avewright/proofwiki-math huggingface dataset:
Theorem: Structure is Group iff Semigroup and Quasigroup

Let $\struct {S, \circ}$ be an algebraic structure.
Then:
:$\struct {S, \circ}$ is a group
{{iff}}
:$\struct {S, \circ}$ is both a semigroup and a quasigroup.

Proof:
=== Sufficient Condition ===
Let $\struct {S, \circ}$ be a group.
Then a fortiori $\struct {S, \circ}$ is a semigroup.
From Regular Representations in Group are Permutations:
:for all $a \in S$, the left regular representation and the rightt regular representation are permutations of $S$.
Hence by definition $\struct {S, \circ}$ is a quasigroup.
{{qed|lemma}}
Theorem: Condition for Group given Semigroup with Idempotent Element

Let $\struct {S, \circ}$ be a semigroup.
Let there exist an idempotent element $e$ of $S$ such that for all $a \in S$:
:there exists at least one element $x$ of $S$ satisfying $x \circ a = e$
:there exists at most one element $y$ of $S$ satisfying $a \circ y = e$.
Then $\struct {S, \circ}$ is a group.

Proof:
Let $a$ be arbitrary.
We have:
{{begin-eqn}}
{{eqn | q = \exists x \in S
      | l = x \circ a
      | r = e
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \paren {x \circ a} \circ \paren {x \circ a}
      | r = e \circ e
      | c = 
}}
{{eqn | r = e
      | c = {{hypothesis}}: $e$ is idempotent
}}
{{eqn | ll= \leadsto
      | l = x \circ \paren {a \circ x \circ a}
      | r = e
      | c = {{Semigroup-axiom|1}}
}}
{{eqn | ll= \leadsto
      | l = a \circ x \circ a
      | r = a
      | c = both $a \circ x \circ a$ and $a$ are $y \in S$ such that $x \circ y = e$
}}
{{eqn | ll= \leadsto
      | l = a \circ e
      | r = a
      | c = as $x \circ a = e$
}}
{{end-eqn}}
So $e$ is a right identity.
Again, let $a$ be arbitrary.
Let $x \in S$ be such that $x \circ a = e$.
We have:
{{begin-eqn}}
{{eqn | l = e
      | r = x \circ a
      | c = 
}}
{{eqn | r = \paren {x \circ e} \circ a
      | c = {{Defof|Right Identity}}
}}
{{eqn | r = x \circ \paren {e \circ a}
      | c = {{Semigroup-axiom|1}}
}}
{{eqn | ll= \leadsto
      | l = e \circ a
      | r = a
      | c = both $e \circ a$ and $a$ are $y \in S$ such that $x \circ y = e$
}}
{{end-eqn}}
So $e$ is a left identity.
We note the meaning of the criterion:
:there exists at least one element $x$ of $S$ satisfying $x \circ a = e$
As we now know that $e$ is a left identity, the above means that $x$ is a left inverse for $a$ in $S$.
To summarise, we have an algebraic structure $\struct {S, \circ}$:
:$(1): \quad$ which is closed, from {{Semigroup-axiom|0}}
:$(2): \quad$ which is associative, from {{Semigroup-axiom|1}}
:$(3): \quad$ which has a left identity
:$(4): \quad$ for which every element has a left inverse.
That is, $\struct {S, \circ}$ fulfils all the left group axioms.
Hence the result.
{{qed}}
Theorem: Existence and Uniqueness of Distributional Primitive

Let $T \in \map {\DD'} \R$ be a Schwartz distribution.
Then there exists a Schwartz distribution $S \in \map {\DD'} \R$ such that in the distributional sense:
:$S' = T$
Furthermore, $S$ is unique up to an arbitrary constant.

Proof:
Let $\mathbf 0 : \R \to 0$ be the zero mapping.
Let $\phi_0 \in \map \DD \R \setminus \set {\mathbf 0}$ be a test function.
Let $\psi \in \map \DD \R$ be a test function.
Let $\phi : \R \to \R$ be a real function such that:
:$\ds \phi := \psi - \frac {\int_{-\infty}^\infty \map \psi x \rd x} {\int_{-\infty}^\infty \map {\phi_0} x \rd x} \phi_0$
Theorem: Conditions under which Commutative Semigroup is Group

Suppose the following:
{{:Conditions under which Commutative Semigroup is Group/Statement of Conditions}}
Then $\struct {S, \circ}$ is a group.

