nyuuzyou/wikis huggingface dataset:
Valid Patterns of Categorical Syllogism

== Theorem ==
The following categorical syllogisms are valid:
:$\begin{array}{rl}
\text{I} & AAA \\
\text{I} & AII \\
\text{I} & EAE \\
\text{I} & EIO \\
* \text{I} & AAI \\
* \text{I} & EAO \\
\end{array}
\qquad
\begin{array}{rl}
\text{II} & EAE \\
\text{II} & AEE \\
\text{II} & AOO \\
\text{II} & EIO \\
* \text{II} & EAO \\
* \text{II} & AEO \\
\end{array}
\qquad
\begin{array}{rl}
\dagger \text{III} & AAI \\
\text{III} & AII \\
\text{III} & IAI \\
\dagger \text{III} & EAO \\
\text{III} & EIO \\
\text{III} & OAO \\
\end{array}
\qquad
\begin{array}{rl}
\S \text{IV} & AAI \\
\text{IV} & AEE \\
\dagger \text{IV} & EAO \\
\text{IV} & EIO \\
\text{IV} & IAI \\
* \text{IV} & AEO \\
\end{array}$
In the above:
:$\text{I}, \text{II}, \text{III}, \text{IV}$ denote the four figures of the categorical syllogisms.
:$A, E, I, O$ denote the universal affirmative, universal negative, particular affirmative and particular negative respectively: see Shorthand for Categorical Syllogism.
:Syllogisms marked $*$ require the assumption that $\exists x: \map S x$, that is, that there exists an object fulfilling the secondary predicate.
:Syllogisms marked $\dagger$ require the assumption that $\exists x: \map M x$, that is, that there exists an object fulfilling the middle predicate.
:Syllogisms marked $\S$ require the assumption that $\exists x: \map P x$, that is, that there exists an object fulfilling the primary predicate.
== Proof ==
From Elimination of all but 24 Categorical Syllogisms as Invalid, all but these $24$ patterns have been shown to be invalid.
It remains to be shown that these remaining syllogisms are in fact valid.
== Sources ==
* : Chapter $4$: The Predicate Calculus $2$: $4$ The Syllogism
* 
* 
Category:Categorical Syllogisms
Valid Syllogism in Figure III needs Particular Conclusion and if Negative then Negative Major Premise

== Theorem ==
Let $Q$ be a valid categorical syllogism in Figure $\text {III}$.
Then it is a necessary condition that:
:The conclusion of $Q$ be a particular categorical statement
and:
:If the conclusion of $Q$ be a negative categorical statement, then so is the major premise of $Q$.
== Proof ==
Consider Figure $\text {III}$:
Let the major premise of $Q$ be denoted $\text{Maj}$.
Let the minor premise of $Q$ be denoted $\text{Min}$.
Let the conclusion of $Q$ be denoted $\text{C}$.
$M$ is:
:the subject of $\text{Maj}$
:the subject of $\text{Min}$.
So, in order for $M$ to be distributed, either:
:$(1): \quad$ From Universal Categorical Statement Distributes its Subject: $\text{Maj}$ must be universal
or:
:$(2): \quad$ From Universal Categorical Statement Distributes its Subject: $\text{Min}$ must be universal.
Suppose $\text{Min}$ to be a negative categorical statement.
Then by No Valid Categorical Syllogism contains two Negative Premises:
:$\text{Maj}$ is an affirmative categorical statement.
From Conclusion of Valid Categorical Syllogism is Negative iff one Premise is Negative:
:$\text{C}$ is a negative categorical statement.
From Negative Categorical Statement Distributes its Predicate:
:$P$ is distributed in $\text{C}$.
From Negative Categorical Statement Distributes its Predicate:
:$P$ is undistributed in $\text{Maj}$.
From Distributed Term of Conclusion of Valid Categorical Syllogism is Distributed in Premise:
:$P$ is distributed in $\text{Maj}$.
That is, $P$ is both distributed and undistributed in $\text{Maj}$.
From this Proof by Contradiction it follows that $\text{Min}$ is an affirmative categorical statement.
Thus from Conclusion of Valid Categorical Syllogism is Negative iff one Premise is Negative:
:if $\text{C}$ is a negative categorical statement, then so is $\text{Maj}$
We have that $\text{Min}$ is an affirmative categorical statement.
Hence from Negative Categorical Statement Distributes its Predicate:
:$S$ is undistributed in $\text{Min}$.
From Distributed Term of Conclusion of Valid Categorical Syllogism is Distributed in Premise:
:$S$ is undistributed in $\text{C}$.
So from Universal Categorical Statement Distributes its Subject:
:$\text{C}$ is a particular categorical statement.
Hence, in order for $Q$ to be valid:
:$\text{C}$ must be a particular categorical statement
:if $\text{C}$ is a negative categorical statement, then so is $\text{Maj}$.
== Sources ==
* : Chapter $4$: The Predicate Calculus $2$: $4$ The Syllogism: Exercise $\text{(f)}$
Category:Categorical Syllogisms
Valid Syllogism in Figure II needs Negative Conclusion and Universal Major Premise

