nyuuzyou/wikis huggingface dataset:
Seventeen Horses

== Classic Problem ==
A man dies, leaving $17$ horses to be divided among his heirs.
They are to be distributed in the ratio $\dfrac 1 2 : \dfrac 1 3 : \dfrac 1 9$.
How can this be done?
=== General Problem: $1$ ===
=== General Problem: $2$ ===
== Solution ==
One son gets $9$ horses, another gets $6$, and the final son gets $2$.
== Proof ==
We note that:
:$\dfrac 1 2 + \dfrac 1 3 + \dfrac 1 9 = \dfrac 9 {18} + \dfrac 6 {18} + \dfrac 2 {18} = \dfrac {17} {18}$
The fact that the ratios do not add up to $1$ tells us straight away that those ratios are not in fact the actual fractions of the inheritance that each will receive.
It is noted that the question carefully does not state that one son receives $\dfrac 1 2$, the next son $\dfrac 1 3$ and the third $\dfrac 1 9$, because that is not what actually happens here.
The solution is usually couched in the form of a story, along the following lines:
:A stranger passing through on horseback heard the sons squabbling amongst their string of horses, and stops and listens to their argument.
:Dismounting, he says to them, "Calm down, gentlemen! Let us see what we have here."
:He leaves his horse to go and mingle with the other horses, and says to the youngest son:
:"Go ahead and select one ninth of the horses here, but please leave my horse alone."
:And the youngest son does so, and puts a halter over the neck of each of two of the finest horses.
:The stranger continues to the second son: "Go ahead and select one third of the horses here, but not those that your brother has chosen, and again, please leave my horse alone."
:The second son does so, carefully counting, and puts a halter over the neck of each of six of the remaining horses.
:The stranger continues to the first son: "The remaining horses are yours, but please leave my horse alone."
:And as the first son puts a halter over the neck of each of the nine remaining horses of his father, the stranger remounted his own horse and went on his way.
== Historical Note ==
== Sources ==
* 
* : Exchanging the Knights: $103$
Category:Classic Problems
Category:Seventeen Horses
Seventeen Horses/General Problem 1

== Theorem ==
A man dies, leaving $n$ indivisible and indistinguishable objects to be divided among $3$ heirs.
They are to be distributed in the ratio $\dfrac 1 a : \dfrac 1 b : \dfrac 1 c$.
Let $\dfrac 1 a + \dfrac 1 b + \dfrac 1 c < 1$.
Then there are $7$ possible values of $\tuple {n, a, b, c}$ such that the required shares are:
:$\dfrac {n + 1} a, \dfrac {n + 1} b, \dfrac {n + 1} c$
These values are:
:$\tuple {7, 2, 4, 8}, \tuple {11, 2, 4, 6}, \tuple {11, 2, 3, 12}, \tuple {17, 2, 3, 9}, \tuple {19, 2, 4, 5}, \tuple {23, 2, 3, 8}, \tuple {41, 2, 3, 7}$
leading to shares, respectively, of:
:$\tuple {4, 2, 1}, \tuple {6, 3, 2}, \tuple {6, 4, 1}, \tuple {9, 6, 2}, \tuple {10, 5, 4}, \tuple {12, 8, 3}, \tuple {21, 14, 6}$
== Proof ==
It is taken as a condition that $a \ne b \ne c \ne a$.
We have that:
:$\dfrac 1 a + \dfrac 1 b + \dfrac 1 c + \dfrac 1 n = 1$
and so we need to investigate the solutions to the above equations.
From Sum of 4 Unit Fractions that equals 1, we have that the only possible solutions are:
From these, we can eliminate the following, because it is not the case that $a \ne b \ne c \ne a$:
Then we can see by inspection that the following are indeed solutions to the problem:
The remaining tuple we have is:
:$\dfrac 1 2 + \dfrac 1 3 + \dfrac 1 {10} + \dfrac 1 {15} = 1$
But we note that:
:$\dfrac 1 2 + \dfrac 1 3 + \dfrac 1 {10} = \dfrac {28} {30}$
which is not in the correct form.
Hence the $7$ possible solutions given.
== Sources ==
* 
* : Exchanging the Knights: $103$ (solution)
Category:Seventeen Horses
Category:Unit Fractions
Seventeen Horses/General Problem 2

