avewright/proofwiki-math huggingface dataset:
Theorem: Subtraction on Numbers is Anticommutative/Integral Domains

The operation of subtraction on the numbers is anticommutative.
That is:
:$a - b = b - a \iff a = b$

Proof:
Let $a, b$ be elements of one of the standard number sets: $\Z, \Q, \R, \C$.
Each of those systems is an integral domain, and so is closed under the operation of subtraction.
=== Necessary Condition ===
Let $a = b$.
Then $a - b = 0 = b - a$.
{{qed|lemma}}
=== Sufficient Condition ===
Let $a - b = b - a$.
Then:
{{begin-eqn}}
{{eqn | l = a - b
      | r = a + \paren {-b}
      | c = {{Defof|Subtraction}}
}}
{{eqn | r = \paren {-b} + a
      | c = Commutative Law of Addition
}}
{{eqn | r = \paren {-b} + \paren {-\paren {-a} }
      | c = 
}}
{{eqn | r = -\paren {b - a}
      | c = {{Defof|Subtraction}}
}}
{{end-eqn}}
We have that:
:$a - b = b - a$
So from the above:
:$b - a = - \paren {b - a}$
That is:
:$b - a = 0$
and so:
:$a = b$
{{qed}}
Theorem: Natural Number Subtraction is not Closed

The operation of subtraction on the natural numbers is not closed.

Proof:
By definition of natural number subtraction:
:$n - m = p$
where $p \in \N$ such that $n = m + p$.
However, when $m > n$ there exists no $p \in \N$ such that $n = m + p$.
{{qed}}
Theorem: Integers under Addition form Semigroup

The set of integers under addition $\struct {\Z, +}$ forms a semigroup.

Proof:
=== Closure ===
Integer Addition is Closed.
{{qed|lemma}}
Theorem: Natural Numbers under Multiplication form Subsemigroup of Integers

Let $\struct {\N, \times}$ denote the set of natural numbers under multiplication.
Let $\struct {\Z, \times}$ denote the set of integers under multiplication.
Then $\struct {\N, \times}$ is a subsemigroup of $\struct {\Z, \times}$.

Proof:
We have from Natural Numbers under Multiplication form Semigroup that $\struct {\N, \times}$ forms a semigroup.
We have from Integers under Multiplication form Semigroup that $\struct {\Z, \times}$ forms a semigroup.
From Natural Numbers are Non-Negative Integers, we have that $\N \subseteq \Z$.
From the definition of integer multiplication it follows that $\times: \Z \to \Z$ is an extension of $\times: \N \to \N$.
Hence the result.
{{qed}}
Theorem: Cauchy Sequence Is Eventually Bounded Away From Non-Limit

Let $\struct {R, \norm {\, \cdot \,} }$ be a normed division ring.
Let $\sequence {x_n}$ be a Cauchy sequence in $R$.
Suppose $\sequence {x_n}$ does not converge to $l \in R$.
Then:
:$\exists K \in \N$ and $C \in \R_{>0}: \forall n > K: C < \norm {x_n - l}$

Proof:
Because $\sequence {x_n}$ does not converge to $l$:
:$\exists \epsilon \in \R_{>0}: \forall n \in \N: \exists m \ge n: \norm {x_m - l} \ge \epsilon$
Because $\sequence {x_n}$ is a Cauchy sequence:
:$\exists K \in \N: \forall n, m \ge K: \norm {x_n - x_m} < \dfrac \epsilon 2$
Let $M \ge K: \norm {x_M - l} \ge \epsilon$.
Then $\forall n > K$:
{{begin-eqn}}
{{eqn | l = \epsilon
      | o = \le
      | r = \norm {x_M - l}
      | c = 
}}
{{eqn | r = \norm {x_M - x_n + x_n - l}
      | c = 
}}
{{eqn | o = \le
      | r = \norm {x_M - x_n} + \norm {x_n - l}
      | c = {{Norm-axiom-mult|3}}
}}
{{eqn | o = <
      | r = \dfrac \epsilon 2 + \norm {x_n - l}
      | c = because $n, M \ge K$
}}
{{eqn | ll= \leadsto
      | l = \dfrac \epsilon 2
      | o = <
      | r = \norm {x_n - l}
      | c = subtracting $\dfrac \epsilon 2$ from both sides of the equation
}}
{{end-eqn}}
Let $C = \dfrac \epsilon 2$ and the result follows.
{{qed}}
Category:Cauchy Sequence in Normed Division Ring is Bounded
rpx42lf45x8h3iu8t67fj7997bld965
Theorem: Embedding Division Ring into Quotient Ring of Cauchy Sequences