== Theorem ==
Let $Q$ be a valid categorical syllogism in Figure $\text{II}$.
Then it is a necessary condition that:
:The major premise of $Q$ be a universal categorical statement
and
:The conclusion of $Q$ be a negative categorical statement.
== Proof ==
Consider Figure $\text{II}$:
Let the major premise of $Q$ be denoted $\text{Maj}$.
Let the minor premise of $Q$ be denoted $\text{Min}$.
Let the conclusion of $Q$ be denoted $\text{C}$.
$M$ is:
:the predicate of $\text{Maj}$
:the predicate of $\text{Min}$.
So, in order for $M$ to be distributed, either:
:$(1): \quad$ From Negative Categorical Statement Distributes its Predicate: $\text{Maj}$ must be negative
or:
:$(2): \quad$ From Negative Categorical Statement Distributes its Predicate: $\text{Min}$ must be negative.
Note that from No Valid Categorical Syllogism contains two Negative Premises, it is not possible for both $\text{Maj}$ and $\text{Min}$ to be negative.
From Conclusion of Valid Categorical Syllogism is Negative iff one Premise is Negative:
:$\text{C}$ is a negative categorical statement.
From Negative Categorical Statement Distributes its Predicate:
:$P$ is distributed in $\text{C}$.
From Distributed Term of Conclusion of Valid Categorical Syllogism is Distributed in Premise:
:$P$ is distributed in $\text{Maj}$.
From Universal Categorical Statement Distributes its Subject:
:$\text{Maj}$ is a universal categorical statement.
Hence, in order for $Q$ to be valid:
:$\text{Maj}$ must be a universal categorical statement
:Either $\text{Maj}$ or $\text{Min}$, and therefore $\text{C}$, must be a negative categorical statement.
== Sources ==
* : Chapter $4$: The Predicate Calculus $2$: $4$ The Syllogism: Exercise $\text{(e)}$
Category:Categorical Syllogisms
Valid Syllogism in Figure I needs Affirmative Minor Premise and Universal Major Premise

== Theorem ==
Let $Q$ be a valid categorical syllogism in Figure $\text I$.
Then it is a necessary condition that:
:The major premise of $Q$ be a universal categorical statement
and
:The minor premise of $Q$ be an affirmative categorical statement.
== Proof ==
Consider Figure $\text I$:
Let the major premise of $Q$ be denoted $\text{Maj}$.
Let the minor premise of $Q$ be denoted $\text{Min}$.
Let the conclusion of $Q$ be denoted $\text{C}$.
$M$ is:
:the subject of $\text{Maj}$
:the predicate of $\text{Min}$.
So, in order for $M$ to be distributed, either:
:$(1): \quad$ From Universal Categorical Statement Distributes its Subject: $\text{Maj}$ must be universal
or:
:$(2): \quad$ From Negative Categorical Statement Distributes its Predicate: $\text{Min}$ must be negative.
Suppose $\text{Min}$ is a negative categorical statement.
Then by Conclusion of Valid Categorical Syllogism is Negative iff one Premise is Negative:
:$\text{C}$ is a negative categorical statement.
From $(2)$:
:$P$ is distributed in $\text{C}$.
From Distributed Term of Conclusion of Valid Categorical Syllogism is Distributed in Premise:
:$P$ is distributed in $\text{Maj}$.
From Negative Categorical Statement Distributes its Predicate:
:$\text{Maj}$ is a negative categorical statement.
Thus both:
:$\text{Min}$ is a negative categorical statement
:$\text{Maj}$ is a negative categorical statement.
But from No Valid Categorical Syllogism contains two Negative Premises, this means $Q$ is invalid.
Thus $\text{Min}$ is not a negative categorical statement in Figure $\text I$.
As $\text{Min}$ needs to be an affirmative categorical statement, $M$ is not distributed in $\text{Min}$.
From Middle Term of Valid Categorical Syllogism is Distributed at least Once, this means $M$ must be distributed in $\text{Maj}$.
As $M$ is the subject of $\text{Maj}$ in Figure $\text I$, it follows from $(1)$ that:
:$\text{Maj}$ is a universal categorical statement.
Hence, in order for $Q$ to be valid:
:$\text{Maj}$ must be a universal categorical statement
:$\text{Min}$ must be an affirmative categorical statement.
== Sources ==
* : Chapter $4$: The Predicate Calculus $2$: $4$ The Syllogism: Exercise $\text{(d)}$
Category:Categorical Syllogisms
Valid Syllogisms in Figure IV

