nyuuzyou/wikis huggingface dataset:
Units of Ring of Polynomial Forms over Commutative Ring

== Theorem ==
Let $\struct {R, +, \circ}$ be a non-null commutative ring with unity whose zero is $0_R$ and whose unity is $1_R$.
Let $R \sqbrk X$ be the ring of polynomial forms in an indeterminate $X$ over $R$.
Let $\map P X = a_0 + a_1 X + \cdots + a_n X^n \in R \sqbrk X$.
Then:
:$\map P X$ is a unit of $R \sqbrk X$
:
:$a_0$ is a unit of $R$
Also, for $i = 1, \ldots, n$, $a_i$ is nilpotent in $R$.
=== Corollary ===
Let $R$ be a reduced ring.
Let $R \sqbrk X$ be the ring of polynomial forms in an indeterminate $X$ over $R$.
The group of units of $R \sqbrk X$ is precisely the group of elements of $R \sqbrk X$ of degree zero that are units of $R$.
== Proof ==
=== Necessary condition ===
Let $a_0$ be a unit of $R$.
For $i = 1, \ldots, n$, let $a_i$ be nilpotent in $R$.
Because the nilradical is an ideal of $R$, it follows that:
:$Q = -a_1 X + \dotsb + -a_n X^n$
is nilpotent.
Moreover, multiplying through by $a_0^{-1}$ we may as well assume that $a_0 = 1_R$.
Then:
:$P = 1_R - Q$
and from Unity plus Negative of Nilpotent Ring Element is Unit $P$ is a unit of $R \sqbrk X$.
=== Sufficient condition ===
Let $\map P X$ be a unit of $R \sqbrk X$.
That is, there exists:
:$Q = b_0 + b_1 X + \dotsb + b_m X^m \in R \sqbrk X$
such that $P Q = 1$.
By the definition of polynomial multiplication the degree zero term of $P Q$ is $a_0 b_0$.
Therefore $a_0 b_0 = 1_R$.
So $a_0$ is a unit of $R$.
Next we show that $a_1, \dotsc, a_n$ are nilpotent.
By Spectrum of Ring is Nonempty, $R$ has at least one prime ideal $\mathfrak p$.
By Prime Ideal iff Quotient Ring is Integral Domain:
:$R / \mathfrak p$ is an integral domain.
By Ring of Polynomial Forms over Integral Domain is Integral Domain:
:$R / \mathfrak p \sqbrk X$ is also an integral domain.
For any polynomial $T \in R \sqbrk X$ let $\overline T$ denote the image of $T$ under the Induced Homomorphism of Polynomial Forms defined by the quotient mapping $R \to R / \mathfrak p$.
Now we have:
:$\overline P \cdot \overline Q = 1_{R / \mathfrak p}$
By Units of Ring of Polynomial Forms over Integral Domain this implies that $\overline P$ has degree zero.
In particular for $i = 1, \dotsc, n$, the image of $a_i$ in $R / \mathfrak p$ is $0_{R / \mathfrak p}$.
By definition, this means that $a_i \in \mathfrak p$.
But this is true for every prime ideal $\mathfrak p$.
Thus by definition:
:$a_i \in \Nil R$
where $\Nil R$ denotes the nilradical of $R$.
Category:Polynomial Theory
Units of Ring of Polynomial Forms over Field

== Theorem ==
Let $\struct {F, +, \circ}$ be a field whose zero is $0_F$ and whose unity is $1_F$.
Let $F \sqbrk X$ be the ring of polynomial forms in an indeterminate $X$ over $F$.
Then the units of $F \sqbrk X$ are all the elements of $F \sqbrk X$ whose degree is $0$.
== Proof ==
An element of $F \sqbrk X$ whose degree is $0$ is merely an element of $F$.
But note that $0_F$, considered as an element of $F \sqbrk X$, has a degree which is not defined, so the null polynomial is seen to be excluded.
Any element $a$ of $F$ has an inverse $1_F / a$.
So all the elements of $F \sqbrk X$ whose degree is $0$ are units of $F$ and hence of $F \sqbrk X$.
Now suppose $\map a X \in F \sqbrk X$ is a unit of $F \sqbrk X$.
Then $\map a X \, \map q X = 1_F$ for some $\map q X \in F \sqbrk X$.
Neither $\map a X$ nor $\map q X$ can be null.
So by Properties of Degree $\map \deg {\map a X} + \map \deg {\map q X} = 0$.
So $\map \deg {\map a X} = 0$.
== Sources ==
* : Chapter $6$: Polynomials and Euclidean Rings: $\S 26$. Divisibility: Example $48$
Category:Polynomial Theory
Category:Units of Rings
Units of Ring of Polynomial Forms over Integral Domain

