nyuuzyou/wikis huggingface dataset:
Volume of Spherical Segment

== Theorem ==
Let $S$ be a spherical segment.
The volume of $S$ is given by the formula:
:$V = \dfrac {\pi h \paren {3 {r_1}^2 + 3 {r_2}^2 + h^2} } 6$
where:
:$r_1$ and $r_2$ are the radii of the bases of $S$
:$h$ is the height of the zone formed by the bases of $S$.
== Proof ==
== Sources ==
* 
* 
Category:Spherical Segments
Category:Volume Formulas
Volume of Spherical Segment of One Base

== Theorem ==
Let $S$ be a spherical segment of one base.
The volume of $S$ is given by the formula:
:$V = \dfrac {\pi h \paren {3 r^2 + h^2} } 6$
where:
:$r$ is the radius of the base of $S$
:$h$ is the height of the zone formed by the base of $S$.
== Proof ==
== Sources ==
* 
* 
Category:Spherical Segments
Category:Volume Formulas
Volume of Spherical Wedge

== Theorem ==
Let $W$ be a spherical wedge on the surface of a sphere $S$.
The volume $V$ enclosed by $W$ is given by:
:$V = \dfrac {\pi r^3 \theta} {270}$
where:
:$r$ is the radius of $S$
:$\theta$ is the inclination of one of the planes to the other, measured in degrees.
== Proof ==
== Sources ==
* 
* 
Category:Spherical Wedges
Category:Volume Formulas
Volume of Torus/Formulation 1

== Theorem ==
Let $\TT$ be a torus.
Let $r$ be the radius of the generating circle of $\TT$.
Let $R$ be the distance of the center of the generating circle from the axis of revolution of $\TT$.
Then the volume $\VV$ enclosed by $\TT$ is given by:
:$\VV = 2 \pi^2 r^2 R$
== Proof ==
Recall Pappus's Centroid Theorem for Volume:
In this context:
:$C$ is the generating circle of $\TT$, which has radius $r$
:the centroid of $C$ is the center of $C$, which is at a distance $R$ from the axis of revolution.
From Area of Circle, the area of $C$ is $\pi r^2$.
From Perimeter of Circle, the distance travelled by the center of $C$ is $2 \pi R$.
Hence:
:$\VV = \paren {\pi r^2} \paren {2 \pi R}$
Hence the result.
== Sources ==
* 
* 
Category:Volume of Torus
Volume of Torus/Formulation 2

== Theorem ==
Let $\TT$ be a torus.
Let $a$ and $b$ be the inner radius and outer radius respectively of $\TT$.
Then the volume $\VV$ enclosed by $\TT$ is given by:
:$\VV = \dfrac {\pi^2 \paren {a + b} \paren {b - a}^2} 4$
== Proof ==
== Sources ==
* : $\S 4$: Geometric Formulas: Torus of Inner Radius $a$ and Outer Radius $b$: $4.45$
* : $\S 7$: Geometric Formulas: Torus of Inner Radius $a$ and Outer Radius $b$: $7.45.$
Category:Volume of Torus
Volume of Unit Hypersphere

== Theorem ==
The volume of the unit sphere in $n$-dimensional space increases as $n$ goes up to $5$, but decreases thereafter.
== Proof ==
=== Sequence of Volumes of Unit Hyperspheres ===
== Also see ==
* Maximum Volume of Unit Radius Sphere in Fractional Dimensions
== Sources ==
* : $5$
* : $5$
Category:Geometry
Category:Spheres
Volume of Unit Hypersphere/Sequence

== Sequence of Volumes of Unit Hyperspheres ==
The sequence of volumes of the unit sphere in $n$-dimensional space begins as follows:
== Sources ==
* : $5 \cdotp 256 \, 946 \, 404 \, 860 \ldots$
* : $5 \cdotp 25694 \, 64048 \, 60 \ldots$
Category:Spheres
Volume of Zone of One Base

== Theorem ==
Let $\ZZ$ be a zone of one base of a sphere $\SS$.
The volume $\VV$ of $\ZZ$ is given by:
:$\VV = \dfrac {\pi h^2 \paren {3 R - h} } 3$
where:
:$R$ is the radius of $\SS$
:$h$ is the height of $\ZZ$.
== Proof ==
== Sources ==
* : $\S 4$: Geometric Formulas: Spherical Cap of Radius $r$ and Height $h$: $4.40$
* : $\S 7$: Geometric Formulas: Spherical Cap of Radius $r$ and Height $h$: $7.40.$
Category:Zones of One Base
Category:Volume Formulas
Volumes of Cones or Cylinders on Equal Bases are in Same Ratio as Heights

