avewright/proofwiki-math huggingface dataset:
Theorem: Restriction of Canonical Surjection to Restricted Dipper Semigroup is Isomorphism

Let $m, n \in \N_{>0}$ be non-zero natural numbers.
Let $\N_{< \paren {m \mathop + n} }$ denote the initial segment of the natural numbers:
:$\N_{< \paren {m \mathop + n} } := \set {0, 1, \ldots, m + n - 1}$
Let $\N^*_{< \paren {m \mathop + n} }$ denote the set defined as $\N_{< \paren {m \mathop + n} } \setminus \set 0$:
:$\N^*_{< \paren {m \mathop + n} } := \set {1, 2, \ldots, m + n - 1}$
Let $+_{m, n}$ be the '''dipper operation''' on $\N_{< \paren {m \mathop + n} }$:
:$\forall a, b \in \N_{< \paren {m \mathop + n} }: a +_{m, n} b = \begin{cases} a + b & : a + b < m \\ a + b - k n & : a + b \ge m \end{cases}$
where $k$ is the largest integer satisfying:
:$m + k n \le a + b$
Let $+^*_{m, n}$ be the '''restricted dipper operation''' on $\N^*_{< \paren {m \mathop + n} }$ defined as:
:$\forall a, b \in \N^*_{< \paren {m \mathop + n} }: a +^*_{m, n} b = \begin{cases} a + b & : a + b < m \\ a + b - k n & : a + b \ge m \end{cases}$
where $k$ is the largest integer satisfying:
:$m + k n \le a + b$
Consider the algebraic structure $\struct {\N^*_{< \paren {m \mathop + n} }, +^*_{m, n} }$.
Let $\phi^*_{m, n}$ be the restriction of the canonical surjection from $\N_{>0}$ to the restricted dipper semigroup $\struct {\map {D^*} {m, n}, \oplus^*_{m, n} }$.
Then $\phi^*_{m, n}$ is an isomorphism from $\struct {\N^*_{< \paren {m \mathop + n} }, +^*_{m, n} }$ to $\struct {\map {D^*} {m, n}, \oplus^*_{m, n} }$.

Proof:
{{ProofWanted|Formalise the notion that everything is a restriction of everything else}}
Theorem: Inductive Semigroup whose Inductive Elements Commute is Commutative Semigroup

Let $\struct {S, \circ}$ be a semigroup.
Let there exist $\alpha, \beta \in S$ which fulfil the condition for $\struct {S, \circ}$ to be an inductive semigroup:
:the only subset of $S$ containing both $\alpha$ and $x \circ \beta$ whenever it contains $x$ is $S$ itself.
Let $\alpha$ and $\beta$ commute.
Then $\struct {S, \circ}$ is a commutative semigroup.

Proof:
Suppose $\struct {S, \circ}$ is a semigroup.
Suppose there exist $\alpha, \beta \in S$ such that $\struct {S, \circ}$ is an inductive semigroup. 
That is, suppose there exist $\alpha, \beta \in S$ such that the only subset of $S$ containing both $\alpha$ and $x \circ \beta$ whenever it contains $x$ is $S$ itself.
That is:
:$\exists \alpha, \beta \in S: \forall A \subseteq S: \paren {\alpha \in A \land \paren {\forall x \in A: x \circ \beta \in A} } \implies A = S$
By Form of Elements of Inductive Semigroup, the elements of $S$ are of the form:
:$\alpha \circ \beta \circ \beta \circ \cdots \circ \beta$
{{ProofWanted}}
Then:
:$\forall x, y \in S: x \circ y = y \circ x$
Hence, $\struct {S, \circ}$ is a commutative semigroup, as required.
Theorem: Strictly Inductive Semigroup is Inductive Semigroup

Let $\struct {S, \circ}$ be a strictly inductive semigroup.
Then $\struct {S, \circ}$ is an inductive semigroup.