== Theorem ==
Let $\struct {D, +, \circ}$ be an integral domain.
Let $D \sqbrk X$ be the ring of polynomial forms in an indeterminate $X$ over $D$.
Then the group of units of $D \sqbrk X$ is precisely the group of elements of $D \sqbrk X$ of degree zero that are units of $D$.
== Proof ==
It is immediate that a unit of $D$ is also a unit of $D \sqbrk X$.
Let $P$ be a unit of $D \sqbrk X$.
Then there exists $Q \in D \sqbrk X$ such that $P Q = 1$.
By Corollary 2 to Degree of Product of Polynomials over Ring we have:
:$0 = \map \deg 1 = \map \deg P + \map \deg Q$
Therefore:
:$\map \deg P = \map \deg Q = 0$
That is, $P \in R$ and $Q \in R$.
Moreover $P Q = 1$ in $R$, so it follows that $P$ is a unit of $R$.
Category:Polynomial Theory
Unity Divides All Elements

== Theorem ==
Let $\struct {D, +, \circ}$ be an integral domain whose unity is $1_D$.
Then unity is a divisor of every element of $D$:
:$\forall x \in D: 1_D \divides x$
Also:
:$\forall x \in D: -1_D \divides x$
== Proof 1 ==
== Proof 2 ==
Category:Integral Domains
Category:Unity Divides All Elements
Category:Divisibility
Unity Divides All Elements/Proof 1

== Theorem ==
== Proof ==
The element $1_D$ is the unity of $\struct {D, +, \circ}$, and so:
:$1_D \in D: x = 1_D \circ x$
Similarly, from Product of Ring Negatives:
:$-1_D \in D: x = \paren {-1_D} \circ \paren {-x}$
The result follows from the definition of divisor.
Category:Unity Divides All Elements
Unity Divides All Elements/Proof 2

== Theorem ==
== Proof ==
This is a special case of Unit of Integral Domain divides all Elements, as Unity is Unit.
Furthermore, from Unity and Negative form Subgroup of Units we also have that $-1_D$ is a unit of $D$.
Hence the result.
== Sources ==
* : $\S 62.1$ Factorization in an integral domain: $\text{(ii)}$
Category:Unity Divides All Elements
Unity Function is Completely Multiplicative

== Theorem ==
Let $f_1: \Z_{> 0} \to \Z_{> 0}$ be the constant function:
:$\forall n \in \Z_{> 0}: f_1 \left({n}\right) = 1$
Then $f_1$ is completely multiplicative.
== Proof ==
:$\forall m, n \in \Z_{> 0}: f_1 \left({m n}\right) = 1 = f_1 \left({m}\right) f_1 \left({n}\right)$
Category:Completely Multiplicative Functions
Unity and Negative form Subgroup of Units

== Theorem ==
Let $\struct {R, +, \circ}$ be a ring with unity.
Then:
:$\struct {\set {1_R, -1_R}, \circ} \le U_R$
That is, the set consisting of the unity and its negative forms a subgroup of the group of units.
== Proof ==
From Unity is Unit:
:$1_R \in U_R$
It remains to be shown that $-1_R \in U_R$.
From , $\struct {R, +}$ is an abelian group.
Therefore:
:$1_R \in R \implies -1_R \in R$.
From Product of Ring Negatives:
:$-1_R \circ -1_R = 1_R \circ 1_R = 1_R$
Thus $-1_R$ has a ring product inverse (itself) and therefore $-1_R \in U_R$.
We have:
This exhausts all the ways we can form a ring product between $1_R$ and $-1_R$.
That is:
:$\forall x, y \in \set {1_R, -1_R}: x \circ y \in \set {1_R, -1_R}$
Hence the result from the Two-Step Subgroup Test.
Category:Rings with Unity
Category:Subgroups
Unity is Unity in Ring of Idempotents

