hcju/proofwikiqa huggingface dataset:
By the definition of finite set, there is an $n \in \N$ such that $S \sim \N_n$. Suppose $m \in \N$ and $S \sim \N_m$. It follows from Set Equivalence is Equivalence Relation that $\N_n \sim \N_m$. Thus by Equality of Natural Numbers, $n = m$.   Therefore the cardinality of a finite set is well-defined. {{qed}}
We must show that the commutative and unitary ring axioms are satisfied: {{:Definition:Commutative and Unitary Ring Axioms}} === Proof of the additive axioms === '''A1:''' This is shown by Polynomials Closed under Addition. '''A2-A5:''' According to the formal definition, a polynomial is a map from the free commutative monoid to $R$. Now observe that addition of polynomial forms is induced by addition in $R$. Therefore: :'''A2''' is shown by Structure Induced by Associative Operation is Associative :'''A3''' is shown by Induced Structure Identity :'''A4''' is shown by Induced Structure Inverse :'''A5''' is shown by Structure Induced by Commutative Operation is Commutative === Proof of the multiplicative axioms === '''M1:''' This is shown by Polynomials Closed under Ring Product. Multiplication of polynomial forms is not induced by multiplication in $R$, so we must show the multiplicative axioms by hand. '''M2:''' This is shown by Multiplication of Polynomials is Associative. '''M3:''' This is shown by Polynomials Contain Multiplicative Identity. '''M4:''' This is shown by Multiplication of Polynomials is Commutative. '''D:''' This is shown by Multiplication of Polynomials Distributes over Addition. Therefore, all of the axioms of a commutative ring with unity are satisfied. {{qed}} Category:Polynomial Rings ld9w9xriw4vtobh8neg5cpkgnx0kh3k
By Subsemigroup Closure Test we need only show that: :For all $x, y \in I$: $x \circ y \in I$. That is: :$\paren {x \circ y} \circ \paren {x \circ y} = x \circ y$ We reason as follows: {{begin-eqn}} {{eqn | l = \paren {x \circ y} \circ \paren {x \circ y}       | r = \paren {x \circ y} \circ \paren {y \circ x}       | c = $\circ$ is commutative }} {{eqn | r = \paren {x \circ \paren {y \circ x} } \circ y       | c = $\circ$ is associative }} {{eqn | r = \paren {x \circ \paren {x \circ y} } \circ y       | c = $\circ$ is commutative }} {{eqn | r = \paren {x \circ x} \circ \paren {y \circ y}       | c = $\circ$ is associative }} {{eqn | r = x \circ y       | c = $x$ and $y$ are idempotent }} {{end-eqn}} Hence the result. {{qed}}
By Subsemigroup Closure Test we need only show that: : For all $x, y \in I$: $x \circ y \in I$. As $x, y \in I$, they are idempotent. We have that $\circ$ is commutative. Thus, by definition, $x$ and $y$ commute. From Product of Commuting Idempotent Elements is Idempotent, $\left({x \circ y}\right)$ is idempotent. That is: :$x \circ y \in I$ {{qed}}
{{begin-eqn}} {{eqn | l = \paren {a \circ b} \circ \paren {a \circ b}       | r = \paren {a \circ \paren {b \circ a} } \circ b       | c = $\circ$ is associative by definition of semigroup }} {{eqn | r = \paren {a \circ \paren {a \circ b} } \circ b       | c = $a$ and $b$ commute }} {{eqn | r = \paren {a \circ a} \circ \paren {b \circ b}       | c = $\circ$ is associative }} {{eqn | r = a \circ b       | c = $a$ and $b$ are idempotent by the premise }} {{end-eqn}} Thus $a \circ b$ is idempotent. {{qed}} Category:Semigroups Category:Idempotence 5hcn0hrqp0efy8tk9dzdvy04zje69lm
