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$$
R\subseteq A^2 \text{ anti-symmetric }\stackrel{\text{def}}{\Longleftrightarrow} \forall x, y \in A. x\ne y \Rightarrow (x, y)\notin R \vee (y, x) \notin R\\
\Longleftrightarrow (x, y) \in R \wedge (y, x) \in R \Rightarrow x=y
$$
$$
\forall x\ \forall y ((x, y \in A) \wedge (x, y) \in R \wedge (y, x) \in R \to x = y), \text{ 则称 } R \text{ 在 } A \text{ 上是反对称的 }
$$
Transitivity/传递性
$$
R\subseteq A^2 \text{ is transitive }\stackrel{\text{def}}{\Longleftrightarrow}\forall x, y, z\in A. (x, y) \in R \wedge (y, z)\in R \Rightarrow (x, z) \in R
$$
$$
R, S \subseteq A^2\\
R {}_{;} S\stackrel{\text{def}}{=}\left{(x, z)\in A^2 | \exist y \in A. (x, y)\in R \wedge (y, z)\in S\right}
$$
$_;$ means composed with
$$
\forall x\ \forall y\ \forall z\ ((x, y, z \in A) \wedge (y, z) \in R \wedge (y, z) \in R \to (x, z) \in R) \text{ 则称 } R \text{ 在 } A \text{ 上是传递的 }
$$
Transitivity-closure/传递性闭包
$$
\text{trans-closure}(R)\stackrel{\text{def}}{=}R\cup R {}{;} R \cup R {}{;} R {}{;} R\cup R {}{;} R{}{;}R{}{;}R\cup {...}\
= \text{"all R-paths"}
$$
$$
R \subseteq A^2 \text{ is an order relation if it is relexive, anti-symmetric and transitive.}
$$
$$
偏序 = 自反 + 反对称 + 传递
$$
$$
a. b \in \N, a\ |\ b\\
\text{1. }\ a\ |\ a, a = 1 \times a\\
\text{2. }\ a\ |\ b \wedge b\ |\ a\Rightarrow a=b\\
\text{3. }\ a\ |\ b \wedge b \ |\ c \Rightarrow a\ |\ c
$$
Hasse Diagram
Equivalence Relations/等价关系
$$
R \subseteq A^2 \text{ is an order reflation if it is reflexive, symmetric and transitive.}
$$
$$
等价 = 自反 + 对称 + 传递
$$
Only difference is symmetric.
Equivalence is like quality disregarding some details.
Equivalence class/等价类
$$
\approx\text{ equivalence relationship on }A\qquad a\in A\
[a]{\approx}\stackrel{\text{def}}{=}\left{x\in A| x \approx a\right}\subseteq A\
[a]{\approx} \text{ will never be empty, at least } a \in [a]_{\approx}
$$
定义: 设 $R$ 为非空集合 $A$ 上的等价关系,$\forall x \in A$,令
$$
[x]_R=\left{y \mid y \in A \wedge xRy \right}
$$
如果 $R$ 为非空集合 $A$ 上的等价关系,则:
$\forall x \in A, [x] \text{ 是 } A \text{ 的非空子集}$
$\forall x, y \in A, \text{如果 } xRy \text{ 则 } [x]=[y]$
$\forall x, y \in A, \text{如果 } \neg xRy \text{ 则 } [x]\text{与} [y] \text{ 不交}$
$\cup\left{[x] \mid x \in A\right}=A$
Theorem:$\approx$ equ. relation on A, $a, b \in A$, then either $[a]{\approx}\cap[b]{\approx}=\empty$ or $[a]{\approx}=[b]{\approx}$