nyuuzyou/wikis huggingface dataset:
Truth Table/Examples/((not p) and q) implies ((not q) and r)

== Example of Truth Table ==
The truth table for the WFF of propositional logic:
:$\paren {\paren {\lnot p} \land q} \implies \paren {\paren {\lnot q} \land r}$:
can be presented as:
:$\begin{array}{cccc|c|cccc}
((\lnot & p) & \land & q) & \implies & ((\lnot & q) & \land & r) \\
\hline
\T & \F & \F & \F & \T & \T & \F & \F & \F \\
\T & \F & \F & \F & \T & \T & \F & \T & \T \\
\T & \F & \T & \T & \F & \F & \T & \F & \F \\
\T & \F & \T & \T & \T & \F & \T & \T & \T \\
\F & \T & \F & \F & \T & \T & \F & \T & \F \\
\F & \T & \F & \F & \T & \T & \F & \T & \T \\
\F & \T & \F & \T & \T & \F & \T & \F & \F \\
\F & \T & \F & \T & \T & \F & \T & \F & \T \\
\end{array}$
== Sources ==
* : $\S 1$: Informal statement calculus: $\S 1.2$: Truth functions and truth tables: Exercises $3 \ \text{(g)}$
Category:Truth Tables
Truth Table/Examples/(not p) and (not q)

== Example of Truth Table ==
The truth table for the WFF of propositional logic:
:$\paren {\lnot p} \land \paren {\lnot q}$:
can be presented as:
:$\begin{array}{cc|c|cc}
(\lnot & p) & \land & (\lnot & q) \\
\hline
\T & \F & \T & \T & \F \\
\T & \F & \F & \F & \T \\
\F & \T & \F & \T & \F \\
\F & \T & \F & \F & \T \\
\end{array}$
== Sources ==
* : $\S 1$: Informal statement calculus: $\S 1.2$: Truth functions and truth tables: Exercises $3 \ \text{(a)}$
Category:Truth Tables
Truth Table/Examples/(p and q) implies r

== Example of Truth Table ==
The truth table for the WFF of propositional logic:
:$\paren {p \land q} \implies r$:
can be depicted as:
:$\begin{array}{ccc|c|c}
(p & \land & q) & \implies & r \\
\hline
\F & \F & \F & \T & \F \\
\F & \F & \F & \T & \T \\
\F & \F & \T & \T & \F \\
\F & \F & \T & \T & \T \\
\T & \F & \F & \T & \F \\
\T & \F & \F & \T & \T \\
\T & \T & \T & \F & \F \\
\T & \T & \T & \T & \T \\
\end{array}$
== Sources ==
* : $\S 1$: Informal statement calculus: $\S 1.2$: Truth functions and truth tables: Exercises $3 \ \text{(d)}$
Category:Truth Tables
Truth Table/Examples/(p and q) or (r and s)

== Example of Truth Table ==
The truth table for the WFF of propositional logic:
:$\paren {p \land q} \lor \paren {r \land s}$:
can be presented as:
:$\begin{array}{ccc|c|ccc}
(p & \land & q) & \lor & (r & \land & s) \\
\hline
\F & \F & \F & \F & \F & \F & \F \\
\F & \F & \F & \F & \F & \F & \T \\
\F & \F & \F & \F & \T & \F & \F \\
\F & \F & \F & \T & \T & \T & \T \\
\F & \F & \T & \F & \F & \F & \F \\
\F & \F & \T & \F & \F & \F & \T \\
\F & \F & \T & \F & \T & \F & \F \\
\F & \F & \T & \T & \T & \T & \T \\
\T & \F & \F & \F & \F & \F & \F \\
\T & \F & \F & \F & \F & \F & \T \\
\T & \F & \F & \F & \T & \F & \F \\
\T & \F & \F & \T & \T & \T & \T \\
\T & \T & \T & \T & \F & \F & \F \\
\T & \T & \T & \T & \F & \F & \T \\
\T & \T & \T & \T & \T & \F & \F \\
\T & \T & \T & \T & \T & \T & \T \\
\end{array}$
== Sources ==
* : $\S 1$: Informal statement calculus: $\S 1.2$: Truth functions and truth tables: Exercises $3 \ \text{(f)}$
Category:Truth Tables
Truth Table/Examples/(p iff (not q)) or q