Let $f, g \in S^S$. As the domain of $g$ and codomain of $f$ are the same, the composition $f \circ g$ is defined. By the definition of composition, $f \circ g$ is a mapping from the domain of $g$ to the codomain of $f$. Thus $f \circ g: S \to S$, so $f \circ g \in S^S$. Since this holds for all $f, g \in S^S$,  $\struct {S^S, \circ}$ is closed. By Composition of Mappings is Associative, $\circ$ is associative. Since $\struct {S^S, \circ}$ is closed and $\circ$ is associative: :$\struct {S^S, \circ}$ is a semigroup. {{qed}}
We have that Integers form Subdomain of Rationals. We have that Rational Numbers form Subfield of Real Numbers. Hence the result, from the definition of subdomain. {{qed}} Category:Integers Category:Real Numbers k7dkktib40yaper07azh5367u9i60v8
By the definition of an identity element: :$\forall x \in S: e \circ x = x$ Thus in particular: :$e \circ e = e$ Therefore $e$ is idempotent under $\circ$. {{qed}}
Let $x \in T$. By the definition of subset, $x \in S$. Since $e$ is an identity of $\struct {S, \circ}$: :$e \circ x = x \circ e = x$ Since this holds for all $x \in T$, $e$ is an identity of $\struct {T, \circ}$. {{qed}} Category:Identity Elements n2ie5wgzy61jxufp2t0pdsatrdnts8o
By Idempotent Elements form Subsemigroup of Commutative Semigroup, $\left({I, \circ}\right)$ is a subsemigroup of $\left({S, \circ}\right)$. By Identity Element is Idempotent, $e \in I$. By Identity of Algebraic Structure is Preserved in Substructure, $e$ is an identity of $(I, \circ)$. Since $\left({T, \circ}\right)$ is a semigroup and has an identity, $\left({T, \circ}\right)$ is a monoid. Since $T \subseteq S$, $\left({T, \circ}\right)$ is a submonoid of $\left({S, \circ}\right)$ with identity $e$. {{qed}} Category:Monoids hf714gnpsngl4li3spzhqmiu6kvh0p7
Let $x \in \map {\phi^{-1} } {\map \phi p^{\succ'} }$. By the definition of inverse image: :$\map \phi x \in \map \phi p^{\succ'}$ By the definition of strict upper closure: :$\map \phi p \prec' \map \phi x$ Since $\phi$ is an order embedding: :$p \prec x$ Thus by the definition of strict upper closure: :$x \in p^\succ$ and so: :$\map {\phi^{-1} } {\map \phi p^{\succ'} } \subseteq p^\succ$ Let $x \in p^\succ$. By the definition of strict upper closure: :$p \prec x$ Since $\phi$ is an order embedding: :$\map \phi p \prec' \map \phi x$ Thus by the definition of strict upper closure: :$\map \phi x \in \map \phi p^{\succ'}$ Thus by the definition of inverse image: :$x \in \map {\phi^{-1} } {\map \phi p^{\succ'} }$ and so: :$p^\succ \subseteq \map {\phi^{-1} } {\map \phi p^{\succ'} }$ Thus by definition of set equality: :$\map {\phi^{-1} } {\map \phi p^{\succ'} } = p^\succ$ {{qed}} Category:Order Embeddings Category:Upper Closures io0oltfff5r2qj4nelhv9cohyuvf1sj
Let $\struct {S, \preceq_1}$ and $\struct {T, \preceq_2}$ be ordered sets. Let $\phi: S \to T$ be a mapping. === Definition 1 implies Definition 2 === Let $\phi$ be an order embedding by definition 1. Then by definition: :$\forall x, y \in S: x \preceq_1 y \iff \map \phi x \preceq_2 \map \phi y$ It follows from Order Embedding is Injection that $\phi$ is an injection. Thus $\phi$ is an order embedding by definition 2. {{qed|lemma}} === Definition 2 implies Definition 1 === Let $\phi$ be an order embedding by definition 2. Then condition $(2)$ of that definition directly states: :$\forall x, y \in S: x \preceq_1 y \iff \map \phi x \preceq_2 \map \phi y$ Thus $\phi$ is an order embedding by definition 1. {{qed|lemma}} === Definition 