hcju/proofwikiqa huggingface dataset:
=== Definition 1 implies Definition 2 === Let $\text {SR}$ be a ring of sets which is closed under countable unions. We have: {{begin-axiom}} {{axiom | n = \text {RS} 1_2         | lc= Empty Set:         | m = \O \in \text {SR} }} {{axiom | n = \text {RS} 2_2         | lc= Closure under Set Difference:         | q = \forall A, B \in \text {SR}         | m = A \setminus B \in \text {SR} }} {{end-axiom}} which are exactly $\text {SR} 1$ and $\text {SR} 2$. Then as $\text {SR}$ is closed under countable unions: :$\ds A_1, A_2, \ldots \in \text {SR} \implies \bigcup_{n \mathop = 1}^\infty A_n \in \text {SR}$ and so $\text {SR} 3$ is fulfilled. {{qed|lemma}} === Definition 2 implies Definition 1 === Let $\text {SR}$ be a system of sets such that: {{begin-axiom}} {{axiom | n = \text {SR} 1         | m = \O \in \text {SR} }} {{axiom | n = \text {SR} 2         | q = \forall A, B \in \text {SR}         | m = A \setminus B \in \text {SR} }} {{axiom | n = \text {SR} 3         | q = \forall A_n \in \text {SR}: n = 1, 2, \ldots         | m = \bigcup_{n \mathop = 1}^\infty A_n \in \text {SR} }} {{end-axiom}} As noted above, $\text {SR} 1$ and $\text {SR} 2$ are exactly $\text {RS} 1_2$ and $\text {RS} 2_2$. Let $A, B \in \text {SR}$. Let $A_1 = A, A_2 = B$ and $A_n = \O$ for all $n = 3, 4, \ldots$ Then: :$\ds \forall A_n \in \text {SR}: n = 1, 2, \ldots: \bigcup_{n \mathop = 1}^\infty A_n = A \cup B \in \text {SR}$ Thus criterion $(\text {RS} 3_2)$ is fulfilled. So $\text {SR}$ is a ring of sets which is closed under countable unions. {{qed}} {{ProofWanted|Add equivalence to Definition 3}}
=== Definition 1 implies Definition 3 === Let $\Sigma$ be a system of sets on a set $X$ such that: :$(1): \quad X \in \Sigma$ :$(2): \quad \forall A, B \in \Sigma: \relcomp X A \in \Sigma$ :$(3): \quad \ds \forall A_n \in \Sigma: n = 1, 2, \ldots: \bigcup_{n \mathop = 1}^\infty A_n \in \Sigma$ Let $A, B \in \Sigma$. From the definition: :$\forall A \in \Sigma: A \subseteq X$. Hence from Intersection with Subset is Subset: :$\forall A \in \Sigma: A \cap X = A$ Hence $X$ is the unit of $\Sigma$. So by definition 2 of $\sigma$-ring it follows that $\Sigma$ is a $\sigma$-ring with a unit. Thus $\Sigma$ is a $\sigma$-algebra by definition 3. {{qed|lemma}} === Definition 3 implies Definition 1 === Let $\Sigma$ be a $\sigma$-ring with a unit $X$. By definition, $X \in \Sigma$. From definition 2 of $\sigma$-ring, $\Sigma$ is: :$(1) \quad$ closed under set difference. :$(2) \quad$ closed under countable union From Unit of System of Sets is Unique, we have that: :$\forall A \in \Sigma: A \subseteq X$ from which we have that $X \setminus A = \relcomp X A$. So $\Sigma$ is a $\sigma$-algebra by definition 1. {{qed|lemma}} === Definition 1 implies Definition 2 === Follows directly from the definitions, as a disjoint union is a type of union. {{qed|lemma}} === Definition 2 implies Definition 1 === Let $\Sigma$ be a system of sets on a set $X$ such that: :$(1): \quad X \in \Sigma$ :$(2): \quad \forall A, B \in \Sigma: \relcomp X A \in \Sigma$ :$(3): \quad \ds \forall A_n \in \Sigma: n = 1, 2, \ldots: \bigsqcup_{n \mathop = 1}^\infty A_n \in \Sigma$ Conditions $(1)$ and $(2)$ in definition 2 are identical to that of conditions $(1)$ and $(2)$ in definition 1.  