Proof:
In accordance with our assertion, let $\struct {S, \circ}$ be a strictly inductive semigroup.
By definition of strictly inductive semigroup, there exists $\beta \in S$ such that the only subset of $S$ containing both $\beta$ and $x \circ \beta$ whenever it contains $x$ is $S$ itself.
That is:
:$\exists \beta \in S: \forall A \subseteq S: \paren {\beta \in S \land \paren {\forall x \in A: x \circ \beta \in A} } \implies A = S$
Let $\alpha = \beta$.
Then we may say that there exist $\alpha, \beta \in S$ such that the only subset of $S$ containing both $\alpha$ and $x \circ \beta$ whenever it contains $x$ is $S$ itself.
That is:
:$\exists \alpha, \beta \in S: \forall A \subseteq S: \paren {\alpha \in A \land \paren {\forall x \in A: x \circ \beta \in A} } \implies A = S$
Hence, by identifying $\alpha$ with $\beta$, $\struct {S, \circ}$ fulfils the conditions of being an inductive semigroup.
{{qed}}
Theorem: Natural Numbers under Addition form Inductive but not Strictly Inductive Semigroup

Let $\struct {\N, +}$ denote the algebraic structure consisting of the set of natural numbers $\N$ under addition $+$.
Then $\struct {\N, +}$ forms an inductive semigroup, but not a strictly inductive semigroup

Proof:
Recall the definition of inductive semigroup:
{{:Definition:Inductive Semigroup}}
Recall the definition of strictly inductive semigroup:
{{:Definition:Strictly Inductive Semigroup/Definition 1}}
The natural numbers $\N$ can be considered as a naturally ordered semigroup.
From Naturally Ordered Semigroup is Unique, $\struct {\N, +}$ is unique up to isomorphism.
{{explain|For some reason I can't find the exact result which performs that identification}}
Hence from Naturally Ordered Semigroup forms Peano Structure, $\N$ is a Peano structure.
Let $S$ be a subset of $\N$.
Then by {{PeanoAxiom|5}}:
:$\paren {0 \in S \land \paren {\forall z \in S: z + 1 \in S} } \implies S = \N$
The latter condition is that which defines an inductive semigroup, where $0$ is identified with $\alpha$ and $1$ with $\beta$.
However, there is no element in $\struct {\N, +}$ which can be identified with $\beta$ in the definition of strictly inductive semigroup.
Indeed, the element $0$ can be expressed in the form $x \circ 1$ for any $x$.
{{qed}}
Theorem: Ideals of P-adic Integers

Let $\Z_p$ be the $p$-adic integers for some prime $p$.
Then the ideals of $\Z_p$ are the principal ideals:
:$\text a) \quad \set 0$
:$\text b) \quad \forall k \in \N: p^k \Z_p$

Proof:
Let $\nu_p: \Q_p \to \Z \cup \set {+\infty}$ denote the $p$-adic valuation on the $p$-adic numbers.
Let $\Z_p^\times$ denote the $p$-adic units.
Let $I \ne \set 0$ be a non-null ideal of $\Z_p$.
Hence:
:$\exists j \in I : \map {\nu_p} j < \infty$
Let:
:$k = \inf \set {\map {\nu_p} i : i \in I}$
Hence:
:$k \le j < \infty$
Let:
:$a \in I : a \ne 0 \land \map {\nu_p} a = k$
From P-adic Number is Power of p Times P-adic Unit:
:$\exists u \in \Z_p^\times : a = p^k u$
We have:
{{begin-eqn}}
{{eqn | l = p^k 
      | r = u^{-1} a
}}
{{eqn | o = \in 
      | r = I
      | c = {{Defof|Ideal of Ring}}
}}
{{eqn | ll= \leadsto
      | l = \paren {p^k}
      | r = p^k\Z_p
      | c = {{Defof|Principal Ideal}}
}}
{{eqn | o = \subseteq
      | r = I
      | c = {{Defof|Ideal of Ring}}
}}
{{end-eqn}}
Let $b \in I$.
Theorem: Closed Subgroups of P-adic Integers

Let $\Z_p$ be the $p$-adic integers for some prime $p$.
Then the closed subgroups of $\Z_p$ are the principal ideals:
:$\text a) \quad \set 0$
:$\text b) \quad \forall k \in \N : p^k \Z_p$

