nyuuzyou/wikis huggingface dataset:
Circuits of Matroid iff Matroid Circuit Axioms/Formulation 2 implies Circuits of Matroid

== Theorem ==
Let $S$ be a finite set.
Let $\mathscr C$ be a non-empty set of subsets of $S$ that satisfies the circuit axioms:
Then:
:$\mathscr C$ is the set of circuits of a matroid $M = \struct{S, \mathscr I}$ on $S$
== Proof ==
We will define a mapping $\rho$ associated with $\mathscr C$.
It will be shown that $\rho$ is the rank function of a matroid $M$ which has $\mathscr C$ as the set of circuits.
For any ordered tuple $\tuple{x_1, \ldots, x_q}$ of elements of $S$, let $\map \theta {x_1, \ldots, x_q}$ be the ordered tuple defined by:
:$\forall i \in \set{1, \ldots, q} : \map \theta {x_1, \ldots, x_q}_i = \begin{cases} 0 & : \exists C \in \mathscr C : x_i \in C \subseteq \set{x_1, \ldots, x_i}\\ 1 & : \text {otherwise} \end{cases}$
Let $t$ be the mapping from the set of ordered tuple of $S$ defined by:
:$\map t {x_1, \ldots, x_q} = \ds \sum_{i = 1}^q \map \theta {x_1, \ldots, x_q}_i$
==== Lemma 1 ====
Let $\rho : \powerset S \to \Z$ be the mapping defined by:
:$\forall A \subseteq S$:
::$\map \rho A = \begin{cases} 0 & : \text{if } A = \O \\ \map t {x_1, \ldots, x_q } & : \text{if } A = \set{x_1, \ldots, x_q} \end{cases}$
From Lemma 1:
:$\rho$ is well-defined
We now show that $\rho$ satisfies the rank function axioms:
==== $\rho$ satisfies $(\text R 1)$ ====
Follows from the definition of $\rho$.
==== $\rho$ satisfies $(\text R 2)$ ====
Let:
:$X = \set{x_1, \ldots, x_q} \subseteq S$ and $y \in S$
We have:
==== Lemma 2 ====
==== $\rho$ satisfies $(\text R 3)$ ====
Let:
:$X \subseteq S$ and $y, z \in S$
Let:
:$\map \rho {X \cup \set y} = \map \rho {X \cup \set z} = \map \rho X$
===== Case 1 : $z \in X$ =====
Let $z \in X$.
Then:
:$\map \rho {X \cup \set {y,z}} = \map \rho {X \cup \set y}$
Since $\map \rho {X \cup \set y} = \map \rho X$, then:
:$\map \rho {X \cup \set {y,z}} = \map \rho X$
===== Case 2 : $z = y$ =====
Let $z = y$.
Then
:$\map \rho {X \cup \set {y,z}} = \map \rho {X \cup \set y}$
Since $\map \rho {X \cup \set y} = \map \rho X$, then:
:$\map \rho {X \cup \set {y,z}} = \map \rho X$
===== Case 3 : $z \neq y$ and $z \notin X$ =====
Let $z \neq y$ and $z \notin X$.
Then $z \notin X \cup \set y$.
From Lemma 2:
:$\exists C_z \in \mathscr C : z \in C_z \subseteq X \cup \set{z} \subseteq X \cup \set{y,z}$
From Lemma 2:
:$\map \rho {X \cup \set{y, z}} = \map \rho {X \cup \set y} = \map \rho X$ 
From Matroid Rank Function Iff Matroid Rank Axioms:
:$\rho$ is the rank function of a matroid $M = \struct{S, \mathscr I}$ on $S$
==== Lemma 3 ====
== Sources ==
* Chapter $1.$ $\S 9.$ Circuits, Proof of Theorem $2.5$
Category:Circuits of Matroid iff Matroid Circuit Axioms
Circuits of Matroid iff Matroid Circuit Axioms/Lemma 2