== Example of Truth Table ==
The truth table for the WFF of propositional logic:
:$\paren {p \iff \paren {\lnot q} } \lor q$:
can be presented as:
:$\begin{array}{cccc|c|c}
(p & \iff & (\lnot & q)) & \lor & q \\
\hline
\F & \F & \T & \F & \F & \F \\
\F & \T & \F & \T & \T & \T \\
\T & \T & \T & \F & \T & \F \\
\T & \F & \F & \T & \T & \T \\
\end{array}$
== Sources ==
* : $\S 1$: Informal statement calculus: $\S 1.2$: Truth functions and truth tables: Exercises $3 \ \text{(e)}$
Category:Truth Tables
Truth Table/Examples/not ((p implies q) implies (not (q implies p)))

== Example of Truth Table ==
The truth table for the WFF of propositional logic:
:$\lnot \paren {\paren {p \implies q} \implies \paren {\lnot \paren {q \implies p} } }$:
can be presented as:
:$\begin{array}{c|ccc|c|cccc}
\lnot & ((p & \implies & q) & \implies & (\lnot & (q & \implies & p))) \\
\hline
\T & \F & \T & \F & \F & \F & \F & \T & \F \\
\F & \F & \T & \T & \T & \T & \T & \F & \F \\
\F & \T & \F & \F & \T & \F & \F & \T & \T \\
\T & \T & \T & \T & \F & \F & \T & \T & \T \\
\end{array}$
== Sources ==
* : $\S 1$: Informal statement calculus: $\S 1.2$: Truth functions and truth tables: Exercises $3 \ \text{(b)}$
Category:Truth Tables
Truth Table/Examples/p implies (q implies r)

== Example of Truth Table ==
The truth table for the WFF of propositional logic:
:$p \implies \paren {q \implies r}$:
can be presented as:
:$\begin{array}{c|c|ccc}
p & \implies & (q & \implies & r) \\
\hline
\F & \T & \F & \T & \F \\
\F & \T & \F & \T & \T \\
\F & \T & \T & \F & \F \\
\F & \T & \T & \T & \T \\
\T & \T & \F & \T & \F \\
\T & \T & \F & \T & \T \\
\T & \F & \T & \F & \F \\
\T & \T & \T & \T & \T \\
\end{array}$
== Sources ==
* : $\S 1$: Informal statement calculus: $\S 1.2$: Truth functions and truth tables: Exercises $3 \ \text{(c)}$
Category:Truth Tables
Truth Table/Examples/p implies (q or r)

== Example of Truth Table ==
The truth table for the WFF of propositional logic:
:$p \implies \paren {q \lor r}$:
can be presented as:
:$\begin{array}{c|c|c|c|c}
p & q & r & q \lor r & p \implies \paren {q \lor r} \\
\hline
\F & \F & \F & \F & \T \\
\F & \F & \T & \T & \T \\
\F & \T & \F & \T & \T \\
\F & \T & \T & \T & \T \\
\T & \F & \F & \F & \F \\
\T & \F & \T & \T & \T \\
\T & \T & \F & \T & \T \\
\T & \T & \T & \T & \T \\
\end{array}$
== Sources ==
* : $\S 1$: Informal statement calculus: $\S 1.2$: Truth functions and truth tables: Example $1.4 \ \text{(b)}$
Category:Truth Tables
Tschirnhaus Transformation yields Depressed Polynomial

== Theorem ==
Let $\map f x$ be a polynomial of order $n$:
:$a_n x^n + a_{n - 1} x^{n - 1} + \cdots + a_1 x + a_0$
Then the Tschirnhaus transformation:
$y = x + \dfrac {a_{n - 1} } {n a_n}$
converts $f$ into a depressed polynomial:
:$b_n y^n + b_{n - 1} y^{n - 1} + \cdots + b_1 y + b_0$
where $b_{n - 1} = 0$.
== Proof ==
Substituting $y = x + \dfrac {a_{n - 1} } {n a_n}$ gives us:
:$x = y - \dfrac {a_{n - 1} } {n a_n}$
By the Binomial Theorem:
:$a_n x^n = a_n \paren {y^n - \dfrac {a_{n - 1} } {a_n} y^{n - 1} + \map {f'_{n - 2} } y}$
where $\map {f'_{n - 2} } y$ is a polynomial in $y$ of order $n - 2$.
Now we note that:
:$a_{n - 1} x^{n- 1} = a_{n - 1} y^{n - 1} - \map {f' '_{n - 2} } y$
where $\map {f' '_{n - 2} } y$ is another polynomial in $y$ of order $n - 2$.
The terms in $y^{n - 1}$ cancel out.
Hence the result.
Category:Polynomial Theory
Category:Tschirnhaus Transformations
Tubular Neighborhood Theorem