Proof:
From Metric Space is Hausdorff:
:$\Z_p$ is a Hausdorff space
From Finite Subspace of Hausdorff Space is Closed:
:$\set 0$ is closed
From Cosets Form Local Basis of P-adic Number:
:$\forall k \in \N : p^k \Z_p = 0 + p^k \Z_p$ is closed
Hence the ideals:
:$\text a) \quad \set 0$
:$\text b) \quad \forall k \in \N : p^k \Z_p$
are closed subgroups.
It remains to show that an arbitrary closed subgroups is indeed one of these ideals.
Let $H$ be a closed subgroup of the additive group of $\Z_p$.
From Correspondence between Abelian Groups and Z-Modules:
:$\Z H \subseteq H$
Let $a \in \Z_p$.
Let $h \in H$.
From Integers are Dense in P-adic Integers, there exists a sequence $\sequence {a_n}$:
:$\forall n \in \N: a_n \in Z$
:$\ds \lim_{n \mathop \to \infty} a_n = a$
From Multiple Rule for Sequences in Normed Division Ring:
:$\ds \lim_{n \mathop \to \infty} a_n h = a h$
We have:
:$a_n h \in \Z H \subseteq H$
From Subset of Metric Space contains Limits of Sequences iff Closed:
:$a h = \ds \lim_{n \mathop \to \infty} a_n h \in H$
Since $a \in \Z_p$ and $h \in H$ were arbitrary, it follows that:
:$\Z_p H \subseteq H$
By definition of ideal:
:$H$ is an ideal of $\Z_p$
From Ideals of P-adic Integers, $H$ is one of the principal ideals:
:$\text a) \quad \set 0$
:$\text b) \quad \forall k \in \N : p^k \Z_p$
{{qed}}
Category:P-adic Integers
2qv8tuqco2f3i9rbsq7pzdw3uo89nl3
Theorem: Ideals of P-adic Integers/Corollary

:$\Z_p$ is a principal ideal domain

Proof:
From Ideals of P-adic Integers, all ideals of $\Z_p$ are the principal ideals:
:* $\quad\set 0$
:* $\quad\forall n \in \N : p^n \Z_p$
Hence $\Z_p$ is a principal ideal domain by definition.
{{qed}}
Theorem: Form of Elements of Inductive Semigroup

Let $\struct {S, \circ}$ be an inductive semigroup.
Then the elements of $S$ are of the form:
:$\alpha \circ \beta \circ \beta \circ \cdots \circ \beta$

Proof:
Recall the definition of inductive semigroup:
{{:Definition:Inductive Semigroup}}
It follows from the definition that all elements of the form:
:$\alpha \circ \beta \circ \beta \circ \cdots \circ \beta$
are indeed elements of $S$.
{{AimForCont}} $x \in S$ is not of the above form.
Let $A \subseteq S$ be such that $\alpha, x \in A$.
Then $A$ contains elements of the form:
:$x, x \circ \beta, x \circ \beta \circ \beta, \ldots$
Now let $B \subseteq S$ be such that $\alpha \in B$ but $x \notin B$.
Then $B$ is of the form:
:$\exists \alpha, \beta \in S: \paren {\alpha \in B \land \paren {\forall x \in B: x \circ \beta \in B} }$
but because $x \in S$ while $x \notin B$ it follows that:
:$B \ne S$
and so $S$ is not inductive.
From this contradiction it follows that all elements of $S$ are of the form:
:$\alpha \circ \beta \circ \beta \circ \cdots \circ \beta$
{{qed}}
Category:Inductive Semigroups
pqsh1lp4gbhs1c0krx1p5c0igj33ca4
Theorem: Left Self-Distributive Operation with Right Identity is Idempotent

Let $\struct {S, \circ}$ be an algebraic structure.
Let $\circ$ be left self-distributive.
Let $\struct {S, \circ}$ have a right identity.
Then $\circ$ is an idempotent operation.