Proof:
Some lemmata:
Theorem: Operation is Right Operation iff Anticommutative with Left Cancellable Element

Let $\struct {S, \circ}$ be a semigroup.
Then:
:$\circ$ is the right operation
{{iff}}:
:$\circ$ is anticommutative and has a left cancellable element.

Proof:
=== Sufficient Condition ===
Let $\circ$ be the right operation.
Then from Right Operation is Anticommutative we have that $\circ$ is anticommutative.
Let $x \in S$ be arbitrary.
Let $y, z \in S$ such that:
:$x \circ z = x \circ y$
Then:
{{begin-eqn}}
{{eqn | l = x \circ z
      | r = z
      | c = {{Defof|Right Operation}}
}}
{{eqn | l = x \circ y
      | r = y
      | c = {{Defof|Right Operation}}
}}
{{eqn | ll= \leadsto
      | l = z
      | r = y
      | c = 
}}
{{end-eqn}}
That is, $x$ is a left cancellable element for all $x \in S$.
Thus:
:$\circ$ is anticommutative and has a left cancellable element.
{{qed|lemma}}
Theorem: Operation is Left Operation iff Anticommutative with Right Cancellable Element

Let $\struct {S, \circ}$ be an semigroup.
Then:
:$\circ$ is the left operation
{{iff}}:
:$\circ$ is anticommutative and has a right cancellable element.

Proof:
=== Sufficient Condition ===
Let $\circ$ be the left operation.
Then from Left Operation is Anticommutative we have that $\circ$ is anticommutative.
Let $x \in S$ be arbitrary.
Let $y, z \in S$ such that:
:$z \circ x = y \circ x$
Then:
{{begin-eqn}}
{{eqn | l = z \circ x
      | r = z
      | c = {{Defof|Left Operation}}
}}
{{eqn | l = y \circ x
      | r = y
      | c = {{Defof|Left Operation}}
}}
{{eqn | ll= \leadsto
      | l = z
      | r = y
      | c = 
}}
{{end-eqn}}
That is, $x$ is a right cancellable element for all $x \in S$.
Thus:
:$\circ$ is anticommutative and has a right cancellable element.
{{qed|lemma}}
Theorem: Count of Operations on Finite Set which are Closed on Every Subset

Let $S$ be a finite set with $n$ elements.
There are $2^{n \paren {n - 1} }$ binary operations on $S$ which are closed on all subsets of $S$.

Proof:
Suppose $\circ$ is a binary operation on $S$ such that $a \circ b = c$ for $a \ne b \ne c \ne a$.
Then $\circ$ is not closed on $\set {a, b} \subseteq S$.
Similarly, suppose $\circ$ is a binary operation on $S$ such that $a \circ a = b$ for $a \ne b$.
Then $\circ$ is not closed on $\set a \subseteq S$.
Hence for $\circ$ to be closed on all subsets of $S$:
:$\forall a \in S: a \circ a = a$
and:
:$\forall a, b \in S: \paren {a \circ b = a} \lor \paren {a \circ b = b}$.
So for every ordered pair $\tuple {x, y}$ of elements of $S$ such that $a \ne b$ there are two possible products: $x$ or $y$.
For an algebraic structure of order $n$, there are $n^2$ such ordered pairs.
However, of these $n^2$ such ordered pairs $\tuple {x, y}$, $n$ of these are such that $x = y$.
There remain $n^2 - n = n \paren {n - 1}$ ordered pairs $\tuple {x, y}$, such that $x \ne y$.
Each one contributes $2$ different products
It follows by the Product Rule for Counting that there are $2^{n \paren {n - 1} }$ binary operations on $S$ which are closed on all subsets of $S$.
{{qed}}
Theorem: Count of Commutative Operations on Finite Set which are Closed on Every Subset

Let $S$ be a finite set with $n$ elements.
There are $2^{n \paren {n - 1} / 2}$ commutative binary operations on $S$ which are closed on all subsets of $S$.