1 implies Definition 3 === {{:Equivalence of Definitions of Order Embedding/Definition 1 implies Definition 3}}{{qed|lemma}} === Definition 3 implies Definition 1 === {{:Equivalence of Definitions of Order Embedding/Definition 3 implies Definition 1}}{{qed|lemma}} === Definition 2 implies Definition 4 === Let $\phi$ be an order embedding by definition 2. Then by definition: :$\phi$ is an injection :$\forall x, y \in S: x \preceq_1 y \iff \map \phi x \preceq_2 \map \phi y$ Let $T' = \Img S$ be the image of $S$ under $\phi$. From Injection to Image is Bijection: :$\phi {\restriction_{S \times T'} }: S \to T'$ is a bijection. It follow by definition that $\phi {\restriction_{S \times T'} }: S \to T'$ is an order isomorphism. Thus $\phi$ is an order embedding by definition 4. {{qed|lemma}} === Definition 4 implies Definition 1 === Let $\phi$ be an order embedding by definition 4. Then by definition: :the restriction of $\phi$ to $S \times T'$ is an order isomorphism between $\struct {S, \preceq_1}$ and $\struct {T', \preceq_2 \restriction_{T' \times T'} }$. By definition of order isomorphism: :$\forall x, y \in S: x \preceq_1 y \iff \map {\phi {\restriction_{S \times T'} } } x \preceq_2 \map {\phi {\restriction_{S \times T'} } } y$ But by definition of restriction: :$\forall z \in S: \map {\phi {\restriction_{S \times T'} } } z = \map \phi z$ Thus: :$\forall x, y \in S: x \preceq_1 y \iff \map \phi x \preceq_2 \map \phi y$ and so $\phi$ is an order embedding by definition 1. {{qed}}
Let $\phi$ be an order embedding by definition 1. Then by definition: :$\forall x, y \in S: x \preceq_1 y \iff \map \phi x \preceq_2 \map \phi y$ $\phi$ is injective by Order Embedding is Injection. It remains to be shown that: : $x \prec_1 y \iff \map \phi x \prec_2 \map \phi y$ Suppose first that $x \prec_1 y$. Then $x \preceq_1 y$ and $x \ne y$. Thus by the premise: :$\map \phi x \preceq_2 \map \phi y$ Since $\phi$ is injective: :$\map \phi x \ne \map \phi y$ Therefore: :$\map \phi x \prec_2 \map \phi y$ Suppose instead that $\map \phi x \prec_2 \map \phi y$ Then: :$\map \phi x \preceq_2 \map \phi y$ and: :$\map \phi x \ne \map \phi y$ By the premise: :$x \preceq_1 y$ By the substitutive property of equality: :$x \ne y$ Thus: :$x \prec_1 y$ Thus $\phi$ is an order embedding by definition 3.
Let $\phi$ be an order embedding by definition 3. Then by definition: :$(1): \quad \phi$ is injective :$(2): \quad \forall x, y, \in S: x \prec_1 y \iff \map \phi x \prec_2 \map \phi y$ Let $x \preceq_1 y$. Then $x \prec_1 y$ or $x = y$. If $x \prec_1 y$, then by hypothesis: :$\map \phi x \prec_2 \map \phi y$ Thus: :$\map \phi x \preceq_2 \map \phi y$ If $x = y$, then: :$\map \phi x = \map \phi y$ Thus: :$\map \phi x \preceq_2 \map \phi y$ Thus it has been shown that: :$x \preceq_1 y \implies \map \phi x \preceq_2 \map \phi y$ {{qed|lemma}} Let $\map \phi x \preceq_2 \map \phi y$. Then: :$\map \phi x \prec_2 \map \phi y$ or: :$\map \phi x = \map \phi y$ Suppose $\map \phi x \prec_2 \map \phi y$. Then by hypothesis: :$x \prec_1 y$ and so: :$x \preceq_1 y$ Suppose $\map \phi x = \map \phi y$. Then since $\phi$ is injective: :$x = y$ and so: :$x \preceq_1 y$ Thus in both cases: :$x \preceq_1 y$ and so: :$\map \phi x \preceq_2 \map \phi y \implies x \preceq_1 y$ {{qed|lemma}} Hence the result: :$x \preceq_1 y \iff \map \phi x \preceq_2 \map \phi y$ and so $\phi$ is an order embedding by definition 1.