Proof:
Let the right identity of $\struct {S, \circ}$ be $e_R$.
We have:
{{begin-eqn}}
{{eqn | q = \forall a, b, c \in S
      | l = a \circ \paren {b \circ c}
      | r = \paren {a \circ b} \circ \paren {a \circ c}
      | c = {{Defof|Left Self-Distributive Operation}}
}}
{{eqn | ll= \leadsto
      | q = \forall a \in S
      | l = a \circ \paren {e_R \circ e_R}
      | r = \paren {a \circ e_R} \circ \paren {a \circ e_R}
      | c = In particular, it holds for $e_R$
}}
{{eqn | ll= \leadsto
      | q = \forall a \in S
      | l = a
      | r = a \circ a
      | c = {{Defof|Right Identity}}
}}
{{end-eqn}}
The result follows by definition of idempotent operation.
{{qed}}
Theorem: Right Self-Distributive Operation with Left Identity is Idempotent

Let $\struct {S, \circ}$ be an algebraic structure.
Let $\circ$ be right self-distributive.
Let $\struct {S, \circ}$ have a left identity.
Then $\circ$ is an idempotent operation.

Proof:
Let the left identity of $\struct {S, \circ}$ be $e_L$.
We have:
{{begin-eqn}}
{{eqn | q = \forall a, b, c \in S
      | l = \paren {a \circ b} \circ c
      | r = \paren {a \circ c} \circ \paren {b \circ c}
      | c = {{Defof|Right Self-Distributive Operation}}
}}
{{eqn | ll= \leadsto
      | q = \forall c \in S
      | l = \paren {e_L \circ e_L} \circ c
      | r = \paren {e_L \circ c} \circ \paren {e_L \circ c}
      | c = In particular, it holds for $e_L$
}}
{{eqn | ll= \leadsto
      | q = \forall c \in S
      | l = c
      | r = c \circ c
      | c = {{Defof|Left Identity}}
}}
{{end-eqn}}
The result follows by definition of idempotent operation.
{{qed}}
Theorem: Product on Left with Idempotent Element under Left Self-Distributive Operation is Idempotent

Let $\struct {S, \circ}$ be an algebraic structure.
Let $\circ$ be left self-distributive.
Let $a \in S$ be an idempotent element of $\struct {S, \circ}$.
Then for all $b \in S$, $b \circ a$ is an idempotent element of $\struct {S, \circ}$

Proof:
Let $a \in S$ be an idempotent element of $\struct {S, \circ}$.
We have:
{{begin-eqn}}
{{eqn | q = \forall b \in S
      | l = \paren {b \circ a} \circ \paren {b \circ a}
      | r = b \circ \paren {a \circ a}
      | c = {{Defof|Left Self-Distributive Operation}}
}}
{{eqn | r = b \circ a
      | c = {{Defof|Idempotent Element}}
}}
{{end-eqn}}
The result follows by definition of idempotent element.
{{qed}}
Theorem: Product on Right with Idempotent Element under Right Self-Distributive Operation is Idempotent

Let $\struct {S, \circ}$ be an algebraic structure.
Let $\circ$ be right self-distributive.
Let $a \in S$ be an idempotent element of $\struct {S, \circ}$.
Then for all $b \in S$, $a \circ b$ is an idempotent element of $\struct {S, \circ}$

Proof:
Let $a \in S$ be an idempotent element of $\struct {S, \circ}$.
We have:
{{begin-eqn}}
{{eqn | q = \forall b \in S
      | l = \paren {a \circ b} \circ \paren {a \circ b}
      | r = \paren {a \circ a} \circ b
      | c = {{Defof|Right Self-Distributive Operation}}
}}
{{eqn | r = a \circ b
      | c = {{Defof|Idempotent Element}}
}}
{{end-eqn}}
The result follows by definition of idempotent element.
{{qed}}
Theorem: Condition for Operation to be Left Distributive over Constant Operation

Let $\struct {S, \circ}$ be an algebraic structure.
Let $\sqbrk c$ be the constant operation for some $c \in S$.
Then:
:$\circ$ is left distributive over $\sqbrk c$
{{iff}}:
:$\forall x \in S: x \circ c = c$

Proof:
=== Sufficient Condition ===
Let $\circ$ be left distributive over $\sqbrk c$.
{{begin-eqn}}
{{eqn | q = \forall x, y, z \in S
      | l = c
      | r = \paren {x \circ y} \sqbrk c \paren {x \circ z}
      | c = {{Defof|Constant Operation}}
}}
{{eqn | r = x \circ \paren {y \sqbrk c z}
      | c = {{Defof|Left Distributive Operation}}
}}
{{eqn | r = x \circ c
      | c = {{Defof|Constant Operation}}
}}
{{end-eqn}}
That is:
:$\forall x \in S: x \circ c = c$
{{qed|lemma}}
Theorem: Condition for Operation to be Right Distributive over Constant Operation

