hcju/proofwikiqa huggingface dataset:
=== Sufficient Condition === Let $\mathcal R$ be an equivalence relation. Then by definition it is reflexive. It is also transitive and symmetric. So, let $x, y, z \in \mathcal R$ such that $x \mathcal R y$ and $x \mathcal R z$. From symmetry, we have $y \mathcal R x$, and by transitivity it follows that $y \mathcal R z$. Hence $\mathcal R$ is right-Euclidean. Now let $x, y, z \in \mathcal R$ such that $x \mathcal R z$ and $y \mathcal R z$. From symmetry, we have $z \mathcal R y$, and by transitivity it follows that $x \mathcal R y$. Hence $\mathcal R$ is left-Euclidean. So $\mathcal R$ is '''both''' left-Euclidean and right-Euclidean. We know by definition that it is reflexive. Thus an equivalence relation is both left-Euclidean and right-Euclidean and also reflexive. === Necessary Condition === Let $\mathcal R$ be a relation which is both left-Euclidean and reflexive. Checking in turn each of the criteria for equivalence: ==== Reflexive ==== By definition: $\mathcal R$ has been defined as a relation which is reflexive. ==== Symmetric ==== Suppose $x \mathcal R y$. We have that $y \mathcal R y$ from the reflexivity of $\mathcal R$. So we have $y \mathcal R y$ and $x \mathcal R y$, and so $y \mathcal R x$ from the fact that $\mathcal R$ is left-Euclidean. Hence $\mathcal R$ is symmetric. ==== Transitive ==== Suppose $x \mathcal R y$ and $y \mathcal R z$. As $\mathcal R$ is symmetric (from above), we have that $z \mathcal R y$. So we have $x \mathcal R y$ and $z \mathcal R y$, and by the fact that $\mathcal R$ is left-Euclidean, we have that $x \mathcal R z$. So, if $\mathcal R$ be a relation which is both left-Euclidean and reflexive, it is also an equivalence relation. A similar argument shows that if $\mathcal R$ be a relation which is both right-Euclidean and reflexive, it is also an equivalence relation. {{Qed}} Category:Reflexive Relations Category:Euclidean Relations Category:Equivalence Relations r01dhclv0emixbnwmamuu2yy61ioyi8
Let $\mathcal R \subseteq S \times S$ be a relation in $S$. We have that $\mathcal R$ is serial {{iff}}: :$\forall x \in S: \exists y \in S: \left({x, y}\right) \in \mathcal R$ That is, {{iff}} every element relates to at least one element. We have that $\mathcal R$ is reflexive {{iff}}: :$\forall x \in S: \left({x, x}\right) \in \mathcal R$ Hence if $\mathcal R$ is reflexive, every $x$ is related to itself, thereby fulfilling the criterion for $\mathcal R$ to be serial. {{qed}} Category:Reflexive Relations Category:Serial Relations odv59hgjd19v9b124olsw94o6qf7wcz
By definition, the complete bipartite graph $K_{p, q}$ consists of two sets of vertices: $A$ of cardinality $p$, and $B$ of cardinality $q$, such that: :Every vertex in $A$ is adjacent to every vertex in $B$ :No vertex in $A$ is adjacent to any other vertex in $A$ :No vertex in $B$ is adjacent to any other vertex in $B$. The complement of $K_{p, q}$ therefore must be a graph $G$ such that: :No vertex in $A$ is adjacent to any vertex in $B$ :Every vertex in $A$ is adjacent to every other vertex in $A$ :Every vertex in $B$ is adjacent to every other vertex in $B$. The second and third of these conditions describes the complete graphs $K_p$ and $K_q$. From the first of these conditions, it follows that $G$ comes in two disconnected graph components. Hence the result. {{qed}} Category:Bipartite Graphs 4a3yu1at7zsk2bdzjhseodlc0qv808t
