nyuuzyou/wikis huggingface dataset:
Straight Lines which make Equal Angles with Perpendicular to Straight Line are Equal

== Theorem ==
Let $AB$ be a straight line.
Let $C$ be a point which is not on $AB$.
Let $D$ be a point on $AB$ such that $CD$ is perpendicular to $AB$.
Let $E, F$ be points on $AB$ such that $\angle DCE = \angle DCF$.
Then $CE = CF$. 
== Proof ==
$\triangle CDE$ and $\triangle CDF$ are right triangle where $CE$ and $CF$ are the hypotenuses.
We have:
:$\angle CDE = \angle CDF$ as both are right angles.
:$\angle DCE = \angle DCF$ by hypothesis.
:$CD$ is common.
Thus by Triangle Angle-Side-Angle Congruence, $\triangle CDE$ and $\triangle CDF$ are congruent.
Hence the result.
== Sources ==
* : Chapter $1$: Triangles and Polygons: Theorems and Corollaries $1.21 \ \text{(ii)}$
Category:Lines
Category:Angles
Strassen Algorithm/Examples/Order 2

== Example of Use of the Strassen Algorithm ==
Let $\mathbf A$ and $\mathbf B$ be square matrices of order $2$:
Let $\mathbf C = \mathbf {A B}$.
Then:
:$C = \begin {pmatrix} p_1 + p_4 - p_5 + p_7 & p_3 + p_5 \\ p_2 + p_4 & p_1 + p_3 - p_2 + p_6 \end {pmatrix}$
where:
It is seen that the Strassen algorithm uses only $7$ multiplication operations as opposed to the usual $8$.
== Proof ==
By definition of matrix multiplication:
:$\mathbf {A B} = \begin {pmatrix} a_{11} b_{11} + a_{12} b_{21} & a_{11} b_{12} + a_{12} b_{22} \\ a_{21} b_{11} + a_{22} b_{21} & a_{21} b_{12} + a_{22} b_{22} \end {pmatrix}$
Hence to show that $\mathbf C = \mathbf {A B}$, we need to show that:
So:
Hence the result.
== Sources ==
* 
* 
Category:Examples of Use of Strassen Algorithm
Stratified Sample/Examples/Arbitrary Example 1

== Example of Stratified Sample ==
Suppose a survey is to be carried out in a school to determine attitudes towards the possible banning of nuclear weapons.
It is proposed that different proportions of boys and girls may or may not favour such a ban.
Hence a two-strata sample may be used, with the gender of the pupils forming the strata.
Suppose there were $400$ boys and $200$ girls at the school in question.
Then a stratified sample of $40$ boys and $20$ girls might be taken, giving a total sample size of $60$.
This adheres to the guideline that the sizes of the samples are proportional to the stratum sizes.
== Sources ==
* 
* 
Category:Examples of Stratified Samples
Strawson Leap

== Philosophical Argument ==
Let the following statements be assumed:
:$(1): \quad$ God exists or God doesn't exist.
:$(2): \quad$ If God does exist, it would be beneficial to believe this.
:$(3): \quad$ If there is no God then there can be no Free Will.
:$(4): \quad$ Either $\textrm a)$ we have Free Will or $\textrm b)$ we do not have Free Will, and our perception of what we believe to be our Free Will is in fact an illusion.
:$(5): \quad$ It is impossible to test whether or not we have Free Will.
$(1)$ and $(4)$ are dependent upon the Law of Excluded Middle.
$(3)$ is an adaptation of 's work, which can be found at http://www.bbc.co.uk/programmes/b00z5y9z
Then either:
: Situation $\textrm A$ pertains: There is free will
or:
: Situation $\textrm B$ pertains: There is no free will.
We cannot know in which situation we are living.
In situation $\textrm A$, free will can only be a result of the existence of God.
God exists.
We benefit from believing in God.
Therefore the most beneficial action is to believe in God.
In situation $\textrm B$, since free will does not exist, we have no choice in what we believe.
Therefore if we end up believing in God, it was going to happen anyway.
We have no say in our choice.
In this situation it doesn't matter what our choice is.
Let us now make our choice.
If I have chosen to believe, then if situation $\textrm A$ pertains I have chosen well.
If situation $\textrm B$ pertains, nothing is lost, as I was destined to believe anyway.
If I have chosen not to believe, then if situation $\textrm A$ pertains, I have chosen badly.
If situation $\textrm B$ pertains, nothing is lost, as I was destined not to believe anyway.
Even if I think there's the slimmest possibility of genuine free will, there is still nothing to be lost in ending up believing.
The reason that this argument is called a "Leap" is that even if we accept the argument, for some of us it still may take a "leap of faith" to act as though we accept it.
== Also see ==
* Pascal's Wager
Category:Philosophical Questions
Strict Lower Closure in Restricted Ordering