Let $\struct {S, \circ}$ be an algebraic structure.
Let $\sqbrk c$ be the constant operation for some $c \in S$.
Then:
:$\circ$ is right distributive over $\sqbrk c$
{{iff}}:
:$\forall x \in S: c \circ x = c$

Proof:
=== Sufficient Condition ===
Let $\circ$ be right distributive over $\sqbrk c$.
{{begin-eqn}}
{{eqn | q = \forall x, y, z \in S
      | l = c
      | r = \paren {y \circ x} \sqbrk c \paren {z \circ x}
      | c = {{Defof|Constant Operation}}
}}
{{eqn | r = \paren {y \sqbrk c z} \circ x
      | c = {{Defof|Right Distributive Operation}}
}}
{{eqn | r = c \circ x
      | c = {{Defof|Constant Operation}}
}}
{{end-eqn}}
That is:
:$\forall x \in S: c \circ x = c$
{{qed|lemma}}
Theorem: Condition for Constant Operation to be Distributive over Another Operation

Let $\struct {S, \circ}$ be an algebraic structure.
Let $\sqbrk c$ denote the constant operation for some $c \in S$.
Then:
:$\sqbrk c$ is distributive over $\circ$
{{iff}}:
:$c \circ c = c$

Proof:
=== Sufficient Condition ===
Let $\sqbrk c$ be distributive over $\circ$
{{begin-eqn}}
{{eqn | q = \forall x, y, z \in S
      | l = \paren {x \sqbrk c y} \circ \paren {x \sqbrk c z}
      | r = c circ c
      | c = {{Defof|Constant Operation}}
}}
{{eqn | q = \forall x, y, z \in S
      | l = x \sqbrk c \paren {y \circ z}
      | r = c
      | c = {{Defof|Constant Operation}}
}}
{{end-eqn}}
:$c \circ c = c$
{{qed|lemma}}
Theorem: Operation which is Left Distributive over Every Commutative Associative Operation is Right Operation

Let $\struct {S, \circ}$ be an algebraic structure.
Let $\circ$ have the property that:
:for every arbitrary operation $*$ on $S$ which is both commutative and associative, $\circ$ is left distributive over $*$.
Then $\circ$ is the right operation $\to$:
:$\forall a, b \in S: a \to b = b$

Proof:
First recall from Right Operation is Left Distributive over All Operations that the right operation is indeed  left distributive over all operations, whether commutative or associative.
Let $*$ be an arbitrary operation on $S$ which is both commutative and associative.
As asserted, let $\circ$ be left distributive over $*$.
Let $c \in S$ be arbitrary.
Consider the constant operation $\sqbrk c$:
:$\forall a, b \in S: a \sqbrk c b := c$
We have from Constant Operation is Commutative and Constant Operation is Associative that $\sqbrk c$ is both commutative and associative.
Hence $\circ$ must be left distributive over $\sqbrk c$.
But then from Condition for Operation to be Left Distributive over Constant Operation:
:$x \circ c = c$
As $c$ is arbitrary:
:$\forall a, b \in S: a \circ b = b$
so $\circ$ has to be the right operation.
{{qed}}
Theorem: Operation which is Right Distributive over Every Commutative Associative Operation is Left Operation

Let $\struct {S, \circ}$ be an algebraic structure.
Let $\circ$ have the property that:
:for every arbitrary operation $*$ on $S$ which is both commutative and associative, $\circ$ is right distributive over $*$.
Then $\circ$ is the left operation $\gets$:
:$\forall a, b \in S: a \gets b = a$