We need to show that $\not \prec$ is a weak total ordering. First we check in turn each of the criteria for an ordering: ==== Reflexivity ==== By Complement of Reflexive Relation, it follows directly that $\not \prec$ is reflexive. ==== Transitivity ==== Let $a \not \prec b$ and $b \not \prec c$. Now, suppose that $a \prec c$. As $b \not \prec c$, from the Trichotomy Law, one of the following is the case: * $b = c$ and so $a \prec b$; * $c \prec b$ and $a \prec b$ from transitivity of $\prec$. Both are an immediate contradiction. So $a \not \prec c$ and so $\not \prec$ is transitive. ==== Antisymmetry ==== Suppose $a \not \prec b$. Then from the Trichotomy Law, either $a = b$ or $b \prec a$. Similarly, if $b \not \prec a$ we have that either $a = b$ or $a \prec b$. As $\prec$ is asymmetric, $b \prec a \iff \neg a \prec b$. So if $a \not \prec b$ and $b \not \prec a$, it follows that $a = b$. So $\not \prec$ is antisymmetric. The fact that $\not \prec$ is connected follows directly from the Trichotomy Law. {{qed}} Category:Total Orderings khi6brweos2nicv39pie5ig2uzlqg6a
From the definition of strict weak ordering, we define the symbol $\Bumpeq$ as: :$a \Bumpeq b := \neg a \prec b \land \neg b \prec a$ that is, $a \Bumpeq b$ means "$a$ and $b$ are non-comparable". Checking in turn each of the criteria for equivalence: === Reflexive === As $\prec$ is antireflexive, by definition $\forall a \in S: \neg a \prec a$. Hence by the Rule of Idempotence $\neg a \prec a \land \neg a \prec a$ and so $\forall a \in S: a \Bumpeq a$. === Symmetric === We have that $a \Bumpeq b$ is defined as being $\neg a \prec b \land \neg b \prec a$. It follows from the Rule of Commutation that $\neg b \prec a \land \neg a \prec b$, and so $b \Bumpeq a$. === Transitive === The relation $a \Bumpeq b$ is ''defined'' as being transitive in a strict weak ordering. {{qed}} Category:Order Theory Category:Equivalence Relations i1kgsbvg0574xpdfd3xky75lomjjmgj
{{begin-eqn}} {{eqn | l = \prod_{k \mathop = 1}^n \paren {\frac {x_k} {x_k + y_k} }^{1/n}       | o = \le       | r = \frac 1 n \sum_{k \mathop = 1}^n \frac {x_k} {x_k + y_k}       | c = Cauchy's Mean Theorem }} {{eqn | l = \prod_{k \mathop = 1}^n \paren {\frac {y_k} {x_k + y_k} }^{1/n}       | o = \le       | r = \frac 1 n \sum_{k \mathop = 1}^n \frac {y_k} {x_k + y_k}       | c = Cauchy's Mean Theorem }} {{eqn | ll= \leadsto       | l = \prod_{k \mathop = 1}^n \paren {\frac {x_k} {x_k + y_k} }^{1/n} + \prod_{k \mathop = 1}^n \paren {\frac {y_k} {x_k + y_k} }^{1/n}       | o = \le       | r = \frac 1 n \sum_{k \mathop = 1}^n \frac {x_k} {x_k + y_k} + \frac 1 n \sum_{k \mathop = 1}^n \frac {y_k} {x_k + y_k}       | c = adding them together }} {{eqn | r = \frac 1 n \sum_{k \mathop = 1}^n \frac {x_k + y_k} {x_k + y_k}       | c =  }} {{eqn | r = \frac 1 n n       | c =  }} {{eqn | r = 1       | c =  }} {{end-eqn}} This leads to: {{begin-eqn}} {{eqn | l = \prod_{k \mathop = 1}^n \paren {\frac {x_k} {x_k + y_k} }^{1/n} + \prod_{k \mathop = 1}^n \paren {\frac {y_k} {x_k + y_k} }^{1/n}       | o = \le       | r = 1       | c =  }} {{eqn | ll= \leadsto       | l = \frac {\displaystyle \prod_{k \mathop = 1}^n \paren {x_k}^{1/n} } {\displaystyle \prod_{k \mathop = 1}^n \paren {x_k + y_k}^{1/n} } + \frac {\displaystyle \prod_{k \mathop = 1}^n \paren {y_k}^{1/n} } {\displaystyle \prod_{k \mathop = 1}^n \paren {x_k + y_k}^{1/n} }       | o = \le       | r = 1       | c =  }} {{eqn | ll= \leadsto       | l = \frac {\displaystyle \prod_{k \mathop = 1}^n \paren {x_k}^{1/n} + \prod_{k \mathop = 1}^n \paren {y_k}^{1/n} } {\displaystyle \prod_{k \mathop = 1}^n \paren {x_k + y_k}^{1/n} }       | o = \le       | r = 1       | c =  }} {{eqn | ll= \leadsto       | l = \prod_{k \mathop = 1}^n \paren {x_k}^{1/n} + \prod_{k \mathop = 1}^n \paren {y_k}^{1/n}       | o = \le       | r = \prod_{k \mathop = 1}^n \paren {x_k + y_k}^{1/n}       | c =  }} {{end-eqn}} {{qed}} {{Namedfor|Kurt Mahler|cat = Mahler}} Category:Algebra Category:Measures of Central Tendency kumfwqaoihlbglrwl519pylp3tptki7
Let $\mathcal R_1$ and $\mathcal R_2$ be transitive relations (on what sets is immaterial for this argument). Let $\left({s_1, s_2}\right) \in \mathcal R_1 \cap \mathcal R_2$ and $\left({s_2, s_3}\right) \in \mathcal R_1 \cap \mathcal R_2$. Then by definition of intersection: : $\left({s_1, s_2}\right) \in \mathcal R_1$ and $\left({s_1, s_2}\right) \in \mathcal R_2$ : $\left({s_2, s_3}\right) \in \mathcal R_1$ and $\left({s_2, s_3}\right) \in \mathcal R_2$ Then as $\mathcal R_1$ and $\mathcal R_2$ are both transitive: : $\left({s_1, s_3}\right) \in \mathcal R_1$ and  $\left({s_1, s_3}\right) \in \mathcal R_2$ and by definition of intersection: : $\left({s_1, s_3}\right) \in \mathcal R_1 \cap \mathcal R_2$ hence $\mathcal R_1 \cap \mathcal R_2$ is transitive. {{qed}} Category:Transitive Relations Category:Set Intersection qut6uwlh9coyuh5mdhceoa68klqt8c8
Let $\mathcal R_1$ and $\mathcal R_2$ be reflexive relations on a set $S$. From Relation Contains Diagonal Relation iff Reflexive, we have that: : $\Delta_S \subseteq \mathcal R_1$ : $\Delta_S \subseteq \mathcal R_2$ Hence from Intersection is Largest Subset: : $\Delta_S \subseteq \mathcal R_1 \cap \mathcal R_2$ Hence the result, from Relation Contains Diagonal Relation iff Reflexive. {{qed}} Category:Reflexive Relations Category:Set Intersection i9e8b6o95p3em0typc41tvb8nuilqom
Let $\mathcal R_1$ and $\mathcal R_2$ be symmetric relations on a set $S$. Let $\mathcal R_3 = \mathcal R_1 \cap \mathcal R_2$. Then: {{begin-eqn}} {{eqn | o=       | r=\left({x, y}\right) \in \mathcal R_3       | c= }} {{eqn | o=\implies       | r=\left({x, y}\right) \in \mathcal R_1 \cap \mathcal R_2       | c=by definition of $\mathcal R_3$ }} {{eqn | o=\implies       | r=\left({x, y}\right) \in \mathcal R_1 \land \left({x, y}\right) \in \mathcal R_2       | c=Definition of Intersection }} {{eqn | o=\implies       | r=\left({y, x}\right) \in \mathcal R_1 \land \left({y, x}\right) \in \mathcal R_2       | c=$\mathcal R_1$ and $\mathcal R_2$ are symmetric }} {{eqn | o=\implies       | r=\left({y, x}\right) \in \mathcal R_1 \cap \mathcal R_2       | c=Definition of Intersection }} {{eqn | o=\implies       | r=\left({y, x}\right) \in \mathcal R_3       | c=by definition of $\mathcal R_3$ }} {{end-eqn}} {{qed}} Category:Symmetric Relations Category:Set Intersection aq7s9vtktv87ycyqbmvvnp0zk43tisc
Let $\mathcal R_1$ and $\mathcal R_2$ be reflexive relations on a set $S$. From Relation Contains Diagonal Relation iff Reflexive, we have that: : $\Delta_S \subseteq \mathcal R_1$ : $\Delta_S \subseteq \mathcal R_2$ Hence from Subset Relation is Transitive: : $\Delta_S \subseteq \mathcal R_1 \cup \mathcal R_2$ Hence the result, by definition of reflexive relation. {{qed}} Category:Reflexive Relations Category:Set Union tqikddu7bqes9eonhgo0a4aoms9jy7n