== Theorem ==
Let $\left({S, \preceq}\right)$ be an ordered set.
Let $T \subseteq S$ be a subset of $S$, and let $\preceq \restriction_T$ be the restricted ordering on $T$.
Then for all $t \in T$:
:$t^{\prec T} = T \cap t^{\prec S}$
where:
: $t^{\prec T}$ is the strict lower closure of $t$ in $\left({T, \preceq \restriction_T}\right)$
: $t^{\prec S}$ is the strict lower closure of $t$ in $\left({S, \preceq}\right)$.
== Proof ==
Let $t \in T$, and suppose that $t' \in t^{\prec T}$.
By definition of strict lower closure, this is equivalent to:
:$t' \preceq \restriction_T t \land t \ne t'$
By definition of $\preceq \restriction_T$, the first condition comes down to:
:$t' \preceq t \land t' \in T$
as it is assumed that $t \in T$.
In conclusion, $t' \in t^{\prec T}$ is equivalent to:
:$t' \in T \land t' \preceq t \land t \ne t'$
These last two conjuncts precisely express that $t' \in t^{\prec S}$.
By definition of set intersection, it also holds that:
:$t' \in T \cap t^{\prec S}$
 $t' \in T$ and $t' \in t^{\prec S}$.
Thus, it follows that the following are equivalent:
:$t' \in t^{\prec T}$
:$t' \in T \cap t^{\prec S}$
and hence the result follows, by definition of set equality.
Category:Lower Closures
Strict Lower Closure is Dual to Strict Upper Closure

== Theorem ==
Let $\left({S, \preceq}\right)$ be an ordered set.
Let $a, b \in S$.
The following are dual statements:
:$b \in a^\prec$, the strict lower closure of $a$
:$b \in a^\succ$, the strict upper closure of $a$
== Proof ==
By definition of strict lower closure:
:$b \in a^\prec$
:$b$ strictly precedes $a$
The dual of this statement is:
:$b$ strictly succeeds $a$
by Dual Pairs (Order Theory).
By definition of strict upper closure, this means:
: $b \in a^\succ$
The converse follows from Dual of Dual Statement (Order Theory).
== Also see ==
* Duality Principle (Order Theory)
Category:Lower Closures
Category:Upper Closures
Category:Dual Pairs (Order Theory)
Strict Lower Closure is Lower Section

== Theorem ==
Let $\struct {S, \preceq}$ be an ordered set.
Let $p \in S$.
Let $p^\prec$ denote the strict lower closure of $p$.
Then $p^\prec$ is a lower section.
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Lower Closure is Lower Section
* Strict Upper Closure is Upper Section
Category:Lower Sections
Category:Lower Closures
Category:Strict Lower Closure is Lower Section
Strict Lower Closure is Lower Section/Proof 1

== Theorem ==
== Proof ==
Let $l \in p^\prec$.
Let $s \in S$ with $s \preceq l$.
Then by the definition of strict lower closure:
:$l \prec p$
Thus by Extended Transitivity:
:$s \prec p$
So by the definition of strict lower closure:
:$s \in p^\prec$
Since this holds for all such $l$ and $s$, $p^\prec$ is a lower section.
Category:Strict Lower Closure is Lower Section
Strict Lower Closure of Element is Proper Lower Section

== Theorem ==
Let $A$ be a class under an ordering $\preccurlyeq$.
Let $a \in A$.
Let $a^\prec$ denote the strict lower closure of $a$ in $A$.
Then $a^\prec$ is a proper lower section of $A$.
== Proof ==
By definition of strict lower closure:
:$a^\prec := \set {b \in S: b \prec a}$
By definition of $\prec$:
:$a \not \prec a$
and so:
:$a \notin a^\prec$
and so:
:$a \in A \setminus a^\prec$
Let $x \in a^\prec$.
Then by definition of $a^\prec$:
:$x \prec a$
and so:
:$\exists a \in A \setminus a^\prec: x \prec a$
and so $x$ is in a proper lower section of $A$.
As $x$ is arbitrary, it follows that every element of $a^\prec$ is in a proper lower section of $A$.
Hence $a^\prec$ is a proper lower section of $A$.
== Sources ==
* : Chapter $6$: Order Isomorphism and Transfinite Recursion: $\S 1$ A few preliminaries
Category:Lower Closures
Category:Lower Sections
Strict Lower Closure of G-Tower is Set of Elements which are Proper Subsets