Let $\family {E_n}_{n \mathop \in \N}$ be a countable indexed family of sets in $\Sigma$. By Union of Indexed Family of Sets Equal to Union of Disjoint Sets: :$\ds \bigsqcup_{n \mathop \in \N} F_n = \bigcup_{n \mathop \in \N} E_n$ for an appropriately constructed countable indexed family of disjoint sets in $\Sigma$. By the hypotheses of definition $2$: :$\ds \bigsqcup_{k \mathop \in \N}^\infty F_k$  is measurable. Thus $\displaystyle \bigcup_{n \mathop \in \N} E_n$ is also measurable by $(\text {SA} 2)$ of definition $1$. {{qed|lemma}} {{explain|Either rewrite the aboveby removing references to "measurable", or provide a link to a RIGOROUSLY IDENTICAL definition of "measurable" as is meant in the above.}} === Definition 1 implies Definition 4 === Immediate from the definition of algebra along with the added condition of closure under countable unions. {{qed|lemma}} === Definition 4 implies Definition 1 === By definition $1$ of algebra of sets, an algebra has the properties: {{begin-axiom}} {{axiom | n = \text {AS} 1         | lc= Unit:         | m = X \in \Sigma }} {{axiom | n = \text {AS} 2         | lc= Closure under Union:         | q = \forall A, B \in \Sigma         | m = A \cup B \in \Sigma }} {{axiom | n = \text {AS} 3         | lc= Closure under Complement Relative to $X$:         | q = \forall A \in \Sigma         | m = \relcomp X A \in \Sigma }} {{end-axiom}} Replacing $(\text {AS} 2)$ with closure under countable unions immediately yields the first definition. {{qed}}
{{begin-eqn}} {{eqn | l = \complement \left({\liminf_{n \mathop \to \infty} \ E_n}\right)       | r = \complement \left({\bigcup_{n \mathop = 0}^\infty \bigcap_{i \mathop = n}^\infty E_n}\right)       | c = {{Defof|Limit Inferior of Sequence of Sets|index = 1}} }} {{eqn | r = \bigcap_{n \mathop = 0}^\infty \complement \left({\bigcap_{i \mathop = n}^\infty E_n}\right)       | c = De Morgan's Laws }} {{eqn | r = \bigcap_{n \mathop = 0}^\infty \bigcup_{i \mathop = n}^\infty \complement \left({E_n}\right)       | c = De Morgan's Laws }} {{eqn | r = \limsup_{n \mathop \to \infty} \ \complement \left({E_n}\right)       | c = {{Defof|Limit Superior of Sequence of Sets|index = 1}} }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn|l = \complement \left({\limsup_{n \to \infty} \ E_n}\right)      |r = \complement \left({\bigcap_{n \mathop = 0}^\infty \bigcup_{i \mathop = n}^\infty E_n}\right)      |c = Definition of limit superior }} {{eqn|r = \bigcup_{n \mathop = 0}^\infty \complement \left({\bigcup_{i \mathop = n}^\infty E_n}\right)      |c = De Morgan's Laws }} {{eqn|r = \bigcup_{n \mathop = 0}^\infty \bigcap_{i \mathop = n}^\infty \complement \left({E_n}\right)      |c = De Morgan's Laws }} {{eqn|r = \liminf_{n \to \infty} \ \complement \left({E_n}\right)      |c = Definition of limit inferior }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | ll= \forall n \in \N_{>0}       | l = A_1, A_n \in \mathcal R       | o = \implies       | r = A_1 \setminus A_n \in \mathcal R       | c = Axiom $(SR2)$ for $\sigma$-rings }} {{eqn | o = \implies       | r = \bigcup_{n \mathop = 2}^\infty \left({A_1 \setminus A_n}\right) \in \mathcal R       | c = Axiom $(SR3)$ for $\sigma$-rings }} {{eqn | o = \implies       | r = A_1 \setminus \left({\bigcup_{n \mathop = 2}^\infty \left({A_1 \setminus A_n}\right) }\right) \in \mathcal R       | c = Axiom $(SR2)$ for $\sigma$-rings }} {{end-eqn}} From De Morgan's laws: Difference with Intersection: :$\displaystyle \bigcup_{n \mathop = 2}^\infty \left({A_1 \setminus A_n}\right) = A_1 \setminus \left({\bigcap_{n \mathop = 2}^\infty A_n}\right)$ From Set Difference with Set Difference: {{begin-eqn}} {{eqn | l = A_1 \setminus \left({A_1 \setminus \left({\displaystyle \bigcap_{n \mathop = 2}^\infty A_n}\right) }\right)       | r = A_1 \cap \left({\displaystyle \bigcap_{n \mathop = 2}^\infty A_n}\right) }} {{eqn | r = {\bigcap_{n \mathop = 1}^\infty A_n} }} {{end-eqn}} Combining the previous equalities, it follows that: : $\displaystyle \bigcap_{n \mathop = 1}^\infty A_n \in \mathcal R$ {{qed}}
A plane contains an infinite number of points. Not all these points are collinear. Let $A$, $B$ and $C$ be points in a plane $P$. From Propositions of Incidence: Line in Plane, any two of these points determine a line. Consider the lines $AB$, $AC$ and $BC$, all of which are distinct. Let $X$ be one of the infinite number of points on $BC$ which is not $B$ or $C$. Then $AX$ is a line in $P$ which is distinct from both $AB$ and $AC$. As there is are infinite number of points on $BC$, there are an infinite number of lines incident to $A$ and $BC$. All these lines are in $P$. Hence the result. {{qed}}
Take two distinct lines in plane $1$. From Propositions of Incidence: Plane and Line, they each meet plane $2$ in one point each, say at $A$ and $B$. Thus $A$ and $B$ both lie in both planes. Thus the line defined by $A$ and $B$ lies in both planes. {{qed}} {{Handwaving}}
Let $A$, $B$ and $C$ be the three planes in question. From Two Planes have Line in Common, $A$ and $B$ share a line, $p$ say. From Propositions of Incidence: Plane and Line, $p$ meets $C$ in one point. {{qed}}
:500px Let $\triangle ABC$ and $\triangle A'B'C'$ be in different planes $\pi$ and $\pi'$ respectively. Since $BB'$ and $CC'$ intersect in $O$, it follows that $B$, $B'$, $C$ and $C'$ lie in a plane. Thus $BC$ must meet $B'C'$ in a point $L$. By the same argument, $CA$ meets $C'A'$ in a point $M$ and $AB$ meets $A'B'$ in a point $N$. These points $L, M, N$ are in each of the planes $\pi$ and $\pi'$. By Two Planes have Line in Common they are therefore collinear on the line where $\pi$ and $\pi'$ meet. Now let $\triangle ABC$ and $\triangle A'B'C'$ be in the same plane $\pi$. Let $OPP'$ be any line through $O$ which does not lie in $\pi$. Then since $PP'$ meets $AA'$ in $O$, the four points $P, P', A, A$ are coplanar. Thus $PA$ meets $P'A'$ at a point $A''$. Similarly $PB$ meets $P'B'$ at a point $B''$, and $PC$ meets $P'C'$ at a point $C''$. The lines $BC, B'C'$ and $B''C''$ are the three lines of intersection of the three planes $PBC$, $P'B'C'$ and $\pi$ taken in pairs. So $BC$, $B'C'$ and $B''C''$ meet in a point $L$. Similarly $CA$, $C'A'$ and $C''A''$ meet in a point $M$ and $AB$, $A'B'$ and $A''B''$ meet in a point $N$. The two triangles $\triangle ABC$ and $\triangle A''B''C''$ are in different planes, and $AA''$, $BB''$ and $CC''$ meet in $P$. Thus $L$, $M$ and $N$ are collinear by the first part of this proof. {{qed}}
