Contributed by Mario Carneiro, 26-Feb-2016.) |
⊢ 〈“𝐴”〉 = 〈“( I ‘𝐴)”〉 | ||
Theorem | s1val 14606 | 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 14607 | The range of a singleton word. (Contributed by Mario Carneiro, 18-Jul-2016.) |
⊢ (𝐴 ∈ 𝑉 → ran 〈“𝐴”〉 = {𝐴}) | ||
Theorem | s1eq 14608 | Equality theorem for a singleton word. (Contributed by Mario Carneiro, 26-Feb-2016.) |
⊢ (𝐴 = 𝐵 → 〈“𝐴”〉 = 〈“𝐵”〉) | ||
Theorem | s1eqd 14609 | Equality theorem for a singleton word. (Contributed by Mario Carneiro, 26-Feb-2016.) |
⊢ (𝜑 → 𝐴 = 𝐵) ⇒ ⊢ (𝜑 → 〈“𝐴”〉 = 〈“𝐵”〉) | ||
Theorem | s1cl 14610 | 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 14611 | A singleton word is a word. (Contributed by Mario Carneiro, 26-Feb-2016.) |
⊢ (𝜑 → 𝐴 ∈ 𝐵) ⇒ ⊢ (𝜑 → 〈“𝐴”〉 ∈ Word 𝐵) | ||
Theorem | s1prc 14612 | Value of a singleton word if the symbol is a proper class. (Contributed by AV, 26-Mar-2022.) |
⊢ (¬ 𝐴 ∈ V → 〈“𝐴”〉 = 〈“∅”〉) | ||
Theorem | s1cli 14613 | A singleton word is a word. (Contributed by Mario Carneiro, 26-Feb-2016.) |
⊢ 〈“𝐴”〉 ∈ Word V | ||
Theorem | s1len 14614 | Length of a singleton word. (Contributed by Stefan O'Rear, 15-Aug-2015.) (Revised by Mario Carneiro, 26-Feb-2016.) |
⊢ (♯‘〈“𝐴”〉) = 1 | ||
Theorem | s1nz 14615 | 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 14616 | The domain of a singleton word is a singleton. (Contributed by AV, 9-Jan-2020.) |
⊢ dom 〈“𝐴”〉 = {0} | ||
Theorem | s1dmALT 14617 | Alternate version of s1dm 14616, 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 14618 | 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 14619 | The last symbol of a singleton word is its symbol. (Contributed by AV, 22-Oct-2018.) |
⊢ (𝐴 ∈ 𝑉 → (lastS‘〈“𝐴”〉) = 𝐴) | ||
Theorem | eqs1 14620 | 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 14621* | 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 14622 | 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 14623 | 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 14624 | 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 14625 | 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 14626 | 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 14627 | 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 14628 | 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 14629 | 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 14630 | 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 14631 | 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 14632 | 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 14633 | 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 14634 | 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 14635 | 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 14636 | 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 14637 | 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 14638 | 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 14639 | 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 14640 | 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 14641 | 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 14642 | 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 14643 | 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 14644 | 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 14645 | 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 14646 | 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 14647 | 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 14648 | Syntax for the subword operator. |