== Theorem ==
Let $Q$ be a valid categorical syllogism in Figure $\text {IV}$.
Then it is a necessary condition that:
:$(1): \quad$ Either:
:: the major premise of $Q$ be a negative categorical statement
:or:
:: the minor premise of $Q$ be a universal categorical statement
:or both.
:$(2): \quad$ If the conclusion of $Q$ be a negative categorical statement, then the major premise of $Q$ be a universal categorical statement.
:$(3): \quad$ If the conclusion of $Q$ be a universal categorical statement, then the minor premise of $Q$ be a negative categorical statement.
== Proof ==
Consider Figure $\text {IV}$:
Let the major premise of $Q$ be denoted $\text{Maj}$.
Let the minor premise of $Q$ be denoted $\text{Min}$.
Let the conclusion of $Q$ be denoted $\text{C}$.
$M$ is:
:the predicate of $\text{Maj}$
:the subject of $\text{Min}$.
We have:
:Middle Term of Valid Categorical Syllogism is Distributed at least Once.
So, in order for $M$ to be distributed, either:
:From Negative Categorical Statement Distributes its Predicate: $\text{Maj}$ must be negative
or:
:From Universal Categorical Statement Distributes its Subject: $\text{Min}$ must be universal.
Both may be the case.
Thus $(1)$ is seen to hold.
Let $\text{C}$ be a negative categorical statement.
From Negative Categorical Statement Distributes its Predicate:
:$P$ is distributed in $\text{C}$.
From Distributed Term of Conclusion of Valid Categorical Syllogism is Distributed in Premise:
:$P$ is distributed in $\text{Maj}$.
So from Universal Categorical Statement Distributes its Subject:
:$\text{Maj}$ is a universal categorical statement.
Thus $(2)$ is seen to hold.
Let $\text{C}$ be a universal categorical statement.
From Universal Categorical Statement Distributes its Subject:
:$S$ is distributed in $\text{C}$.
From Distributed Term of Conclusion of Valid Categorical Syllogism is Distributed in Premise:
:$S$ is distributed in $\text{Min}$.
From Negative Categorical Statement Distributes its Predicate:
:$S$ is a negative categorical statement.
Thus $(3)$ is seen to hold.
== Sources ==
* : Chapter $4$: The Predicate Calculus $2$: $4$ The Syllogism
Category:Categorical Syllogisms
Valuation Ideal is Maximal Ideal of Induced Valuation Ring