Let $\mathcal R_1$ and $\mathcal R_2$ be symmetric relations on a set $S$. Let $\mathcal R_3 = \mathcal R_1 \cup \mathcal R_2$. Then: {{begin-eqn}} {{eqn | o=       | r=\left({x, y}\right) \in \mathcal R_3       | c= }} {{eqn | o=\implies       | r=\left({x, y}\right) \in \mathcal R_1 \cup \mathcal R_2       | c=Definition of $\mathcal R_3$ }} {{eqn | o=\implies       | r=\left({x, y}\right) \in \mathcal R_1 \lor \left({x, y}\right) \in \mathcal R_2       | c=Definition of Union }} {{eqn | o=\implies       | r=\left({y, x}\right) \in \mathcal R_1 \lor \left({y, x}\right) \in \mathcal R_2       | c=$\mathcal R_1$ and $\mathcal R_2$ are symmetric }} {{eqn | o=\implies       | r=\left({y, x}\right) \in \mathcal R_1 \cup \mathcal R_2       | c=Definition of Union }} {{eqn | o=\implies       | r=\left({y, x}\right) \in \mathcal R_3       | c=Definition of $\mathcal R_3$ }} {{end-eqn}} {{qed}} Category:Symmetric Relations Category:Set Union e9ys5s8ei652ugbe8sgl36s5bcyulbr
Proof by counterexample: Let $S = \set {a, b, c, d}$. Let $\RR_1$ be the transitive relation $\set {\tuple {a, b}, \tuple {b, c}, \tuple {a, c} }$. Let $\RR_2$ be the transitive relation $\set {\tuple {b, c}, \tuple {c, d}, \tuple {b, d} }$. Then we have that $\tuple {a, b} \in \RR_1 \cup \RR_2$ and $\tuple {b, d} \in \RR_1 \cup \RR_2$. However, $\tuple {a, d} \notin \RR_1 \cup \RR_2$, and so $\RR_1 \cup \RR_2$ is not transitive. {{qed}} Category:Transitive Relations Category:Set Union ja9x5q1i1pnxh46bnj8d1jv6ykh6bip
=== Closure === By definition (above), $\left({\mathcal S, \cap}\right)$ is closed. === Associativity === The operation $\cap$ is associative from Intersection is Associative. === Commutativity === The operation $\cap$ is commutative from Intersection is Commutative. Hence, by definition, the result. {{qed}}
===  Closure === By definition (above), $\struct {\mathcal S, \cup}$ is closed. === Associativity === The operation $\cup$ is associative from Union is Associative. === Commutativity === The operation $\cup$ is commutative from Union is Commutative. Hence, by definition, the result. {{qed}}
=== $\text G 0$: Closure === By presupposition on $\SS$, $\struct {\SS, *}$ is closed. {{qed|lemma}} === $\text G 1$: Associativity === :$\forall A, B, C \in \SS: \paren {A * B} * C = A * \paren {B * C}$ as Symmetric Difference is Associative. So $*$ is associative. {{qed|lemma}} === $\text G 2$: Identity === From Symmetric Difference with Self is Empty Set, we have that: :$\forall A \in \SS: A * A = \O$ So it is clear that $\O$ is in $\SS$, from the fact that $\struct {\SS, *}$ is closed. Then we have: :$\forall A \in \SS: A * \O = A = \O * A$ from Symmetric Difference with Empty Set and Symmetric Difference is Commutative. Thus $\O$ acts as an identity. {{qed|lemma}} === $\text G 3$: Inverses === From the above, we know that $\O$ is the identity element of $\struct {\SS, *}$. We also noted that :$\forall A \in \SS: A * A = \O$ From Symmetric Difference with Self is Empty Set. Thus each $A \in \SS$ is self-inverse. {{qed|lemma}} === Commutativity === :$\forall A, B \in \SS: A * B = B * A$ as Symmetric Difference is Commutative. So $*$ is commutative. {{qed|lemma}} We see that $\struct {\SS, *}$ is closed, associative, commutative, has an identity element $\O$, and each element has an inverse (itself), so it satisfies the criteria for being an abelian group. {{Qed}}