class substr | ||
Definition | df-substr 14649* | 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 14650 | 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 6936). (Contributed by AV, 2-Dec-2022.) (New usage is discouraged.) |
⊢ (¬ (𝐹 ∈ ℤ ∧ 𝐿 ∈ ℤ) → (𝑆 substr 〈𝐹, 𝐿〉) = ∅) | ||
Theorem | swrdval 14651* | 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 14652 | A zero length substring. (Contributed by Stefan O'Rear, 27-Aug-2015.) |
⊢ (𝑆 substr 〈𝑋, 𝑋〉) = ∅ | ||
Theorem | swrdcl 14653 | 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 14654* | 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 14655 | 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 14656 | 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 14657 | 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 14658 | 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 14659 | 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 14660 | 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 14661 | 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 14662 | 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 14663 | 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 14664 | 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 14665 | 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 14666 | 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 14667 | 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 14668 | 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 14669 | 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 14670 | 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 14671 | 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 14672 | 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 14673* | 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 14674 | 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 14675 | 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 14676 | 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 14677 | 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 14678 | Syntax for the prefix operator. |
class prefix | ||
Definition | df-pfx 14679* | 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 14680 | 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 6936). (Contributed by AV, 3-Dec-2022.) (New usage is discouraged.) |
⊢ (¬ (𝑆 ∈ V ∧ 𝐿 ∈ ℕ0) → (𝑆 prefix 𝐿) = ∅) | ||
Theorem | pfxval 14681 | Value of a prefix operation. (Contributed by AV, 2-May-2020.) |
⊢ ((𝑆 ∈ 𝑉 ∧ 𝐿 ∈ ℕ0) → (𝑆 prefix 𝐿) = (𝑆 substr 〈0, 𝐿〉)) | ||
Theorem | pfx00 14682 | The zero length prefix is the empty set. (Contributed by AV, 2-May-2020.) |
⊢ (𝑆 prefix 0) = ∅ | ||
Theorem | pfx0 14683 | A prefix of an empty set is always the empty set. (Contributed by AV, 3-May-2020.) |
⊢ (∅ prefix 𝐿) = ∅ | ||
Theorem | pfxval0 14684 | Value of a prefix operation. This theorem should only be used in proofs if 𝐿 ∈ ℕ0 is not available. Otherwise (and usually), pfxval 14681 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 14685 | Closure of the prefix extractor. (Contributed by AV, 2-May-2020.) |
⊢ (𝑆 ∈ Word 𝐴 → (𝑆 prefix 𝐿) ∈ Word 𝐴) | ||
Theorem | pfxmpt 14686* | 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))) | ||