== Theorem ==
== Proof ==
:
Let $EB$ and $FD$ be cylinders on equal bases, the circles $AB$ and $CD$.
It is to be shown that:
:$EB : FD = GH : KL$
where $GH$ and $KL$ are the axes of $GH$ and $KL$.
Let the axis $KL$ be produced to the point $N$.
Let $LN = GH$.
Let the cylinder $CM$ be described about the axis $LN$.
We have that:
:$EB$ and $CM$ have the same height
and
:$EB$ and $CM$ are on equal bases.
From :
:$EB$ and $CM$ are equal.
We have that the cylinder $FM$ has been cut by the plane $CD$ which is parallel to its bases.
Therefore by :
:$CM : FD = LN : KL$
But:
:$CM = EB$
and:
:$LN = GH$
Therefore:
:$EB : FD = GH : KL$
But from :
:$EB : FD = ABG : CDK$
where $ABG$ and $CDK$ are cones whose bases are $AB$ and $CD$ and whose apices are $G$ and $K$.
Therefore:
:$GH : KL = ABG : CDK$
and:
:$GH : KL = EB : FD$
== Sources ==
* : Book $\text{XII}$. Propositions
Category:Right Circular Cylinders
Category:Cones
Volumes of Parts of Cylinder cut by Plane Parallel to Opposite Planes are as Parts of Axis

== Theorem ==
== Proof ==
:
Let the cylinder $AD$ be cut by the plane $GH$ which is parallel to the bases $AB$ and $CD$ of $AD$.
Let $GH$ meet the axis of $AD$ at the point $K$.
It is to be demonstrated that:
:$BG : GD = EK : FK$
where:
:$BG$ and $GD$ are the cylinder whose parts together make the cylinder $AD$
:$EK$ and $FK$ are the axes of $BG$ and $GD$.
Let $EF$ be produced in both directions to $L$ and $M$.
Let there be set out:
:any number of axes $EN, NL$ equal to $EK$
and
:any number of axes $FO, OM$ equal to $FK$.
Let $PW$ be the cylinder on the axis $LM$.
Let $PQ$ and $VW$ be the bases of $PW$.
Let planes be defined parallel to $AB$ and $CD$ through the points $N$ and $O$.
Let the circles $RS$ and $TU$ be drawn in these planes with centers $N, O$ and with circumferences on $PV$ and $QW$.
We have that:
:$LN = NE = EK$
Therefore from :
:$QR : RB : BG = PQ : RS : AB$
But the bases $PQ, RS, AB$ are all equal.
Therefore the cylinders $QR, RB, BG$ are all equal.
So as:
:the axes $LN, NE, EK$ are all equal
and:
:the cylinders $QR, RB, BG$ are all equal
and:
:the number of axes equals the number of cylinders
it follows that:
:whatever multiple the axis $KL$ is of axis $EK$, the same multiple cylinder $QG$ will be of cylinder $GB$.
For the same reason:
:whatever multiple the axis $MK$ is of axis $KF$, the same multiple cylinder $WG$ will be of cylinder $GD$.
If the axis is greater than the axis, the cylinder will also be greater than the cylinder.
If the axis is less than the axis, the cylinder will also be less than the cylinder.
The conditions of are seen to be fulfilled.
Hence the result.
== Sources ==
* : Book $\text{XII}$. Propositions
Category:Right Circular Cylinders
Volumes of Similar Parallelepipeds are in Triplicate Ratio to Length of Corresponding Sides