Let $p \in S$. {{begin-eqn}} {{eqn |o =       |r = x \in \mathcal R_S (p) }} {{eqn |lo= \iff |o=       |r = p \mathrel{\mathcal R_S} x       |c = Definition of the image of $p$ under $\mathcal R_S$ }} {{eqn |lo= \iff |o=       |r = \phi(p) \mathrel{\mathcal R_T} \phi(x)       |c = Premise }} {{eqn |lo= \iff |o=       |r = \phi(x) \in \mathcal R_T \left({ \phi(p) }\right)       |c = Definition of the image of $\phi(x)$ under $\mathcal R_T$ }} {{eqn |lo = \iff |o=       |r = x \in \phi^{-1}\left({\mathcal R_T \left({ \phi(p) }\right) }\right)       |c = Definition of inverse image }} {{end-eqn}} Thus by the Axiom of Extension: : $\mathcal R_S (p) = \phi^{-1}\left({\mathcal R_T \left({ \phi(p) }\right) }\right)$ {{qed}} Category:Relation Theory p01vhur6jk7qpkalxpz316iw3q5m1rh
Put $\gamma_k = \gamma {\restriction_{ \left[{a_{k-1} \,.\,.\, a_k}\right] } } : \left[{a_{k-1} \,.\,.\, a_k}\right] \to \C$. By definition, it follows that there exists a directed smooth curve $C_k$ with parameterization $\gamma_k$. For all $k \in \left\{ {1, \ldots, n-1}\right\}$, we have $\gamma_k \left({a_k}\right) = \gamma_{k + 1} \left({a_k}\right)$. Define the contour $C$ as the concatenation $C_1 \cup C_2 \cup \ldots \cup C_n$. Then, it follows by definition that $\gamma$ is a possible parameterization of $C$. {{qed|lemma}} Suppose that $\gamma$ is a closed path. Then $\gamma \left({a}\right) = \gamma_1 \left({a_0}\right) = \gamma_n \left({a_n}\right)$, so $C_1$ has start point $\gamma \left({a}\right)$, and $C_n$ has end point $\gamma \left({a}\right)$. By definition, it follows that $C$ is a closed contour. {{qed|lemma}} Suppose that $\gamma$ is a Jordan arc. Let $k_1, k_2 \in \left\{ {1, \ldots, n}\right\}$, and $t_1 \in \left[{a_{k_1 - 1} \,.\,.\, a_{k_1} }\right), t_2 \in \left[{a_{k_2 - 1} \,.\,.\, a_{k_2} }\right)$. Then $\gamma \left({t_1}\right) \ne \gamma \left({t_2}\right)$ by the definition of Jordan arc, so $\gamma_{k_1} \left({t_1}\right) \ne \gamma_{k_2} \left({t_2}\right)$. Let instead $k \in \left\{ {1, \ldots, n}\right\}$ and $t \in \left[{a_{k-1} \,.\,.\, a_k}\right)$ with $t \ne a_1$. Then $\gamma \left({t}\right) \ne \gamma \left({a_n}\right)$ by the definition of Jordan arc, so $\gamma_k \left({t}\right) \ne \gamma_n \left({a_n}\right)$. By definition, it follows that $C$ is a simple contour. {{qed|lemma}} Suppose that $\gamma$ is a Jordan curve. As a Jordan curve by definition is both a Jordan arc and a closed path, it follows from what is shown above that $C$ is a simple closed contour. {{qed}} Category:Contour Integration oy9jy2rtbhg9g4e65kfchpvs7rx5u52
Let $P \left(X\right) = a_0 + a_1 X + \cdots + a_n X^n \in R \left[{X}\right]$. Suppose first that $\phi \left({a_i}\right) = 0$ for $i = 0, \ldots, n$. We have by definition that: :$\bar\phi \left({a_0 + a_1 X + \cdots + a_n X^n}\right) = \phi \left({a_0}\right) + \phi \left({a_1}\right) X + \cdots + \phi \left({a_n}\right) X^n = 0$ That is to say, $P \left({X}\right) \in \ker \bar\phi$. Conversely, suppose that $P \left({X}\right) \in \ker \bar\phi$. That is, $\bar\phi \left({P \left({X}\right)}\right) = \phi \left({a_0}\right) + \phi \left({a_1}\right) X + \cdots + \phi \left({a_n}\right) X^n$ is the null polynomial. This by definition means that for $i = 0, \ldots, n$ we have $\phi \left({a_i}\right) = 0$. Hence, $P \left({X}\right) \in \left({\ker \phi}\right) \left[{X}\right]$. This concludes the proof. {{qed}} Category:Polynomial Theory gn2m9numjm5k02aoqfmo9vc114v0qsk