Proof:
From Count of Operations on Finite Set which are Closed on Every Subset, there are $2^{n \paren {n - 1} }$ binary operations on $S$ which are closed on all subsets of $S$.
For each $a, b \in S$, we have that
:there are $2$ possible products for $a \circ b$
:there are $2$ possible products for $b \circ a$
and so on the set $\set {a, b}$ there are $4$ binary operations which are closed on all subsets of $\set {a, b}$.
Let $\circ$ be commutative.
Then:
:$a \circ b = b \circ a$
and so now there are only $2$ possible such products.
Hence the number of commutative binary operations on $S$ which are closed on all subsets of $S$ equals the number of distinct doubleton subsets of $S$.
There are half as many of these as there are ordered pairs.
That is, there are $\dfrac {n^2 - n} 2 = \dfrac {n \paren {n - 1} } 2$ doubletons $\set {x, y} \subseteq S$, such that $x \ne y$.
The result follows.
{{qed}}
Theorem: Canonical P-adic Expansion of Rational is Eventually Periodic/Lemma 4

:$\forall n \in \N: r_n = d_{n + 1} b + p r_{n + 1}$

Proof:
We have:
{{begin-eqn}}
{{eqn | q = \forall n \in \N 
      | l = \dfrac a b
      | r = A_n + p^{n + 1} \dfrac {r_n} b
      | c = {{hypothesis}}
}}
{{eqn | r = A_{n + 1} + p^{n + 2} \dfrac {r_{n + 1} } b
      | c = {{hypothesis}}
}}
{{eqn | ll= \leadsto
      | q = \forall n \in \N 
      | l = A_{n + 1} + p^{n + 2} \dfrac {r_{n + 1} } b
      | r = A_n + p^{n + 1} \dfrac {r_n} b
}}
{{eqn | ll= \leadsto
      | q = \forall n \in \N 
      | l = A_{n + 1} - A_n
      | r = p^{n + 1} \paren {\dfrac {r_n - p r_{n + 1} } b}
      | c = rearranging terms
}}
{{eqn | ll = \leadsto
      | q = \forall n \in \N 
      | l = b \paren {A_{n + 1} - A_n}
      | r = p^{n + 1} \paren {r_n - p r_{n + 1} }
      | c = rearranging terms
}}
{{eqn | ll = \leadsto
      | q = \forall n \in \N 
      | l = b 
      | o = \divides
      | r = p^{n + 1} \paren {r_n - p r_{n + 1} }
}}
{{end-eqn}}
From Characterization of Rational P-adic Integer:
:$p \nmid b$
From Prime not Divisor implies Coprime:
:$b, p$ are coprime
From Integer Coprime to all Factors is Coprime to Whole:
:$b, p^{n+1}$ are coprime
Hence:
$b \nmid p^{n + 1}$
From Euclid's Lemma:
:$b \divides \paren {r_n - p r_{n + 1} }$
Then:
:$\dfrac {r_n - p r_{n + 1} } b \in \Z$
Hence:
:$A_{n + 1} \equiv A_n \pmod {p^{n + 1} }$
By definition of coherent sequence:
:the sequence $\sequence{A_n}$ is a coherent sequence
From Coherent Sequence is Partial Sum of P-adic Expansion:
:the sequence $\sequence{A_n}$ is the sequence of partial sums of a $p$-adic expansion $\ds \sum_{i \mathop = 0}^n e_i p^i$
Hence:
:$y = \dfrac a b = \ds \sum_{n \mathop = 0}^\infty e_n p^n$
From P-adic Number is Limit of Unique P-adic Expansion:
:$\forall n \in \N: e_n = d_n$
By definition of partial sums:
:$\forall n \in \N: A_{n + 1} = A_n + d_{n + 1} p^{n + 1}$
Hence:
{{begin-eqn}}
{{eqn | q = \forall n \in \N
      | l = d_{n + 1} p^{n + 1}
      | r = A_{n + 1} - A_n
}}
{{eqn | r = p^{n + 1} \paren {\dfrac {r_n - p r_{n + 1} } b}
      | c = {{apriori}}
}}
{{eqn | ll= \leadsto
      | q = \forall n \in \N
      | l = d_{n + 1}
      | r = \paren {\dfrac {r_n - p r_{n + 1} } b}
      | c = dividing both sides by $p^{n + 1}$
}}
{{eqn | ll = \leadsto
      | q = \forall n \in \N
      | l = r_n
      | r = d_{n + 1} b + p r_{n + 1}
      | c = rearranging terms
}}
{{end-eqn}}
{{qed}}
Category:Canonical P-adic Expansion of Rational is Eventually Periodic
pdg52rqqykr469br6lz0drznmmvkdgv
Theorem: Unit Cylinder as Surface of Revolution

