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Type | Label | Description |
---|---|---|
Statement | ||
Theorem | 0wrd0 13601 | The empty word is the only word over an empty alphabet. (Contributed by AV, 25-Oct-2018.) |
⊢ (𝑊 ∈ Word ∅ ↔ 𝑊 = ∅) | ||
Theorem | ffz0iswrd 13602 | A sequence with zero-based indices is a word. (Contributed by AV, 31-Jan-2018.) (Proof shortened by AV, 13-Oct-2018.) |
⊢ (𝑊:(0...𝐿)⟶𝑆 → 𝑊 ∈ Word 𝑆) | ||
Theorem | wrdsymb 13603 | A word is a word over the symbols it consists of. (Contributed by AV, 1-Dec-2022.) |
⊢ (𝑆 ∈ Word 𝐴 → 𝑆 ∈ Word (𝑆 “ (0..^(♯‘𝑆)))) | ||
Theorem | nfwrd 13604 | Hypothesis builder for Word 𝑆. (Contributed by Mario Carneiro, 26-Feb-2016.) |
⊢ Ⅎ𝑥𝑆 ⇒ ⊢ Ⅎ𝑥Word 𝑆 | ||
Theorem | csbwrdg 13605* | Class substitution for the symbols of a word. (Contributed by Alexander van der Vekens, 15-Jul-2018.) |
⊢ (𝑆 ∈ 𝑉 → ⦋𝑆 / 𝑥⦌Word 𝑥 = Word 𝑆) | ||
Theorem | wrdnval 13606* | Words of a fixed length are mappings from a fixed half-open integer interval. (Contributed by Alexander van der Vekens, 25-Mar-2018.) (Proof shortened by AV, 13-May-2020.) |
⊢ ((𝑉 ∈ 𝑋 ∧ 𝑁 ∈ ℕ0) → {𝑤 ∈ Word 𝑉 ∣ (♯‘𝑤) = 𝑁} = (𝑉 ↑𝑚 (0..^𝑁))) | ||
Theorem | wrdmap 13607 | Words as a mapping. (Contributed by Thierry Arnoux, 4-Mar-2020.) |
⊢ ((𝑉 ∈ 𝑋 ∧ 𝑁 ∈ ℕ0) → ((𝑊 ∈ Word 𝑉 ∧ (♯‘𝑊) = 𝑁) ↔ 𝑊 ∈ (𝑉 ↑𝑚 (0..^𝑁)))) | ||
Theorem | hashwrdn 13608* | If there is only a finite number of symbols, the number of words of a fixed length over these sysmbols is the number of these symbols raised to the power of the length. (Contributed by Alexander van der Vekens, 25-Mar-2018.) |
⊢ ((𝑉 ∈ Fin ∧ 𝑁 ∈ ℕ0) → (♯‘{𝑤 ∈ Word 𝑉 ∣ (♯‘𝑤) = 𝑁}) = ((♯‘𝑉)↑𝑁)) | ||
Theorem | wrdnfi 13609* | If there is only a finite number of symbols, the number of words of a fixed length over these symbols is also finite. (Contributed by Alexander van der Vekens, 25-Mar-2018.) |
⊢ ((𝑉 ∈ Fin ∧ 𝑁 ∈ ℕ0) → {𝑤 ∈ Word 𝑉 ∣ (♯‘𝑤) = 𝑁} ∈ Fin) | ||
Theorem | wrdsymb0 13610 | A symbol at a position "outside" of a word. (Contributed by Alexander van der Vekens, 26-May-2018.) (Proof shortened by AV, 2-May-2020.) |
⊢ ((𝑊 ∈ Word 𝑉 ∧ 𝐼 ∈ ℤ) → ((𝐼 < 0 ∨ (♯‘𝑊) ≤ 𝐼) → (𝑊‘𝐼) = ∅)) | ||
Theorem | wrdlenge1n0 13611 | A word with length at least 1 is not empty. (Contributed by AV, 14-Oct-2018.) |
⊢ (𝑊 ∈ Word 𝑉 → (𝑊 ≠ ∅ ↔ 1 ≤ (♯‘𝑊))) | ||
Theorem | len0nnbi 13612 | The length of a word is a positive integer iff the word is not empty. (Contributed by AV, 22-Mar-2022.) |
⊢ (𝑊 ∈ Word 𝑆 → (𝑊 ≠ ∅ ↔ (♯‘𝑊) ∈ ℕ)) | ||
Theorem | wrdlenge2n0 13613 | A word with length at least 2 is not empty. (Contributed by AV, 18-Jun-2018.) (Proof shortened by AV, 14-Oct-2018.) |
⊢ ((𝑊 ∈ Word 𝑉 ∧ 2 ≤ (♯‘𝑊)) → 𝑊 ≠ ∅) | ||
Theorem | wrdsymb1 13614 | The first symbol of a nonempty word over an alphabet belongs to the alphabet. (Contributed by Alexander van der Vekens, 28-Jun-2018.) |
⊢ ((𝑊 ∈ Word 𝑉 ∧ 1 ≤ (♯‘𝑊)) → (𝑊‘0) ∈ 𝑉) | ||