== Theorem ==
Let $M$ be a class.
Let $g: M \to M$ be a progressing mapping on $M$.
Let $M$ be a $g$-tower.
Let $x \in M$.
Then the strict lower closure of $x$ is the set of all elements of $M$ that are proper subsets of $M$.
== Proof ==
Follows directly from the definitions.
== Sources ==
* : Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text I$ -- Superinduction and Well Ordering: $\S 3$ The well ordering of $g$-towers
Category:Lower Closures
Category:G-Towers
Strict Lower Closure of Limit Element is Infinite

== Theorem ==
Let $A$ be a class.
Let $\preccurlyeq$ be a well-ordering on $A$.
Let $x \in A$ be a limit element of $A$ under $\preccurlyeq$.
Let $x^\prec$ denotes the strict lower closure of $x$ in $A$ under $\preccurlyeq$.
Then $x^\prec$ is an infinite set.
== Proof ==
Let $x \in A$ be a limit element of $A$ under $\preccurlyeq$.
From Characterisation of Limit Element under Well-Ordering it follows that $x^\prec$ has no greatest element $\preccurlyeq$.
The result follows.
== Sources ==
* : Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text I$ -- Superinduction and Well Ordering: $\S 1$ Introduction to well ordering: Proposition $1.7$
Category:Well-Orderings
Category:Lower Closures
Category:Limit Elements
Strict Lower Closure of Sum with One

== Theorem ==
Let $\struct {S, \circ, \preceq}$ be a naturally ordered semigroup.
Then:
:$\forall n \in \struct {S, \circ, \preceq}: \paren {n \circ 1}^\prec = n^\prec \cup \set n$
where $n^\prec$ is defined as the strict lower closure of $n$, that is, the set of elements strictly preceding $n$.
== Proof ==
First note that as $\struct {S, \circ, \preceq}$ is well-ordered and hence totally ordered, the Trichotomy Law applies.
Thus:
So:
Similarly:
So:
:$p \notin n^\prec \cup \set n \iff p \notin \paren {n \circ 1}^\prec$
Thus:
:$\relcomp S {\paren {n \circ 1}^\prec} = \relcomp S {n^\prec \cup \set n}$
from the definition of relative complement.
So:
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 16$: The Natural Numbers: Corollary $16.4.1$
Category:Naturally Ordered Semigroup
Strict Negativity is equivalent to Strict Positivity of Negative

== Theorem ==
Let $\struct {D, +, \times}$ be an ordered integral domain, whose (strict) positivity property is denoted $P$.
Let $N$ be the (strict) negativity property on $D$:
:$\forall a \in D: \map N a \iff \map P {-a}$
Then for all $a \in D$:
:$\map P a \iff \map N {-a}$
== Proof ==
== Sources ==
* : Chapter $2$: Ordered and Well-Ordered Integral Domains: $\S 7$. Order: Theorem $10 \ \text {(ii)}$
Category:Ordered Integral Domains
Strict Negativity is equivalent to Strictly Preceding Zero

== Theorem ==
Let $\struct {D, +, \times}$ be an ordered integral domain, whose (strict) positivity property is denoted $P$.
Let $\le$ be the total ordering induced by $P$, and let $<$ be its strict total ordering counterpart.
Let $N$ be the (strict) negativity property on $D$:
:$\forall a \in D: \map N a \iff \map P {-a}$
Then for all $a \in D$:
:$\map N a \iff a < 0$
== Proof ==
== Sources ==
* : Chapter $2$: Ordered and Well-Ordered Integral Domains: $\S 7$. Order: Theorem $10 \ \text {(i)}$
Category:Ordered Integral Domains
Strict Ordering Preserved under Cancellability in Totally Ordered Semigroup