== Theorem ==
Let $S$ be a finite set.
Let $\mathscr C$ be a non-empty set of subsets of $S$ that satisfies the circuit axioms:
For any ordered tuple $\tuple{x_1, \ldots, x_q}$ of elements of $S$, let $\map \theta {x_1, \ldots, x_q}$ be the ordered tuple defined by:
:$\forall i \in \set{1, \ldots, q} : \map \theta {x_1, \ldots, x_q}_i = \begin{cases} 0 & : \exists C \in \mathscr C : x_i \in C \subseteq \set{x_1, \ldots, x_i}\\ 1 & : \text {otherwise} \end{cases}$
Let $t$ be the mapping from the set of ordered tuple of $S$ defined by:
:$\map t {x_1, \ldots, x_q} = \ds \sum_{i = 1}^q \map \theta {x_1, \ldots, x_q}_i$
Let $\rho : \powerset S \to \Z$ be the mapping defined by:
:$\forall A \subseteq S$:
::$\map \rho A = \begin{cases} 0 & : \text{if } A = \O \\ \map t {x_1, \ldots, x_q } & : \text{if } A = \set{x_1, \ldots, x_q} \end{cases}$
Let $X \subseteq S$ and $y \in S \setminus X$.
Then:
:$\map \rho {X \cup \set y} = \map \rho X$ $\exists C \in \mathscr C : y \in C \subseteq X \cup \set y$
== Proof ==
Let $X = \set{x_1, \ldots, x_q}$
We have:
Hence:
Category:Circuits of Matroid iff Matroid Circuit Axioms
Circuits of Matroid iff Matroid Circuit Axioms/Lemma 3

== Theorem ==
Let $S$ be a finite set.
Let $\mathscr C$ be a non-empty set of subsets of $S$ that satisfies the circuit axioms:
For any ordered tuple $\tuple{x_1, \ldots, x_q}$ of elements of $S$, let $\map \theta {x_1, \ldots, x_q}$ be the ordered tuple defined by:
:$\forall i \in \set{1, \ldots, q} : \map \theta {x_1, \ldots, x_q}_i = \begin{cases} 0 & : \exists C \in \mathscr C : x_i \in C \subseteq \set{x_1, \ldots, x_i}\\ 1 & : \text {otherwise} \end{cases}$
Let $t$ be the mapping from the set of ordered tuple of $S$ defined by:
:$\map t {x_1, \ldots, x_q} = \ds \sum_{i = 1}^q \map \theta {x_1, \ldots, x_q}_i$
Let $\rho : \powerset S \to \Z$ be the mapping satisfying the matroid rank axioms defined by:
:$\forall A \subseteq S$:
::$\map \rho A = \begin{cases} 0 & : \text{if } A = \O \\ \map t {x_1, \ldots, x_q } & : \text{if } A = \set{x_1, \ldots, x_q} \end{cases}$
Let $M = \struct{S, \mathscr I}$ be the matroid corresponding to the rank function $\rho$.
Then:
:$\mathscr C$ is the set of circuits of $M$.
== Proof ==
Let $\mathscr C_M$ be the set of circuit of the matroid $M$.
===== Lemma 4 =====
===== Lemma 5 =====
==== $\mathscr C_M$ is a subset of $\mathscr C$ ====
Let $C \in \mathscr C_M$.
From Lemma 4:
:$\exists C_1 \in \mathscr C : C_1 \subseteq C$
From Lemma 5:
:$\exists C' \in \mathscr C_M : C' \subseteq C_1$
From Subset Relation is Transitive:
:$C' \subseteq C$
By the minimality of a circuit:
:$C' = C$
By definition of set equality:
:$C = C_1$
Hence:
:$C \in \mathscr C$
It follows that $\mathscr C_M \subseteq \mathscr C$.
===== $\mathscr C$ is a subset of $\mathscr C_M$ =====
Let $C \in \mathscr C$.
From Lemma 5:
:$\exists C_1 \in \mathscr C_M : C_1 \subseteq C$
From Lemma 4:
:$\exists C' \in \mathscr C : C' \subseteq C_1$
From Subset Relation is Transitive:
:$C' \subseteq C$
By circuit axiom $(\text C 2)$:
:$C' = C$
By definition of set equality:
:$C = C_1$
Hence:
:$C \in \mathscr C_M$.
It follows that $\mathscr C \subseteq \mathscr C_M$.
By definition of set equality:
:$\mathscr C = \mathscr C_M$
Category:Circuits of Matroid iff Matroid Circuit Axioms
Circuits of Matroid iff Matroid Circuit Axioms/Lemma 4