By definition, the operations $\cap$ and $*$ are closed in $\mathcal R$. Hence we can apply the following results: * Set System Closed under Symmetric Difference is Abelian Group: $\left({\mathcal R, *}\right)$ is an abelian group. * Set System Closed under Intersection is Commutative Semigroup: $\left({\mathcal R, \cap}\right)$ is a commutative semigroup. * Intersection Distributes over Symmetric Difference. * The identity of $\left({\mathcal R, *}\right)$ is $\varnothing$, and this, by definition, is the zero. So $\left({\mathcal R, *, \cap}\right)$ is a commutative ring whose zero is $\varnothing$. {{qed}} Category:Commutative Rings Category:Rings of Sets h7sb8yadj1jae9l5jrhgq9z1996f488
From the definition of symmetric difference: :$\left({A * B}\right) * \left({A \cap B}\right) = \left({\left({A * B}\right) \cup \left({A \cap B}\right)}\right) \setminus \left({\left({A * B}\right) \cap \left({A \cap B}\right)}\right)$ Also from the definition of Symmetric Difference: {{explain}} :$\left({A * B}\right) \cap \left({A \cap B}\right) = \left({\left({A \cup B}\right) \setminus \left({A \cap B}\right)}\right) \cap \left({A \cup B}\right)$ From Set Difference Intersection with Second Set is Empty Set: : $\left({S \setminus T}\right) \cap T = \varnothing$ Hence: :$\left({\left({A \cup B}\right) \setminus \left({A \cup B}\right)}\right) \cap \left({A \cup B}\right) = \varnothing$ This leaves: {{begin-eqn}} {{eqn | o =        | r = \left({\left({A * B}\right) \cup \left({A \cap B}\right)}\right) \setminus \left({\left({A * B}\right) \cap \left({A \cap B}\right)}\right)       | c =  }} {{eqn | r = \left({\left({A * B}\right) \cup \left({A \cap B}\right)}\right) \setminus \varnothing       | c =  }} {{eqn | r = \left({A * B}\right) \cup \left({A \cap B}\right)       | c = Set Difference with Empty Set is Self }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | o =        | r = \left({A * B}\right) \cup \left({A \cap B}\right)       | c =  }} {{eqn | r = \left({A \setminus B}\right) \cup \left({B \setminus A}\right) \cup \left({A \cap B}\right)       | c = {{Defof|Symmetric Difference|index = 1}} }} {{eqn | r = \left({\left({A \setminus B}\right) \cup \left({A \cap B}\right)}\right) \cup \left({\left({B \setminus A}\right) \cup \left({A \cap B}\right)}\right)       | c = Union is Idempotent, Union is Commutative and Union is Associative }} {{eqn | r = A \cup B       | c = Set Difference Union Intersection }} {{end-eqn}} Hence the result. {{qed}} Category:Set Union Category:Set Intersection Category:Symmetric Difference jt1vzab812qhbkdi2qo6a712pebijwp
{{begin-eqn}} {{eqn | l = S * \paren {S \cap T}       | r = \paren {S \setminus \paren {S \cap T} } \cup \paren {\paren {S \cap T} \setminus S}       | c = {{Defof|Symmetric Difference}} }} {{eqn | r = \paren {S \setminus \paren {S \cap T} } \cup \O       | c = Set Difference of Intersection with Set is Empty Set }} {{eqn | r = \paren {S \setminus T} \cup \O       | c = Set Difference with Intersection is Difference }} {{eqn | r = S \setminus T       | c = Union with Empty Set }} {{end-eqn}} {{qed}} Category:Set Intersection Category:Set Difference Category:Symmetric Difference drzwebx8qpcfykkbpdvlcb82bvpd2ym
Let $\mathcal S$ be a system of sets. Suppose $U$ and $U'$ are both units of $\mathcal S$. Then, by definition: * $\forall A \in \mathcal S: A \cap U = A$ * $\forall A \in \mathcal S: A \cap U' = A$ This applies to both $U$ and $U'$, of course. So $U \cap U' = U$ and $U' \cap U = U'$. From Intersection with Subset is Subset it follows that $U \subseteq U'$ and $U' \subseteq U$. By definition of set equality: : $U = U'$ We also see that from Intersection with Subset is Subset, $A \cap U = A \iff A \subseteq U$, which shows that: :$\forall A \in \mathcal S: A \subseteq U$. {{qed}} Category:Set Systems 644wke5stggq0p0947wxsp8knrgxee7