== Theorem ==
Let $\left({R, +, \circ}\right)$ be a commutative and unitary ring whose unity is $1_R$.
Let $\left({A, \oplus, \circ}\right)$ be the ring of idempotents of $R$.
Then $1_R$ is also a unity for $\left({A, \oplus, \circ}\right)$.
== Proof ==
From Unity of Ring is Idempotent, $1_R$ is an idempotent element of $R$.
Hence $1_R \in A$.
Recall that the ring product of $A$ is a restriction from that of $R$.
Hence, for each $x \in A$:
:$x \circ 1_R = x = 1_R \circ x$
so that $1_R$ is a unity for $A$, as desired.
Category:Ring Theory
Unity of Integral Domain is Unique

== Theorem ==
Let $\struct {D, +, \times}$ be an integral domain.
Then the unity of $\struct {D, +, \times}$ is unique.
== Proof ==
From the definition of an integral domain, $\struct {D, +, \times}$ is a commutative ring such that $\struct {D^*, \circ}$ is a monoid.
The result follows from Identity of Monoid is Unique.
== Sources ==
* : Chapter $1$: Integral Domains: $\S 4$. Elementary Properties: Theorem $1 \ \text{(ii)}$
Category:Integral Domains
Unity of Ordered Integral Domain is Strictly Positive

== Theorem ==
Let $\struct {D, +, \times \le}$ be an ordered integral domain whose unity is $1_D$.
Then:
:$\map P {1_D}$
where $P$ is the (strict) positivity property.
== Proof ==
We have by definition of the unity that:
:$\forall a \in D: 1_D \times a = a = a \times 1_D$
This particularly applies to $1_D$ itself:
:$1_D = 1_D \times 1_D$
But then by Square of Non-Zero Element of Ordered Integral Domain is Strictly Positive:
:$\map P {1_D \times 1_D} \implies \map P {1_D}$
== Sources ==
* : Chapter $2$: Ordered and Well-Ordered Integral Domains: $\S 7$. Order: Theorem $7 \ \text{(ii)}$
Category:Ordered Integral Domains
Unity of Ring is Idempotent

== Theorem ==
Let $\left({R, +, \circ}\right)$ be a ring with unity whose unity is $1_R$.
Then $1_R$ is an idempotent element of $R$ under the ring product $\circ$:
:$1_R \circ 1_R = 1_R$
== Proof ==
By definition of ring with unity, $\left({R, \circ}\right)$ is a monoid whose identity element is $1_R$.
From Identity Element is Idempotent (applied to $1_R$):
:$1_R \circ 1_R = 1_R$
which was to be proven. 
Category:Ring Theory
Unity of Ring is Unique

== Theorem ==
A ring can have no more than one unity.
== Proof ==
Let $\struct {R, +, \circ}$ be a ring.
If $\struct {R, \circ}$ has an identity, then it is a monoid.
From Identity of Monoid is Unique, it follows that such an identity is unique.
== Also see ==
* Identity is Unique
== Sources ==
* : Chapter $1$: Rings - Definitions and Examples: $3$: Some special classes of rings
Category:Rings with Unity
Unity of Ring of Bounded Continuous Real-Valued Functions