By definition, $f \cup g$ is a relation whose domain is $X = A \cup B$. Let $\tuple {x, y_1} \in f \cup g$ and $\tuple {x, y_2} \in f \cup g$. At least one of the following must be true: :$(1): \quad \tuple {x, y_1} \in f, \tuple {x, y_2} \in f$ :$(2): \quad \tuple {x, y_1} \in g, \tuple {x, y_2} \in g$ :$(3): \quad \tuple {x, y_1} \in f, \tuple {x, y_2} \in g$ :$(4): \quad \tuple {x, y_1} \in g, \tuple {x, y_2} \in f$ Because $f$ and $g$ are mappings, $(1)$ and $(2)$ imply that $y_1 = y_2$. If $(3)$ holds, then $y_1 = \map f x$ and $y_2 = \map g x$. But then $x \in A \cup B$. So by hypothesis: :$y_1 = \map f x = \map g x = y_2$ Similarly if $(4)$ holds. Thus in all cases: :$\tuple {x, y_1}, \tuple {x, y_2} \in f \cup g \implies y_1 = y_2$ and so by definition $f \cup g: X \to Y$ is a mapping. {{qed}}
By definition of composition of mappings: :$f_2 \circ f_1 := \set {\tuple {x, z} \in S_1 \times S_3: \exists y \in S_2: \tuple {x, y} \in f_1 \land \tuple {y, z} \in f_2}$ {{explain|Not quite what that definition says any more}} Let $x \in \Dom {f_2 \circ f_1}$. Then: :$\exists z \in S_3: \tuple {x, z} \in S_1 \times S_3$ and: :$\exists y \in S_2: \tuple {x, y} \in f_1$ That is: :$x \in \Dom {f_1}$ Thus by definition of subset: :$\Dom {f_2 \circ f_1} \subseteq \Dom {f_1}$ Now suppose $x \in \Dom {f_1}$. By the definition of mapping: :$\exists y \in S_2: \tuple {x, y} \in f_1$ As $f_2$ is likewise a mapping: :$\exists z \in S_3: \tuple {y, z} \in f_2$ and so by definition of composition of mappings: :$\tuple {x, z} \in S_1 \times S_3$ and so: :$x \in \Dom {f_2 \circ f_1}$ Thus by definition of subset: :$\Dom {f_1} \subseteq \Dom {f_2 \circ f_1}$ By definition of set equality: :$\Dom {f_1} = \Dom {f_2 \circ f_1}$ {{qed}}
From Identity Mapping is Left Identity: :$I_S \circ f = f$ for all mappings $f: S \to S$. From Identity Mapping is Right Identity: :$f \circ I_S = f$ for all mappings $f: S \to S$. Substituting $I_S$ for $f$ in either one and the result follows. {{qed}}
By definition, $\displaystyle g = \bigcup \set {g_i: i \in \N}$ is a relation whose domain is $X$. {{AimForCont}} $g$ is not a mapping. Then for some $x \in X$ and $i, h \in \N$: :$(1): \quad x \in X_i, \map {g_i} x \ne \map {g_{i + h} } x$ Let $k \in \N$ be the smallest such that: :$\map {g_i} x \ne g_{i + k}$ where $x$ and $i$ are the same as in $(1)$. Then: :$\map {g_{i + k - 1} } x = g_i$ But then: :$g_{i + k} \restriction X_{i + k - 1} = g_{i + k - 1}$ From this contradiction it follows that our supposition that $g$ is not a mapping must be false. That is:  :$\displaystyle \bigcup \set {g_i: i \in \N}$ is a mapping from $X$ to $Y$. {{qed}}
Let $\mathcal R \subseteq S \times S$ be antireflexive and transitive. From Antireflexive and Transitive Relation is Asymmetric it follows that $\mathcal R$ is asymmetric. The result follows from Asymmetric Relation is Antisymmetric. {{qed}}
Let $\left({T, \mathbf H, \Phi}\right)$ be a tableau confutation of $\mathbf H$. Suppose that $v$ were a boolean interpretation model for $\mathbf H$, i.e.: :$v \models_{\mathrm{BI}} \mathbf H$ By Model of Root of Propositional Tableau is Model of Branch, it follows that: :$v \models_{\mathrm{BI}} \Phi \left[{\Gamma}\right]$ for some branch $\Gamma$ of $T$. Since $T$ is a tableau confutation, there is some WFF $\mathbf A$ such that: :$\mathbf A, \neg\mathbf A \in \Phi \left[{\Gamma}\right]$ Hence $v \models_{\mathrm{BI}} \mathbf A$, i.e.: :$v \left({\mathbf A}\right) = T$ But by the truth table for $\neg$, this means: :$v \left({\neg\mathbf A}\right) = F$ which contradicts that $v \models_{\mathrm{BI}} \neg\mathbf A$. Hence, no such boolean interpretation can exist. That is, $\mathbf H$ is unsatisfiable for boolean interpretations. {{qed}}