== Theorem ==
Let $\struct {M, g}$ be a Riemannian manifold.
Let $S$ be a submanifold of $M$.
Suppose, $S$ is embedded.
Then $S$ has a tubular neighborhood in $M$.
== Proof ==
== Sources ==
* : $\S 5$: The Levi-Civita Connection. Tubular Neighborhoods and Fermi Coordinates
Category:Tubular Neighborhoods
Category:Named Theorems
Tukey's Lemma/Formulation 1

== Theorem ==
Let $S$ be a non-empty set of finite character.
Then $S$ has an element which is maximal with respect to the subset relation.
== Proof ==
Let $C \subseteq S$ be a chain.
We will show that $\ds \bigcup C \in S$.
Let $x$ be a finite subset of $\ds \bigcup C$.
By the definitions of subset and of union, each element of $x$ is an element of at least one element of $C$.
By the Principle of Finite Choice, there is a mapping $c: x \to C$ such that:
:$\forall a \in x: a \in \map c x$
Then $\map c x$ is a finite subset of $C$.
From Finite Totally Ordered Set is Well-Ordered, $\map c x$ has a greatest element $m \in C \subseteq S$.
Then:
:$x$ is a finite subset of $m$
and:
:$m \in S$
Since $S$ has finite character:
:$x \in S$
We have thus shown that every finite subset of $\ds \bigcup C$ is in $S$.
Since $S$ is of finite character:
:$\ds \bigcup C \in S$
Thus by Zorn's Lemma, $S$ has a maximal element.
== Also known as ==
Category:Tukey's Lemma
Tukey's Lemma/Formulation 2

== Theorem ==
Let $S$ be a non-empty set of finite character.
Then every element of $S$ is a subset of a maximal element of $S$ under the subset relation.
== Proof ==
== Also known as ==
== Sources ==
* : Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text {II}$ -- Maximal principles: $\S 5$ Maximal principles
Category:Tukey's Lemma
Tukey's Lemma implies Zorn's Lemma

== Theorem ==
Let Tukey's Lemma be accepted as true.
Then Zorn's Lemma holds.
== Proof ==
Recall Tukey's Lemma:
Recall Zorn's Lemma:
So, let us assume Tukey's Lemma.
Let $S$ be a non-empty ordered set, with $T$ as defined.
From Property of being Totally Ordered is of Finite Character:
:$T$ is of finite character.
From Ordering on Singleton is Total Ordering it follows trivially that $T$ is non-empty.
Then by Tukey's Lemma:
:every element of $T$ is a subset of a maximal element of $T$ under the subset relation.
Thus it is seen that Zorn's Lemma likewise holds.
== Sources ==
* : Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text {II}$ -- Maximal principles: $\S 5$ Maximal principles: Proposition $5.7$
Category:Tukey's Lemma
Category:Zorn's Lemma
Tuning Fork Delta Sequence

== Theorem ==
Let $\sequence {\map {\delta_n} x}$ be a sequence such that:
:$\map {\delta_n} x := \begin{cases}
-n & : \size x < \dfrac 1 {2n} \\ \\
2n & : \dfrac 1 {2n} \le \size x \le \dfrac 1 n \\ \\
0 & : \size x > \dfrac 1 n \end{cases}$
Then $\sequence {\map {\delta_n} x}_{n \mathop \in {\N_{>0} } }$ is a delta sequence.
That is, in the distributional sense it holds that:
:$\ds \lim_{n \mathop \to \infty} \map {\delta_n} x = \map \delta x$
or
:$\ds \lim_{n \mathop \to \infty} \int_{-\infty}^\infty \map {\delta_n} x \map \phi x \rd x = \map \delta \phi$
where $\phi \in \map \DD \R$ is a test function, $\delta$ is the Dirac delta distribution, and $\map \delta x$ is the abuse of notation, usually interpreted as an infinitely thin and tall spike with its area equal to $1$.
== Proof 1 ==
== Proof 2 ==
Category:Tuning Fork Delta Sequence
Category:Examples of Delta Sequences
Category:Dirac Delta Distribution
Tuning Fork Delta Sequence/Proof 2