== Theorem ==
=== Porism ===
== Proof ==
:
Let $AB$ and $CD$ be similar parallelepipeds.
Let $AE$ be the side of $AB$ corresponding to the side $CF$ of $CD$.
It is to be demonstrated that the parallelepiped $AB$ has to the parallelepiped $CD$ the ratio triplicate of that which $AE$ has to $CF$.
That is:
:$AB : CD = AE^3 : CF^3$
Let $EK, EL, EM$ be produced in a straight line with $AE, GE, HE$.
Let $EK = CF, EL = FN, EM = FR$.
Let the parallelogram $KL$ and parallelepiped $KP$ be completed.
We have that the two sides $KE$ and $EL$ equal the two sides $CF$ and $FN$.
Also $\angle KEL = \angle CFN$ because of similarity of $AB$ and $CD$.
Therefore the parallelogram $KL$ is equal and similar to the parallelogram $CN$.
For the same reason:
:the parallelogram $KM$ is equal and similar to the parallelogram $CR$
and:
:the parallelogram $EP$ is equal and similar to the parallelogram $DF$.
Therefore from:
::
and:
::
the whole parallelepiped $KP$ is equal and similar to the parallelepiped $CD$.
Let the parallelogram $GK$ be completed.
On the parallelograms $GK$ and $KL$ as bases, let the parallelepipeds $EO$ and $LQ$ be constructed with the same height as $AB$.
We have that $AB$ and $CD$ are similar.
So:
:$AE : CF = EG : FN = EH : FR$
and:
:$CF = EK$
:$FN : EL$
:$FR : EM$
Therefore:
:$AE : EK = GE : EL = HE : EM$
But from :
:$AE : EK = AG : GK$
:$GE : EL = GK : KL$
:$HE : EM = QE : KM$
Therefore:
:$AG : GK = GK : KL = QE : LM$
But from :
:$AG : GK = AB : EO$
:$GK : KL = OE : QL$
:$QE : KM = QL : KP$
Therefore:
:$AB : EO = EO : QL = QL : KP$
So from :
:the parallelepipeds $AB$ has to the parallelepipeds $KP$ the ratio triplicate of that which $AE$ has to $EO$.
But from :
:$AB : EO = AG : GK = AE : EK$
Hence:
:$AB : KP = AE^3 : EK^3$
But $KP = CD$ and $EK = CF$.
Therefore the parallelepiped $AB$ has to the parallelepiped $CD$ the ratio triplicate of that which $AE$ has to $CF$.
== Sources ==
* : Book $\text{XI}$. Propositions
Category:Parallelepipeds
Volumes of Similar Tetrahedra are in Triplicate Ratio of Corresponding Sides

== Theorem ==
=== Porism ===
== Proof ==
:
Let $ABCG$ be a tetrahedron whose base is $\triangle ABC$ and whose apex is $G$.
Let $DEFH$ be a tetrahedron whose base is $\triangle DEF$ and whose apex is $H$.
Let $ABCG$ and $DEFH$ be similar and similarly situated.
It is to be demonstrated that the ratio of $ABCG$ to $DEFH$ is triplicate of the ratio of the sides $BC$ and $EF$.
Let the parallelepipeds $BGML$ and $EHQP$ be completed.
We have that $ABCG$ is similar to $DEFH$.
Therefore:
:$\angle ABC = \angle DEF$
:$\angle GBC = \angle HEF$
:$\angle ABG = \angle DEH$
and:
:$AB : DE = BC : EF = BG : EH$
It follows that the parallelogram $BM$ is similar to the parallelogram $EQ$.
For the same reason:
:the parallelogram $BN$ is similar to the parallelogram $ER$
and:
:the parallelogram $BK$ is similar to the parallelogram $EO$
We also have that:
:$BM, BK, BN$ are similar to $EQ, EO, ER$.
But by :
:the three parallelograms $BM, BK, BN$ are equal and similar to their opposites
and
:the three parallelograms $EQ, EO, ER$ are equal and similar to their opposites.
Therefore the parallelepipeds $BGML$ and $EHQP$ are contained by equal numbers of similar planes.
Therefore $BGML$ is similar to $EHQP$.
So by :
:$BGML : EQHP = BC^3 : EF^3$
But from:
:
and
:
it follows that:
:$BGML : EQHP = ABCG : DEFH$
Hence the result.
== Sources ==
* : Book $\text{XII}$. Propositions
Category:Tetrahedra
Volumes of Similar Tetrahedra are in Triplicate Ratio of Corresponding Sides/Porism