== Theorem ==
Let $\struct {R, \norm {\,\cdot\,} }$ be a non-Archimedean normed division ring with zero $0_R$ and unity $1_R$.
Let $\OO$ be the valuation ring induced by the non-Archimedean norm $\norm {\,\cdot\,}$, that is:
:$\OO = \set{x \in R : \norm x \le 1}$
Let $\PP$ be the valuation ideal induced by the non-Archimedean norm $\norm {\,\cdot\,}$, that is:
:$\PP = \set{x \in R : \norm x < 1}$
Then $\PP$ is an ideal of $\OO$:
:$(a):\quad \PP$ is a maximal left ideal
:$(b):\quad \PP$ is a maximal right ideal
:$(c):\quad$ the quotient ring $\OO / \PP$ is a division ring.
=== Corollary ===
== Proof ==
First it is shown that $\PP$ is an ideal of $\OO$ by applying Test for Ideal.
That is, it is shown that:
:$(1): \quad \PP \ne \O$
:$(2): \quad \forall x, y \in \PP: x + \paren {-y} \in \PP$
:$(3): \quad \forall x \in \PP, y \in \OO: x y \in \PP$
By Maximal Left and Right Ideal iff Quotient Ring is Division Ring the statements (a), (b) and (c) above are equivalent.
So then it is shown:
:$(4): \quad \PP$ is a maximal left ideal
==== $(1): \quad \PP \ne \O$ ====
By :
:$\norm {0_R} = 0$
Hence:
:$0_R \in \PP \ne \O$
==== $(2): \quad \forall x, y \in \PP: x + \paren {-y} \in \PP$ ====
Let $x, y \in \PP$.
Then:
Hence:
:$x + \paren {-y} \in \PP$
==== $(3): \quad \forall x \in \PP, y \in \OO: x y \in \PP$ ====
Let $x \in \PP, y \in \OO$.
Then:
Hence:
:$x y \in \PP$
By Test for Ideal it follows that $\PP$ is an ideal of $\OO$.
==== $(4):\quad \PP$ is a maximal left ideal ====
Let $J$ be a left ideal of $\OO$:
:$\PP \subsetneq J \subset \OO$
Let $x \in J \setminus \PP$, then:
:$\norm x = 1$
By Norm of Inverse then:
:$\norm {x^{-1} } = 1 / \norm x = 1 / 1 = 1$ 
Hence:
:$x^{-1} \in \OO$
Since $J$ is a left ideal then:
:$x^{-1} x = 1_R \in J$
Thus:
:$\forall y \in \OO: y \cdot 1_R = y \in J$
That is, $J = \OO$
Hence $\PP$ is a maximal left ideal.
The result follows.
== Sources ==
* : $\S 2.4$ Algebra: Proposition $2.4.1$
Category:Normed Division Rings
Category:Non-Archimedean Norms
Valuation Ideal is Maximal Ideal of Induced Valuation Ring/Corollary 1

== Theorem ==
Let $\struct {R, \norm {\,\cdot\,} }$ be a non-Archimedean normed division ring with zero $0_R$ and unity $1_R$.
Let $\OO$ be the valuation ring induced by the non-Archimedean norm $\norm {\,\cdot\,}$, that is:
:$\OO = \set {x \in R : \norm x \le 1}$
Let $\PP$ be the valuation ideal induced by the non-Archimedean norm $\norm {\,\cdot\,}$, that is:
:$\PP = \set {x \in R : \norm x < 1}$
Then:
:$(a): \quad \OO$ is a local ring.
:$(b): \quad \PP$ is the unique maximal left ideal of $\OO$
:$(c): \quad \PP$ is the unique maximal right ideal of $\OO$
== Proof ==
By Valuation Ideal is Maximal Ideal of Induced Valuation Ring then:
:$\PP$ is a maximal left ideal of $\OO$.
Let $J \subsetneq \OO$ be any maximal left ideal of $\OO$.
Let $x \in \OO \setminus \PP$.
 $x \in J$.
By Norm of Inverse then:
:$\norm {x^{-1}} = 1 / \norm x = 1 / 1 = 1$ 
Hence:
:$x^{-1} \in \OO$
Since $J$ is a left ideal then:
:$x^{-1} x = 1_R \in J$
Thus:
:$\forall y \in \OO: y \cdot 1_R = y \in J$
That is: 
:$J = \OO$
This contradicts the assumption that $J \ne \OO$.
So:
:$x \notin J$
Hence: 
:$\paren {\OO \setminus \PP} \cap J = \O$
That is:
:$J \subseteq \PP$
Since $J$ and $\PP$ are both maximal left ideals then:
:$J = \PP$
Hence:
:$\PP$ is the unique maximal left ideal of $\OO$.
By :
:$\OO$ is a local ring
From Maximal Left and Right Ideal iff Quotient Ring is Division Ring:
:every maximal right ideal of $\OO$ is a maximal left ideal
Hence:
:$\PP$ is the unique maximal right ideal of $\OO$.
The result follows.
== Sources ==
* : $\S 2.4$ Algebra, Proposition $2.4.1$
Category:Normed Division Rings
Category:Non-Archimedean Norms
Valuation Ideal of P-adic Norm on Rationals