== Theorem ==
Let $\struct {S, \tau}$ be a topological space.
Let $\R$ denote the real number line.
Let $\struct {\map {C^*} {S, \R}, +, *}$ be the ring of bounded continuous real-valued functions from $S$.
Then:
:the unity of $\struct{\map {C^*} {S, \R}, +, *}$ is the constant mapping $1_{\R^S} : S \to \R$ defined by:
::$\forall s \in S : \map {1_{\R^S}} s = 1$.
== Proof ==
Let $\struct{\map C {S, \R}, +, *}$ be the ring of continuous real-valued functions from $S$.
From Ring of Bounded Continuous Functions is Subring of Continuous Real-Valued Functions:
:$\struct{\map {C^*} {S, \R}, +, *}$ is a subring of $\struct{\map C {S, \R}, +, *}$
From Unity of Ring of Continuous Real-Valued Functions:
:the unity of $\struct{\map C {S, \R}, +, *}$ is the constant mapping $1_{\R^S} : S \to R$ defined by:
::$\forall s \in S : \map {1_{\R^S}} s = 1$
From Constant Mapping is Continuous and Constant Real-Valued Function is Bounded:
:$1_{\R^S} \in \map {C^*} {S, \R}$
From Subring Containing Ring Unity has Unity:
:$1_{\R^S}$ is the unity of $\struct{\map {C^*} {S, \R}, +, *}$
== Also see ==
* Ring of Bounded Continuous Functions is Subring of Continuous Real-Valued Functions
* Zero of Ring of Bounded Continuous Real-Valued Functions
* Additive Inverse in Ring of Bounded Continuous Real-Valued Functions
* Ring of Bounded Continuous Real-Valued Functions is Commutative
== Sources ==
: Chapter $1$: Functions on a Topological Space, $\S 1.4$
Category:Rings of Bounded Continuous Real-Valued Functions
Unity of Ring of Continuous Mappings

== Theorem ==
Let $\struct {S, \tau_{_S} }$ be a topological space.
Let $\struct {R, +, *, \tau_{_R} }$ be a topological ring with unity $1_R$.
Let $\struct{\map C {S, R}, +, *}$ be the ring of continuous mappings from $S$ to $R$.
Then:
:the unity of $\struct{\map C {S, R}, +, *}$ is the constant mapping $1_{R^S} : S \to R$ defined by:
::$\forall s \in S : \map {1_{R^S}} s = 1_R$.
== Proof ==
Let $\struct {R^S, +, *}$ be the ring of mappings from $S$ to $R$.
From Ring of Continuous Mappings is Subring of All Mappings:
:$\struct{\map C {S, R}, +, *}$ is a subring of $\struct {R^S, +, *}$
From Induced Structure Identity:
:the unity of $\struct {R^S, +, *}$ is the constant mapping $1_{R^S} : S \to R$ defined by:
::$\forall s \in S : \map {1_{R^S}} s = 1_R$
From Constant Mapping is Continuous:
:$1_{R^S} \in \map C {S, R}$
From Subring Containing Ring Unity has Unity:
:$1_{R^S}$ is the unity of $\map C {S, R}$
== Also see ==
* Ring of Continuous Mappings is Subring of All Mappings
* Zero of Ring of Continuous Mappings
* Additive Inverse in Ring of Continuous Mappings
* Commutativity of Ring of Continuous Mappings
Category:Rings of Continuous Mappings
Unity of Ring of Continuous Real-Valued Functions

== Theorem ==
Let $\struct {S, \tau}$ be a topological space.
Let $\R$ denote the real number line.
Let $\struct {\map C {S, \R}, +, *}$ be the ring of continuous real-valued functions from $S$.
Then:
:the unity of $\struct{\map C {S, \R}, +, *}$ is the constant mapping $1_{\R^S} : S \to \R$ defined by:
::$\forall s \in S : \map {1_{\R^S}} s = 1$.
== Proof ==
By definition of ring of continuous real-valued functions:
:$\struct {\map C {S, \R}, +, *}$ is the ring of continuous mappings from $S$ to $\R$.
From Unity of Ring of Continuous Mappings:
:the unity of $\struct{\map C {S, \R}, +, *}$ is the constant mapping $1_{\R^S} : S \to \R$ defined by:
::$\forall s \in S : \map {1_{\R^S}} s = 1$.
== Also see ==
* Ring of Continuous Real-Valued Functions is Ring
* Zero of Ring of Continuous Real-Valued Functions
* Ring of Continuous Real-Valued Functions is Commutative
== Sources ==
: Chapter $1$: Functions on a Topological Space, $\S 1.3$
Category:Rings of Continuous Real-Valued Functions
Unity of Subfield is Unity of Field