== Porism to Volumes of Similar Tetrahedra are in Triplicate Ratio of Corresponding Sides ==
== Proof ==
Let them be divided into the tetrahedra, by virtue of the fact that the polygons forming their bases can be divided into triangles.
It follows from :
:the triangles forming the bases of these tetrahedra are similar.
So from :
:the ratio of one tetrahedron in the one complete pyramid to the corresponding tetrahedron in the other complete pyramid, so is the ratio of all the tetrahedra together in the one complete pyramid to all the tetrahedra together in the other complete pyramid.
But from :
:the ratio of one tetrahedron to another tetrahedron is in triplicate ratio to their edges.
Hence the result.
== Sources ==
* : Book $\text{XII}$. Propositions
Category:Pyramids
Volumes of Spheres are in Triplicate Ratio of Diameters

== Theorem ==
== Proof ==
:
Let the spheres $ABC$ and $DEF$ be described.
Let their diameters be $BC$ and $EF$.
It is to be demonstrated that:
:$ABC : DEF = BC^3 : EF^3$
Suppose it is not the case that:
:$ABC : DEF = BC^3 : EF^3$
Then:
:$ABC : O = BC^3 : EF^3$
where $O$ is either less than or greater than $DEF$.
Let $O$ be equal to a sphere $GHK$ which is less than $DEF$.
Let $DEF$ be described about $GHK$ with the same center as $GHK$.
From :
:let a polyhedron $X$ be inscribed inside $DEF$ which does not touch $GHK$.
Let a polyhedron $Y$ be inscribed inside $ABC$ which is similar to the one inscribed inside $DEF$.
From :
:$X : Y = BC^3 : EF^3$
But we have that:
:$ABC : GHK = BC^3 : EF^3$
Therefore:
:$ABC : GHK = X : Y$
So from :
:$ABC : X = GHK : Y$
But $ABC > X$, as $ABC$ completely encloses $X$.
Therefore $GHK > Y$.
But we have that $GHK < Y$, because $Y$ completely encloses $GHK$.
From this impossibility it follows that it is not the case that:
:$ABC : O = BC^3 : EF^3$
where $O < DEF$.
Similarly it can be shoen that it is not the case that:
:$DEF : O = EF^3 : BC^3$
where $O < ABC$.
Now it is to be shown that it is not the case that:
:$ABC : O = BC^3 : EF^3$
where $O$ is greater than $DEF$.
Suppose that:
:$ABC : LMN = BC^3 : EF^3$
where $LMN$ is a sphere which is greater than $DEF$.
Then:
:$LMN : ABC = EF^3 : BC^3$
We have that:
:$LMN > DEF$
From :
:$LMN : ABC = DEF : Z$
where $Z$ is some sphere which is less than $ABC$.
Therefore:
:$DEF : Z = EF^3 : BC^3$
where $Z$ is less than $ABC$.
This has been shown to be impossible.
Therefore it is not the case that:
:$ABC : LMN : BC^3 : EF^3$
where $LMN > DEF$.
It follows that:
:$ABC : DEF = BC^3 : EF^3$
== Sources ==
* : Book $\text{XII}$. Propositions
Category:Volume of Sphere
Category:Spheres
Von Mangoldt Equivalence

== Theorem ==
For $n \in \N_{>0}$, let $\map \Lambda n$ be the von Mangoldt function.
Then:
:$\ds \lim_{N \mathop \to \infty} \frac 1 N \sum_{n \mathop = 1}^N \map \Lambda n = 1$
is logically equivalent to the Prime Number Theorem.
== Proof ==
Observe:
Notice this sum will have:
:as many $\map \ln 2$ terms as there are powers of $2$ less than or equal to $N$
:as many $\map \ln 3$ terms as there are powers of $3$ less than or equal to $N$
and in general, if $p$ is a prime less than $N$, $\map \ln p$ will occur in this sum $\floor {\map {\log_p} N}$ times.
But:
:$\map \ln p \floor {\map {\log_p} N} \sim \map \ln p \map {\log_p} N = \map \ln N$
so:
:$\ds \sum_{p \text{ prime} \mathop \le N} \map \ln p \floor {\map {\log_p} N} \sim \sum_{p \text{ prime} \mathop \le N} \map \ln N = \map \pi N \map \ln N$
Therefore:
:$\ds \sum_{n \mathop = 1}^N \map \Lambda n \sim \map \pi N \map \ln N$
and so if:
:$\ds \lim_{N \mathop \to \infty} \frac 1 N \sum_{n \mathop = 1}^N \map \Lambda n = 1$ 
then:
:$\ds \lim_{N \mathop \to \infty} \frac 1 N \map \pi N \map \ln N = 1$
and vice versa.
But this last equation is precisely the Prime Number Theorem.
Hence our statement regarding the von Mangoldt function is logically equivalent to the Prime Number Theorem.
Category:Prime Numbers
Von Neumann-Bounded Set in Weak-* Topology of Normed Dual of Banach Space is Norm Bounded