== Theorem ==
Let $S$ be a finite set.
Let $\mathscr C$ be a non-empty set of subsets of $S$ that satisfies the circuit axioms:
For any ordered tuple $\tuple{x_1, \ldots, x_q}$ of elements of $S$, let $\map \theta {x_1, \ldots, x_q}$ be the ordered tuple defined by:
:$\forall i \in \set{1, \ldots, q} : \map \theta {x_1, \ldots, x_q}_i = \begin{cases} 0 & : \exists C \in \mathscr C : x_i \in C \subseteq \set{x_1, \ldots, x_i}\\ 1 & : \text {otherwise} \end{cases}$
Let $t$ be the mapping from the set of ordered tuple of $S$ defined by:
:$\map t {x_1, \ldots, x_q} = \ds \sum_{i = 1}^q \map \theta {x_1, \ldots, x_q}_i$
Let $\rho : \powerset S \to \Z$ be the mapping satisfying the matroid rank axioms defined by:
:$\forall A \subseteq S$:
::$\map \rho A = \begin{cases} 0 & : \text{if } A = \O \\ \map t {x_1, \ldots, x_q } & : \text{if } A = \set{x_1, \ldots, x_q} \end{cases}$
Let $M = \struct{S, \mathscr I}$ be the matroid corresponding to the rank function $\rho$.
Let $\mathscr C_M$ be the set of circuits of $M$
Then:
:$\forall C \in \mathscr C_M : \exists C' \in \mathscr C : C' \subseteq C$
== Proof ==
Let $C \in \mathscr C_M$.
By definition of circuit:
:$C$ is dependent
By matroid axiom $(\text I 1)$:
:$C \neq \O$
Let $x \in C$.
From Proper Subset of Matroid Circuit is Independent:
:$C \setminus \set x$ is independent
We have:
=== Lemma 2 ===
From Lemma 2:
:$\exists C' \in \mathscr C : x \in C' \subseteq C$
Category:Circuits of Matroid iff Matroid Circuit Axioms
Circuits of Matroid iff Matroid Circuit Axioms/Lemma 5