== Theorem ==
A man dies, leaving $n$ indivisible and indistinguishable objects to be divided among $m$ heirs.
They are to be distributed in the ratio $\dfrac 1 {a_1} : \dfrac 1 {a_2} : \cdots : \dfrac 1 {a_m}$.
Let $t = \dfrac q r = \ds \sum_{k \mathop = 1}^m \dfrac 1 {a_k}$ expressed in canonical form.
Let $t \ne 1$.
Then it is possible to achieve the required share by adding $s$ objects to the existing $n$ such that:
:$s + q = r$
when $q = n$.
This still works whether $q$ is positive or negative.
== Proof ==
== Examples ==
== Sources ==
* : Exchanging the Knights: $103$ (solution)
Category:Seventeen Horses
Seventeen Horses/General Problem 2/Examples/31

== Example of Seventeen Horses Problem ==
Suppose $31$ horses are to be divided between $3$ sons in the ratio $\dfrac 1 2 : \dfrac 1 3 : \dfrac 1 5$.
We have that:
:$\dfrac 1 2 + \dfrac 1 3 + \dfrac 1 5 = \dfrac {15 + 10 + 6} {30} = \dfrac {31} {30}$
of which the numerator is seen to equal the total number of horses to be divided.
So we lend $1$ horse to someone, leaving $30$ horses.
The first son takes $\dfrac 1 2$ of these $30$, that is, $15$.
The second son takes $\dfrac 1 3$ of these $30$, that is, $10$.
The third son takes the remaining $5$ horses, then goes and recovers the $1$ horse that was lent, and so has his share of $6$ horses.
== Sources ==
* : Exchanging the Knights: $103$ (solution)
Category:Seventeen Horses
Seventeen Horses/General Problem 2/Examples/77

== Example of Seventeen Horses Problem ==
Suppose $77$ horses are to be divided between $4$ sons in the ratio $\dfrac 1 2 : \dfrac 1 3 : \dfrac 1 4 : \dfrac 1 5$.
We have that:
:$\dfrac 1 2 + \dfrac 1 3 + \dfrac 1 4 + \dfrac 1 5 = \dfrac {30 + 20 + 14 + 12} {60} = \dfrac {77} {60}$
of which the numerator is seen to equal the total number of horses to be divided.
So we lend $17$ horses to someone, leaving $60$ horses.
The fourth son takes $\dfrac 1 5$ of these $60$, that is, $12$.
The third son takes $\dfrac 1 4$ of these $60$, that is, $15$.
The second son takes $\dfrac 1 3$ of these $60$, that is, $20$.
The first son takes the remaining $13$ horses, then goes and hunts down the $17$ horses that were lent, and so has his share of $30$ horses.
== Sources ==
* : Exchanging the Knights: $103$ (solution)
Category:Seventeen Horses
Seventeen Horses/Historical Note

== Historical Note on Seventeen Horses ==
 tells us in his of $1992$ that the Seventeen Horses riddle appears in one of 's works: either of $1546$ or of $1556$.
, however, does not tell us which one of those two it appears in.
It has since appeared in countless anthologies of puzzles and riddles.
== Sources ==
* : Exchanging the Knights: $103$
Category:Historical Notes
Sextuple Angle Formulas

== Theorem ==
=== Sextuple Angle Formula for Sine ===
=== Sextuple Angle Formula for Cosine ===
=== Sextuple Angle Formula for Tangent ===
where $\sin, \cos, \tan$ denote sine, cosine and tangent respectively.
== Also see ==
* Quintuple Angle Formulas
Category:Sextuple Angle Formulas
Category:Multiple Angle Formulas
Category:Trigonometric Identities
Sextuple Angle Formulas/Cosine

== Theorem ==
:$\cos 6 \theta = 32 \cos^6 \theta - 48 \cos^4 \theta + 18 \cos^2 \theta - 1$
where $\cos$ denotes cosine.
== Proof ==
Hence:
== Sources ==
* : $\S 2$. Geometrical Representations: Exercise $9$
Category:Cosine Function
Category:Sextuple Angle Formulas
Category:Sextuple Angle Formula for Cosine
Sextuple Angle Formulas/Sine