== Theorem ==
Let $\struct {X, \norm {\, \cdot \,} }$ be a Banach space. 
Let $\struct {X^\ast, \norm {\, \cdot \,}_{X^\ast} }$ be the normed dual space of $\struct {X, \norm {\, \cdot \,} }$.
Let $w^\ast$ be the weak-$\ast$ topology on $X^\ast$. 
Let $E$ be a von Neumann-bounded subset of $\struct {X^\ast, w^\ast}$.
Then $E$ is a von Neumann-bounded subset of $\struct {X^\ast, \norm {\, \cdot \,}_{X^\ast} }$.
== Proof ==
Let $x \in X$. 
Then from Open Sets in Weak-* Topology of Topological Vector Space:
:$V = \set {f \in X^\ast : \cmod {\map f x} < 1}$ is an open neighborhood of $\mathbf 0_{X^\ast}$ in $\struct {X^\ast, w^\ast}$.
Then there exists $r > 0$ such that: 
:$E \subseteq r V$ 
Then for each $f \in E$, we have $f = r g$ for some $g \in V$. 
Then $\cmod {\map f x} = r \cmod {\map g x} < r$.
So:
:$\ds \sup_{f \in E} \cmod {\map f x} < \infty$
for each $x \in X$.
Since $X$ is a Banach space, we may apply the Banach-Steinhaus Theorem to obtain: 
:$\ds \sup_{f \in E} \norm f_{X^\ast} < \infty$
That is, there exists $M > 0$ such that: 
:$\norm f_{X^\ast} \le M$ for all $f \in E$. 
From Characterization of von Neumann-Boundedness in Normed Vector Space, we may conclude that $E$ is von Neumann-bounded in $\struct {X^\ast, \norm {\, \cdot \,}_{X^\ast} }$. 
Category:Weak-* Topologies
Category:Von Neumann-Bounded Subsets of Topological Vector Spaces
Von Neumann-Bounded Set in Weak Topology is Norm Bounded

== Theorem ==
Let $\struct {X, \norm {\, \cdot \,} }$ be a normed vector space. 
Let $w$ be the weak topology on $\struct {X, \norm {\, \cdot \,} }$. 
Let $B$ be a von Neumann-bounded subset of $\struct {X, w}$. 
Then $B$ is a von Neumann-bounded subset of $\struct {X, \norm {\, \cdot \,} }$
== Proof ==
Let $X^\ast$ be the normed dual space of $X$.
Let $X^{\ast \ast}$ be the second normed dual of $X$.
Let $f \in X^\ast$. 
Then from Open Sets in Weak Topology of Topological Vector Space:
:$V = \set {y \in X : \cmod {\map f y} < 1}$ is an open neighborhood of $\mathbf 0_X$ in $\struct {X, w}$. 
So there exists $r > 0$ such that: 
:$B \subseteq r V$
That is, if $x \in B$ we have $x = r y$ for some $y \in V$. 
That is: 
:$\cmod {\map f x} = r \cmod {\map f y} < r$ 
giving: 
:$\ds \cmod {\map f x} < r$ 
for all $x \in B$.
Hence: 
:$\ds \sup_{x \mathop \in B} \cmod {\map f x} < \infty$ 
for each $f \in X^\ast$.
That is: 
:$\ds \sup_{x \mathop \in B} \cmod {\map {x^\wedge} f} < \infty$ 
for each $f \in X^\ast$. 
Since $X^\ast$ is a Banach space, we can apply the Banach-Steinhaus Theorem to the maps $\family {x^\wedge : X^\ast \to \Bbb F}_{x \in X}$.
Doing that, we obtain:
:$\ds \sup_{x \mathop \in B} \norm {x^\wedge}_{X^{\ast \ast} } < \infty$
From Evaluation Linear Transformation on Normed Vector Space is Linear Isometry, we deduce that: 
:$\ds \sup_{x \mathop \in B} \norm x < \infty$
So there exists $M \in \R_{>0}$ such that: 
:$\norm x \le M$ for each $x \in B$.
From Characterization of von Neumann-Boundedness in Normed Vector Space, we may conclude that $B$ is von Neumann-bounded in $\struct {X, \norm {\, \cdot \,} }$.
Category:Weak Topologies on Topological Vector Spaces
Category:Von Neumann-Bounded Subsets of Topological Vector Spaces
Von Neumann Conjecture