== Theorem ==
Let $\struct {S, \circ, \preceq}$ be a totally ordered semigroup.
If either:
:$x \circ z \prec y \circ z$
or
:$z \circ x \prec z \circ y$
then $x \prec y$.
== Proof ==
Let $x \circ z \prec y \circ z$.
 $x \succeq y$.
As $\struct {S, \circ, \preceq}$ is an ordered semigroup, $\preceq$ is compatible with $\circ$.
Hence we have:
:$x \succeq y \implies x \circ z \succeq y \circ z$
which contradicts $x \circ z \prec y \circ z$.
We have that $\preceq$ is a total ordering, and that it is not the case that $x \succeq y$.
Hence by the Trichotomy Law:
:$x \prec y$
Similarly for $z \circ x \prec z \circ y$.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 15$: Ordered Semigroups: Theorem $15.1: \ 2^\circ$
Category:Ordered Semigroups
Strict Ordering Preserved under Product with Cancellable Element

== Theorem ==
Let $\struct {S, \circ, \preceq}$ be an ordered semigroup.
Let $x, y, z \in S$ be such that:
:$(1): \quad z$ is cancellable for $\circ$
:$(2): \quad x \prec y$
Then:
:$x \circ z \prec y \circ z$
:$z \circ x \prec z \circ y$
== Proof ==
Let $z$ be cancellable and $x \prec y$.
Then by the definition of ordered semigroup:
:$x \circ z \preceq y \circ z$
From the fact that $z$ is cancellable:
:$x \circ z = y \circ z \iff x = y$
Thus as $x \circ z \ne y \circ z$ it follows from Strictly Precedes is Strict Ordering that:
:$x \circ z \prec y \circ z$
Similarly, $z \circ x \prec z \circ y$ follows from $z \circ x \preceq z \circ y$.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 15$: Ordered Semigroups: Theorem $15.1: \ 1^\circ$
Category:Ordered Semigroups
Strict Ordering Preserved under Product with Invertible Element

== Theorem ==
Let $\struct {S, \circ, \preceq}$ be an ordered semigroup.
Let $z \in S$ be invertible.
Suppose that either $x \circ z \prec y \circ z$ or $z \circ x \prec z \circ y$.
Then $x \prec y$.
== Proof ==
Suppose $x \circ z \prec y \circ z$.
By Invertible Element of Monoid is Cancellable, $z^{-1}$ is cancellable.
Then from Strict Ordering Preserved under Product with Cancellable Element:
:$x = \paren {x \circ z} \circ z^{-1} \prec \paren {y \circ z} \circ z^{-1} = y$
Likewise, if $z \circ x \prec z \circ y$:
:$x = z^{-1} \circ \paren {z \circ x} \prec z^{-1} \circ \paren {z \circ y} = y$
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 15$: Ordered Semigroups: Theorem $15.1: \ 2^\circ$
Category:Ordered Semigroups
Strict Ordering can be Expanded to Compare Additional Pair

== Theorem ==
Let $\struct {S, \prec}$ be an ordered set.
Let $a$ and $b$ be distinct, $\prec$-incomparable elements of $S$.
That is, let:
:$a \nprec b$ and $b \nprec a$.
Let $\prec' = {\prec} \cup \set {\tuple {a, b} }$.
Define a relation $\prec'^+$ by letting $p \prec'^+ q$ :
:$p \prec q$
or:
:$p \preceq a$ and $b \preceq q$
where $\preceq$ is the reflexive closure of $\prec$.
Then:
:$\prec'^+$ is a strict ordering
:$\prec^+$ is the transitive closure of $\prec'$.
== Proof 1 ==
Category:Strict Orderings
Category:Strict Ordering can be Expanded to Compare Additional Pair
Strict Ordering of Naturally Ordered Semigroup is Strongly Compatible

== Theorem ==
Let $\struct {S, \circ, \preceq}$ be a naturally ordered semigroup.
Then $\prec$ is strongly compatible with $\circ$:
:$\forall m, n, p \in S: m \prec n \iff m \circ p \prec n \circ p$
== Proof ==
By , all $n \in S$ are cancellable.
Hence from Strict Ordering Preserved under Product with Cancellable Element:
:$\forall m, n, p \in S: m \prec n \implies m \circ p \prec n \circ p$
By , $\preceq$ is a total ordering.
Therefore, the contrapositive of:
:$\forall m, n, p \in S: m \circ p \prec n \circ p \implies m \prec n$
is:
:$\forall m, n, p \in S: m \preceq n \implies m \circ p \preceq n \circ p$
which we know to be true by virtue of .
The result follows.
== Also see ==
* Ordering of Naturally Ordered Semigroup is Strongly Compatible
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 16$: The Natural Numbers: Theorem $16.3$
Category:Naturally Ordered Semigroup
Strict Ordering of Ordinals is Equivalent to Membership Relation