== Theorem ==
:$\dfrac {\sin 6 \theta} {\sin \theta} = 32 \cos^5 \theta - 32 \cos^3 \theta + 6 \cos \theta$
where $\cos$ denotes cosine and $\sin$ denotes sine.
=== Corollary ===
== Proof ==
Hence:
== Sources ==
* : $\S 2$. Geometrical Representations: Exercise $9$
Category:Sine Function
Category:Sextuple Angle Formula for Sine
Category:Sextuple Angle Formulas
Sextuple Angle Formulas/Sine/Corollary

== Theorem ==
:$\sin 6 \theta = 6 \sin \theta \cos \theta - 32 \sin^3 \theta \cos \theta + 32 \sin^5 \theta \cos \theta$
where $\sin$ denotes sine and $\cos$ denotes cosine.
== Proof ==
Category:Sextuple Angle Formula for Sine
Sextuple Angle Formulas/Tangent

== Theorem ==
:$\tan 6 \theta = \dfrac { 6 \tan \theta - 20 \tan^3 \theta + 6 \tan^5 \theta } { 1 - 15 \tan^2 \theta + 15 \tan^4 \theta - \tan^6 \theta }$
where $\tan$ denotes tangent.
== Proof ==
Category:Tangent Function
Category:Sextuple Angle Formulas
Shape of Cosecant Function

== Theorem ==
The nature of the cosecant function on the set of real numbers $\R$ is as follows:
:$(1): \quad$ strictly decreasing on the intervals $\hointr {-\dfrac \pi 2} 0$ and $\hointl 0 {\dfrac \pi 2}$
:$(2): \quad$ strictly increasing on the intervals $\hointr {\dfrac \pi 2} \pi$ and $\hointl \pi {\dfrac {3 \pi} 2}$
:$(3): \quad$ $\csc x \to +\infty$ as $x \to 0^+$
:$(4): \quad$ $\csc x \to +\infty$ as $x \to \pi^-$
:$(5): \quad$ $\csc x \to -\infty$ as $x \to \pi^+$
:$(6): \quad$ $\csc x \to -\infty$ as $x \to 2 \pi^-$
== Proof ==
From Derivative of Cosecant Function::
:$\map {D_x} {\csc x} = -\dfrac {\cos x} {\sin^2 x}$
From Sine and Cosine are Periodic on Reals: Corollary:
:$\forall x \in \openint {-\dfrac \pi 2} {\dfrac {3 \pi} 2} \setminus \set {0, \pi}: \sin x \ne 0$
Thus, from Square of Non-Zero Element of Ordered Integral Domain is Strictly Positive:
:$\forall x \in \openint {-\dfrac \pi 2} {\dfrac {3 \pi} 2} \setminus \set {0, \pi}: \sin^2 x > 0$
From Sine and Cosine are Periodic on Reals: Corollary:
:$\cos x > 0$ on the open interval $\openint {-\dfrac \pi 2} {\dfrac \pi 2}$
It follows that:
:$\forall x \in \openint {-\dfrac \pi 2} {\dfrac \pi 2} \setminus \set 0: -\dfrac {\cos x} {\sin^2 x} < 0$
From Sine and Cosine are Periodic on Reals: Corollary:
:$\cos x < 0$ on the open interval $\openint {\dfrac \pi 2} {\dfrac {3 \pi} 2}$
It follows that:
:$\forall x \in \openint {\dfrac \pi 2} {\dfrac {3 \pi} 2} \setminus \set \pi: -\dfrac {\cos x} {\sin^2 x} > 0$
Thus, $(1)$ and $(2)$ follow from Derivative of Monotone Function.
From Zeroes of Sine and Cosine: $\sin 0 = \sin \pi = \sin 2 \pi = 0$.
From Sine and Cosine are Periodic on Reals: Corollary:
:$\sin x > 0$ on the open interval $\openint 0 \pi$
From the same source:
:$\sin x < 0$ on the open interval $\openint \pi {2 \pi}$
Thus, $(3)$, $(4)$, $(5)$ and $(6)$ follow from Infinite Limit Theorem.
=== Graph of Cosecant Function ===
== Also see ==
* Shape of Sine Function
* Shape of Cosine Function
* Shape of Tangent Function
* Shape of Cotangent Function
* Shape of Secant Function
== Sources ==
* : $\S 5$: Trigonometric Functions: Signs and Variations of Trigonometric Functions
Category:Cosecant Function
Shape of Cosine Function