Theorem | wrdlen1 13615* | A word of length 1 starts with a symbol. (Contributed by AV, 20-Jul-2018.) (Proof shortened by AV, 19-Oct-2018.) |
⊢ ((𝑊 ∈ Word 𝑉 ∧ (♯‘𝑊) = 1) → ∃𝑣 ∈ 𝑉 (𝑊‘0) = 𝑣) | ||
Theorem | fstwrdne 13616 | The first symbol of a nonempty word is element of the alphabet for the word. (Contributed by AV, 28-Sep-2018.) (Proof shortened by AV, 14-Oct-2018.) |
⊢ ((𝑊 ∈ Word 𝑉 ∧ 𝑊 ≠ ∅) → (𝑊‘0) ∈ 𝑉) | ||
Theorem | fstwrdne0 13617 | The first symbol of a nonempty word is element of the alphabet for the word. (Contributed by AV, 29-Sep-2018.) (Proof shortened by AV, 14-Oct-2018.) |
⊢ ((𝑁 ∈ ℕ ∧ (𝑊 ∈ Word 𝑉 ∧ (♯‘𝑊) = 𝑁)) → (𝑊‘0) ∈ 𝑉) | ||
Theorem | eqwrd 13618* | Two words are equal iff they have the same length and the same symbol at each position. (Contributed by AV, 13-Apr-2018.) |
⊢ ((𝑈 ∈ Word 𝑉 ∧ 𝑊 ∈ Word 𝑉) → (𝑈 = 𝑊 ↔ ((♯‘𝑈) = (♯‘𝑊) ∧ ∀𝑖 ∈ (0..^(♯‘𝑈))(𝑈‘𝑖) = (𝑊‘𝑖)))) | ||
Theorem | elovmpt2wrd 13619* | Implications for the value of an operation defined by the maps-to notation with a class abstration of words as a result having an element. Note that 𝜑 may depend on 𝑧 as well as on 𝑣 and 𝑦. (Contributed by Alexander van der Vekens, 15-Jul-2018.) |
⊢ 𝑂 = (𝑣 ∈ V, 𝑦 ∈ V ↦ {𝑧 ∈ Word 𝑣 ∣ 𝜑}) ⇒ ⊢ (𝑍 ∈ (𝑉𝑂𝑌) → (𝑉 ∈ V ∧ 𝑌 ∈ V ∧ 𝑍 ∈ Word 𝑉)) | ||
Theorem | elovmptnn0wrd 13620* | Implications for the value of an operation defined by the maps-to notation with a function of nonnegative integers into a class abstraction of words as a result having an element. Note that 𝜑 may depend on 𝑧 as well as on 𝑣 and 𝑦 and 𝑛. (Contributed by AV, 16-Jul-2018.) (Revised by AV, 16-May-2019.) |
⊢ 𝑂 = (𝑣 ∈ V, 𝑦 ∈ V ↦ (𝑛 ∈ ℕ0 ↦ {𝑧 ∈ Word 𝑣 ∣ 𝜑})) ⇒ ⊢ (𝑍 ∈ ((𝑉𝑂𝑌)‘𝑁) → ((𝑉 ∈ V ∧ 𝑌 ∈ V) ∧ (𝑁 ∈ ℕ0 ∧ 𝑍 ∈ Word 𝑉))) | ||
Theorem | wrdred1 13621 | A word truncated by a symbol is a word. (Contributed by AV, 29-Jan-2021.) |
⊢ (𝐹 ∈ Word 𝑆 → (𝐹 ↾ (0..^((♯‘𝐹) − 1))) ∈ Word 𝑆) | ||
Theorem | wrdred1hash 13622 | The length of a word truncated by a symbol. (Contributed by Alexander van der Vekens, 1-Nov-2017.) (Revised by AV, 29-Jan-2021.) |
⊢ ((𝐹 ∈ Word 𝑆 ∧ 1 ≤ (♯‘𝐹)) → (♯‘(𝐹 ↾ (0..^((♯‘𝐹) − 1)))) = ((♯‘𝐹) − 1)) | ||
Syntax | clsw 13623 | Extend class notation with the Last Symbol of a word. |
class lastS | ||
Definition | df-lsw 13624 | Extract the last symbol of a word. May be not meaningful for other sets which are not words. The name lastS (as abbreviation of "lastSymbol") is a compromise between usually used names for corresponding functions in computer programs (as last() or lastChar()), the terminology used for words in set.mm ("symbol" instead of "character") and brevity ("lastS" is shorter than "lastChar" and "lastSymbol"). Labels of theorems about last symbols of a word will contain the abbreviation "lsw" (Last Symbol of a Word). (Contributed by Alexander van der Vekens, 18-Mar-2018.) |
⊢ lastS = (𝑤 ∈ V ↦ (𝑤‘((♯‘𝑤) − 1))) | ||
Theorem | lsw 13625 | Extract the last symbol of a word. May be not meaningful for other sets which are not words. (Contributed by Alexander van der Vekens, 18-Mar-2018.) |