== Theorem ==
Let $\On$ denote the class of all ordinals.
Let $<$ denote the (strict) usual ordering of $\On$.
Then:
:$\forall \alpha, \beta \in \On: \alpha < \beta \iff \alpha \in \beta$
== Proof ==
=== Necessary Condition ===
Let $\alpha \in \beta$.
Then from Ordinal is Transitive:
:$\alpha \subseteq \beta$
But if $\alpha = \beta$ we would have $\alpha \in \alpha$.
This is contrary to Ordinal is not Element of Itself.
Hence we have:
:$\alpha \subseteq \beta$
and:
:$\alpha \ne \beta$
That is:
:$\alpha \subsetneqq \beta$
Hence by definition of the (strict) usual ordering of $\On$:
:$\alpha < \beta$
=== Sufficient Condition ===
Let $\alpha < \beta$.
Then by Sandwich Principle for $g$-Towers: Corollary 1:
:$\alpha^+ \subseteq \beta$
where $\alpha^+$ denotes the successor set of $\alpha$.
By definition of successor set:
:$\alpha^+ = \alpha \cup \set \alpha$
and so:
:$\alpha \in \alpha^+$
That is:
:$\alpha \in \beta$
== Sources ==
* : Chapter $5$: Ordinal Numbers: $\S 1$ Ordinal numbers: Theorem $1.12$
Category:Ordinals
Strict Ordering on Integers is Asymmetric

== Theorem ==
Let $\eqclass {a, b} {}$ denote an integer, as defined by the formal definition of integers.
Then:
That is, strict ordering on the integers is asymmetric.
== Proof ==
By the formal definition of integers, we have that $a, b, c, d, e, f$ are all natural numbers.
To eliminate confusion between integer ordering and the ordering on the natural numbers, let $a \prec b$ denote that the natural number $a$ is less than the natural number $b$.
We have:
== Sources ==
* : Introduction $\S 5$: The system of integers
Category:Orderings on Integers
Strict Ordering on Integers is Transitive

== Theorem ==
Let $\eqclass {a, b} {}$ denote an integer, as defined by the formal definition of integers.
Then:
That is, strict ordering on the integers is transitive.
== Proof ==
By the formal definition of integers, we have that $a, b, c, d, e, f$ are all natural numbers.
To eliminate confusion between integer ordering and the ordering on the natural numbers, let $a \prec b$ denote that the natural number $a$ is less than the natural number $b$.
We have:
== Sources ==
* : Introduction $\S 5$: The system of integers
Category:Orderings on Integers
Strict Ordering on Integers is Trichotomy

== Theorem ==
Let $\eqclass {a, b} {}$ and $\eqclass {c, d} {}$ be integers, as defined by the formal definition of integers.
Then exactly one of the following holds:
That is, strict ordering is a trichotomy.
== Proof ==
By the formal definition of integers, we have that $a, b, c, d, e, f$ are all natural numbers.
To eliminate confusion between integer ordering and the ordering on the natural numbers, let $a \preccurlyeq b$ denote that the natural number $a$ is less than or equal to the natural number $b$.
We have:
Then:
Similarly:
Then:
and:
This demonstrates that $<$, $=$ and $>$ are mutually exclusive.
Now:
Similarly:
and:
demonstrating that either $<$, $=$ or $>$ must hold.
== Sources ==
* : Introduction $\S 5$: The system of integers
Category:Orderings on Integers
Strict Ordering on Integers is Well-Defined

== Theorem ==
Let $\eqclass {a, b} {}$ denote an integer, as defined by the formal definition of integers.
Let:
Then:
== Proof ==
This is a direct application of the Extension Theorem for Total Orderings.
== Sources ==
* : Introduction $\S 5$: The system of integers
Category:Orderings on Integers
Strict Positivity Property induces Total Ordering