Let $n \in \N$ be the number of sides of $P$. Denote the vertices of $P$ as $A_1, \ldots, A_n$. From Complex Plane is Metric Space, it follows that $\C$ is homeomorphic to $\R^2$. Then, we can consider $\partial P$ as a subset of $\R^2$. From Boundary of Polygon is Jordan Curve, it follows that there exists a Jordan curve $\gamma: \left[{0 \,.\,.\, 1}\right] \to \R^2$ such that $\operatorname{Im} \left({\gamma}\right) = \partial P$. The same theorem shows that $\gamma$ is a concatenation of $n$ line segments, parameterized as: :$\gamma_k \left({t}\right) = \left({1 - t}\right) A_k + tA_{k+1}$ where $k \in \left\{ {1, \ldots, n}\right\}$. We identify $A_{n + 1}$ with $A_1$. Then, $\gamma_k: \left[{a_{k - 1} \,.\,.\, a_k}\right] \to \R^2$, where $a_{k - 1}, a_k \in \left[{0 \,.\,.\, 1}\right]$. As $\gamma$ is a concatenation of $\gamma_1, \ldots, \gamma_n$, it follows that $\left\{ {a_0, a_1, \ldots, a_n}\right\}$ is a subdivision of $\left[{0 \,.\,.\, 1}\right]$. We have: :$\dfrac \d {\d t} \gamma_k \left({t}\right) = A_{k + 1} - A_k \ne \mathbf 0$ as $A_k$ and $A_{k + 1}$ are two different vertices. As $\C$ is homeomorphic to $\R^2$, we can consider $\gamma$ as a continuous complex function $\gamma: \left[{0 \,.\,.\, 1}\right] \to \C$. Then $\gamma_k$ is complex-differentiable for all values of $t \in \left({a_{k - 1} \,.\,.\, a_k}\right)$ with its derivative $\gamma_k'$ defined by: :$\gamma_k' \left({t}\right) = x \left({\dfrac \d {\d t} \gamma_k \left({t}\right) }\right) + i y \left({\dfrac \d {\d t} \gamma_k \left({t}\right) }\right)$ where: :$x \left({\dfrac \d {\d t} \gamma_k \left({t}\right) }\right)$ is the real part of $\dfrac \d {\d t} \gamma_k \left({t}\right)$ :$y \left({\dfrac \d {\d t} \gamma_k \left({t}\right) }\right)$ is the imaginary part of $\dfrac \d {\d t} \gamma_k \left({t}\right)$ As shown above, $\gamma_k' \left({t}\right) \ne 0$. By definition, it follows that $\gamma_k = \gamma {\restriction_{\left[{a_{k-1} \,.\,.\, a_k}\right] } }$ is a smooth path. From Path as Parameterization of Contour, it follows that there exists a simple closed contour $C$ with $\operatorname{Im} \left({C}\right) = \partial P$. {{qed}} Category:Contour Integration s08ifxybz9kg9dn79jt3a8qti4hr4mb
To show that $\sim$ is an equivalence relation, we must show that it is reflexive, transitive, and symmetric. By the definition of preordering, $\precsim$ is transitive and reflexive. === Transitive === Let $p, q, r \in S$. Suppose that $p \sim q$ and $q \sim r$. Then $p \precsim q$, $q \precsim r$, $r \precsim q$, and $q \precsim p$. Since $\precsim$ is transitive: :$p \precsim r$ and $r \precsim p$. Thus by the definition of $\sim$, $p \sim r$. Since this holds for all $p$, $q$, and $r$, $\sim$ is transitive. {{qed|lemma}} === Reflexive === Let $p \in S$. Since $\precsim$ is reflexive: : $p \precsim p$ Thus by the definition of $\sim$: : $p \sim p$ As this holds for all $p$, $\sim$ is reflexive. {{qed|lemma}} === Symmetric === Let $p, q \in S$ with $p \sim q$. Then $p \precsim q$ and $q \precsim p$. Thus $q \sim p$. Since this holds for all $p$ and $q$, $\sim$ is symmetric. {{qed}}