== Theorem ==
== Proof ==
We have that:
Let $\map g x = \map \phi x - \map \phi 0$
Then:
Furthermore:
We have that $\map g 0 = 0$.
By definition, $\phi$ is a smooth real function on $\R$.
By Differentiable Function is Continuous, $\map g x$ is continuous at $x = 0$.
Then:
:$\ds \forall \epsilon' \in \R_{> 0} : \exists \delta \in \R_{> 0} : 0 < \size x < \delta \implies \size {\map g x} < \epsilon'$
Let $\ds \delta = \frac 1 N$ where $N \in \R_{> 0}$.
Let $\ds \epsilon' = \frac \epsilon 3$ where $\epsilon \in \R_{> 0}$.
Then
:$\ds \forall \epsilon > 0 : \exists N \in \R_{> 0} : \size x < \frac 1 N \implies \size {\map g x} < \frac \epsilon 3$
Suppose:
:$\ds n \in \N : \size x \le \frac 1 n < \frac 1 N$
That is, suppose:
:$\ds n \in \N : \frac 1 {\size {x} } \ge n > N$
Then:
Since $n$ was arbitrary:
:$\ds \forall \epsilon \in \R_{>0} : \exists N \in \R_{>0} : \forall n \in \N_{>0} : n > N \implies \size {\int_{-\infty}^\infty \map g x \map {\delta_n} x \rd x} < \epsilon$
By definition of the limit of a real sequence:
:$\ds \lim_{n \mathop \to \infty} \int_{-\infty}^\infty \map g x \map {\delta_n} x \rd x = 0$
However:
== Sources ==
* : Chapter $1$ Mathematical Preliminaries $1.11$ Dirac Delta Function
Category:Tuning Fork Delta Sequence
Turing Machine/Examples/Natural Number Addition

== Example of Turing Machine ==
Let $\TT$ be a Turing machine whose task is to implement addition on the non-zero natural numbers.
Let $a, b \in \N_{>0}$ be expressed in unary notation:
$a$ and $b$ are to be presented to $\TT$ on a tape in the form:
:$0000 \underbrace {111 \ldots 1}_{\text {$a$ times} } 0 \underbrace {111 \ldots 1}_{\text {$b$ times} } 000$
That is, $a$ and $b$ are presented on the tape with one $0$ between them.
The program of $\TT$ is of the form:
The elements of each tuple are interpreted as follows:
$\TT$ starts in state $1$.
== Proof ==
When $\TT$ is in state $1$:
:If the symbol read is $0$, leave it where it is, read the next symbol to the right, and leave $\TT$ in state $1$.
:If the symbol read is $1$, replace it with $0$, read the next symbol to the right, and move $\TT$ into state $2$.
When $\TT$ is in state $2$:
:If the symbol read is $0$, replace it with $1$, and go into halt state. The direction to move is irrelevant.
:If the symbol read is $1$, leave it where it is, read the next symbol to the right, and leave $\TT$ in state $2$.
If $a = 2$ and $b = 3$, the tape starts as:
:$000110111000$
After $\TT$ encounters the first $1$, the tape looks like:
:$000010111000$
After $\TT$ encounters the next $0$, the tape looks like:
:$000011111000$
revealing $5$ as the sum of $2$ and $3$.
At this stage, $\TT$ has halted.
</onlyinclude>
== Sources ==
* 
* 
Category:Examples of Turing Machines
Tusi Couple is Diameter of Deferent

== Theorem ==
A Tusi couple is a degenerate case of the hypocycloid whose form is a straight line that forms a diameter of the deferent.
== Proof ==
Let $C_1$ be a circle of radius $b$ rolling without slipping around the inside of a circle $C_2$ of radius $a$.
Let $C_2$ be embedded in a cartesian plane with its center $O$ located at the origin.
Let $P$ be a point on the circumference of $C_1$.
Let $C_1$ be initially positioned so that $P$ is its point of tangency to $C_2$, located at point $A = \tuple {a, 0}$ on the $x$-axis.
Let $H$ be the hypocycloid formed by the locus of $P$.
From Number of Cusps of Hypocycloid from Integral Ratio of Circle Radii we have that $H$ will have $2$ cusps :
:$a = 2 b$
By Equation of Hypocycloid a hypocycloid can be expressed in parametric form as:
:$x = \paren {a - b} \cos \theta + b \map \cos {\paren {\dfrac {a - b} b} \theta}$
:$y = \paren {a - b} \sin \theta - b \map \sin {\paren {\dfrac {a - b} b} \theta}$
Hence:
Thus the $x$ coordinate of the $2$ cusp hypocycloid has a range $\closedint {-b} b$.
Similarly:
Thus the $y$ coordinate of the $2$ cusp hypocycloid is fixed at $y = 0$.
Thus the $2$ cusp hypocycloid consists of the line segment:
:$x \in \closedint {-b} b, y = 0$.
which is a diameter of the containing circle.
== Sources ==
* : Chapter $\text {B}.21$: The Cycloid: Problem $9$
Category:Tusi Couples
Tutte's Wheel Theorem