== Theorem ==
The cosine function is:
:$(1): \quad$ strictly decreasing on the interval $\closedint 0 \pi$
:$(2): \quad$ strictly increasing on the interval $\closedint \pi {2 \pi}$
:$(3): \quad$ concave on the interval $\closedint {-\dfrac \pi 2} {\dfrac \pi 2}$
:$(4): \quad$ convex on the interval $\closedint {\dfrac \pi 2} {\dfrac {3 \pi} 2}$
== Proof ==
From the discussion of Real Cosine Function is Periodic, we know that:
:$\cos x \ge 0$ on the closed interval $\closedint {-\dfrac \pi 2} {\dfrac \pi 2}$
and:
:$\cos x > 0$ on the open interval $\openint {-\dfrac \pi 2} {\dfrac \pi 2}$
From the same discussion, we have that:
:$\map \sin {x + \dfrac \pi 2} = \cos x$
So immediately we have that $\sin x \ge 0$ on the closed interval $\closedint 0 \pi$, $\sin x > 0$ on the open interval $\openint 0 \pi$.
But $\map {\dfrac \d {\d x} } {\cos x} = -\sin x$ from Derivative of Cosine Function.
Thus from Derivative of Monotone Function, $\cos x$ is strictly decreasing on $\closedint 0 \pi$.
From Derivative of Sine Function it follows that:
:$\map {\dfrac {\d^2} {\d x^2} } {\cos x} = -\cos x$
On $\closedint {-\dfrac \pi 2} {\dfrac \pi 2}$ where $\cos x \ge 0$, therefore, $\map {\dfrac {\d^2} {\d x^2} } {\cos x} \le 0$.
From Second Derivative of Concave Real Function is Non-Positive it follows that $\cos x$ is concave on $\closedint {-\dfrac \pi 2} {\dfrac \pi 2}$.
The rest of the result follows similarly.
=== Graph of Cosine Function ===
== Also see ==
* Properties of Real Cosine Function 
* Shape of Sine Function
* Shape of Tangent Function
* Shape of Cotangent Function
* Shape of Secant Function
* Shape of Cosecant Function
== Sources ==
* : $\S 5$: Trigonometric Functions: Signs and Variations of Trigonometric Functions
* : $\S 16.5 \ (1)$
Category:Cosine Function
Shape of Cotangent Function