Let $\struct {\R^3, d}$ be the Euclidean space.
Let $S_C \subseteq \R^3$ be the surface of revolution.
Let $C$ be a straight line in the open upper half-plane.
Let the smooth local parametrization of $C$ be:
:$\map \gamma t = \tuple {t, 1}$
Then the induced metric on $S_C$ is:
:$g = d t^2 + d \theta^2$

Proof:
We have that:
:$\map {\gamma'} t = \tuple {1, 0}$
Hence, $\map \gamma t$ is a unit-speed curve.
By the corollary of the induced metric on the surface of revolution:
:$g = d t^2 + d \theta^2$
{{qed}}
Theorem: Canonical P-adic Expansion of Rational is Eventually Periodic/Lemma 2

:$\exists n_0 \in \N : \forall n \ge n_0 : -b \le r_n \le 0$

Proof:
=== Lemma 11 ===
{{:Canonical P-adic Expansion of Rational is Eventually Periodic/Lemma 11}}{{qed|lemma}}
Theorem: Canonical P-adic Expansion of Rational is Eventually Periodic/Lemma 3

:$\ds \lim_{n \mathop \to \infty} A_n = \dfrac a b$

Proof:
Let $\epsilon \in \R_{> 0}$.
Let $M = \max \set {\norm{r_0}_p, \norm{r_1}_p, \ldots, \norm{r_{n_0}}_p, \norm{-1}_p, \norm{-2}_p, \ldots, \norm{-b}_p}$
From Power Function is Unbounded Above:
:$\exists N \in \N: p^{N+1} >  \dfrac  M {\epsilon \norm b_p}$
We have:
{{begin-eqn}}
{{eqn | q = \forall n \in \N: n \ge N
      | l = \norm{\dfrac a b - A_n}_p
      | r = \norm{p^{n+1} \dfrac {r_n} b} _p
      | c = {{hypothesis}}
}}
{{eqn | r = \norm p_p^{n+1} \dfrac {\norm{r_n}_p} {\norm b_p} 
      | c = {{Norm-axiom-mult|2}}
}}
{{eqn | o = \le
      | r = \norm p_p^{n+1} \dfrac M {\norm b_p} 
      | c = {{Defof|Max Operation}}
}}
{{eqn | o = \le
      | r = \dfrac 1 {p^{n+1} } \dfrac M {\norm b_p} 
      | c = {{Defof|P-adic Norm}}
}}
{{eqn | o = \le
      | r = \dfrac 1 {p^{N+1} } \dfrac M {\norm b_p}
      | c = Power Function on Base between Zero and One is Strictly Decreasing
}}
{{eqn | o = <
      | r = \dfrac {\epsilon  \norm b_p } M \dfrac M {\norm b_p}
      | c = choice of $N$
}}
{{eqn | r = \epsilon
      | c = cancelling terms
}}
{{end-eqn}}
By definition of convergence in $\struct {\Q_p, \norm {\,\cdot\,}_p}$:
:$\ds \lim_{n \mathop \to \infty} A_n = \dfrac a b$
{{qed}}
Category:Canonical P-adic Expansion of Rational is Eventually Periodic
eeh9wp5i7ujb5jfm3hpknuzu8w2w2s5
Theorem: Powerset of Subset is Closed under Union

Let $S$ be a set.
Let $T \subseteq S$ be a subset of $S$.
Let $\powerset S$ denote the power set of $S$.
Then $\powerset T$ is a closed subset of $\powerset S$ under set union:
:$\forall A, B \in \powerset T: A \cup B \in \powerset T$

Proof:
A direct application of Power Set is Closed under Union.
{{qed}}
Theorem: Powerset of Subset is Closed under Intersection

Let $S$ be a set.
Let $T \subseteq S$ be a subset of $S$.
Let $\powerset S$ denote the power set of $S$.
Then $\powerset T$ is a closed subset of $\powerset S$ under set intersection:
:$\forall A, B \in \powerset T: A \cap B \in \powerset T$

Proof:
A direct application of Power Set is Closed under Intersection.
{{qed}}
Theorem: Powerset of Subset is Closed under Symmetric Difference

Let $S$ be a set.
Let $T \subseteq S$ be a subset of $S$.
Let $\powerset S$ denote the power set of $S$.
Then $\powerset T$ is a closed subset of $\powerset S$ under set intersection:
:$\forall A, B \in \powerset T: A \symdif B \in \powerset T$
where $\symdif$ denotes symmetric difference.