== Theorem ==
Let $\struct {F, +, \times}$ be a field whose unity is $1$.
Let $\struct {K, +, \times}$ be a subfield of $F$.
The unity of $\struct {K, +, \times}$ is also $1$.
== Proof ==
By definition, $\struct {K, +, \times}$ is a subset of $F$ which is a field.
By definition of field, $\struct {K^*, \times}$ and $\struct {F^*, \times}$ are groups such that $K \subseteq F$.
So $\struct {K^*, \times}$ is a subgroup of $\struct {F^*, \times}$.
By Identity of Subgroup, the identity of $\struct {F^*, \times}$, which is $1$, is also the identity of $\struct {K^*, \times}$.
== Sources ==
* : $\S 56.3$ Subrings and subfields
Category:Subfields
Unity of Subring is not necessarily Unity of Ring

== Theorem ==
Let $\struct {S, +, \circ}$ be a ring with unity whose unity is $1_S$.
Let $\struct {T, + \circ}$ be a subring of $\struct {S, + \circ}$ whose unity is $1_T$.
Then it is not necessarily the case that $1_T = 1_S$.
== Proof ==
Let $\struct {S, + \times}$ be the ring formed by the set of order $2$ square matrices over the real numbers $R$ under (conventional) matrix addition and (conventional) matrix multiplication.
Let $T$ be the subset of $S$ consisting of the matrices of the form $\begin{bmatrix} x & 0 \\ 0 & 0 \end{bmatrix}$ for $x \in \R$.
We have that:
:$\begin{bmatrix} x & 0 \\ 0 & 0 \end{bmatrix} + \begin{bmatrix} -y & 0 \\ 0 & 0 \end{bmatrix} = \begin{bmatrix} x - y & 0 \\ 0 & 0 \end{bmatrix}$
and so $T$ is closed under subtraction.
From Matrices of the Form $\begin{bmatrix} x & 0 \\ 0 & 0 \end{bmatrix}$ we have that $\struct {T, \times}$ is a subsemigroup of $\struct {S, \times}$.
It follows from the Subring Test that $\struct {T, + \circ}$ be a subbring of $\struct {S, + \circ}$.
From (some result somewhere) $\struct {S, \times}$ has an identity $\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ which is not in $\struct {T, \times}$.
However, note that:
demonstrating that $\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}$ is the identity of $\struct {T, \times}$.
The result follows.
== Also see ==
* Identity of Submonoid is not necessarily Identity of Monoid
== Sources ==
* : $\S 56.3$ Subrings and subfields
Category:Subrings
Unity plus Negative of Nilpotent Ring Element is Unit

== Theorem ==
Let $\struct {R, +, \circ}$ be a ring with unity whose zero is $0_R$ and whose unity is $1_R$.
Let $x \in R$ be nilpotent.
Then $1_R - x$ is a unit of $R$.
== Proof ==
By definition of nilpotent element:
:$x^n = 0_R$
for some $n \in \Z_{>0}$.
From Difference of Two Powers:
for $a, b \in R$.
Putting $a = 1_R$ and $b = x$, we have:
Thus by definition $1_R - x$ has a product inverse $1_R + x + x^2 + \dotsb + x^{n - 2} + x^{n - 1}$.
Hence by definition $1_R - x$ is a unit of $R$.
== Sources ==
* : Chapter $9$: Rings: Exercise $11 \ \text {(i)}$
Category:Nilpotent Ring Elements
Universal Affirmative and Negative are both False iff Particular Affirmative and Negative are both True