By Diagonal Relation is Equivalence it follows a fortiori that $\Delta_S$ is reflexive. The result follows from Reflexive Relation is Serial. {{qed}}
As $S$ is not the empty set: : $\exists x: x \in S$ As $\mathcal R$ be a serial relation on $S$: :$\exists y \in S: \left({x, y}\right) \in \mathcal R$ That is: :$\mathcal \ne \varnothing$ Hence the result by definition of null relation. {{qed}}
=== Necessary Condition === Let $\mathcal R \subseteq S \times S$ be antireflexive. Then by Antireflexive and Transitive Relation is Asymmetric it follows that $\mathcal R$ is asymmetric. {{qed|lemma}} === Sufficient Condition === Let $\mathcal R$ be asymmetric. Then from Asymmetric Relation is Antireflexive it follows directly that $\mathcal R$ is antireflexive. {{qed}}
Let $\tuple {x, y}, \tuple {y, z} \in \RR$. By Relation is Symmetric and Antisymmetric iff Coreflexive: :$x = y, y = z$ and so trivially: :$\tuple {x, z} = \tuple {x, x} \in \RR$ Thus $\RR$ is transitive. {{qed}}
=== Necessary Condition === Let $\RR$ is reflexive, symmetric and antisymmetric. By definition of reflexive: :$\Delta_S \subseteq \RR$ From Relation is Symmetric and Antisymmetric iff Coreflexive: :$\RR \subseteq \Delta_S$ By definition of set equality: :$\RR = \Delta_S$ {{qed}} === Sufficient Condition === Let $\RR = \Delta_S$. From Relation is Reflexive and Coreflexive iff Diagonal: :$\RR$ is reflexive and :$\RR$ is coreflexive. From Relation is Symmetric and Antisymmetric iff Coreflexive it follows that $\RR$ is both symmetric and antisymmetric. Hence the result. {{qed}}
=== Definition 1 implies Definition 2 === Let $\mathcal R$ be a relation which fulfils the condition: :$\forall x, y \in S: \left({x, y}\right) \in \mathcal R \implies x = y$ Then: {{begin-eqn}} {{eqn | l = \left({x, y}\right)       | o = \in       | r = \mathcal R       | c =  }} {{eqn | ll= \implies       | l = \left({x, y}\right)       | r = \left({x, x}\right)       | c = by hypothesis }} {{eqn | ll= \implies       | l = \left({x, y}\right)       | o = \in       | r = \Delta_S       | c = {{Defof|Diagonal Relation}} }} {{eqn | ll= \implies       | l = \mathcal R       | o = \subseteq       | r = \Delta_S       | c = {{Defof|Subset}} }} {{end-eqn}} Hence $\mathcal R$ is coreflexive by definition 2. {{qed|lemma}} === Definition 2 implies Definition 1 === Let $\mathcal R$ be a relation which fulfils the condition: : $\mathcal R \subseteq \Delta_S$ Then: {{begin-eqn}} {{eqn | l = \left({x, y}\right)       | o = \in       | r = \mathcal R       | c =  }} {{eqn | ll= \implies       | l = \left({x, y}\right)       | o = \in       | r = \Delta_S       | c = {{Defof|Subset}} }} {{eqn | ll= \implies       | l = x       | r = y       | c = {{Defof|Diagonal Relation}} }} {{end-eqn}} Hence $\mathcal R$ is coreflexive by definition 1. {{Qed}} Category:Reflexive Relations bvumkuu4y62kdv5f55m2gy3ihm71yx6
=== Necessary Condition === Let $\mathcal R \subseteq S \times S$ be reflexive and coreflexive. By definition of reflexive: :$\Delta_S \subseteq \mathcal R$ By definition of coreflexive: :$\mathcal R \subseteq \Delta_S$ By definition of set equality: :$\mathcal R = \Delta_S$ {{qed|lemma}} === Necessary Condition === Let $\mathcal R = \Delta_S$ By definition of set equality: :$\Delta_S \subseteq \mathcal R$ and :$\mathcal R \subseteq \Delta_S$ From $\Delta_S \subseteq \mathcal R$ it follows by definition that $\mathcal R$ is reflexive. From $\mathcal R \subseteq \Delta_S$ it follows by definition that $\mathcal R$ is coreflexive. {{qed}} Category:Reflexive Relations 8g8ji62qcfanauvepmqi6n2ynegwx7q