== Theorem ==
Let $\norm {\,\cdot\,}_p$ be the $p$-adic norm on the rationals $\Q$ for some prime $p$.
The induced valuation ideal on $\struct {\Q,\norm {\,\cdot\,}_p}$ is the set:
:$\PP = p \Z_{\ideal p} = \set {\dfrac a b \in \Q : p \nmid b, p \divides a}$
where $\Z_{\ideal p}$ is the induced valuation ring on $\struct {\Q,\norm {\,\cdot\,}_p}$
== Proof ==
Let $\nu_p: \Q \to \Z \cup \set {+\infty}$ be the $p$-adic valuation on $\Q$.
Then:
Let $\dfrac a b \in \Q$ be in canonical form. 
Then $a \perp b$
Suppose $p \divides a$. 
Then $p \nmid b$.
Hence:
:$\map {\nu_p} a > 0 = \map {\nu_p} b$
Suppose $p \nmid a$. 
Then:
:$\map {\nu_p} b \ge 0 = \map {\nu_p} a$
So: 
:$\map {\nu_p} a > \map {\nu_p} b$ $p \nmid b$ and $p \divides a$ 
Hence:
:$\PP = \set {\dfrac a b \in \Q : p \nmid b, p \divides a}$
So:
Hence:
:$\PP = p \Z_{\ideal p}$
== Sources ==
* : $\S 2.4$ Algebra: Proposition $2.4.3$
Category:P-adic Number Theory
Valuation Ideal of P-adic Numbers

== Theorem ==
Let $p$ be a prime number.
Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers.
Then the valuation ideal induced by norm $\norm {\,\cdot\,}_p$ is the principal ideal:
:$p \Z_p = \set {x \in \Q_p: \norm x_p < 1}$
where $\Z_p$ denotes the $p$-adic integers.
== Proof ==
From P-adic Integers is Local Ring, $\Z_p$ is a local ring.
From Principal Ideal from Element in Center of Ring, $p \Z_p$ is a principal ideal.
Now:
Hence:
:$p \Z_p = \set {x \in \Q_p: \norm x_p < 1}$
== Sources ==
* : $\S 3.3$ Exploring $\Q_p$: Lemma $3.3.4$
Category:P-adic Numbers
Valuation Ring is Local

== Theorem ==
Let $R$ be a valuation ring.
Then $R$ is a local ring.
== Proof ==
By definition of local ring:
:$R$ is a local ring :
::$R$ is non-trivial
::the sum of two arbitrary non-units is a non-unit.
First we recall that as a valuation ring is an integral domain, then :
:$R$ is non-trivial
:$R$ has no (proper) zero divisors
:$R$ is a commutative and unitary ring.
Let $0$ and $1$ be the zero and unity of $R$ respectively.
Let $x, y \in R$ be non-units.
We shall show that $x + y$ is non-unit.
, let $x = 0$.
Then:
:$x + y = y$
and so $x + y$ is a non-unit .
Similarly for $y = 0$.
Let $x \ne 0$ and $y \ne 0$.
By definition of non-unit, neither $x$ nor $y$ has an inverse in $R$.
Let $K$ be the field of quotients of $R$.
In particular, $x^{-1} \in K$ and $y^{-1} \in K$.
First note that:
By definition of valuation ring, either:
:$x y^{-1} \in R$
or:
, let $x y^{-1} \in R$.
 there exists a $u \in R$ such that:
:$u \paren {x + y} = 1$
Then:
That is, the inverse of $y$ is in $R$.
This contradicts the fact that $y$ is a non-unit.
Therefore $x + y$ is a non-unit.
Category:Valuation Rings
Category:Local Rings
Valuation Ring of Non-Archimedean Division Ring is Clopen