== Theorem ==
The nature of the cotangent function on the set of real numbers $\R$ is as follows:
:$\cot x$ is continuous and strictly decreasing on the interval $\openint 0 \pi$
:$\cot x \to +\infty$ as $x \to 0^+$
:$\cot x \to -\infty$ as $x \to \pi^-$
:$\cot x$ is not defined on $\forall n \in \Z: x = n \pi$, at which points it is discontinuous
:$\forall n \in \Z: \map \cot {n + \dfrac 1 2} \pi = 0$
== Proof ==
$\cot x$ is continuous and strictly decreasing on $\openint 0 \pi$:
Continuity follows from the Quotient Rule for Continuous Real Functions:
:$(1): \quad$ Both $\sin x$ and $\cos x$ are continuous on $\openint 0 \pi$ from Real Sine Function is Continuous and Cosine Function is Continuous
:$(2): \quad \sin x > 0$ on this interval.
The fact of $\cot x$ being strictly decreasing on this interval has been demonstrated in the discussion on Cotangent Function is Periodic on Reals.
$\cot x \to + \infty$ as $x \to 0^+$:
From Sine and Cosine are Periodic on Reals, we have that both $\sin x > 0$ and $\cos x > 0$ on $\openint 0 {\dfrac \pi 2}$.
We have that:
:$(1): \quad \cos x \to 1$ as $x \to 0^+$
:$(2): \quad \sin x \to 0$ as $x \to 0^+$
Thus it follows that $\cot x = \dfrac {\cos x} {\sin x} \to + \infty$ as $x \to 0^+$.
* $\tan x \to - \infty$ as $x \to \pi^-$:
From Sine and Cosine are Periodic on Reals, we have that $\sin x > 0$ and $\cos x < 0$ on $\openint {\dfrac \pi 2} \pi$.
We have that:
:$(1): \quad \cos x \to -1$ as $x \to \pi^-$
:$(2): \quad \sin x \to 0$ as $x \to \pi^-$
Thus it follows that $\cot x = \dfrac {\cos x} {\sin x} \to - \infty$ as $x \to \pi^-$.
$\cot x$ is not defined and discontinuous at $x = n \pi$:
From the discussion of Sine and Cosine are Periodic on Reals, it was established that $\forall n \in \Z: x = n \pi \implies \sin x = 0$.
As division by zero is not defined, it follows that at these points $\cot x$ is not defined either.
Now, from the above, we have:
:$(1): \quad \cot x \to + \infty$ as $x \to 0^+$
:$(2): \quad \cot x \to - \infty$ as $x \to \pi^-$
As $\map \cot {x + \pi} = \cot x$ from Cotangent Function is Periodic on Reals, it follows that $\cot x \to + \infty$ as $x \to \pi^+$.
Hence the left hand limit and right hand limit at $x = \pi$ are not the same.
From the periodic nature of $\cot x$, it follows that the same applies $\forall n \in \Z: x = n \pi$.
The fact of its discontinuity at these points follows from the definition of discontinuity.
$\map \cot {n + \dfrac 1 2} \pi = 0$:
Follows directly from Sine and Cosine are Periodic on Reals:
:$\forall n \in \Z: \map \cos {n + \dfrac 1 2} \pi = 0$
=== Graph of Cotangent Function ===
== Also see ==
* Shape of Sine Function
* Shape of Cosine Function
* Shape of Tangent Function
* Shape of Secant Function
* Shape of Cosecant Function
== Sources ==
* : $\S 5$: Trigonometric Functions: Signs and Variations of Trigonometric Functions
Category:Cotangent Function
Shape of Secant Function

== Theorem ==
The nature of the secant function on the set of real numbers $\R$ is as follows:
:$(1): \quad \sec x$ is continuous and strictly increasing on the intervals $\hointr 0 {\dfrac \pi 2}$ and $\hointl {\dfrac \pi 2} \pi$
:$(2): \quad \sec x$ is continuous and strictly decreasing on the intervals $\hointr {-\pi} {-\dfrac \pi 2}$ and $\hointl {-\dfrac \pi 2} 0$
:$(3): \quad \sec x \to + \infty$ as $x \to -\dfrac \pi 2^+$
:$(4): \quad \sec x \to + \infty$ as $x \to \dfrac \pi 2^-$
:$(5): \quad \sec x \to - \infty$ as $x \to \dfrac \pi 2^+$
:$(6): \quad \sec x \to - \infty$ as $x \to \dfrac {3 \pi} 2^-$
== Proof ==
From Derivative of Secant Function:
:$\map {D_x} {\sec x} = \dfrac {\sin x} {\cos^2 x}$
From Sine and Cosine are Periodic on Reals: Corollary:
:$\forall x \in \openint {-\pi} \pi \setminus \set {-\dfrac \pi 2, \dfrac \pi 2}: \cos x \ne 0$
Thus, from Square of Non-Zero Element of Ordered Integral Domain is Strictly Positive:
:$\forall x \in \openint {-\pi} \pi \setminus \set {-\dfrac \pi 2, \dfrac \pi 2}: \cos^2 x > 0$
From Sine and Cosine are Periodic on Reals: Corollary:
:$\sin x > 0$ on the open interval $\openint 0 \pi$
It follows that:
:$\forall x \in \openint 0 \pi \setminus \set {\dfrac \pi 2}: \dfrac {\sin x} {\cos^2 x} > 0$
From Sine and Cosine are Periodic on Reals: Corollary::
:$\sin x < 0$ on the open interval $\openint {-\pi} 0$
It follows that:
:$\forall x \in \openint {-\pi} 0 \setminus \set {-\dfrac \pi 2}: \dfrac {\sin x} {\cos^2 x} < 0$
Thus, $(1)$ and $(2)$ follow from Derivative of Monotone Function and Differentiable Function is Continuous.
From Zeroes of Sine and Cosine::
:$\cos - \dfrac \pi 2 = \cos \dfrac \pi 2 = \cos \dfrac {3 \pi} 2 = 0$
From Sine and Cosine are Periodic on Reals: Corollary:
:$\cos x > 0$ on the open interval $\openint {-\dfrac \pi 2} {\dfrac \pi 2}$
From the same source:
:$\cos x < 0$ on the open interval $\openint {\dfrac \pi 2} {\dfrac {3 \pi} 2}$
Thus, $(3)$, $(4)$, $(5)$ and $(6)$ follow from Infinite Limit Theorem.
=== Graph of Secant Function ===
== Also see ==
* Shape of Sine Function
* Shape of Cosine Function
* Shape of Tangent Function
* Shape of Cotangent Function
* Shape of Cosecant Function
== Sources ==
* : $\S 5$: Trigonometric Functions: Signs and Variations of Trigonometric Functions
Category:Secant Function
Shape of Sine Function