Proof:
First recall from Left Operation is Right Distributive over All Operations that the left operation is indeed right distributive over all operations, whether commutative or associative.
Let $*$ be an arbitrary operation on $S$ which is both commutative and associative.
As asserted, let $\circ$ be right distributive over $*$.
Let $c \in S$ be arbitrary.
Consider the constant operation $\sqbrk c$:
:$\forall a, b \in S: a \sqbrk c b := c$
We have from Constant Operation is Commutative and Constant Operation is Associative that $\sqbrk c$ is both commutative and associative.
Hence $\circ$ must be right distributive over $\sqbrk c$.
But then from Condition for Operation to be Right Distributive over Constant Operation:
:$c \circ x = c$
As $c$ is arbitrary:
:$\forall a, b \in S: a \circ b = a$
so $\circ$ has to be the left operation.
{{qed}}
Theorem: Every Operation is Distributive over Right Operation

Let $\struct {S, \circ}$ be an algebraic structure.
Let $\to$ denote the right operation on $S$.
Then $\circ$ is distributive over $\to$.

Proof:
{{begin-eqn}}
{{eqn | q = \forall a, b, c \in S
      | l = a \circ \paren {b \to c}
      | r = a \circ c
      | c = {{Defof|Right Operation}}
}}
{{eqn | q = \forall a, b, c \in S
      | l = \paren {a \circ b} \to \paren {a \circ c}
      | r = a \circ c
      | c = {{Defof|Right Operation}}
}}
{{eqn | ll= \leadsto
      | q = \forall a, b, c \in S
      | l = a \circ \paren {b \to c}
      | r = \paren {a \circ b} \to \paren {a \circ c}
      | c = from above
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | q = \forall a, b, c \in S
      | l = \paren {a \to b} \circ c
      | r = b \circ c
      | c = {{Defof|Right Operation}}
}}
{{eqn | q = \forall a, b, c \in S
      | l = \paren {a \circ c} \to \paren {b \circ c}
      | r = b \circ c
      | c = {{Defof|Right Operation}}
}}
{{eqn | ll= \leadsto
      | q = \forall a, b, c \in S
      | l = \paren {a \to b} \circ c
      | r = \paren {a \circ c} \to \paren {b \circ c}
      | c = from above
}}
{{end-eqn}}
Hence the result by definition of distributive.
{{qed}}
Theorem: Every Operation is Distributive over Left Operation

Let $\struct {S, \circ}$ be an algebraic structure.
Let $\gets$ denote the left operation on $S$.
Then $\circ$ is distributive over $\gets$.

Proof:
{{begin-eqn}}
{{eqn | q = \forall a, b, c \in S
      | l = a \circ \paren {b \gets c}
      | r = a \circ b
      | c = {{Defof|Left Operation}}
}}
{{eqn | q = \forall a, b, c \in S
      | l = \paren {a \circ b} \gets \paren {a \circ c}
      | r = a \circ b
      | c = {{Defof|Left Operation}}
}}
{{eqn | ll= \leadsto
      | q = \forall a, b, c \in S
      | l = a \circ \paren {b \gets c}
      | r = \paren {a \circ b} \gets \paren {a \circ c}
      | c = from above
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | q = \forall a, b, c \in S
      | l = \paren {a \gets b} \circ c
      | r = a \circ c
      | c = {{Defof|Left Operation}}
}}
{{eqn | q = \forall a, b, c \in S
      | l = \paren {a \circ c} \gets \paren {b \circ c}
      | r = a \circ c
      | c = {{Defof|Left Operation}}
}}
{{eqn | ll= \leadsto
      | q = \forall a, b, c \in S
      | l = \paren {a \gets b} \circ c
      | r = \paren {a \circ c} \gets \paren {b \circ c}
      | c = from above
}}
{{end-eqn}}
Hence the result by definition of distributive.
{{qed}}
Theorem: Absolute Continuity of Measures is Transitive Relation

Let $\struct {X, \Sigma}$ be a measurable space.
Let $\mu$, $\nu$ and $\lambda$ be measures on $\struct {X, \Sigma}$ such that: 
:$\mu$ is absolutely continuous {{WRT}} $\nu$
and:
:$\nu$ is absolutely continuous {{WRT}} $\lambda$.
That is: 
:$\mu \ll \nu$
and:
:$\nu \ll \lambda$
Then:
:$\mu$ is absolutely continuous {{WRT}} $\lambda$.
That is: 
:$\mu \ll \lambda$
In other words:
:$\ll$ is a transitive relation.