Proof:
A direct application of Power Set is Closed under Symmetric Difference.
{{qed}}
Theorem: Dipper Semigroup is Commutative Semigroup

The dipper semigroup is a commutative semigroup.

Proof:
Recall the definition of the '''dipper semigroup''':
Let $m, n \in \Z$ be integers such that $m \ge 0, n > 0$.
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 $+_{m, n}$ 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:
:$m + k n \le a + b$
Taking the semigroup axioms in turn:
Theorem: Weak Solution to Dx u = u

thumbright$\map u {x, t} = \map H t e^x$
Let $H$ be the Heaviside step function.
Let $\map u {x, t} = \map H t e^x$
Then $u$ is a weak solution of the partial differential equation $\ds \dfrac {\partial u} {\partial x} = u$.
That is, for the Schwartz distribution $T_u \in \map {\DD'} {\R^2}$ associated with $u$ in the distributional sense it holds that:
:$\ds \dfrac {\partial T_u} {\partial x} = T_u$

Proof:
Let $\phi \in \map \DD {\R^2}$ be a test function.
Then in the distributional sense we have that:
{{begin-eqn}}
{{eqn | l = \paren {\dfrac \partial {\partial x} - 1} \map {T_u} \phi
      | r = -\map {T_u} {\dfrac {\partial \phi} {\partial x} } - \map {T_u} \phi
      | c = {{Defof|Distributional Partial Derivative}}
}}
{{eqn | r = -\iint_{\R^2} \paren {\map H t e^x \dfrac {\partial \map \phi {x, t} } {\partial x} + \map H t e^x \map \phi {x, t} }\rd x \rd t
      | c = {{Defof|Schwartz Distribution}}
}}
{{eqn | r = -\int_0^\infty \int_{-\infty}^\infty \paren {e^x \dfrac {\partial \map \phi {x, t} } {\partial x} + e^x \map \phi {x, t} }\rd x \rd t
      | c = {{Defof|Heaviside Step Function}}
}}
{{eqn | r = -\int_0^\infty \int_{-\infty}^\infty \dfrac \partial {\partial x} \paren {e^x \map \phi {x, t} }\rd x \rd t
      | c = Product Rule for Derivatives
}}
{{eqn | r = -\int_0^\infty \bigintlimits {e^x \map \phi {x, t} } {x \mathop = -\infty} {x \mathop = \infty} \rd t
      | c = Fundamental Theorem of Calculus
}}
{{eqn | r = -\int_0^\infty 0 \rd t
      | c = {{Defof|Test Function}}
}}
{{eqn | r = 0
}}
{{end-eqn}}
{{qed}}
Theorem: Existence of Subgroup of Dipper Semigroup

Let $m, n \in \Z$ be integers such that $m \ge 0, n > 0$.
Let $\struct {N_{< \paren {m \mathop + n} }, +_{m, n} }$ denote the dipper semigroup.
Consider the subset $H \subseteq N_{< \paren {m \mathop + n} }$ defined as:
:$H = \set {k \in \N: m \le k < m + n} = \set {m, m + 1, \ldots, m + n - 1}$
Then the substructure $\struct {H, +_{m, n} }$ is a subgroup of $\struct {N_{< \paren {m \mathop + n} }, +_{m, n} }$.

Proof:
Recall the definition of the '''dipper semigroup''':
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 $+_{m, n}$ 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:
:$m + k n \le a + b$
Taking the group axioms in turn:
Theorem: Canonical P-adic Expansion of Rational is Eventually Periodic/Lemma 5

:$\exists \mathop m, l \in \N : \forall n \ge m: r_n = r_{n + l}$ and $d_n = d_{n + l}$

Proof:
We have {{hypothesis}} that the set $\set {r_n : n \in \N}$ of values of $\sequence {r_n}$ is a subset of:
:$\set {r_0, r_1, \ldots, r_{n_0} } \cup \set {-b, -b + 1, -b + 2, \ldots, 2, 1, 0}$
It follows that $\set {r_n : n \in \N}$ takes only finitely many values.
Hence:
:$\exists m_0, l \in \N : l > 0 : r_{m_0} = r_{ {m_0} + l}$
Theorem: If n is Triangular then so is (2m+1)^2 n + m(m+1)/2

Let $n$ be a triangular number.
Let $m \in \Z_{\ge 0}$ be a positive integer.
Then $\paren {2 m + 1}^2 n + \dfrac {m \paren {m + 1} } 2$ is also a triangular number.