== Theorem ==
Let $S$ be a finite set.
Let $\mathscr C$ be a non-empty set of subsets of $S$ that satisfies the circuit axioms:
For any ordered tuple $\tuple{x_1, \ldots, x_q}$ of elements of $S$, let $\map \theta {x_1, \ldots, x_q}$ be the ordered tuple defined by:
:$\forall i \in \set{1, \ldots, q} : \map \theta {x_1, \ldots, x_q}_i = \begin{cases} 0 & : \exists C \in \mathscr C : x_i \in C \subseteq \set{x_1, \ldots, x_i}\\ 1 & : \text {otherwise} \end{cases}$
Let $t$ be the mapping from the set of ordered tuple of $S$ defined by:
:$\map t {x_1, \ldots, x_q} = \ds \sum_{i = 1}^q \map \theta {x_1, \ldots, x_q}_i$
Let $\rho : \powerset S \to \Z$ be the mapping satisfying the matroid rank axioms defined by:
:$\forall A \subseteq S$:
::$\map \rho A = \begin{cases} 0 & : \text{if } A = \O \\ \map t {x_1, \ldots, x_q } & : \text{if } A = \set{x_1, \ldots, x_q} \end{cases}$
Let $M = \struct{S, \mathscr I}$ be the matroid corresponding to the rank function $\rho$.
Let $\mathscr C_M$ be the set of circuits of $M$
Then:
:$\forall C \in \mathscr C : \exists C' \in \mathscr C_M : C' \subseteq C$
== Proof ==
Let $C \in \mathscr C$.
From circuit axiom $(\text C 1)$:
:$\exists y \in C$
We have:
:$C \subseteq C = \paren{C \setminus \set y} \cup \set y$
=== Lemma 2 ===
We have:
From Rank of Independent Subset Equals Cardinality:
:$C \notin \mathscr I$
That is, $C$ is dependent.
From Dependent Subset Contains a Circuit:
:$\exists C' \in \mathscr C_M$ such that $C' \subseteq C$
Category:Circuits of Matroid iff Matroid Circuit Axioms
Circulant/Examples/Arbitrary Example 1

== Example of Circulant ==
This is an example of a circulant:
:$\begin {pmatrix} a & b & c & d \\ d & a & b & c \\ c & d & a & b \\ b & c & d & a \end {pmatrix}$
== Sources ==
* 
* 
Category:Examples of Circulants
Circumcenter of Triangle is Orthocenter of Medial

== Theorem ==
Let $\triangle ABC$ be a triangle.
Let $\triangle DEF$ be the medial triangle of $\triangle ABC$.
Let $K$ be the circumcenter of $\triangle ABC$.
Then $K$ is the orthocenter of $\triangle DEF$.
== Proof ==
:
Let $FG$, $DH$ and $EJ$ be the perpendicular bisectors of the sides of $AC$, $AB$ and $BC$ respectively.
From Circumscribing Circle about Triangle, the point $K$ where they intersect is the circumcenter of $\triangle ABC$.
From Perpendicular Bisector of Triangle is Altitude of Medial Triangle, $FG$, $DH$ and $EJ$ are the altitudes of $\triangle DEF$.
The result follows by definition of orthocenter.
Category:Circumcircles of Triangles
Category:Orthocenters of Triangles
Category:Medial Triangles
Circumscribing Circle about Regular Pentagon

== Theorem ==
About any given regular pentagon it is possible to circumscribe a circle.
== Construction ==
:
Let $ABCDE$ be the given regular pentagon.
Bisect $\angle BCD$ and $\angle CDE$ by the straight lines $CF$ and $DF$ respectively.
Draw the circle whose center is at $F$ and whose radius is $FC$ (or $FD$).
This circle is the one required.
== Proof ==
From $F$ join the straight lines $FB, FA, FE$.
In a similar manner to Inscribing Circle in Regular Pentagon it can be shown that $\angle CBA, \angle BAE, \angle AED$ have been bisected by the straight lines $FB, FA, FE$ respectively.
Since $\angle BCD = CDE$ and $2 \angle FCD = \angle BCD$ and $2 \angle CDF = \angle CDE$, it follows that $\angle FCD = \angle CDF$.
From Triangle with Two Equal Angles is Isosceles it follows that $FC = FD$.
Similarly it can be shown that $FB = FA = FE = FC = FD$.
Therefore the circle $ABCDE$ circumscribes the regular pentagon $ABCDE$ as required.
== Sources ==
* : Book $\text{IV}$. Propositions
Category:Circles
Category:Pentagons
Circumscribing Circle about Square