== Conjecture ==
Let $G$ be a group.
Then $G$ is non-amenable $G$ contains a free subgroup of rank $2$.
== Resolution ==
This conjecture was answered in the negative, by .
He demonstrated that the Tarski monster is a non-amenable group with no such free subgroup.
== Also see ==
* Tarski Monster Group is Non-Amenable with no Free Subgroup of Two Generators
Category:Free Groups
Category:Famous False Conjectures
Von Neumann Construction of Natural Numbers is Minimally Inductive

== Theorem ==
Let $\omega$ denote the set of natural numbers as defined by the von Neumann construction.
$\omega$ is a minimally inductive class under the successor mapping.
== Proof ==
Consider Peano's axioms:
From Inductive Construction of Natural Numbers fulfils Peano's Axioms, $\omega$ fulfils Peano's axioms.
We note that from :
:$\O \in \omega$
We acknowledge from :
:the successor mapping defines that $n^+ := n \cup \set n$
and from the result follows.
== Sources ==
* : Chapter $3$: The Natural Numbers: $\S 4$ A double induction principle and its applications
Category:Natural Numbers
Category:Minimally Inductive Classes
Von Neumann Hierarchy Comparison

== Theorem ==
Let $x$ and $y$ be ordinals such that $x < y$.
Then:
:$\map V x \in \map V y$
:$\map V x \subset \map V y$
== Proof ==
The proof shall proceed by Transfinite Induction on $y$.
=== Basis for the Induction ===
If $y = 0$, then $x \not < y$.
This proves the basis for the induction.
=== Induction Step ===
Let $x < y \implies \map V x \in \map V y$.
Then:
In either case:
:$\map V x \in \map V {y^+}$
This proves the induction step.
=== Limit Case ===
Let $y$ be a limit ordinal.
Let:
:$\map V x \in \map V z$
for all $z \in y$ such that $x < z$.
Since $x < y$, it follows that $x < z$ for some $z \in y$ by Limit Ordinal Equals its Union.
By the inductive hypothesis:
:$\map V x \in \map V z$
But by Set is Subset of Union of Family:
:$\map V z \subseteq \map V y$
Therefore:
:$\map V x \in \map V y$
This proves the limit case.
:$\map V x \subset \map V y$ follows by Von Neumann Hierarchy is Supertransitive.
== Sources ==
* : $\S 9.10$
Category:Von Neumann Hierarchy
Von Neumann Hierarchy is Cumulative

== Theorem ==
For any two ordinals $x$ and $y$,
If $x < y$ then $V(x) \subsetneqq V(y)$.
== Proof ==
By Von Neumann Hierarchy Comparison, $V(x) \in V(y)$. (1)
By (1) and the Axiom:Axiom of Foundation, $V(x) \ne V(y)$.
Furthermore, by (1) and Von Neumann Hierarchy is Supertransitive,
Von Neumann Hierarchy is Supertransitive

== Theorem ==
Let $V$ denote the Von Neumann Hierarchy.
Let $x$ be an ordinal.
Then $\map V x$ is supertransitive.
== Proof ==
The proof shall proceed by Transfinite Induction on $x$.
=== Basis for the Induction ===
We have that:
:$V_0 = \O$
and $\O$ is supertransitive by the very fact that it has no elements.
This proves the basis for the induction.
=== Induction Step ===
Let $\map V x$ be supertransitive.
First, to prove transitivity.
Let $\map V x$ be transitive.
Then:
Next, to prove supertransitivity:
This proves the induction step.
=== Limit Case ===
Let $x$ be a limit ordinal.
Furthermore, let $\map V y$ be transitive for all $y \in x$.
Then:
This proves transitivity.
Now, to prove supertransitivity:
This proves the limit case.
== Sources ==
* : $\S 9.10$
Category:Von Neumann Hierarchy
WFF of PropLog is Balanced