== Theorem ==
Let $\struct {R, \norm {\,\cdot\,} }$ be a non-Archimedean normed division ring with zero $0_R$.
Let $\OO$ be valuation ring induced by $\norm{\,\cdot\,}$.
Then $\OO$ is a both open and closed in the metric induced by $\norm{\,\cdot\,}$.
=== Corollary ===
== Proof ==
The valuation ring $\OO$ Is the open ball $\map {B_1^-} {0_R}$ by definition.
By Open Balls of Non-Archimedean Division Rings are Clopen then $\OO$ is both open and closed in the metric induced by $\norm {\,\cdot\,}$.
Category:Normed Division Rings
Category:Valuation Ring of Non-Archimedean Division Ring is Clopen
Valuation Ring of Non-Archimedean Division Ring is Clopen/Corollary 1

== Theorem ==
Let $p$ be a prime number.
Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers.
Then the $p$-adic integers $\Z_p$ is both open and closed in the $p$-adic metric.
== Proof ==
The $p$-adic integers $\Z_p$ is the valuation ring induced by $\norm {\,\cdot\,}_p$ by definition.
By Valuation Ring of Non-Archimedean Division Ring is Clopen then the $p$-adic integers $\Z_p$ is both open and closed in the $p$-adic metric.
Category:Valuation Ring of Non-Archimedean Division Ring is Clopen
Valuation Ring of Non-Archimedean Division Ring is Subring

== Theorem ==
Let $\struct {R, \norm{\,\cdot\,}}$ be a non-Archimedean normed division ring with zero $0_R$ and unity $1_R$.
Let $\OO$ be the valuation ring induced by the non-Archimedean norm $\norm {\,\cdot\,}$, that is:
:$\OO = \set {x \in R : \norm{x} \le 1}$
Then $\OO$ is a subring of $R$:
:with a unity: $1_R$
:in which there are no (proper) zero divisors, that is:
:::$\forall x, y \in \OO: x \circ y = 0_R \implies x = 0_R \text{ or } y = 0_R$
== Proof ==
To show that $\OO$ is a subring the Subring Test is used by showing:
:$(1): \quad \OO \ne \O$
:$(2): \quad \forall x, y \in \OO: x + \paren {-y} \in \OO$
:$(3): \quad \forall x, y \in \OO: x y \in \OO$
(1)
By Norm of Unity,
:$\norm{1_R} = 1$
Hence:
:$1_R \in \OO \ne \O$
(2)
Let $x, y \in \OO$.
Then:
Hence:
:$x + \paren {-y} \in \OO$
(3)
Let $x, y \in \OO$.
Then:
Hence:
:$x y \in \OO$
By Subring Test it follows that $\OO$ is a subring of $R$.
Since $1_R \in S$ and $1_R$ is the unity of $R$ then $1_R$ is the unity of $\OO$.
By Division Ring has No Proper Zero Divisors then $R$ has no proper zero divisors.
Hence $\OO$ has no proper zero divisors.
== Sources ==
* : $\S 2.4$ Algebra: Proposition $2.4.1$
Category:Normed Division Rings
Category:Non-Archimedean Norms
Valuation Ring of P-adic Norm is Subring of P-adic Integers

== Theorem ==
Let $p$ be a prime number.
Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers.
Let $\Z_p$ be the $p$-adic integers.
Let $\Z_{\ideal p}$ be the induced valuation ring on $\struct {\Q,\norm {\,\cdot\,}_p}$.
Then:
:$(1): \quad \Z_{\ideal p} = \Q \cap \Z_p$.
:$(2): \quad \Z_{\ideal p}$ is a subring of $\Z_p$.
=== Corollary ===
== Proof ==
The $p$-adic integers is defined as:
:$\Z_p = \set {x \in \Q_p: \norm x_p \le 1}$
The induced valuation ring on $\struct {\Q,\norm {\,\cdot\,}_p}$ is defined as:
:$\Z_{\ideal p} = \set {x \in \Q: \norm x_p \le 1}$
From Rational Numbers are Dense Subfield of P-adic Numbers:
:the $p$-adic norm $\norm {\,\cdot\,}_p$ on $\Q_p$ is an extension of the $p$-adic norm $\norm {\,\cdot\,}_p$ on $\Q$.
It follows that $\Z_{\ideal p} = \Q \cap \Z_p$.
This proves $(1)$ above.
By Valuation Ring of Non-Archimedean Division Ring is Subring then $\Z_p$ is a subring of $\Q_p$.
By definition of $p$-adic integers then $\Q$ is a subring of $\Q_p$.
By Intersection of Subrings is Largest Subring Contained in all Subrings then $\Z_{\paren p}$ is a subring of $\Z_p$.
This proves $(2)$ above.
== Sources ==
* : $\S 3.3$ Exploring $\Q_p$, Proposition $3.3.4$
Category:P-adic Number Theory
Category:Valuation Ring of P-adic Norm is Subring of P-adic Integers
Valuation Ring of P-adic Norm is Subring of P-adic Integers/Corollary 1