== Theorem ==
== Construction ==
:
Let $\Box ABCD$ be the given square.
Join $AC$ and $BD$ and let their point of intersection be $E$.
Draw the circle whose center is $E$ and whose radius is $AE$.
The circle $ABCD$ is the required circle.
== Proof ==
We have that $DA = AB$, $DC = BC$ and $AC$ is common.
So from Triangle Side-Side-Side Congruence $\triangle DAB = \triangle DCB$ and so $\angle DAC = \angle BAC$.
So $\angle DAB$ is bisected by $AC$.
Similarly each of $\angle ABC, \angle BCD, \angle DCA$ are bisected by $AC$ or $DB$.
We have that $\angle DAB = \angle ABC$, $2 \angle EAB = \angle DAB$, $2 \angle EBA = \angle ABC$.
Then $\angle EAB = \angle EBA$.
From Triangle with Two Equal Angles is Isosceles, $EA = EB$.
Similarly, $EA = EB = EC = ED$.
So the circle whose center is $E$ and whose radius is $AE$ passes through each of $B, C, D$.
Hence the result.
== Sources ==
* : Book $\text{IV}$. Propositions
Category:Circles
Category:Squares
Circumscribing Circle about Triangle

== Theorem ==
== Construction ==
Let $\triangle ABC$ be the given triangle.
Let the straight lines $AB$ and $AC$ be bisected at $D$ and $E$.
Let $DF$ and $EF$ be drawn perpendicular to $AB$ and $AC$ respectively.
Let a circle be drawn with center $F$ and radius $AF$.
This is the circle required.
== Proof ==
The point $F$ will be either inside, outside or on the edge $BC$ of $\triangle ABC$.
:
First suppose $F$ is inside $\triangle ABC$.
Join $FB$ and $FC$.
We have that $AD = DB$ and $DF$ is common and at right angles to $AB$.
So from Triangle Side-Angle-Side Congruence, $\triangle ADF = \triangle BDF$, and so $AF = BF$.
Similarly we can prove that $BF = CF$.
So $AF = BF = CF$.
Thus the circle with center $F$ and radius $AF$ also passes through $B$ and $C$.
That is, it circumscribes $\triangle ABC$.
:
Secondly, suppose $F$ lies on $BC$.
We have that $AD = DB$ and $DF$ is common and at right angles to $AB$.
So from Triangle Side-Angle-Side Congruence, $\triangle ADF = \triangle BDF$, and so $AF = BF$.
Similarly we can prove that $BF = CF$.
So $AF = BF = CF$.
Thus the circle with center $F$ and radius $AF$ also passes through $B$ and $C$.
That is, it circumscribes $\triangle ABC$.
:
Thirdly, suppose $F$ lies outside $\triangle ABC$.
Join $FB$ and $FC$.
We have that $AD = DB$ and $DF$ is common and at right angles to $AB$.
So from Triangle Side-Angle-Side Congruence, $\triangle ADF = \triangle BDF$, and so $AF = BF$.
Similarly we can prove that $BF = CF$.
So $AF = BF = CF$.
Thus the circle with center $F$ and radius $AF$ also passes through $B$ and $C$.
That is, it circumscribes $\triangle ABC$.
== Also see ==
* Perpendicular Bisectors of Triangle Meet at Point
== Sources ==
* : Book $\text{IV}$. Propositions
* : $\text {IV}$. Pure Geometry: Plane Geometry: The circumcentre
Category:Circumcircles of Triangles
Category:Circles
Category:Triangles
Circumscribing Regular Pentagon about Circle