By definition, a preordering on $S$ is a relation on $S$ which is: :$(1): \quad$ reflexive and: :$(2): \quad$ transitive. Thus $\RR$ is a relation on $S$ which is reflexive, transitive and symmetric. Thus by definition $\RR$ is an equivalence relation on $S$. {{qed}}
By definition, a preordering on $S$ is a relation on $S$ which is: :$(1): \quad$ reflexive and: :$(2): \quad$ transitive. Thus $\mathcal R$ is a relation on $S$ which is reflexive, transitive and antisymmetric. Thus by definition $\mathcal R$ is an ordering on $S$. {{qed}}
=== Definition 1 implies Definition 2 === Let $\RR$ be an ordering which is also connected. By definition 2 of ordering: :$(1): \quad \RR \circ \RR = \RR$ :$(2): \quad \RR \cap \RR^{-1} = \Delta_S$ By Relation is Connected and Reflexive iff Total, $\RR$ is a total relation. Thus by Relation is Total iff Union with Inverse is Trivial Relation: :$(3): \quad \RR \cup \RR^{-1} = S \times S$ Hence $\RR$ is a total ordering by definition 2. {{qed|lemma}} === Definition 2 implies Definition 1 === Let $\RR$ be a relation which fulfils the conditions: :$(1): \quad \RR \circ \RR = \RR$ :$(2): \quad \RR \cap \RR^{-1} = \Delta_S$ :$(3): \quad \RR \cup \RR^{-1} = S \times S$ By Relation is Total iff Union with Inverse is Trivial Relation it follows that $\RR$ is a total relation. From $(1)$ and $(2)$, $\RR$ is an ordering by definition 2. Thus by definition 1 of ordering, $\RR$ is a reflexive relation. From Relation is Connected and Reflexive iff Total, $\RR$ is a connected relation. Hence $\RR$ is a total ordering by definition 1. {{Qed}} Category:Total Orderings 81npba7z2cil8max0rmmi6rsthkayp1
=== Definition 1 implies Definition 2 === Let $\RR$ be a relation on $S$ satisfying: {{begin-axiom}} {{axiom | n = 1         | lc= $\RR$ is reflexive         | q = \forall a \in S         | m = a \mathrel \RR a }} {{axiom | n = 2         | lc= $\RR$ is transitive         | q = \forall a, b, c \in S         | m = a \mathrel \RR b \land b \mathrel \RR c \implies a \mathrel \RR c }} {{axiom | n = 3         | lc= $\RR$ is antisymmetric         | q = \forall a \in S         | m = a \mathrel \RR b \land b \mathrel \RR a \implies a = b }} {{end-axiom}} ==== Condition $(1)$ ==== Let $\tuple {x, y} \in \RR \circ \RR$. Then there exists a $z \in \RR$ such that: :$\tuple {x, z}, \tuple {z, y} \in \RR$ By $\RR$ being transitive: :$\tuple {x, y} \in \RR$ Hence: :$\RR \circ \RR \subseteq \RR$ Now let $\tuple {x, y} \in \RR$. By $\RR$ being reflexive: :$\tuple {y, y} \in \RR$ Hence by the definition of relation composition: :$\tuple {x, y} \in \RR \circ \RR$ Hence: :$\RR \subseteq \RR \circ \RR$ ==== Condition $(2)$ ==== Follows immediately from Relation is Antisymmetric iff Intersection with Inverse is Coreflexive and $\RR$ being reflexive. Thus $\RR$ is an ordering by definition 2. {{qed|lemma}} === Definition 2 implies Definition 1 === Let $\RR$ be a relation which fulfils the conditions: :$(1): \quad \RR \circ \RR = \RR$ :$(2): \quad \RR \cap \RR^{-1} = \Delta_S$ ==== Reflexivity ==== By Intersection is Subset the condition: :$\RR \cap \RR^{-1} = \Delta_S$ implies: :$\Delta_S \subseteq \RR$ Thus $\RR$ is reflexive by definition. ==== Antisymmetry ==== By Relation is Antisymmetric iff Intersection with Inverse is Coreflexive the condition: :$\RR \cap \RR^{-1} = \Delta_S$ implies that $\RR$ is antisymmetric. ==== Transitivity ==== Let $\tuple {x, y}, \tuple {y, z} \in \RR$. Then by the definition of relation composition: :$\tuple {x, z} \in \RR \circ \RR$ But by the condition: :$\RR \circ \RR = \RR$ It follows that: :$\tuple {x, z} \in \RR$ Hence $\RR$ is transitive. Thus $\RR$ is an ordering by definition 1. {{qed}}