⊢ (𝑊 ∈ 𝑋 → (lastS‘𝑊) = (𝑊‘((♯‘𝑊) − 1))) | ||
Theorem | lsw0 13626 | The last symbol of an empty word does not exist. (Contributed by Alexander van der Vekens, 19-Mar-2018.) (Proof shortened by AV, 2-May-2020.) |
⊢ ((𝑊 ∈ Word 𝑉 ∧ (♯‘𝑊) = 0) → (lastS‘𝑊) = ∅) | ||
Theorem | lsw0g 13627 | The last symbol of an empty word does not exist. (Contributed by Alexander van der Vekens, 11-Nov-2018.) |
⊢ (lastS‘∅) = ∅ | ||
Theorem | lsw1 13628 | The last symbol of a word of length 1 is the first symbol of this word. (Contributed by Alexander van der Vekens, 19-Mar-2018.) |
⊢ ((𝑊 ∈ Word 𝑉 ∧ (♯‘𝑊) = 1) → (lastS‘𝑊) = (𝑊‘0)) | ||
Theorem | lswcl 13629 | Closure of the last symbol: the last symbol of a not empty word belongs to the alphabet for the word. (Contributed by AV, 2-Aug-2018.) (Proof shortened by AV, 29-Apr-2020.) |
⊢ ((𝑊 ∈ Word 𝑉 ∧ 𝑊 ≠ ∅) → (lastS‘𝑊) ∈ 𝑉) | ||
Theorem | lswlgt0cl 13630 | The last symbol of a nonempty word is element of the alphabet for the word. (Contributed by Alexander van der Vekens, 1-Oct-2018.) (Proof shortened by AV, 29-Apr-2020.) |
⊢ ((𝑁 ∈ ℕ ∧ (𝑊 ∈ Word 𝑉 ∧ (♯‘𝑊) = 𝑁)) → (lastS‘𝑊) ∈ 𝑉) | ||
Syntax | cconcat 13631 | Syntax for the concatenation operator. |
class ++ | ||
Definition | df-concat 13632* | Define the concatenation operator which combines two words. Definition in Section 9.1 of [AhoHopUll] p. 318. (Contributed by FL, 14-Jan-2014.) (Revised by Stefan O'Rear, 15-Aug-2015.) |
⊢ ++ = (𝑠 ∈ V, 𝑡 ∈ V ↦ (𝑥 ∈ (0..^((♯‘𝑠) + (♯‘𝑡))) ↦ if(𝑥 ∈ (0..^(♯‘𝑠)), (𝑠‘𝑥), (𝑡‘(𝑥 − (♯‘𝑠)))))) | ||
Theorem | ccatfn 13633 | The concatenation operator is a two-argument function. (Contributed by Mario Carneiro, 27-Sep-2015.) (Proof shortened by AV, 29-Apr-2020.) |
⊢ ++ Fn (V × V) | ||
Theorem | ccatfval 13634* | Value of the concatenation operator. (Contributed by Stefan O'Rear, 15-Aug-2015.) |
⊢ ((𝑆 ∈ 𝑉 ∧ 𝑇 ∈ 𝑊) → (𝑆 ++ 𝑇) = (𝑥 ∈ (0..^((♯‘𝑆) + (♯‘𝑇))) ↦ if(𝑥 ∈ (0..^(♯‘𝑆)), (𝑆‘𝑥), (𝑇‘(𝑥 − (♯‘𝑆)))))) | ||
Theorem | ccatcl 13635 | 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 13636 | The length of a concatenated word. (Contributed by Stefan O'Rear, 15-Aug-2015.) |
⊢ ((𝑆 ∈ Word 𝐵 ∧ 𝑇 ∈ Word 𝐵) → (♯‘(𝑆 ++ 𝑇)) = ((♯‘𝑆) + (♯‘𝑇))) | ||
Theorem | ccat0 13637 | The concatenation of two words is empty iff the two words are empty. (Contributed by AV, 4-Mar-2022.) |
⊢ ((𝑆 ∈ Word 𝐵 ∧ 𝑇 ∈ Word 𝐵) → ((𝑆 ++ 𝑇) = ∅ ↔ (𝑆 = ∅ ∧ 𝑇 = ∅))) | ||
Theorem | ccatval1 13638 | 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.) |
⊢ ((𝑆 ∈ Word 𝐵 ∧ 𝑇 ∈ Word 𝐵 ∧ 𝐼 ∈ (0..^(♯‘𝑆))) → ((𝑆 ++ 𝑇)‘𝐼) = (𝑆‘𝐼)) | ||