== Theorem ==
About a given circle, it is possible to circumscribe a regular pentagon.
== Construction ==
:
Let $ABCDE$ be the given circle.
Find the center $F$ of the circle $ABCDE$.
Let $A, B, C, D, E$ be the vertices of a regular pentagon which has been inscribed within the circle $ABCDE$.
Draw tangents $GH, HK, KL, LM, MG$ to $ABCDE$ at the points $A, B, C, D, E$.
Then $GHKLM$ is the required regular pentagon.
== Proof ==
Join $FB, FK, FC, FL, FD$.
From Radius at Right Angle to Tangent, $HK, KL$ etc. are perpendicular to the radii they touch.
So $\angle KCF$ and $\angle LCF$ are right angles.
For the same reason, $\angle KBF$ and $\angle LDF$ are right angles.
By Pythagoras's Theorem, $FK^2 = FC^2 + CK^2$
For the same reason, $FK^2 = FB^2 + BK^2$.
As $FB = FC$ it follows that $BK = CK$.
From Triangle Side-Side-Side Congruence it follows that $\triangle FCK = \triangle FBK$ and so $\angle FKC = \angle FKB$ and $\angle KFC = \angle KFB$.
So $\angle BFC = 2 \angle KFC$ and $\angle BKC = 2 \angle FKC$.
For the same reason $\angle CFD = 2 \angle CFL$ and $\angle DLC = 2 \angle FLC$.
We have that the arc $BC$ equals the arc $CD$.
So from Angles on Equal Arcs are Equal, $\angle BFC = \angle CFD$.
As $\angle BFC = 2 \angle KFC$ and $\angle CFD = 2 \angle CFL$, it follows that $\angle KFC = \angle CFL$.
But $\angle FCK = \angle FCL$.
So from Triangle Angle-Side-Angle Congruence:
:$\triangle FCK = \triangle FLC$
So $KC = CL$ and $\angle FKC = \angle FLC$.
Since $KC = CL$ it follows that $KL = 2 KC$.
For the same reason $HK = 2 BK$ and $BK = KC$.
So $HK = KL$.
Similarly each of the straight lines $HG, GM, ML$ are all equal to $HK$ and $KL$.
So the pentagon $GHKLM$ is equilateral.
We have that:
:$\angle FKC = \angle FLC$
:$\angle HKL = 2 \angle FKC$
:$\angle KLM = 2 \angle FLC$
So:
:$\angle HKL = \angle KLM$
Similarly each of $\angle KHG, \angle HGM, \angle GML$ are equal to $\angle HKL$ and $\angle KLM$.
So the pentagon $GHKLM$ is equiangular.
== Sources ==
* : Book $\text{IV}$. Propositions
Category:Circles
Category:Pentagons
Circumscribing Square about Circle

== Theorem ==
== Construction ==
:
Let $ABCD$ be the given circle.
Let two diameters$AC, BD$ be drawn at right angles to one another.
Through $A, B, C, D$ draw $FG, GH, HK, KF$ perpendicular to $AC, BD$.
Then $\Box FGHK$ is the required square.
== Proof ==
From Line at Right Angles to Diameter of Circle, $FG, GH, HK, KF$ will be tangent to $ABCD$.
Since $\angle AEB$ is a right angle, $\angle EBG$ is also a right angle.
So $GH$ is parallel to $AC$.
For the same reason $FK$ is parallel to $AC$.
From Parallelism is Transitive Relation, $GH$ is parallel to $FK$.
Similarly, each of $GF, HK$ is parallel to $BD$.
So $GK, GC, AK, FB, BK$ are all parallelograms.
Therefore from Opposite Sides and Angles of Parallelogram are Equal $GF = HK$ and $GH = FK$.
We have that $AC = BD, AC = GH, AC = FK$.
Also $BD = GF = HK$.
So $\Box FGHK$ is equilateral.
Since $GBEA$ is a parallelogram, $\angle AEB$ is a right angle.
Therefore $\angle AGB$ is also a right angle.
Similarly the angles at $H, K, F$ are also right angles.
So $\Box FGHK$ is a square.
== Sources ==
* : Book $\text{IV}$. Propositions
Category:Circles
Category:Squares
Circumscribing about Circle Triangle Equiangular with Given