=== Necessary Condition === Let $\mathcal R$ be both antisymmetric and reflexive. From Relation is Antisymmetric iff Intersection with Inverse is Coreflexive: :$\mathcal R \cap \mathcal R^{-1} \subseteq \Delta_S$ By definition of reflexive relation: :$\Delta_S \subseteq \mathcal R$ By Inverse of Reflexive Relation is Reflexive: :$\Delta_S \subseteq \mathcal R^{-1}$ By Intersection is Largest Subset: :$\Delta_S \subseteq \mathcal R \cap \mathcal R^{-1}$ Thus by definition of set equality: :$\mathcal R \cap \mathcal R^{-1} = \Delta_S$ {{qed|lemma}} === Sufficient Condition === Let $\mathcal R$ be such that: :$\mathcal R \cap \mathcal R^{-1} = \Delta_S$ Then by definition of set equality: :$\mathcal R \cap \mathcal R^{-1} \subseteq \Delta_S$ and so by Relation is Antisymmetric iff Intersection with Inverse is Coreflexive, $\mathcal R$ is antisymmetric. Also by definition of set equality: :$\Delta_S \subseteq \mathcal R \cap \mathcal R^{-1}$ By Intersection is Subset: :$\mathcal R \cap \mathcal R^{-1} \subseteq \mathcal R$ By Subset Relation is Transitive: :$\Delta_S \subseteq \mathcal R$ So, by definition, $\mathcal R$ is reflexive. {{qed}}} Category:Reflexive Relations Category:Symmetric Relations i26nwfe8dndy43nft7zfeti9ij9cgox
Let $\RR$ be both reflexive and  transitive. By definition of transitive relation: :$\RR \circ \RR \subseteq \RR$ Let $\tuple {x, y} \in \RR$. By definition of reflexive relation: :$\tuple {y, y} \in \RR$ By definition of composition of relations: :$\tuple {x, y} \in \RR \circ \RR$ Hence: :$\RR \subseteq \RR \circ \RR$ Thus by definition of set equality: :$\RR \circ \RR = \RR$ {{qed}}
=== Definition 1 implies Definition 2 === Let $\RR$ be a relation on $S$ satisfying: {{begin-axiom}} {{axiom | n = 1         | lc= $\RR$ is reflexive         | q = \forall a \in S         | m = a \mathrel \RR a }} {{axiom | n = 2         | lc= $\RR$ is transitive         | q = \forall a, b, c \in S         | m = a \mathrel \RR b \land b \mathrel \RR c \implies a \mathrel \RR c }} {{axiom | n = 3         | lc= $\RR$ is antisymmetric         | q = \forall a \in S         | m = a \mathrel \RR b \land b \mathrel \RR a \implies a = b }} {{end-axiom}} By Reflexive and Transitive Relation is Idempotent: :$\RR \circ \RR = \RR$ By Relation is Antisymmetric and Reflexive iff Intersection with Inverse equals Diagonal Relation: :$\RR \cap \RR^{-1} = \Delta_S$ Thus $\RR$ is an ordering by definition 2. {{qed|lemma}} === Definition 2 implies Definition 1 === Let $\RR$ be a relation which fulfils the conditions: :$(1): \quad \RR \circ \RR = \RR$ :$(2): \quad \RR \cap \RR^{-1} = \Delta_S$ By definition of set equality, it follows from $(1)$: :$\RR \circ \RR \subseteq \RR$ Thus, by definition, $\RR$ is transitive. By Relation is Antisymmetric and Reflexive iff Intersection with Inverse equals Diagonal Relation, it follows from $(2)$ that: :$\RR$ is reflexive :$\RR$ is antisymmetric. Thus $\RR$ is an ordering by definition 1. {{qed}}