Proof:
Let $A \in \Sigma$ be such that $\map \lambda A = 0$. 
Then, since $\nu \ll \lambda$ we have $\map \nu A = 0$ from the definition of absolute continuity.
Since $\mu \ll \nu$ we similarly have $\map \mu A = 0$ again applying the definition of absolute continuity.
So, whenever $A \in \Sigma$ is such that $\map \lambda A = 0$ we have $\map \mu A = 0$.
So $\mu$ is absolutely continuous {{WRT}} $\lambda$.
{{qed}}
Category:Absolutely Continuous Measures
Category:Transitive Relations
fe9y6awsjxg8j051vfshms696nhs8a3
Theorem: P-adic Integers Form Integral Domain

Let $\Q_p$ be the $p$-adic numbers for some prime $p$.
Let $\Z_p$ be the $p$-adic integers.
Then:
:$\Z_p$ is an integral domain

Proof:
From Field is Integral Domain:
:$\Q_p$ is an  integral domain
From {{Corollary|P-adic Integers is Valuation Ring Induced by P-adic Norm}}:
:$\Z_p$ is a local ring
By the definition of local ring:
:$\Z_p$ is a ring with unity
From Subdomain Test:
:$\Z_p$ is an integral domain
{{qed}}
Theorem: Operation over which Every Commutative Associative Operation is Distributive is either Left or Right Operation/Lemma

Let $S$ be a set.
Let $A \subsetneqq S$ be a proper subset of $S$.
Let $a \in A$ and $b \in \relcomp S A$.
Let $\odot$ be the operation on $S$ defined as:
:$\forall x, y \in S: x \odot y = \begin {cases} a & : \set {x, y} \subseteq A \\ b & : \set {x, y} \not \subseteq A \end {cases}$
Then $\odot$ is commutative and associative.

Proof:
=== Commutativity ===
Trivially:
{{begin-eqn}}
{{eqn | q = \forall x, y \in S
      | l = y \odot x
      | r = \begin {cases} a & : \set {y, x} \subseteq A \\ b & : \set {y, x} \not \subseteq A \end {cases}
      | c = Definition of $\odot$
}}
{{eqn | r = \begin {cases} a & : \set {x, y} \subseteq A \\ b & : \set {x, y} \not \subseteq A \end {cases}
      | c = {{Defof|Set}}
}}
{{eqn | r = x \odot y
      | c = {{Defof|Set}}
}}
{{end-eqn}}
{{qed|lemma}}
Theorem: Operation over which Every Commutative Associative Operation is Distributive is either Left or Right Operation

Let $\struct {S, \circ}$ be an algebraic structure.
Let $\circ$ be such that every operation on $S$ which is both commutative and associative is distributive over $\circ$.
Then $\circ$ is either the left operation $\gets$ or the right operation $\to$.

Proof:
Recall from:
:Every Operation is Distributive over Right Operation
and:
:Every Operation is Distributive over Left Operation
that if $\circ$ is either $\gets$ or $\to$, then every operation is distributive over it, whether commutative or associative.
Theorem: Entropic Idempotent Structure is Self-Distributive

Let $\struct {S, \odot}$ be an algebraic structure such that $\odot$ is both idempotent and entropic.
Then $\struct {S, \odot}$ is a self-distributive structure.

Proof:
{{begin-eqn}}
{{eqn | q = \forall a, b, c \in S
      | l = \paren {a \odot b} \odot \paren {a \odot c}
      | r = \paren {a \odot a} \odot \paren {b \odot c}
      | c = {{Defof|Entropic Operation}}
}}
{{eqn  | r = a \odot \paren {b \odot c}
       | c = {{Defof|Idempotent Operation}}
}}
{{end-eqn}}
and:
{{begin-eqn}}
{{eqn | q = \forall a, b, c \in S
      | l = \paren {a \odot c} \odot \paren {b \odot c}
      | r = \paren {a \odot b} \odot \paren {c \odot c}
      | c = {{Defof|Entropic Operation}}
}}
{{eqn  | r = \paren {a \odot b} \odot c
       | c = {{Defof|Idempotent Operation}}
}}
{{end-eqn}}
Hence the result by definition of self-distributive structure.
{{qed}}