== Theorem ==
The sine function is:
:$(1): \quad$ strictly increasing on the interval $\closedint {-\dfrac \pi 2} {\dfrac \pi 2}$
:$(2): \quad$ strictly decreasing on the interval $\closedint {\dfrac \pi 2} {\dfrac {3 \pi} 2}$
:$(3): \quad$ concave on the interval $\closedint 0 \pi$
:$(4): \quad$ convex on the interval $\closedint \pi {2 \pi}$
== Proof ==
From the discussion of Real Sine Function is Periodic, we have that:
:$\sin \paren {x + \dfrac \pi 2} = \cos x$
The result then follows directly from the Shape of Cosine Function.
=== Graph of Sine Function ===
== Also see ==
* Properties of Real Sine Function
* Shape of Cosine Function
* Shape of Tangent Function
* Shape of Cotangent Function 
* Shape of Secant Function
* Shape of Cosecant Function
== Sources ==
* : $\S 5$: Trigonometric Functions: Signs and Variations of Trigonometric Functions
Category:Sine Function
Sheaf Associated to Injective Module over Noetherian Ring is Flasque

== Theorem ==
Let $A$ be a Noetherian commutative ring.
Let $I$ be an injective $A$-module.
Then the sheaf $\tilde I$ associated to $I$ on $\Spec A$ is flasque.
== Proof ==
== Sources ==
* Proposition $\text{III}.3.4$
Category:Schemes
Sheldon Conjecture/Historical Note

== Historical Note on Sheldon Conjecture ==
The Sheldon Conjecture was expounded by the fictional physicist , who expounded upon the properties of the number $73$ in (fittingly) episode $73$ of , by Lee Aronsohn, Jim Reynolds and Maria Ferrari:
The unspoken implication of this statement was that $73$ was the only (prime) number to have this property.
The question was settled by and in $2019$, who proved that the Sheldon Conjecture is indeed true, thereby turning it into a theorem.
However, as of time of writing (May $2019$), this result is still referred to as the Sheldon Conjecture.
== Sources ==
* 
* 
Category:Historical Notes
Shift Mapping is Lower Adjoint iff Appropriate Maxima Exist