Let $\struct {R, \norm {\, \cdot \,} }$ be a normed division ring.
Let $\CC$ be the ring of Cauchy sequences over $R$
Let $\NN = \set {\sequence {x_n}: \ds \lim_{n \mathop \to \infty} x_n = 0}$
Let $\norm {\, \cdot \,}: \CC \, \big / \NN \to \R_{\ge 0}$ be the norm on the quotient ring $\CC \, \big / \NN$ defined by:
:$\ds \forall \sequence {x_n} + \NN: \norm {\sequence {x_n} + \NN} = \lim_{n \mathop \to \infty} \norm {x_n}$
Let $\phi: R \to \CC \, \big / \NN$ be the mapping from $R$ to the quotient ring $\CC \, \big / \NN$ defined by:
:$\forall a \in R: \map \phi a = \sequence {a, a, a, \dotsc} + \NN$
where $\sequence {a, a, a, \dotsc} + \NN$ is the left coset in $\CC \, \big / \NN$ that contains the constant sequence $\sequence {a, a, a, \dotsc}$.
Then:
:$\phi$ is a distance-preserving ring monomorphism.

Proof:
By the definition of a distance-preserving mapping and a ring monomorphism it has to be shown that:
:$(1): \quad \phi$ is a homomorphism.
:$(2): \quad \phi$ is an injection.
:$(3): \quad \phi$ is distance-preserving.
Theorem: Identity of Submonoid is not necessarily Identity of Monoid

Let $\struct {S, \circ}$ be a monoid whose identity is $e_S$.
Let $\struct {T, \circ}$ be a submonoid of $\struct {S, \circ}$ whose identity is $e_T$.
Then it is not necessarily the case that $e_T = e_S$.

Proof:
Let $\struct {S, \times}$ be the semigroup formed by the set of order $2$ square matrices over the real numbers $R$ under (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 $x \in \R$.
From Matrices of the Form $\begin{bmatrix} x & 0 \\ 0 & 0 \end{bmatrix}$ we have that $\struct {T, \times}$ is a subsemigroup of $\struct {S, \times}$.
From (some result somewhere) $\struct {S, \times}$ has an identity $\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ which is not in $\struct {T, \times}$.
{{explain|Find that result}}
However, note that:
{{begin-eqn}}
{{eqn | l = \begin{bmatrix} x & 0 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}
      | r = \begin{bmatrix} x & 0 \\ 0 & 0 \end{bmatrix}
      | c = 
}}
{{eqn | r = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} x & 0 \\ 0 & 0 \end{bmatrix}
      | c = 
}}
{{end-eqn}}
demonstrating that $\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}$ is the identity of $\struct {T, \times}$.
{{qed}}
Theorem: Non-Zero Real Numbers under Multiplication form Group

Let $\R_{\ne 0}$ be the set of real numbers without zero:
:$\R_{\ne 0} = \R \setminus \set 0$
The structure $\struct {\R_{\ne 0}, \times}$ forms a group.

Proof:
<onlyinclude>
Taking the group axioms in turn:
Theorem: Symmetric Group on n Letters is Isomorphic to Symmetric Group

The symmetric group on $n$ letters $\struct {S_n, \circ}$ is isomorphic to the symmetric group on the $n$ elements of any set $T$ whose cardinality is $n$.
That is:
:$\forall T \subseteq \mathbb U, \card T = n: \struct {S_n, \circ} \cong \struct {\Gamma \paren T, \circ}$
where:
:$\map \Gamma T$ denotes the set of permutations on $T$
{{improve|Link to the actual page defining what $\struct {\map \Gamma T, \circ}$ is without going all the way back to first principles}}
:$\struct {\map \Gamma T, \circ}$ denotes an algebraic structure with $1$ operation
:$\circ$ denotes composition of mappings.

Proof:
The fact that $\struct {S_n, \circ}$ is a group is a direct implementation of the result Symmetric Group is Group.
By definition of cardinality, as $\card T = n$ we can find a bijection between $T$ and $\N_n$.
From Number of Permutations, it is immediate that $\order {\paren {\Gamma \paren T, \circ} } = n! = \order {\struct {S_n, \circ} }$.
Again, we can find a bijection $\phi$ between $\struct {\Gamma \paren T, \circ}$ and $\struct {S_n, \circ}$.
The result follows directly from the Transplanting Theorem.
{{qed}}
Theorem: Symmetric Groups of Same Order are Isomorphic

Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Let $T_1$ and $T_2$ be sets whose cardinality $\card {T_1}$ and $\card {T_2}$ are both $n$.
Let $\struct {\map \Gamma {T_1}, \circ}$ and $\struct {\map \Gamma {T_2}, \circ}$ be the symmetric group on $S$ and $T$ respectively.
Then $\struct {\map \Gamma {T_1}, \circ}$ and $\struct {\map \Gamma {T_2}, \circ}$ are isomorphic.