== Theorem ==
Every $3$-connected graph can be obtained by the following procedure:
* Start with $G_0 := K_4$
* Given $G_i$ pick a vertex $v$
* Split into $v'$ and $v' '$ and add edge $\set {v', v' '}$
This procedure directly follows from the theorem: 
:A graph $G$ is $3$-connected (A) there exists a sequence $G_0, G_1, \dotsc, G_n$ of graphs with the following properties (B): 
:* $G_0 = K_4$ and $G_n = G$
:* $G_{i + 1}$ has an edge $e = x y$ with $\map \deg x, \map \deg y \ge 3$ and $G_i = G_{i + 1} / e$ for every $i < n$.
== Proof ==
=== Lemma ===
=== A $\implies$ B ===
Directly follows from the .
=== B $\implies$ A ===
If $G_i$ is $3$-connected then so is $G_{i + 1}$ for every $i < n$. 
Suppose not. Then:
:$G_i = G_{i + 1} / e$ 
where $G_{i + 1}$ is $2$-connected. 
Let $S$ be a vertex cut such that $\card S \le 2$, and $C_1, C_2$ be two components of $G_{i + 1} - S$.
Since $x, y$ are connected, we can assume:
:$\map V {C_1} \cap \set {x, y} = \O$
Then, $C_2$ cannot contain both $x$ and $y$ nor it can contain any $v \notin \set {x, y}$.
(otherwise $v$ or $u_{x y}$ would be separated from $C_1$ in $G_i$ by at most two vertices.)
But now $C_2$ contains only one vertex: either $x$ or $y$. 
This contradicts:
:$\map \deg x, \map \deg y \ge 3$
Category:Tutte's Wheel Theorem
Category:Graph Theory
Tutte's Wheel Theorem/Lemma

== Lemma ==
Let $G = \struct {V, E}$ be a $3$-connected graph such that $\card V > 4$.
Then there exists $e \in E$ such that $G / e$ is also $3$-connected.
== Proof ==
 that for all $e \in E$, $G / e$ contains a vertex cut $S$ such that $\card S \le 2$.
Let $x, y \in V$ be endvertices of $e$.
Since $\map \kappa G \ge 3$, the contracted vertex $v_{x, y}$ of $G / e$ lies in $S$.
That is, $\exists z \in V: z \notin \set {x, y}$.
Let $S = \set {v_{x, y}, z}$.
Then $T = \set {x, y, z}$ is a vertex cut of $G$. 
Thus, every vertex in $T$ has an edge to every component of $G' = G - T$. 
Let $C = \struct {V', E'}$ be the smallest component of $G'$.
Let $\map {\Gamma_G} z$ be the neighborhood of $z$.
Let $u \in \map {\Gamma_G} z \cap V'$.
By assumption, $G / z u$ is again not $3$-connected.
Thus, there exists $w \in V$ such that $\set {w, z, u}$ is a vertex cut of $G$. 
It also follows that every vertex in $\set {w, z, u}$ has an edge to every component of $G' ' = G - \set {w, z, u}$. 
Since $x, y$ are connected, there exists a component $D = \struct {V' ', E' '}$ of $G' '$ such that $V' ' \cap \set {x, y} = \O$.
Then:
:$V' ' \subseteq \map {\Gamma_G} z \cap V' = \O$
Hence $V' ' \subsetneq V'$ by the choice of $D$, which contradicts the assumption that $C$ is the smallest component. 
Category:Tutte's Wheel Theorem
Twin Prime Conjecture