=== Definition 1 implies Definition 2 === Let $\mathcal R$ be a system of sets such that $\forall A, B \in \mathcal R$: :$(1): \quad A \cup B \in \mathcal R$ :$(2): \quad \complement_X \left({A}\right) \in \mathcal R$ Let $A, B \in \mathcal R$. From the definition: :$\forall A \in \mathcal R: A \subseteq X$. Hence from Intersection with Subset is Subset: : $\forall A \in \mathcal R: A \cap X = A$ Hence $X$ is the unit of $\mathcal R$. From Properties of Algebras of Sets, we have that $\mathcal R$ is closed under set intersection. From the definition of symmetric difference: : $A * B = \left({A \setminus B}\right) \cup \left({B \setminus A}\right)$ Since both set union and set difference are closed operations, it follows that symmetric difference is also closed. So by definition 1 of ring of sets, it follows that $\mathcal R$ is indeed a ring of sets. {{qed|lemma}} === Definition 2 implies Definition 1 === Let $\mathcal R$ be a ring of sets with a unit $X$. By definition, $X \in \mathcal R$. From definition 2 of ring of sets, $\mathcal R$ is: :$(1) \quad$ closed under set union :$(2) \quad$ closed under set difference. From Unit of System of Sets is Unique, we have that: :$\forall A \in \mathcal R: A \subseteq X$ from which we have that $X \setminus A = \complement_X \left({A}\right)$. So $\mathcal R$ is an algebra of sets by definition 1. {{qed}}
From Power Set is Closed under Intersection and Power Set is Closed under Symmetric Difference, we have that: :$(1): \quad \forall A, B \in \mathcal P \left({S}\right): A \cap B \in \mathcal P \left({S}\right)$ :$(2): \quad \forall A, B \in \mathcal P \left({S}\right): A * B \in \mathcal P \left({S}\right)$ From the definition of power set: :$\forall A \in \mathcal P \left({S}\right): A \subseteq S$ and so $S$ is the unit of $\mathcal P \left({S}\right)$. Thus we see that $\mathcal P \left({S}\right)$ is a ring of sets with a unit. Hence the result, by definition of an algebra of sets. {{qed}} Category:Power Set Category:Algebras of Sets j6787sli7tm808fipe4ic7obqg8us0c
Let $A, B \in \powerset S$. Then by the definition of power set, $A \subseteq S$ and $B \subseteq S$. From Intersection is Subset we have that $A \cap B \subseteq A$. It follows from Subset Relation is Transitive that $A \cap B \subseteq S$. Thus $A \cap B \in \powerset S$ and closure is proved. {{qed}}
Let $A, B \in \powerset S$. Then by definition of power set: :$A, B \subseteq S$ Then: {{begin-eqn}} {{eqn | l = A * B       | o = \subseteq       | r = A \cup B       | c = Symmetric Difference is Subset of Union }} {{eqn | o = \subseteq       | r = S       | c = Union is Smallest Superset }} {{eqn | ll= \leadsto       | l = A * B       | o = \in       | r = \powerset S       | c = {{Defof|Power Set}} }} {{end-eqn}} {{qed}}
Consider the set $\displaystyle \mathcal S = \bigcup_k \mathcal R_k$. Let $S = set {X \in Y: Y \in \mathcal S}$. This contains all the elements of all the sets contained in all the $\mathcal R_k$. Now consider the power set $\powerset S$ of $S$. By Power Set is Algebra of Sets and the definition of algebra of sets, we have that $\powerset S$ is a ring of sets. Thus $\struct {\powerset S, *, \cap}$ is a ring (in the abstract algebraic sense of the term). From the method of construction of $\powerset S$, it follows that all the $\struct {\mathcal R_k, *, \cap}$ are subrings of $\powerset S$. The result then follows from Intersection of Subrings is Subring. {{qed}} Category:Rings of Sets ip400w6dlrsex1ihyoafqczxj0zo29s