From Rational Polynomial is Content Times Primitive Polynomial, let $\map f X$ and $\map g X$ be expressed as: :$\map f X = \cont f \cdot \map {f^*} X$ :$\map g X = \cont g \cdot \map {g^*} X$ where: :$\cont f, \cont g$ are the content of $f$ and $g$ respectively :$f^*, g^*$ are primitive. We have, by applications of Rational Polynomial is Content Times Primitive Polynomial: :$\cont h \cdot \map {h^*} X = \cont f \cont g \cdot \map {f^*} X \, \map {g^*} X$ By Gauss's Lemma on Primitive Rational Polynomials we have that $\map {f^*} X \, \map {g^*} X$ is primitive. As $\cont f > 0$ and $\cont g > 0$, then so is $\cont f \cont g > 0$. By the uniqueness clause in Rational Polynomial is Content Times Primitive Polynomial, the result follows. {{qed}}
{{explain|Needs to be made explicit as to exactly what is being assumed and what follows as a consequence. As it stands, the consequence of 1 implies 2, for example, still needs to be completed so as to explain exactly how that show $h$ is irreducible -- it relies upon the implicit understanding of what irreducible means.}} === 1 implies 2 === Suppose first that $h$ is not irreducible in $\Z \sqbrk X$. Let $\displaystyle h = a_0 + a_1 X + \cdots + a_n X^n$. If $\deg h = 0$, then the content of $h$ is: :$\cont h = \gcd \set {a_0} = \size {a_0}$ Since $h$ is primitive by hypothesis, we have $h = \pm 1$. Now by Units of Ring of Polynomial Forms over Field, the units of $\Q \sqbrk X$ are the units of $\Q$. Thus $h$ is a unit of $\Q \sqbrk X$. Therefore $h$ is irreducible. If $\deg h \ge 1$, then by Units of Ring of Polynomial Forms over Integral Domain, the units of $\Z \sqbrk X$ are the units of $\Z$. Therefore $h$ is not a unit of $\Z \sqbrk X$. Thus since $h$ is reducible, there is a non-trivial factorization $h = f g$ in $\Z \sqbrk X$, with $f$ and $g$ both not units. If $\deg f = 0$, that is, $f \in \Z$, then $f$ divides each coefficient of $h$. Since $h$ is primitive, this means that $f$ divides $\cont h = 1$. But the divisors of $1$ are $\pm 1$, so $f = \pm 1$. But then $f$ is a unit in $\Z \sqbrk X$, a contradiction. Therefore $\deg f \ge 1$, so $f$ is a non-unit in $\Q \sqbrk X$. Similarly, $g$ is a non-unit in $\Q \sqbrk X$. Therefore $h = fg$ is a non-trivial factorization in $\Q \sqbrk X$. === 2 implies 1 === {{tidy|ok in terms of details, but presentation needs touching up}} Suppose now that $h$ is not irreducible in $\Q \sqbrk X$. That is, $h$ has a non-trivial factorization in $\Q \sqbrk X$. Since the units of $\Q \sqbrk X$ are the units of $\Q$, this means that $h = f g$, with $f$ and $g$ both of positive degree. Let $c_f$ and $c_g$ be the contents of $f$ and $g$ respectively. Define $\tilde f = c_f^{-1} f$ and $\tilde g = c_g^{-1} g$.  By Content of Scalar Multiple, it follows that $\cont {\tilde f} = \cont {\tilde g} = 1$. Moreover by Polynomial has Integer Coefficients iff Content is Integer we have $\tilde f, \tilde g \in \Z \sqbrk X$. Now we have: :$\tilde f \tilde g = \dfrac {f g} {c_f c_g} = \dfrac h {c_f c_g}$ Taking the content, and using Content of Scalar Multiple we have: :$\cont {\tilde f \tilde g} = \dfrac 1 {c_f c_g} \cont h$ By Gauss's Lemma on Primitive Rational Polynomials we know that $\cont {\tilde f \tilde g} = 1$. Moreover, by Irreducible Integer Polynomial is Primitive, $\cont h = 1$. Therefore we must have $c_f c_g = 1$. Thus we have a factorization in $\Z \sqbrk X$: :$\tilde f \tilde g = h$ This is a non-trivial factorization of $h$, as both $f$ and $g$ have positive degree. Thus $h$ is not irreducible in $\Z \sqbrk X$. {{Qed}} {{Namedfor|Carl Friedrich Gauss|cat = Gauss}} Category:Gauss's Lemma (Polynomial Theory) kbf6czd3gif8u3rldfr8pstovy16j0i
