연습장 - 편집 역사

Pinkgo: /* Ex. 4 */

2025-09-23T05:11:22Z

Ex. 4

← 이전 판		2025년 9월 23일 (화) 05:11 판
49번째 줄:		49번째 줄:

	Uniquness)<br>		Uniquness)<br>
	Suppose <math>u,\,\,v</math> are string, and <math>(\|u\|=\|v\|=1)\land (u</math>@<math>0 = v</math>@<math>0 = x, s.t. x \in \Sigma)</math>		Suppose <math>u,\,\,v</math> are string, and <math>(\|u\|=\|v\|=1)\land (u</math>@<math>0 = v</math>@<math>0 = x, s.t. x \in \Sigma)</math><br>
	Then, the domain of both string is <math>\{1\}</math>, and the alphabet corresponding to index 0 is <math>x</math> for both string.		Then, the domain of both string is <math>\{1\}</math>, and the alphabet corresponding to index 0 is <math>x</math> for both string.
	Hence, <math>u = v = \{(1, x)\}</math>.		Hence, <math>u = v = \{(1, x)\}</math>.

2025-09-23T05:09:11Z

Ex. 6

← 이전 판		2025년 9월 23일 (화) 05:09 판
65번째 줄:		65번째 줄:

	(b)<br>		(b)<br>
			<math>(uv)w
	<math></math>		= (\{(i,x)\|(i,x)\in u\} \cup
			\{(</math><math>\|</math><math>u</math><math>\|</math><math>+j,y)\|(j,y)\in v\}) \cup
			\{(</math><math>\|</math><math>uv</math><math>\|</math><math>+k,z)\|(k,z)\in v\}</math><br><math>
			= uvw
			= \{(i,x)\|(i,x)\in u\} \cup
			\{(</math><math>\|</math><math>u</math><math>\|</math><math>+j,y)\|(j,y)\in v\} \cup
			\{(</math><math>\|</math><math>uv</math><math>\|</math><math>+k,z)\|(k,z)\in v\}</math><br><math>
			= u(vw)
			= \{(i,x)\|(i,x)\in u\} \cup (
			\{(</math><math>\|</math><math>u</math><math>\|</math><math>+j,y)\|(j,y)\in v\} \cup
			\{(</math><math>\|</math><math>uv</math><math>\|</math><math>+k,z)\|(k,z)\in v\})</math>
	<math></math>		<math></math>
	<math></math>		<math></math>

2025-09-23T04:53:10Z

Ex. 4

← 이전 판		2025년 9월 23일 (화) 04:53 판
41번째 줄:		41번째 줄:

	==Ex. 4==		==Ex. 4==
	~~</math><br>~~
	Existence)<br>		Existence)<br>
	<math>\|w\| = 1</math> means that the domain of <math>w</math> is <math>\{1\}</math>.<br>		<math>\|w\| = 1</math> means that the domain of <math>w</math> is <math>\{1\}</math>.<br>

2025-09-23T04:51:13Z

Ex. 5

@@ 57번째 줄: / 57번째 줄: @@
 ==Ex. 5==
 <math></math>
 <math></math>
 <math></math>
 <math></math>

2025-09-23T04:22:59Z

Ex. 4

← 이전 판		2025년 9월 23일 (화) 04:22 판
41번째 줄:		41번째 줄:

	==Ex. 4==		==Ex. 4==
	<math></math>		</math><br>
	<math></math>		Existence)<br>
	<math></math>		<math>\|w\| = 1</math> means that the domain of <math>w</math> is <math>\{1\}</math>.<br>
	<math></math>		So, let relation <math>w</math> be <math>w = \{(1, x)\} \subseteq \mathbb{N} \times \Sigma </math>, s.t. <math>x \in \Sigma</math>.<br>
	<math></math>		First, <math>w</math> is a partial function, since for the input 1, there is only one output value.<br>
	<math></math>		Second, the domain of <math>w</math> is <math>\{1\}</math>, which is the finite initial segment of <math>\mathbb{N}</math>.
	<math></math>		Hence, there is <math>w</math> which satisfy given proposition.
	<math></math>
	<math></math>		Uniquness)<br>
	<math></math>		Suppose <math>u,\,\,v</math> are string, and <math>(\|u\|=\|v\|=1)\land (u</math>@<math>0 = v</math>@<math>0 = x, s.t. x \in \Sigma)</math>
	<math></math>		Then, the domain of both string is <math>\{1\}</math>, and the alphabet corresponding to index 0 is <math>x</math> for both string.
	<math></math>		Hence, <math>u = v = \{(1, x)\}</math>.
	<math></math>
	<math></math>		Thus, for every <math>x \in \Sigma</math>, there exists a unique string <math>w</math> s.t. <math>\|w\|=1</math> and <math>w</math>@<math>0=x</math>.
	<math></math>
	<math></math>		==Ex. 5==
	<math></math>
	<math></math>
	<math></math>		<math></math>
	<math></math>		<math></math>
	<math></math>		<math></math>
	<math></math>		<math></math>

2025-09-23T02:57:57Z

Ex. 3

@@ 28번째 줄: / 28번째 줄: @@
 ==Ex. 3==
 <math></math>
 <math></math>

2025-09-23T02:35:03Z

Ex. 2

← 이전 판		2025년 9월 23일 (화) 02:35 판
14번째 줄:		14번째 줄:

	==Ex. 2==		==Ex. 2==
			Existence)<br>
			By definition, the domain of <math>w</math> is the finite initial segment of <math>\mathbb{N}</math>.<br>
			If the domain is <math>\empty</math>, then there is no position <math>i \in \mathbb{N}</math> where the <math>w</math> is defined.<br>
			In this situation, the length of the string <math>w</math> is 0.<br>
			Hence, there is a string <math>w</math> s.t. <math>\|w\| = 0</math>.

	<math></math>		Uniquness)<br>
	<math></math>		Suppose that there is two sets <math>u,\,\, v</math> s.t. <math>\|u\| = 0,\,\, \|v\| = 0</math>.<br>
	<math></math>		Then <math>u,\,\, v</math> are not defined at any index, so both are empty relation, subset of <math>\mathbb{N} \times \Sigma</math>.<br>
			So <math>u,\,\, v</math> are both empty, which means <math>u = v</math>.

			Hence, there is the a unique empty string <math>w</math> s.t. <math>\|w\| = 0</math>, and we can define <math>\epsilon</math> as the empty string <math>w</math>

			==Ex. 3==
	<math></math>		<math></math>
	<math></math>		<math></math>

2025-09-23T02:01:37Z

Ex. 1

← 이전 판		2025년 9월 23일 (화) 02:01 판
12번째 줄:		12번째 줄:

	Hence, there exists a unique natural number <math>n</math> with the desired property.		Hence, there exists a unique natural number <math>n</math> with the desired property.

			==Ex. 2==

	<math></math>		<math></math>
	<math></math>		<math></math>