=== Uniqueness === Suppose there were two such rings of sets $\map \RR \SS$ and $\map \RR \SS'$. Then by definition their intersection $\map \RR \SS \cap \map \RR \SS'$ would also contain $\SS$. By Intersection of Rings of Sets, $\map \RR \SS \cap \map \RR \SS'$ is also a ring of sets. From Intersection is Subset, $\map \RR \SS \cap \map \RR \SS' \subseteq \map \RR \SS$ and $\map \RR \SS \cap \map \RR \SS' \subseteq \map \RR \SS'$. Hence either $\map \RR \SS$ or $\map \RR \SS'$ can not be minimal. So if $\map \RR \SS$ exists, it has to be unique. === Existence === Consider the union: :$\displaystyle X = \bigcap_{A \mathop \in \SS} A$ of all sets in $\SS$. Now consider the power set $\powerset X$ of all subsets of $X$. From Power Set is Algebra of Sets and by definition of algebra of sets, $\powerset X$ is a ring of sets containing $\SS$. Now let $\Sigma$ be the set of all rings of sets contained in $\powerset X$ which also contain $\SS$. Then consider the intersection: :$\displaystyle \map \RR \SS = \bigcap_{\TT \mathop \in \Sigma} \TT$ of all these rings of sets. It is clear that $\map \RR \SS$ has the properties specified. In particular, as all $\TT \in \Sigma$ contain $\SS$, then so does $\map \RR \SS$. Finally, note that if $\RR'$ is a ring of sets containing $\SS$, then: :$\RR^* = \RR' \cap \powerset X$ is a ring of sets in $\Sigma$. Hence $\map \RR \SS \subseteq \RR^* \subseteq \RR'$ as we needed to show. {{qed}} Category:Set Systems 5grq7uqgy6m7m58mygsouhtkiv6rqce
=== Necessary Condition === First we check that the equation works out negative. From Pasch's Axiom, the line $DEF$ must intersect either two sides of $\triangle ABC$: :500px or none of them: :500px That means there is an odd number of negative contributions to the product. Hence the equation works out to be negative. Next, the magnitude can be checked. The method is illustrated using the second of the above options; the first works the same. :500px Construct perpendiculars $AD$, $CH$ and $BI$ from $A$, $B$ and $C$ to $DEF$. The triangles $\triangle AEG$, $\triangle CEH$ are similar. The triangles $\triangle BDI$, $\triangle CDH$ are also similar. The triangles $\triangle BFI$, $\triangle AFG$ are also similar. Hence: :$\dfrac {CE} {EA} = \dfrac {CH} {AG}, \dfrac {BD} {DC} = \dfrac {BI} {CH}, \dfrac {AF} {FB} = \dfrac {AG} {BI}$ Thus: :$\size {\dfrac {AF} {FB} \cdot \dfrac{BD} {DC} \cdot \dfrac{CE} {EA} } = \size {\dfrac {AG} {BI} \cdot \dfrac {BI} {CH} \cdot \dfrac {CH} {AG} } = 1$ Hence the result. {{qed|lemma}} === Sufficient Condition === Suppose $\dfrac {AF} {FB} \cdot \dfrac {BD} {DC} \cdot \dfrac {CE} {EA} = -1$. Let $F'$ be a point on $AB$ distinct from $F$. Let us define the measurements of $AF$, $AF'$, $AB$ as $n, n', s$ respectively. {{WLOG}}, suppose that $F'$ also satisfies the equation. Then: :$\dfrac {AF} {FB} = \dfrac {AF'} {F'B}$ which means: :$\dfrac n {s - n} = \dfrac {n'} {s - n'}$ and so $n = n'$ and so $F = F'$. So only one point on line $AB$ can satisfy the equation. Let us fix $D$ and $E$. Then if $F$ satisfies the equation, then it must be the point collinear with $D$ and $E$. By a similar argument, the same applies to $D$ and $E$. Hence the result. {{qed}}