== Theorem ==
Consider the categorical statements:
Then:
:$\map {\mathbf A} {S, P}$ and $\map {\mathbf E} {S, P}$ are both false
:
:$\map {\mathbf I} {S, P}$ and $\map {\mathbf O} {S, P}$ are both true.
== Proof ==
=== Necessary Condition ===
Let $\map {\mathbf A} {S, P}$ and $\map {\mathbf E} {S, P}$ both be false.
Hence $\map {\mathbf I} {S, P}$ and $\map {\mathbf O} {S, P}$ are both true.
=== Sufficient Condition ===
Let $\map {\mathbf I} {S, P}$ and $\map {\mathbf O} {S, P}$ both be true.
Hence $\map {\mathbf A} {S, P}$ and $\map {\mathbf E} {S, P}$ are both false.
== Sources ==
* : Chapter $4$: The Predicate Calculus $2$: $4$ The Syllogism
Category:Categorical Statements
Universal Affirmative and Particular Negative are Contradictory

== Theorem ==
Consider the categorical statements:
Then $\mathbf A$ and $\mathbf O$ are contradictory.
Using the symbology of predicate logic:
:$\neg \paren {\paren {\forall x: \map S x \implies \map P x} \iff \paren {\exists x: \map S x \land \neg \map P x} }$
== Proof ==
The argument reverses:
The result follows by definition of contradictory.
== Sources ==
* : Chapter $4$: The Predicate Calculus $2$: $4$ The Syllogism
* : $4$: Propositional Functions and Quantifiers: $4.1$: Singular Propositions and General Propositions
Category:Categorical Statements
Universal Affirmative and Universal Negative are Contrary iff First Predicate is not Vacuous

== Theorem ==
Consider the categorical statements:
Then:
:$\mathbf A$ and $\mathbf E$ are contrary
:
:$\exists x: \map S x$
Using the symbology of predicate logic:
:$\exists x: \map S x \iff \neg \paren {\paren {\forall x: \map S x \implies \map P x} \land \paren {\forall x: \map S x \implies \neg \map P x} }$
== Proof ==
=== Sufficient Condition ===
Let $\exists x: \map S x$.
Suppose $\mathbf A$ and $\mathbf E$ are both true.
As $\mathbf A$ is true, then by Modus Ponendo Ponens:
:$\map P x$
As $\mathbf E$ is true, then by Modus Ponendo Ponens:
:$\neg \map P x$
It follows by Proof by Contradiction that $\mathbf A$ and $\mathbf E$ are not both true.
Thus, by definition, $\mathbf A$ and $\mathbf E$ are contrary statements.
=== Necessary Condition ===
Let $\mathbf A$ and $\mathbf E$ be contrary statements.
Suppose:
:$\neg \exists x: \map S x$
that is, $\map S x$ is vacuous.
From Denial of Existence:
:$\forall x: \neg \map S x \dashv \vdash \neg \exists x: \map S x$
it follows that $\forall x: \map S x$ is false.
From False Statement implies Every Statement:
:$\forall x: \map S x \implies \map P x$
is true, and:
:$\forall x: \map S x \implies \neg \map P x$
is also true.
Thus, by definition, $\mathbf A$ and $\mathbf E$ are not contrary statements.
It follows by Proof by Contradiction that $\exists x: \map S x$.
== Also defined as ==
Some sources gloss over the possibility of $\map S x$ being vacuous and merely report that the universal affirmative and universal negative are contrary.
== Sources ==
* : Chapter $4$: The Predicate Calculus $2$: $4$ The Syllogism
* : $4$: Propositional Functions and Quantifiers: $4.1$: Singular Propositions and General Propositions
Category:Categorical Statements
Universal Affirmative implies Particular Affirmative iff First Predicate is not Vacuous