Proof:
Consider the symmetric group on $n$ letters $S_n$.
From Symmetric Group on n Letters is Isomorphic to Symmetric Group we have that:
:$\struct {\map \Gamma {T_1}, \circ}$ is isomorphic to $S_n$
:$\struct {\map \Gamma {T_2}, \circ}$ is isomorphic to $S_n$
and hence from Isomorphism is Equivalence Relation:
:$\struct {\map \Gamma {T_1}, \circ}$ is isomorphic to $\struct {\map \Gamma {T_2}, \circ}$.
{{qed}}
Theorem: Symmetric Groups of Same Order are Isomorphic

Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Let $T_1$ and $T_2$ be sets whose cardinality $\card {T_1}$ and $\card {T_2}$ are both $n$.
Let $\struct {\map \Gamma {T_1}, \circ}$ and $\struct {\map \Gamma {T_2}, \circ}$ be the symmetric group on $S$ and $T$ respectively.
Then $\struct {\map \Gamma {T_1}, \circ}$ and $\struct {\map \Gamma {T_2}, \circ}$ are isomorphic.

Proof:
Let us define a bijection:
:$\alpha: T_1 \to T_2$
Let $\theta: \struct {\map \Gamma {T_1}, \circ} \to \struct {\map \Gamma {T_2}, \circ}$ be defined as:
:$\forall f \in \struct {\map \Gamma {T_1}, \circ}: \map \theta f = \alpha \circ f \circ \alpha^{-1}$
Let $f, g \in \map \Gamma {T_1}$.
We have:
{{begin-eqn}}
{{eqn | l = \map \theta f \circ \map \theta g
      | r = \paren {\alpha \circ f \circ \alpha^{-1} } \circ \paren {\alpha \circ g \circ \alpha^{-1} }
      | c = Definition of $\theta$
}}
{{eqn | r = \alpha \circ f \circ \paren {\alpha^{-1} \circ \alpha} \circ g \circ \alpha^{-1}
      | c = 
}}
{{eqn | r = \alpha \circ \paren {f \circ g} \circ \alpha^{-1}
      | c = 
}}
{{eqn | r = \map \theta {f \circ g}
      | c = Definition of $\theta$
}}
{{end-eqn}}
That is, $\theta$ is a group homomorphism.
Let $f, g \in \map \Gamma {T_1}$ such that $\map \theta f = \map \theta g$.
Then:
{{begin-eqn}}
{{eqn | l = \map \theta f
      | r = \map \theta g
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \alpha \circ f \circ \alpha^{-1}
      | r = \alpha \circ g \circ \alpha^{-1}
      | c = Definition of $\theta$
}}
{{eqn | ll= \leadsto
      | l = \alpha^{-1} \circ \paren {\alpha \circ f \circ \alpha^{-1} } \circ \alpha
      | r = \alpha^{-1} \circ \paren {\alpha \circ g \circ \alpha^{-1} } \circ \alpha
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \paren {\alpha^{-1} \circ \alpha} \circ f \circ \paren {\alpha^{-1} \circ \alpha}
      | r = \paren {\alpha^{-1} \circ \alpha} \circ g \circ \paren {\alpha^{-1} \circ \alpha}
      | c = {{Group-axiom|1}}
}}
{{eqn | ll= \leadsto
      | l = f
      | r = g
      | c = 
}}
{{end-eqn}}
Thus it is seen that $\theta$ is an injection.
Let $g \in \map \Gamma {T_2}$ be arbitrary.
Let $\alpha^{-1} \circ g \circ \alpha = f$
Then:
{{begin-eqn}}
{{eqn | l = \alpha^{-1} \circ g \circ \alpha
      | r = f
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \alpha \circ \paren {\alpha^{-1} \circ g \circ \alpha} \circ \alpha^{-1}
      | r = \alpha \circ f \circ \alpha^{-1}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = g
      | r = \alpha \circ f \circ \alpha^{-1}
      | c = 
}}
{{eqn | r = \map \theta f
      | c = Definition of $\theta$
}}
{{end-eqn}}
and so $g$ is the image of $f$ under $\theta$.
As $g$ is arbitrary, it follows that $\theta$ is a surjection.
As $\theta$ is an injection and a surjection, it follows that $\theta$ is a bijection by definition.
So $\theta$ is a bijective group homomorphism, and so a group isomorphism by definition.
{{qed}}
Theorem: Quotient Ring of Cauchy Sequences is Normed Division Ring