Theorem | ccatval2 13639 | 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 13640 | 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 13641 | 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 13642 | 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 13643 | 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 13644 | 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 13645 | 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 13646 | 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 13647 | 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 13648 | 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 13649 | 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 13650 | 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 13651 | Concatenation of the empty word by the empty word. (Contributed by AV, 26-Mar-2022.) |
⊢ (∅ ++ ∅) = ∅ | ||
Theorem | lswccatn0lsw 13652 | 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 13653 | 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 13654 | 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 13655 | 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 13656 | Syntax for the singleton word constructor. |
class 〈“𝐴”〉 | ||
Definition | df-s1 13657 | 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 13665, and not, as maybe expected, the empty word, see also s1nz 13668. (Contributed by Stefan O'Rear, 15-Aug-2015.) (Revised by Mario Carneiro, 26-Feb-2016.) |
⊢ 〈“𝐴”〉 = {〈0, ( I ‘𝐴)〉} | ||
Theorem | ids1 13658 | Identity function protection for a singleton word. (Contributed by Mario Carneiro, 26-Feb-2016.) |
⊢ 〈“𝐴”〉 = 〈“( I ‘𝐴)”〉 | ||
Theorem | s1val 13659 | 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 13660 | The range of a singleton word. (Contributed by Mario Carneiro, 18-Jul-2016.) |
⊢ (𝐴 ∈ 𝑉 → ran 〈“𝐴”〉 = {𝐴}) | ||
Theorem | s1eq 13661 | Equality theorem for a singleton word. (Contributed by Mario Carneiro, 26-Feb-2016.) |
⊢ (𝐴 = 𝐵 → 〈“𝐴”〉 = 〈“𝐵”〉) | ||
Theorem | s1eqd 13662 | Equality theorem for a singleton word. (Contributed by Mario Carneiro, 26-Feb-2016.) |
⊢ (𝜑 → 𝐴 = 𝐵) ⇒ ⊢ (𝜑 → 〈“𝐴”〉 = 〈“𝐵”〉) | ||
Theorem | s1cl 13663 | 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 13664 | A singleton word is a word. (Contributed by Mario Carneiro, 26-Feb-2016.) |
⊢ (𝜑 → 𝐴 ∈ 𝐵) ⇒ ⊢ (𝜑 → 〈“𝐴”〉 ∈ Word 𝐵) | ||
Theorem | s1prc 13665 | Value of a singleton word if the symbol is a proper class. (Contributed by AV, 26-Mar-2022.) |
⊢ (¬ 𝐴 ∈ V → 〈“𝐴”〉 = 〈“∅”〉) | ||
Theorem | s1cli 13666 | A singleton word is a word. (Contributed by Mario Carneiro, 26-Feb-2016.) |
⊢ 〈“𝐴”〉 ∈ Word V | ||
Theorem | s1len 13667 | Length of a singleton word. (Contributed by Stefan O'Rear, 15-Aug-2015.) (Revised by Mario Carneiro, 26-Feb-2016.) |
⊢ (♯‘〈“𝐴”〉) = 1 | ||
Theorem | s1nz 13668 | 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 13669 | The domain of a singleton word is a singleton. (Contributed by AV, 9-Jan-2020.) |
⊢ dom 〈“𝐴”〉 = {0} | ||
Theorem | s1dmALT 13670 | Alternate version of s1dm 13669, 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 13671 | 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 13672 | The last symbol of a singleton word is its symbol. (Contributed by AV, 22-Oct-2018.) |