== Theorem ==
Let $\struct {D, +, \times}$ be an integral domain whose zero is $0_D$.
Let $D$ be endowed with a (strict) positivity property $P: D \to \set {\T, \F}$.
Then there exists a total ordering $\le$ on $\struct {D, +, \times}$ induced by $P$ which is compatible with the ring structure of $\struct {D, +, \times}$.
== Proof ==
By definition of the strict positivity property:
Let us define a relation $<$ on $D$ as:
:$\forall a, b \in D: a < b \iff \map P {-a + b}$
Setting $a = 0$:
:$\forall b \in D: 0 < b \iff \map P b$
demonstrating that (strictly) positive elements of $D$ are those which are greater than zero.
From Relation Induced by Strict Positivity Property is Compatible with Addition we have that $<$ is compatible with $+$.
From Relation Induced by Strict Positivity Property is Transitive we have that $<$ is transitive.
From Relation Induced by Strict Positivity Property is Asymmetric and Antireflexive we have that $<$ is asymmetric and antireflexive.
Thus by definition, $<$ is a strict ordering.
Let the relation $\le$ be defined as the reflexive closure of $<$.
From Reflexive Closure of Strict Ordering is Ordering we have that $\le$ is an ordering on $D$.
From Relation Induced by Strict Positivity Property is Trichotomy, and from the Trichotomy Law (Ordering), we have that $\le$ is a total ordering.
Hence the result.
== Also see ==
* Definition:Total Ordering Induced by Strict Positivity Property‎
* Definition:Ordering Induced by Positivity Property
* Ordered Integral Domain is Totally Ordered Ring: a direct corollary
== Sources ==
* : Chapter $2$: Ordered and Well-Ordered Integral Domains: $\S 7$. Order
Category:Ordered Integral Domains
Strict Upper Bound for Composition of Ackermann-Péter Functions

== Theorem ==
For all $x, y, z \in \N$:
:$\map A {x + y + 2, z} > \map A {x, \map A {y, z}}$
where $A$ is the Ackermann-Péter function.
== Proof ==
We have:
:$x + y + 1 > y$
giving us:
:$\paren 1 \quad \map A {x + y + 1, z} > \map A {y, z}$
by Ackermann-Péter Function is Strictly Increasing on First Argument.
Therefore:
Category:Ackermann-Péter Function
Strict Upper Bound for Sum of Ackermann-Péter Functions

== Theorem ==
For all $x, y, z \in \N$:
:$\map A {\map \max {x, y} + 4, z} > \map A {x, z} + \map A {y, z}$
where $A$ is the Ackermann-Péter function.
== Proof ==
Category:Ackermann-Péter Function
Strict Upper Closure in Restricted Ordering

== Theorem ==
Let $\struct {S, \preceq}$ be an ordered set.
Let $T \subseteq S$ be a subset of $S$, and let $\preceq \restriction_T$ be the restricted ordering on $T$.
Then for all $t \in T$:
:$t^{\succ T} = T \cap t^{\succ S}$
where:
:$t^{\succ T}$ is the strict upper closure of $t$ in $\struct {T, \preceq \restriction_T}$
:$t^{\succ S}$ is the strict upper closure of $t$ in $\struct {S, \preceq}$.
== Proof ==
Let $t \in T$, and suppose that $t' \in t^{\succ T}$.
By definition of strict upper closure, this is equivalent to:
:$t \preceq \restriction_T t' \land t \ne t'$
By definition of $\preceq \restriction_T$, the first condition comes down to:
:$t \preceq t' \land t' \in T$
as it is assumed that $t \in T$.
In conclusion, $t' \in t^{\succ T}$ is equivalent to:
:$t' \in T \land t \preceq t' \land t \ne t'$
These last two conjuncts precisely express that $t' \in t^{\succ S}$.
By definition of set intersection, it also holds that:
:$t' \in T \cap t^{\succ S}$
 $t' \in T$ and $t' \in t^{\succ S}$.
Thus, it follows that the following are equivalent:
:$t' \in t^{\succ T}$
:$t' \in T \cap t^{\succ S}$
and hence the result follows, by definition of set equality.
Category:Upper Closures
Strict Upper Closure is Upper Section

== Theorem ==
Let $\struct {S, \preceq}$ be an ordered set.
Let $p \in S$.
Let $p^\succ$ denote the strict upper closure of $p$.
Then $p^\succ$ is an upper section.
== Proof ==
Let $u \in p^\succ$.
Let $s \in S$ with $u \preceq s$.
Then by the definition of strict upper closure:
:$p \prec u$
Thus by Extended Transitivity:
:$p \prec s$
So by the definition of strict upper closure:
:$s \in p^\succ$
Since this holds for all such $u$ and $s$, $p^\succ$ is an upper section.
== Also see ==
* Upper Closure is Upper Section
* Strict Lower Closure is Lower Section
Category:Upper Sections
Category:Upper Closures