== Theorem ==
Let $\struct {S, \preceq}$ be a meet semilattice.
Then the following two conditions are equivalent:
:$(1): \quad \forall x \in S, f: S \to S: \paren {\forall s \in S: \map f s = x \wedge s} \implies f$ is lower adjoint
:$(2): \quad \forall x, t \in S: \max \set {s \in S: x \wedge s \preceq t}$ exists.
== Proof ==
=== $(1) \implies (2)$ ===
Assume that
:$\forall x \in S, f: S \to S: \paren {\forall s \in S: \map f s = x \wedge s} \implies f$ is lower adjoint
Let $x, t \in S$.
Define $f: S \to S$:
:$\forall s \in S: \map f s = x \wedge s$
By assumption:
:$f$ is lower adjoint
By definition of lower adjoint:
:$\exists g: S \to S: \struct {g, f}$ is Galois connection
By Galois Connection is Expressed by Maximum:
:$\forall s \in S: \map g s = \map \max {f^{-1} \sqbrk {s^\preceq} }$
Then:
:$\map \max {f^{-1} \sqbrk {t^\preceq} }$ exists.
By definition of image of set:
:$\max \set {s \in S: \map f s \in t^\preceq}$ exists.
By definition of lower closure of element:
:$\max \set {s \in S: x \wedge s \preceq t}$ exists.
=== $(2) \implies (1)$ ===
Assume that:
:$\forall x, t \in S: \max \set {s \in S: x \wedge s \preceq t}$ exists.
Let $x \in S, f: S \to S$ such that:
:$\forall s \in S: \map f s = x \wedge s$
As maxima exist define a mapping $g: S \to S$:
:$\forall s \in S: \map g s = \map \max {f^{-1} \sqbrk {s^\preceq} }$
We will prove that:
:$f$ is an increasing mapping
Let $y, z \in S$ such that:
:$y \preceq z$
By definition of $f$:
:$\map f y = x \wedge y$ and $\map f z = x \wedge z$
Thus by Meet Semilattice is Ordered Structure:
:$\map f y \preceq \map f z$
By Galois Connection is Expressed by Maximum:
:$\struct {g, f}$ is a Galois connection
By definition:
:$f$ is a lower adjoint.
== Sources ==
* 
* 
Category:Galois Connections
Shift Mapping is Lower Adjoint iff Meet is Distributive with Supremum

== Theorem ==
Let $\struct {S, \preceq}$ be a complete lattice.
Then:
:$\forall x \in S, f: S \to S: \paren {\forall y \in S: \map f y = x \wedge y} \implies f$ is lower adjoint of a Galois connection
:
:$\forall x \in S, X \subseteq S: x \wedge \sup X = \sup \set {x \wedge y: y \in X}$
== Proof ==
=== Sufficient Condition ===
Assume that
:$\forall x \in S, f: S \to S: \paren {\forall y \in S: \map f y = x \wedge y} \implies f$ is lower adjoint of a Galois connection
Let $x \in S, X \subseteq S$.
Define a mapping $f: S \to S$:
:$\forall y \in S: \map f y := x \wedge y$
By assumption:
:$f$ is lower adjoint of a Galois connection.
By Lower Adjoint Preserves All Suprema:
:$f$ preserves all suprema.
By definition of mapping preserves all suprema
:$f$ preserves the supremum of $X$.
By definition of complete lattice:
:$X$ admits a supremum.
Thus
=== Necessary Condition === 
Assume that
:$\forall x \in S, X \subseteq S: x \wedge \sup X = \sup \set {x \wedge y: y \in X}$
Let $x \in S$, $f: S \to S$ such that
:$\forall y \in S: \map f y = x \wedge y$
We will prove that
:$f$ preserves all suprema.
Let $X \subseteq S$ such that
:$X$ admits a supremum.
Thus by definition of complete lattice:
:$\map {f^\to} X$ admits a supremum.
Thus
Thus by definition
:$f$ preserves the supremum of $X$.
Thus by All Suprema Preserving Mapping is Lower Adjoint of Galois Connection:
:$f$ is lower adjoint of a Galois connection. 
== Sources ==
* 
* 
Category:Galois Connections
Category:Complete Lattices
Shift of Stopping Time is Stopping Time

== Theorem ==
Let $\struct {\Omega, \Sigma, \sequence {\FF_n}_{n \ge 0}, \Pr}$ be a filtered probability space.
Let $T$ be a stopping time with respect to $\sequence {\FF_n}_{n \ge 0}$.
Let $t$ be a extended natural number.
Then $T + t$ is a stopping time with respect to $\sequence {\FF_n}_{n \ge 0}$.
== Proof ==
By Constant Function is Stopping Time, $t$ is a stopping time with respect to $\sequence {\FF_n}_{n \ge 0}$.
By Sum of Stopping Times is Stopping Time, $T + t$ is a stopping time with respect to $\sequence {\FF_n}_{n \ge 0}$.
Category:Stopping Times