== Theorem ==
Consider the categorical statements:
Then:
:$\map {\mathbf A} {S, P} \implies \map {\mathbf I} {S, P}$
:
:$\exists x: \map S x$
Using the symbology of predicate logic:
:$\exists x: \map S x \iff \paren {\paren {\forall x: \map S x \implies \map P x} \implies \paren {\exists x: \map S x \land \map P x} }$
== Proof ==
=== Sufficient Condition ===
Let $\exists x: \map S x$.
Let $\map {\mathbf A} {S, P}$ be true.
As $\map {\mathbf A} {S, P}$ is true, then by Modus Ponendo Ponens:
:$\map P x$
From the Rule of Conjunction:
:$\map S x \land \map P x$
Thus $\map {\mathbf I} {S, P}$ holds.
So by the Rule of Implication:
:$\map {\mathbf A} {S, P} \implies \map {\mathbf I} {S, P}$
=== Necessary Condition ===
Let $\map {\mathbf A} {S, P} \implies \map {\mathbf I} {S, P}$.
Suppose:
:$\neg \exists x: \map S x$
that is, $\map S x$ is vacuous.
From De Morgan's Laws: Denial of Existence:
:$\forall x: \neg \map S x \dashv \vdash \neg \exists x: \map S x$
it follows that $\forall x: \map S x$ is false.
From False Statement implies Every Statement:
:$\forall x: \map S x \implies \map P x$
is true.
So $\map {\mathbf A} {S, P}$ holds.
Again, $\neg \exists x: \map S x$.
Then by the Rule of Conjunction:
:$\neg \paren {\exists x: \map S x \land \map P x}$
That is, $\mathbf I$ does not hold.
So $\map {\mathbf A} {S, P}$ is true and $\map {\mathbf I} {S, P}$ is false.
This contradicts $\map {\mathbf A} {S, P} \implies \map {\mathbf I} {S, P}$ by definition of implication.
Thus $\exists x: \map S x$ must hold.
== Also defined as ==
Some sources gloss over the possibility of $\map S x$ being vacuous and merely report that the universal affirmative implies the particular affirmative.
== Sources ==
* : Chapter $4$: The Predicate Calculus $2$: $4$ The Syllogism
* : $4$: Propositional Functions and Quantifiers: $4.1$: Singular Propositions and General Propositions
Category:Categorical Statements
Universal Class is Proper/Proof 1

== Theorem ==
== Proof ==
 $V$ is small.
We have that:
:$\operatorname {Ru} \subseteq V$
where $\operatorname {Ru}$ denotes the Russell class.
By Axiom of Subsets Equivalents, $\operatorname {Ru}$ is also small.
This contradicts Russell's Paradox.
== Sources ==
* : $\S 5.23$
Category:Universal Class is Proper
Universal Class is Proper/Proof 3

== Theorem ==
== Proof ==
 $V$ is small.
By Cantor's Theorem, there is no surjection from $V$ to $\powerset V$.
By definition of universal class:
:$\powerset V \subseteq V$
By Injection from Subset to Superset, there exists an injection from $\powerset V$ to $V$.
By Injection has Surjective Left Inverse Mapping, there is a surjection from $V$ to $\powerset V$.
But this contradicts Cantor's Theorem.
Therefore $V$ is proper. 
Category:Universal Class is Proper
Universal Class less Set is not Transitive

== Theorem ==
Let $V$ be a basic universe.
Let $a \in V$ be a set.
Then:
:$V \setminus \set a$ is not a transitive class
where $\setminus$ denotes class difference.
== Proof ==
By definition, $V$ is the class of all sets.
As $a \in V$, by definition of $V$ it follows that $a$ is a set.
Consider the power set $\powerset a$ of $a$.
From the axiom of powers:
:$\powerset a$ is also a set
and:
:$\powerset {\powerset a}$ is also a set.
By definition:
:$a \in \powerset a$
and so:
:$\set a \in \powerset {\powerset a}$
as $\powerset {\powerset a}$ is a set, it follows that:
:$\powerset {\powerset a} \in V$
But we have by definition of class difference that $\set a \notin V$.
Hence we have an element of an element of $V$ which is not itself in $V$.
Hence, by definition, $V \setminus \set a$ is not a transitive class.
== Sources ==
* : Chapter $2$: Some Basics of Class-Set Theory: $\S 10$ Some useful facts about transitivity: Exercise $10.1 \ \text {(b)}$
Category:Examples of Transitive Classes