== Theorem ==
Let $p$ be a prime number.
Let $\Z_p$ be the $p$-adic integers.
The set of integers $\Z$ is a subring of $\Z_p$.
== Proof ==
Let $\Z_{\paren p}$ be the valuation ring induced by $\norm {\,\cdot\,}_p$ on $\Q$.
By Integers form Subring of Valuation Ring of P-adic Norm on Rationals then:
:$\Z$ is a subring of $\Z_{\paren p}$
By Valuation Ring of P-adic Norm is Subring of P-adic Integers then:
:$\Z_{\paren p}$ is a subring of $\Z_p$
The result follows.
== Sources ==
* : $\S 3.3$ Exploring $\Q_p$, Proposition $3.3.4$
Category:Valuation Ring of P-adic Norm is Subring of P-adic Integers
Valuation Ring of P-adic Norm on Rationals

== Theorem ==
Let $\norm {\,\cdot\,}_p$ be the $p$-adic norm on the rationals $\Q$ for some prime $p$.
The induced valuation ring on $\struct {\Q,\norm {\,\cdot\,}_p}$ is the set:
:$\OO = \Z_{\paren p} = \set {\dfrac a b \in \Q : p \nmid b}$
=== Corollary ===
== Proof ==
Let $\nu_p: \Q \to \Z \cup \set {+\infty}$ be the $p$-adic valuation on $\Q$.
Then:
Let $\dfrac a b \in \Q$ be in canonical form. 
Then $a \perp b$
Suppose $p \divides b$. 
Then $p \nmid a$.
Hence:
:$\map {\nu_p} b \gt 0 = \map {\nu_p} a$
Suppose $p \nmid b$. 
Then:
:$\map {\nu_p} a \ge 0 = \map {\nu_p} b$
So: 
:$\map {\nu_p} a \ge \map {\nu_p} b$ $p \nmid b$ 
Hence:
:$\OO = \set {\dfrac a b \in \Q : p \nmid b }$
== Sources ==
* : $\S 2.4$ Algebra: Proposition $2.4.3$
Category:P-adic Number Theory
Category:Valuation Ring of P-adic Norm on Rationals
Valuation Ring of P-adic Norm on Rationals/Corollary 1

== Theorem ==
Let $\norm {\,\cdot\,}_p$ be the $p$-adic norm on the rationals $\Q$ for some prime $p$.
Let $\OO$ be the induced valuation ring on $\struct {\Q,\norm {\,\cdot\,}_p}$. 
The set of integers $\Z$ is a subring of $\OO$.
== Proof ==
By Valuation Ring of P-adic Norm on Rationals, the induced valuation ring $\OO$ is the set:
:$\OO = \Z_{\paren p} = \set {\dfrac a b \in \Q : p \nmid b}$
Since $p \nmid 1$ then for all $a \in \Z$, $a = \dfrac a 1 \in \OO$.
Hence $\Z \subseteq \OO$.
By Valuation Ring of Non-Archimedean Division Ring is Subring then $\OO$ is a subring of $\Q$.
By Integers form Subdomain of Rationals then $\Z$ is a subring of $\Q$.
By Intersection of Subrings is Largest Subring Contained in all Subrings then $\Z \cap \OO = \Z$ is a subring of $\OO$. 
== Sources ==
* : $\S 2.4$ Algebra, Problem $59$
Category:Valuation Ring of P-adic Norm on Rationals