⊢ (𝐴 ∈ 𝑉 → (lastS‘〈“𝐴”〉) = 𝐴) | ||
Theorem | eqs1 13673 | 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 13674* | 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 13675 | 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 13676 | 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 13677 | 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 13678 | 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 13679 | 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 13680 | 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 13681 | 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 13682 | 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 13683 | 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 13684 | The length of the concatenation of two singleton words. (Contributed by Alexander van der Vekens, 22-Sep-2018.) |
⊢ ((𝑋 ∈ 𝑉 ∧ 𝑌 ∈ 𝑉) → (♯‘(〈“𝑋”〉 ++ 〈“𝑌”〉)) = 2) | ||
Theorem | ccatw2s1cl 13685 | 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 13686 | 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 13687 | 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.) |
⊢ ((𝑊 ∈ Word 𝑉 ∧ 𝑆 ∈ 𝑉 ∧ 𝐼 ∈ (0..^(♯‘𝑊))) → ((𝑊 ++ 〈“𝑆”〉)‘𝐼) = (𝑊‘𝐼)) | ||
Theorem | ccats1val2 13688 | 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 13689 | 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 13690 | Extract the first of two concatenated singleton words. (Contributed by Alexander van der Vekens, 22-Sep-2018.) |
⊢ ((𝑋 ∈ 𝑉 ∧ 𝑌 ∈ 𝑉) → ((〈“𝑋”〉 ++ 〈“𝑌”〉)‘0) = 𝑋) | ||
Theorem | ccat2s1p2 13691 | Extract the second of two concatenated singleton words. (Contributed by Alexander van der Vekens, 22-Sep-2018.) |
⊢ ((𝑋 ∈ 𝑉 ∧ 𝑌 ∈ 𝑉) → ((〈“𝑋”〉 ++ 〈“𝑌”〉)‘1) = 𝑌) | ||
Theorem | ccatw2s1ass 13692 | Associative law for a concatenation of a word with two singleton words. (Contributed by Alexander van der Vekens, 22-Sep-2018.) |
⊢ ((𝑊 ∈ Word 𝑉 ∧ 𝑋 ∈ 𝑉 ∧ 𝑌 ∈ 𝑉) → ((𝑊 ++ 〈“𝑋”〉) ++ 〈“𝑌”〉) = (𝑊 ++ (〈“𝑋”〉 ++ 〈“𝑌”〉))) | ||
Theorem | ccatws1n0 13693 | 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 13694 | 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 13695 | 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 13696 | 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 13697 | 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.) |
⊢ (((𝑊 ∈ Word 𝑉 ∧ (♯‘𝑊) = 𝑁) ∧ (𝑋 ∈ 𝑉 ∧ 𝑌 ∈ 𝑉)) → (((𝑊 ++ 〈“𝑋”〉) ++ 〈“𝑌”〉)‘𝑁) = 𝑋) | ||
Theorem | ccatw2s1p2 13698 | 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 13699 | 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.) |
⊢ (((𝑊 ∈ Word 𝑉 ∧ 𝐼 ∈ ℕ0 ∧ 𝐼 < (♯‘𝑊)) ∧ (𝑋 ∈ 𝑉 ∧ 𝑌 ∈ 𝑉)) → (((𝑊 ++ 〈“𝑋”〉) ++ 〈“𝑌”〉)‘𝐼) = (𝑊‘𝐼)) | ||
Theorem | ccat2s1fst 13700 | The first symbol of the concatenation of a word with two single symbols. (Contributed by Alexander van der Vekens, 22-Sep-2018.) |
⊢ (((𝑊 ∈ Word 𝑉 ∧ 0 < (♯‘𝑊)) ∧ (𝑋 ∈ 𝑉 ∧ 𝑌 ∈ 𝑉)) → (((𝑊 ++ 〈“𝑋”〉) ++ 〈“𝑌”〉)‘0) = (𝑊‘0)) |
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