By the definition of equivalence relation, $\sim$ is transitive, reflexive, and symmetric. By the definition of Definition:preordering, $\precsim$ is transitive and reflexive. To show that $\preceq$ is an ordering, we must show that it is transitive, reflexive, and antisymmetric. === Transitive === Let $P, Q, R \in S / {\sim}$. Suppose that $P \preceq Q$ and $Q \preceq R$. Then for some $p \in P$, $q_1, q2 \in Q$, and $r \in R$: :$p \precsim q_1$ and $q_2 \precsim r$. By the definition of quotient set, $q_1 \sim q_2$. By the definition of $\sim$: :$q_1 \precsim q_2$ Since $p \precsim q_1$, $q_1 \precsim q_2$, $q_2 \precsim r$, and $\precsim$ is transitive, Transitive Chaining shows that: :$p \precsim r$ Thus by the definition of $\preceq$: :$P \preceq R$. Since this holds for all such $P$, $Q$, and $R$, $\preceq$ is transitive. {{qed|lemma}} === Reflexive === Let $P \in S / {\sim}$. By the definition of quotient set, $P$ is non-empty. Thus there exists a $p \in P$. Since $\precsim$ is a preordering, it is reflexive, so $p \precsim p$. By definition of the equivalence relation, we have that $q\sim p$ for any other $q\in P$ This gives that: :$q \sim p \precsim p$ which is equivalent to $q \precsim p \precsim p$ by definition of the equivalence relation. It also gives that $p \precsim p \precsim q$ by similar reasoning. Both by transitivity gives $p\precsim q$ and $q\precsim p$. This gives us that $P \preceq P$ As this holds for all $P \in S / {\sim}$, $\preceq$ is reflexive. {{qed|lemma}} === Antisymmetric === Let $P, Q \in S / {\sim}$ such that: :$P \preceq Q$ :$Q \preceq P$ By the definition of $\preceq$, there are elements $p_1, p_2 \in P$ and $q_1, q_2 \in Q$ such that: :$(1)\quad p_1 \precsim q_1$ :$(2)\quad q_2 \precsim p_2$ Let $p \in P$. Then by the definition of quotient set: :$p \sim p_1$ :$p_2 \sim p$ By the definition of $\sim$: :$p \precsim p_1$ :$p_2 \precsim p$ Thus by $(1)$ and $(2)$ and the fact that $\precsim$ is transitive: :$p \precsim q_1$ :$q_2 \precsim p$ By the definition of quotient set: :$q_1 \sim q_2$ Thus by the definition of $\sim$: :$q_1 \precsim q_2$ Since $\precsim$ is transitive: :$p \precsim q_2$ We already know that: :$q_2 \precsim p$ Thus $p \sim q_2$. By the definition of quotient set: :$p \in Q$ The same argument shows that each element of $Q$ is also in $P$. Thus by the Axiom of Extension: :$P = Q$ As this holds for all such $P, Q \in S / {\sim}$, $\preceq$ is antisymmetric. {{qed|lemma}} === Relation between Sets implies all their Elements are Related === Let $P, Q \in S / {\sim}$ with $P \preceq Q$. Then by the definition of $\preceq$, there are $p \in P$ and $q \in Q$ such that $p \precsim q$. Let $p' \in P$ and $q' \in Q$. By the definition of quotient set: :$p' \sim p$ :$q \sim q'$ Thus by the definition of $\sim$: :$p' \precsim p$ :$q \precsim q'$ Since $p \precsim q$ and $\precsim$ is transitive: :$p' \precsim q'$ We have shown that: :$\forall P, Q \in S / {\sim}: \paren {P \preceq Q} \land \paren {p \in P} \land \paren {q \in Q} \implies p \precsim q$ {{qed}}