Proof:
Let $n$ be a triangular number.
Then:
:$\exists k \in \Z: n = \dfrac {k \paren {k + 1} } 2$
So:
{{begin-eqn}}
{{eqn | l = \paren {2 m + 1}^2 n + \dfrac {m \paren {m + 1} } 2
      | r = \paren {2 m + 1}^2 \dfrac {k \paren {k + 1} } 2 + \dfrac {m \paren {m + 1} } 2
      | c = 
}}
{{eqn | r = \dfrac {\paren {2 m + 1}^2 k^2 + \paren {2 m + 1}^2 k + m \paren {m + 1} } 2
      | c = 
}}
{{eqn | r = \dfrac {\paren {\paren {2 m + 1} k + m} \paren {\paren {2 m + 1} k + m + 1} } 2
      | c = 
}}
{{end-eqn}}
which is of the form:
:$\dfrac {t \paren {t + 1} } 2$
where $t = \paren {2 m + 1} k + m$.
{{qed}}
Category:Triangular Numbers
1szwd95ilf6jk06p5hf2b5a5mzqzjai
Theorem: Identity of Subgroup of Dipper Semigroup is not Identity of Dipper

Let $m, n \in \Z$ be integers such that $m, n > 0$.
Let $\struct {N_{< \paren {m \mathop + n} }, +_{m, n} }$ denote the dipper semigroup.
Let $\struct {H, +_{m, n} }$ be the subgroup of $\struct {N_{< \paren {m \mathop + n} }, +_{m, n} }$ where $H = \set {k \in \N: m \le k < m + n}$
Then the identity of $\struct {H, +_{m, n} }$ is not the identity of $\struct {N_{< \paren {m \mathop + n} }, +_{m, n} }$.

Proof:
This is demonstrated by Proof by Counterexample.
First we note that by Existence of Subgroup of Dipper Semigroup:
:$\struct {H, +_{m, n} }$ is indeed a subgroup of $\struct {N_{< \paren {m \mathop + n} }, +_{m, n} }$
:the identity of $\struct {H, +_{m, n} }$ is $n$.
But we note that by definition of $+_{m, n}$:
:$0 +_{m, n} n = n - k n$
where $m \le n - k n$
As we have specified that $m > 0$, it follows that:
:$0 +_{m, n} n > 0$
Thus $n$ is not the identity of $\struct {N_{< \paren {m \mathop + n} }, +_{m, n} }$.
{{qed}}
Theorem: Elements of Subgroup of Dipper Semigroup are not Invertible in Dipper

Let $m, n \in \Z$ be integers such that $m, n > 0$.
Let $\struct {N_{< \paren {m \mathop + n} }, +_{m, n} }$ denote the dipper semigroup.
Let $\struct {H, +_{m, n} }$ be the subgroup of $\struct {N_{< \paren {m \mathop + n} }, +_{m, n} }$ where $H = \set {k \in \N: m \le k < m + n}$
Then the elements of $\struct {H, +_{m, n} }$ are not invertible in $\struct {N_{< \paren {m \mathop + n} }, +_{m, n} }$.

Proof:
From Identity of Subgroup of Dipper Semigroup is not Identity of Dipper, the identity of $\struct {H, +_{m, n} }$ is not an identity of $\struct {N_{< \paren {m \mathop + n} }, +_{m, n} }$.
Hence (indirectly) from Identity of Submagma containing Identity of Magma is Same Identity, $\struct {N_{< \paren {m \mathop + n} }, +_{m, n} }$ can have no identity.
Hence there can be no elements of $\struct {H, +_{m, n} }$ that are invertible, by definition.
{{qed}}
Theorem: De Rham Cohomology of Sphere

Let $S^n$ denote the $n$-Sphere.
Then the de Rham Cohomology of $S^n$ are:
:$\map {H^k} {S^n} = \begin {cases} \Z^2 & : k = 0, n = 0 \\ \Z & : k = 0 \text { or } k = n, n > 0 \\ 0 & : 0 < k < n \end {cases}$
and Higher de Rham Cohomology Vanishes.
{{explain|The condition for the middle case is not uniquely readable, and need to be clarified. If the cases are exhaustive, the middle one might be "otherwise".}}