== Conjecture ==
It is conjectured that there exist infinitely many pairs of twin primes: that is, primes which differ by $2$.
== Landau's Problems ==
This is the $2$nd of Landau's problems.
== Sources ==
* : $\text I$: The Series of Primes: $1.4$ The sequence of primes
* : $1,32032 36316 \ldots$
* : $1 \cdotp 90195 \ldots$
* 
* : Chapter $\text {B}.16$: The Sequence of Primes
* : $1 \cdotp 90216 \, 054 \ldots$
* 
* 
* : Chapter $7$: Patterns in Numbers: Fermat
* 
Category:Twin Primes
Category:Landau's Problems
Category:Unproven Hypotheses
Twin Primes/Examples/2,003,663,613 x 2^195,000 pm 1

== Example of Pair of Twin Primes ==
The integers:
:$2 \, 003 \, 663 \, 613 \times 2^{195 \, 000} \pm 1$
are twin primes.
== Historical Note ==
== Sources ==
* : Chapter $7$: Patterns in Numbers: Fermat
Category:Twin Primes/Examples‎
Twin Primes/Examples/2,003,663,613 x 2^195,000 pm 1/Historical Note

== Historical Note on $2 \, 003 \, 663 \, 613 \times 2^{195 \, 000} \pm 1$ ==
According to in his of $2008$, the numbers:
:$2 \, 003 \, 663 \, 613 \times 2^{195 \, 000} \pm 1$
were the largest pair of twin primes known at the date of writing.
They were discovered in $2007$ by , and .
== Sources ==
* : Chapter $7$: Patterns in Numbers: Fermat
Category:Historical Notes‎
Twin Primes/Sequence

== Integer Sequence ==
The sequence of twin primes begins:
:$3, 5, 7, 11, 13, 17, 19, 29, 31, 41, 43, 59, 61, 71, 73, 101, 103, 107, 109, \ldots$
== Sources ==
* 
* 
Category:Twin Primes
Category:Prime Numbers
Two-Person Zero-Sum Game/Examples

== Examples of Two-Person Zero-Sum Game ==
The two players are $A$ and $B$.
Player $A$ has $m$ strategies: $A_1, A_2, \ldots, A_m$.
Player $B$ has $n$ strategies: $B_1, B_2, \ldots, B_n$.
The game is zero-sum in that a payoff of $p$ to $A$ corresponds to a payoff of $-p$ to $B$.
The game is completely defined by the table specifying the payoff to $A$ for each strategy of $A$ and $B$.
=== Abstract Example 1 ===
=== Abstract Example 2 ===
=== Abstract Example 3 ===
=== Abstract Example 4 ===
=== Abstract Example 5 ===
=== Political Example 1 ===
=== Political Example 2 ===
=== Political Example 3 ===
Category:Examples of Two-Person Zero-Sum Games
Two-Person Zero-Sum Game/Examples/-1, 0, 1

== Example of Two-Person Zero-Sum Game ==
The two players are $A$ and $B$.
$A$ and $B$ each have three possible moves, the values:
:$-1, 0, 1$
Let $A$'s move be $s$.
Let $B$'s move be $t$.
Then the payoff to $A$ is given by:
:$\map p A = s \paren {t - s} + t \paren {t + s}$
As $G$ is zero-sum it follows that the payoff to $A$ is given by:
:$\map p B = s \paren {s - t} - t \paren {t + s}$
=== Payoff Table ===
== Sources ==
* : Chapter $\text{I}$: An Outline of the Theory of Games: $3$
Category:Examples of Two-Person Zero-Sum Games
Two-Person Zero-Sum Game/Examples/-1, 0, 1/Payoff Table

== Payoff Table for $-1$, $0$, $1$ Game ==
The following payoff table specifies the $-1$, $0$, $1$ game whose payoff to $A$ is given by:
:$\map p A = s \paren {t - s} + t \paren {t + s}$
== Sources ==
* : Chapter $\text{I}$: An Outline of the Theory of Games: $3$
Category:Examples of Payoff Tables‎
Two-Person Zero-Sum Game/Examples/Abstract 1