== Theorem ==
== Construction ==
:
Let $ABC$ be the given circle, and $\triangle DEF$ the given triangle.
Produce $EF$ in both directions to $G$ and $H$.
Find the center $K$ of $ABC$, and draw $KB$ at random.
Construct $\angle BKA$ equal to $\angle DEG$.
Construct $\angle BKC$ equal to $\angle DFH$.
Through $A, B, C$ draw tangents $LM, MN, LN$ to $ABC$.
Then $\triangle LMN$ is the required triangle.
== Proof ==
From Radius at Right Angle to Tangent, $LM$ is perpendicular to $AK$, $MN$ is perpendicular to $BK$, and $LN$ is perpendicular to $CK$.
We have that $\angle KAM, \angle KBM$ are right angles.
This leaves us with $\angle AMB + \angle AKB$ equal to two right angles.
We have that $\angle AKB = \angle DEG$ and so $\angle AMB + \angle DEG$ equals two right angles.
But from Two Angles on Straight Line make Two Right Angles $\angle DEF + \angle DEG$ equals two right angles.
It follows that $\angle AMB = \angle DEF$.
In a similar way we can show that $\angle ALC = \angle EDF$ and $\angle BNC = \angle DFE$.
Hence the result.
== Sources ==
* : Book $\text{IV}$. Propositions
Category:Circles
Category:Triangles
Clairaut's Differential Equation

== Theorem ==
Clairaut's differential equation is a first order ordinary differential equation which can be put into the form:
:$y = x y' + \map f {y'}$
Its general solution is:
:$y = C x + \map f C$
where $C$ is a constant.
== Proof ==
We have:
:$y = x y' + \map f {y'}$
Differentiating the equation $x$ we have:
=== Proof for General Solution ===
The first solution is:
By substituting into the original equation, we obtain:
Hence the result:
:$y = C_1 x + \map f {C_1}$
== Also known as ==
== Sources ==
* 
* 
* 
Category:Clairaut's Differential Equation
Category:First Order ODEs
Clairaut's Theorem

== Theorem ==
Let $\map f {x, y}$ be a function of the two independent variables $x$ and $y$.
Let $\map f {x, y}$ be continuous.
Let the partial deriviatives of $f$ also be continuous.
Then:
:$\dfrac {\partial^2 f} {\partial x \partial y} = \dfrac {\partial^2 f} {\partial y \partial x}$
== Proof ==
== Also known as ==
== Sources ==
* : $\S 13$: Derivatives: Partial Derivatives: $13.61$
* : $\S 15$: Derivatives: Partial Derivatives: $15.61.$
Category:Clairaut's Theorem
Category:Partial Differentiation
Class Difference of B with Class Difference of A with B

== Theorem ==
Let $A$ and $B$ be classes.
Then:
:$B \setminus \paren {A \setminus B} = B$
where $S \setminus T$ denotes class difference.
== Proof ==
== Sources ==
* : Chapter $2$: Some Basics of Class-Set Theory: $\S 5$ The union axiom: Exercise $5.6. \ \text {(d)}$
Category:Class Difference
Class Difference of B with Class Difference of A with B/Mistake

== Source Work ==
:
:Chapter $2$: Some Basics of Class-Set Theory:
::$\S 5$ The union axiom
:::Exercise $5.6. \ \text {(d)}$
== Mistake ==
:Show that for any classes $A$, $B$, $\ldots{}$
::$B - \paren {A - B} = \O$
== Correction ==
From Class Difference of B with Class Difference of A with B, the correct result is:
:$B - \paren {A - B} = B$
Perhaps the following exercise might have been meant:
:Show that for any classes $A$, $B$, $\ldots{}$
::$B \cap \paren {A - B} = \O$
the set-theoretical version of which is proved in Set Difference Intersection with Second Set is Empty Set.
== Sources ==
* : Chapter $2$: Some Basics of Class-Set Theory: $\S 5$ The union axiom: Exercise $5.6. \ \text {(d)}$
Category:Mistakes/Set Theory and the Continuum Problem
Class Difference with Class Difference