Proof:
{{ProofWanted|Mayer–Vietoris sequence and Induction}}
Category:de Rham Cohomology
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Theorem: Isomorphisms between Symmetry Groups of Isosceles Triangle and Equilateral Triangle

Let $\TT = ABC$ be an isosceles triangle whose apex is $A$.
Let $\struct {\TT, \circ}$ be the symmetry group of $\TT$, where the symmetries are identified as:
:the identity mapping $e$
:the reflection $d$ in the line through $A$ and the midpoint of $BC$.
:240px
Let $\SS = A'B'C'$ be an equilateral triangle.
We define in cycle notation the following symmetries on $\triangle A'B'C'$:
{{begin-eqn}}
{{eqn | l = e
      | o = :
      | r = \tuple {A'} \tuple {B'} \tuple {C'}
      | c = Identity mapping
}}
{{eqn | l = p
      | o = :
      | r = \tuple {A'B'C'}
      | c = Rotation of $120 \degrees$ anticlockwise about center
}}
{{eqn | l = q
      | o = :
      | r = \tuple {A'C'B'}
      | c = Rotation of $120 \degrees$ clockwise about center
}}
{{eqn | l = r
      | o = :
      | r = \tuple {B'C'}
      | c = Reflection in line $r$
}}
{{eqn | l = s
      | o = :
      | r = \tuple {A'C'}
      | c = Reflection in line $s$
}}
{{eqn | l = t
      | o = :
      | r = \tuple {A'B'}
      | c = Reflection in line $t$
}}
{{end-eqn}}
Then $\struct {\TT, \circ}$ is isomorphic to the $3$ subgroups of $S_3$:
{{begin-eqn}}
{{eqn | o = 
      | r = \set {e, \tuple {12} }
}}
{{eqn | o = 
      | r = \set {e, \tuple {13} }
}}
{{eqn | o = 
      | r = \set {e, \tuple {23} }
}}
{{end-eqn}}

Proof:
We have that $\struct {\TT, \circ}$ is of order $2$.
We also have that:
{{begin-eqn}}
{{eqn | o = 
      | r = \set {e, \tuple {12} }
}}
{{eqn | o = 
      | r = \set {e, \tuple {13} }
}}
{{eqn | o = 
      | r = \set {e, \tuple {23} }
}}
{{end-eqn}}
are also groups of order $2$.
From Parity Group is Only Group with 2 Elements, all these groups are isomorphic.
{{qed}}
Theorem: Set of Associating Elements forms Subsemigroup of Magma

Let $S$ be a set.
Let $\oplus$ be an operation on $S$ such that $\struct {S, \oplus}$ is a magma.
Let $T \subseteq S$ be the subset of $S$ defined as:
:$T = \set {x \in S: \paren {x \oplus y} \oplus z = x \oplus \paren {y \oplus z} }$
Suppose $T \ne \O$.
Then $\struct {T, \oplus {\restriction_T} }$ is a subsemigroup of $\struct {S, \oplus}$.

Proof:
Taking the semigroup axioms in turn:
Theorem: Subsemigroup of Cancellable Mappings is Subgroup of Invertible Mappings

Let $S$ be a set.
Let $S^S$ denote the set of mappings from $S$ to itself.
Let $\CC \subseteq S^S$ denote the set of cancellable mappings on $S$.
Let $\MM \subseteq S^S$ denote the set of invertible mappings on $S$.
Then:
:the subsemigroup $\struct {\CC, \circ}$ of $\struct {S^S, \circ}$ coincides with the subgroup $\struct {\MM, \circ}$ of $\struct {S^S, \circ}$
where $\circ$ denotes composition of mappings.

Proof:
From Set of Invertible Mappings forms Symmetric Group, we have that $\struct {\MM, \circ}$ is a group.
Hence, by definition, $\struct {\MM, \circ}$ is a subgroup of $\struct {S^S, \circ}$.
Recall from Bijection iff Left and Right Inverse that a mapping is invertible {{iff}} it is a bijection.
By definition, a cancellable mapping is a mapping both left cancellable and right cancellable.
From Injection iff Left Cancellable and Surjection iff Right Cancellable, a cancellable mapping is both an injection and a surjection.
That is, a mapping is cancellable mapping {{iff}} it is a bijection.
That is, $\struct {\CC, \circ}$ is exactly the same as $\struct {\MM, \circ}$.
{{qed}}