== Example of Two-Person Zero-Sum Game ==
The two players are $A$ and $B$.
Player $A$ has $m$ strategies: $A_1, A_2, \ldots, A_m$.
Player $B$ has $n$ strategies: $B_1, B_2, \ldots, B_n$.
The game is zero-sum in that a payoff of $p$ to $A$ corresponds to a payoff of $-p$ to $B$.
The following table specifies the payoff to $A$ for each strategy of $A$ and $B$.
This has an equilibrium point at $\tuple {A_2, B_3}$ with a payoff to $A$ of $2$.
== Proof ==
By inspection, strategy $A_2$ dominates strategy $A_1$.
Therefore by Elimination by Domination, $A_1$ can be eliminated.
$A_1$ having been eliminated, it is seen by inspection that $B_3$ dominates both $B_1$ and $B_2$.
$B_1$ and $B_2$ having been eliminated, it is seen by inspection that $A_2$ dominates $A_3$.
The optimum strategies are therefore $A_2$ and $B_3$, and the payoff to $A$ is $2$.
Given $B_3$, every other strategy adopted by $A$ will result in a smaller payoff to $A$.
Given $A_2$, every other strategy adopted by $B$ will result in a larger payoff to $A$.
Hence, by definition, $\tuple {A_2, B_3}$ is an equilibrium point.
== Sources ==
* : $\S 2$: The Two-Person, Zero-Sum Game with Equilibrium Points: Introductory Problems: Figure $2.1$
Category:Examples of Two-Person Zero-Sum Games
Two-Person Zero-Sum Game/Examples/Abstract 2

== Example of Two-Person Zero-Sum Game ==
The two players are $A$ and $B$.
Player $A$ has $m$ strategies: $A_1, A_2, \ldots, A_m$.
Player $B$ has $n$ strategies: $B_1, B_2, \ldots, B_n$.
The game is zero-sum in that a payoff of $p$ to $A$ corresponds to a payoff of $-p$ to $B$.
The following table specifies the payoff to $A$ for each strategy of $A$ and $B$.
This has an equilibrium point at $\tuple {A_1, B_1}$ with a payoff to $A$ of $-2$.
== Proof ==
By inspection, strategy $B_1$ dominates strategy $B_3$.
Therefore by Elimination by Domination, $B_3$ can be eliminated.
$B_3$ having been eliminated, it is seen by inspection that $A_1$ dominates both $A_2$ and $A_3$.
$A_2$ and $A_3$ having been eliminated, it is seen by inspection that $B_1$ dominates $B_2$.
The optimum strategies are therefore $A_1$ and $B_1$, and the payoff to $A$ is $-2$.
Given $B_1$, every other strategy adopted by $A$ will result in a smaller payoff to $A$.
Given $A_1$, every other strategy adopted by $B$ will result in a larger payoff to $A$.
Hence, by definition, $\tuple {A_1, B_1}$ is an equilibrium point.
== Sources ==
* : $\S 2$: The Two-Person, Zero-Sum Game with Equilibrium Points: Introductory Problems: Figure $2.2$
Category:Examples of Two-Person Zero-Sum Games
Two-Person Zero-Sum Game/Examples/Abstract 3

== Example of Two-Person Zero-Sum Game ==
The two players are $A$ and $B$.
Player $A$ has $m$ strategies: $A_1, A_2, \ldots, A_m$.
Player $B$ has $n$ strategies: $B_1, B_2, \ldots, B_n$.
The game is zero-sum in that a payoff of $p$ to $A$ corresponds to a payoff of $-p$ to $B$.
The following table specifies the payoff to $A$ for each strategy of $A$ and $B$.
This has an equilibrium point at $\tuple {A_2, B_3}$ with a payoff to $A$ of $5$.
== Proof ==
By inspection, strategy $B_1$ dominates strategy $B_4$.
Therefore by Elimination by Domination, $B_4$ can be eliminated.
$B_4$ having been eliminated, it is seen by inspection that $A_2$ dominates both $A_3$ and $A_4$.
$A_3$ and $A_4$ having been eliminated, it is seen by inspection that $B_3$ dominates both $B_1$ and $B_2$.
$B_1$ and $B_2$ having been eliminated, it is seen by inspection that $A_2$ dominates $A_1$.
The optimum strategies are therefore $A_2$ and $B_3$, and the payoff to $A$ is $5$.
Given $B_3$, every other strategy adopted by $A$ will result in a smaller payoff to $A$.
Given $A_2$, every other strategy adopted by $B$ will result in a larger payoff to $A$.
Hence, by definition, $\tuple {A_2, B_3}$ is an equilibrium point.
== Sources ==
* : $\S 2$: The Two-Person, Zero-Sum Game with Equilibrium Points: Introductory Problems: Figure $2.3$
Category:Examples of Two-Person Zero-Sum Games