Theorem | pfxres 14687 | Value of the subword extractor for left-anchored subwords. (Contributed by Stefan O'Rear, 24-Aug-2015.) (Revised by AV, 2-May-2020.) |
⊢ ((𝑆 ∈ Word 𝐴 ∧ 𝐿 ∈ (0...(♯‘𝑆))) → (𝑆 prefix 𝐿) = (𝑆 ↾ (0..^𝐿))) | ||
Theorem | pfxf 14688 | A prefix of a word is a function from a half-open range of nonnegative integers of the same length as the prefix to the set of symbols for the original word. (Contributed by AV, 2-May-2020.) |
⊢ ((𝑊 ∈ Word 𝑉 ∧ 𝐿 ∈ (0...(♯‘𝑊))) → (𝑊 prefix 𝐿):(0..^𝐿)⟶𝑉) | ||
Theorem | pfxfn 14689 | Value of the prefix extractor as function with domain. (Contributed by AV, 2-May-2020.) |
⊢ ((𝑆 ∈ Word 𝑉 ∧ 𝐿 ∈ (0...(♯‘𝑆))) → (𝑆 prefix 𝐿) Fn (0..^𝐿)) | ||
Theorem | pfxfv 14690 | A symbol in a prefix of a word, indexed using the prefix' indices. (Contributed by Alexander van der Vekens, 16-Jun-2018.) (Revised by AV, 3-May-2020.) |
⊢ ((𝑊 ∈ Word 𝑉 ∧ 𝐿 ∈ (0...(♯‘𝑊)) ∧ 𝐼 ∈ (0..^𝐿)) → ((𝑊 prefix 𝐿)‘𝐼) = (𝑊‘𝐼)) | ||
Theorem | pfxlen 14691 | Length of a prefix. (Contributed by Stefan O'Rear, 24-Aug-2015.) (Revised by AV, 2-May-2020.) |
⊢ ((𝑆 ∈ Word 𝐴 ∧ 𝐿 ∈ (0...(♯‘𝑆))) → (♯‘(𝑆 prefix 𝐿)) = 𝐿) | ||
Theorem | pfxid 14692 | A word is a prefix of itself. (Contributed by Stefan O'Rear, 16-Aug-2015.) (Revised by AV, 2-May-2020.) |
⊢ (𝑆 ∈ Word 𝐴 → (𝑆 prefix (♯‘𝑆)) = 𝑆) | ||
Theorem | pfxrn 14693 | The range of a prefix of a word is a subset of the set of symbols for the word. (Contributed by AV, 2-May-2020.) |
⊢ ((𝑊 ∈ Word 𝑉 ∧ 𝐿 ∈ (0...(♯‘𝑊))) → ran (𝑊 prefix 𝐿) ⊆ 𝑉) | ||
Theorem | pfxn0 14694 | A prefix consisting of at least one symbol is not empty. (Contributed by Alexander van der Vekens, 4-Aug-2018.) (Revised by AV, 2-May-2020.) |
⊢ ((𝑊 ∈ Word 𝑉 ∧ 𝐿 ∈ ℕ ∧ 𝐿 ≤ (♯‘𝑊)) → (𝑊 prefix 𝐿) ≠ ∅) | ||
Theorem | pfxnd 14695 | The value of a prefix operation for a length argument larger than the word length is the empty set. (This is due to our definition of function values for out-of-domain arguments, see ndmfv 6936). (Contributed by AV, 3-May-2020.) |
⊢ ((𝑊 ∈ Word 𝑉 ∧ 𝐿 ∈ ℕ0 ∧ (♯‘𝑊) < 𝐿) → (𝑊 prefix 𝐿) = ∅) | ||
Theorem | pfxnd0 14696 | The value of a prefix operation for a length argument not in the range of the word length is the empty set. (This is due to our definition of function values for out-of-domain arguments, see ndmfv 6936). (Contributed by AV, 3-Dec-2022.) |
⊢ ((𝑊 ∈ Word 𝑉 ∧ 𝐿 ∉ (0...(♯‘𝑊))) → (𝑊 prefix 𝐿) = ∅) | ||
Theorem | pfxwrdsymb 14697 | A prefix of a word is a word over the symbols it consists of. (Contributed by AV, 3-Dec-2022.) |
⊢ (𝑆 ∈ Word 𝐴 → (𝑆 prefix 𝐿) ∈ Word (𝑆 “ (0..^𝐿))) | ||
Theorem | addlenrevpfx 14698 | The sum of the lengths of two reversed parts of a word is the length of the word. (Contributed by Alexander van der Vekens, 1-Apr-2018.) (Revised by AV, 3-May-2020.) |
⊢ ((𝑊 ∈ Word 𝑉 ∧ 𝑀 ∈ (0...(♯‘𝑊))) → ((♯‘(𝑊 substr 〈𝑀, (♯‘𝑊)〉)) + (♯‘(𝑊 prefix 𝑀))) = (♯‘𝑊)) | ||
Theorem | addlenpfx 14699 | The sum of the lengths of two parts of a word is the length of the word. (Contributed by AV, 21-Oct-2018.) (Revised by AV, 3-May-2020.) |
⊢ ((𝑊 ∈ Word 𝑉 ∧ 𝑀 ∈ (0...(♯‘𝑊))) → ((♯‘(𝑊 prefix 𝑀)) + (♯‘(𝑊 substr 〈𝑀, (♯‘𝑊)〉))) = (♯‘𝑊)) | ||
Theorem | pfxfv0 14700 | The first symbol of a prefix is the first symbol of the word. (Contributed by Alexander van der Vekens, 16-Jun-2018.) (Revised by AV, 3-May-2020.) |
⊢ ((𝑊 ∈ Word 𝑉 ∧ 𝐿 ∈ (1...(♯‘𝑊))) → ((𝑊 prefix 𝐿)‘0) = (𝑊‘0)) |
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