== Theorem ==
Let $A$ and $B$ be classes.
Then:
:$A \setminus \paren {A \setminus B} = A \cap B$
== Proof ==
== Also see ==
* Set Difference with Set Difference
== Sources ==
* : Chapter $2$: Some Basics of Class-Set Theory: $\S 5$ The union axiom: Exercise $5.6. \ \text {(c)}$
Category:Class Intersection
Category:Class Difference
Class Difference with Class Difference with Subclass

== Theorem ==
Let $A$ and $B$ be classes.
Let $B \subseteq A$.
Then:
:$A \setminus \paren {A \setminus B} = B$
== Proof ==
From Class Difference with Class Difference:
:$A \setminus \paren {A \setminus B} = A \cap B$
for all classes $A$ and $B$.
From Intersection with Subclass is Subclass:
:$A \subseteq B \iff A \cap B = A$
The result follows.
== Also see ==
* Relative Complement of Relative Complement
== Sources ==
* : Chapter $2$: Some Basics of Class-Set Theory: $\S 5$ The union axiom: Exercise $5.6. \ \text {(g)}$
Category:Subclasses
Category:Set Difference
Class Equality is Symmetric

== Theorem ==
Let $A$ and $B$ be classes.
Let $=$ denote class equality.
Then:
:$A = B \implies B = A$
== Proof ==
From Biconditional is Commutative:
:$\forall x: \left({ x \in A \iff x \in B }\right) \implies \forall x: \left({ x \in B \iff x \in A }\right)$
Hence the result by definition of class equality.
== Sources ==
* : $\S 4.7 \ (2)$
Category:Class Theory
Class Equality is Transitive

== Theorem ==
Let $A$, $B$, and $C$ be classes.
Let $=$ denote class equality.
Then
:$\left({ A = B \land B = C }\right) \implies B = A$
== Proof ==
By Universal Generalisation and Biconditional is Transitive:
:$\forall x: \left({ \left({ x \in A \iff x \in B }\right) \land \left({ x \in B \iff x \in C }\right) }\right) \implies \forall x: \left({ x \in A \iff x \in C }\right)$ by Universal Generalisation and Biconditional is Transitive
== Sources ==
* : $\S 4.7 \ (3)$
Category:Class Theory
Class Intersection Distributes over Class Union

== Theorem ==
Let $A$, $B$ and $C$ be classes.
Then:
:$A \cap \paren {B \cup C} = \paren {A \cap B} \cup \paren {A \cap C}$
where:
:$A \cap B$ denotes class intersection
:$B \cup C$ denotes class union.
== Proof ==
== Also see ==
* Intersection Distributes over Union
* Class Union Distributes over Class Intersection
== Sources ==
* : Chapter $2$: Some Basics of Class-Set Theory: $\S 5$ The union axiom: Exercise $5.6. \ \text {(a)}$
Category:Class Intersection
Category:Class Union
Category:Examples of Distributive Operations
Class Interval/Examples/Arbitrary Example 1

== Examples of Class Intervals ==
Consider the problem domain of employee wages.
We may count the number of employees earning a wage in each of the following ranges:
:between $\pounds 0.00$ and $\pounds 199.99$
:between $\pounds 200.00$ and $\pounds 299.99$
:between $\pounds 300.00$ and $\pounds 399.99$
Hence the class intervals in question are:
:$\closedint {0.00} {199.99}$
:$\closedint {200.00} {299.99}$
:$\closedint {300.00} {399.99}$
Hence the number of employees in each class interval gives the class frequency of that interval.
== Sources ==
* 
* 
Category:Examples of Class Intervals
Class Mapping has Minimally Superinductive Class

== Theorem ==
Let $g$ be a (class) mapping.
Then there exists a class $M$ that is minimally superinductive under $g$.
== Proof ==
== Sources ==
* : Chapter $6$: Order Isomorphism and Transfinite Recursion: $\S 5$ Transfinite recursion theorems: Exercise $5.2$
Category:Minimally Superinductive Classes