== Theorem ==
Let $\mathbf A$ be a WFF of propositional logic.
Then $\mathbf A$ is a balanced string.
== Proof ==
We will prove by strong induction on $n$ that:
:All WFFs of length $n$ are balanced.
Let $\map l {\mathbf A}$ denote the number of left brackets in a string $\mathbf A$.
Let $\map r {\mathbf A}$ denote the number of right brackets in a string $\mathbf A$.
=== Induction Basis ===
The only WFFs of PropLog of Length 1‎ are letters of the alphabet and the symbols $\bot$ and $\top$.
By definition, none of these consist of brackets.
So every WFF $\mathbf A$ of length $1$ has $\map l {\mathbf A} = \map r {\mathbf A} = 0$.
So every WFF of length $1$ is balanced.
This provides the induction basis.
=== Induction Hypothesis ===
Now, suppose all WFFs of propositional logic of length at most $n$ are balanced.
This provides the induction hypothesis.
=== Induction Step ===
Let $\mathbf A$ be a WFF of propositional logic of length $n + 1$.
The following cases are to be considered (with $\mathbf B$ and $\mathbf C$ WFFs):
:$\mathbf A = \neg \mathbf B$
:$\mathbf A = \paren {\mathbf B \circ \mathbf C}$, where $\circ$ is one of the binary connectives
Suppose $\mathbf A = \neg \mathbf B$.
Then $\mathbf B$ is a WFF of length $n$.
By the induction hypothesis, it is hence balanced.
Since $\mathbf A$ contains the same brackets as $\mathbf B$, it must also be balanced.
Suppose now $\mathbf A = \paren {\mathbf B \circ \mathbf C}$, where $\circ$ is one of the binary connectives.
Then:
:$\map l {\mathbf A} = \map l {\mathbf B} + \map l {\mathbf C} + 1$
:$\map r {\mathbf A} = \map r {\mathbf B} + \map r {\mathbf C} + 1$.
Now, $\mathbf B$ and $\mathbf C$ are both of length strictly less than $n + 1$, and thus balanced, by the induction hypothesis.
That is:
:$\map l {\mathbf B} = \map r {\mathbf B}$
:$\map l {\mathbf C} = \map r {\mathbf C}$
It follows that also:
:$\map l {\mathbf A} = \map r {\mathbf A}$
We also note that the first symbol of $\paren {\mathbf B \circ \mathbf C}$ is $($.
Hence as $\mathbf B$ and $\mathbf C$ are both balanced, every prefix of $\paren {\mathbf B \circ \mathbf C}$ must have strictly more left brackets than right brackets.
This precisely states that $\mathbf A$ is balanced.
We assumed that all WFFs of length $n$ and less are balanced.
We have proved as a consequence all WFFs of length $n + 1$ are balanced.
The result now follows by strong induction.
== Sources ==
* : $\S 1.3$: Induction on Length of Wffs: Proposition $1.3.1$
Category:Language of Propositional Logic
WFFs of PropLog of Length 1

== Theorem ==
The only WFFs of propositional logic of length $1$ are:
* The letters of the formal grammar of propositional logic $\LL_0$
* The tautology symbol $\top$
* The contradiction symbol $\bot$.
== Proof ==
We refer to the rules of formation.
From $\mathbf W: \T \F$, $\top$ and $\bot$ (both of length 1) are WFFs.
From $\mathbf W: \PP_0$, all elements of $\PP_0$ (all of length 1) are WFFs.
Every other rule of formation of the formal grammar of propositional logic consists of an existing WFF in addition to at least one other primitive symbol.
Hence the result.
== Sources ==
* : $\S 1.3$: Induction on Length of Wffs
Category:Language of Propositional Logic
Waiting Time for Poisson Process obeys Exponential Distribution

== Theorem ==
Let $P$ be a Poisson process in which events occur at a rate of $k$ per unit time.
Then the waiting times of $P$ are exponentially distributed with expectation $\dfrac 1 k$.
== Proof ==
== Examples ==
== Sources ==
* 
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* 
Category:Waiting Time for Poisson Process obeys Exponential Distribution
Category:Poisson Processes
Category:Exponential Distribution