Let $\struct {R, \norm {\, \cdot \,} }$ be a normed division ring.
Let $\CC$ be the ring of Cauchy sequences over $R$
Let $\NN$ be the set of null sequences.
For all $\sequence {x_n} \in \CC$, let $\eqclass {x_n} {}$ denote the left coset $\sequence {x_n} + \NN$
Let $\norm {\, \cdot \,}_1: \CC \,\big / \NN \to \R_{\ge 0}$ be defined by:
:$\ds \forall \eqclass {x_n} {} \in \CC \,\big / \NN: \norm {\eqclass {x_n} {} }_1 = \lim_{n \mathop \to \infty} \norm {x_n}$
Then:
:$\struct {\CC \,\big / \NN, \norm {\, \cdot \,}_1 }$ is a normed division ring.

Proof:
By Quotient Ring of Cauchy Sequences is Division Ring then $\CC \,\big / \NN$ is a  division ring.
It remains to be proved that:
:$\norm {\, \cdot \,}_1$ is well-defined
:$\norm {\, \cdot \,}_1$ satisfies the norm axioms.
Theorem: Inequality Rule for Real Sequences

Let $\sequence {x_n}$ and $\sequence {y_n}$ be sequences in $\R$.
Let $\sequence {x_n}$ and $\sequence {y_n}$ be convergent to the following limits:
{{begin-eqn}}
{{eqn | l = \lim_{n \mathop \to \infty} x_n
      | r = l
}}
{{eqn | l = \lim_{n \mathop \to \infty} y_n
      | r = m
}}
{{end-eqn}}
Let there exist $N \in \N$ such that:
:$\forall n \ge N: x_n \le y_n$
Then:
:$l \le m$

Proof:
Suppose $l > m$.
Then:
:$m = \dfrac m 2 + \dfrac m 2 < \dfrac {l + m} 2 < \dfrac l 2 + \dfrac l 2 = l$
Let $\epsilon = \dfrac {l - m} 2$.
Then:
:$\epsilon > 0$
We are given that:
:$\ds \lim_{n \mathop \to \infty} x_n = l$
By definition of the limit of a real sequence, we can find $N_1$ such that:
:$\forall n \ge N_1: \size {x_n - l} < \epsilon$
where $\size {x_n - l}$ denotes the absolute value of $x_n - l$
Suppose $n \ge N_1$
If $x_n \ge l$ then $x_n > \dfrac {l + m} 2$
If $x_n < l$ then:
{{begin-eqn}}
{{eqn | l = \epsilon
      | o = >
      | r = \size {x_n - l}
      | c = as $n > N_1$
}}
{{eqn | r = l - x_n
      | c = as $x_n < l$
}}
{{eqn | ll= \leadsto
      | l = \dfrac {l - m} 2
      | o = >
      | r = l - x_n
      | c = as $\epsilon = \dfrac {l - m} 2$
}}
{{eqn | ll= \leadsto
      | l = x_n
      | o = >
      | r = \dfrac {l + m} 2
      | c = rearranging terms
}}
{{end-eqn}}
In either case $x_n > \dfrac {l+m} 2$
We are also given that:
:$\ds \lim_{n \mathop \to \infty} y_n = m$
Similarly we can find $N_2$ such that:
:$\forall n > N_2: \size {y_n - m} < \epsilon$
Suppose $n \ge N_2$
If $y_n \le m$ then $y_n < \dfrac {l + m} 2$
If $y_n > m$ then:
{{begin-eqn}}
{{eqn | l = \epsilon
      | o = >
      | r = \size {y_n - l}
      | c = as $n > N_2$
}}
{{eqn | r = y_n - m
      | c = as $y_n > m$
}}
{{eqn | ll= \leadsto
      | l = \dfrac {l - m} 2
      | o = >
      | r = y_n - m
      | c = as $\epsilon = \dfrac {l - m} 2$
}}
{{eqn | ll= \leadsto
      | l = \dfrac {l + m} 2
      | o = >
      | r = y_n
      | c = rearranging terms
}}
{{end-eqn}}
In either case $y_n < \dfrac {l + m} 2$
Let $N = \max \set {N_1, N_2}$.
Then if $n > N$, both the above inequalities will be true:
:$n > N_1$
:$n > N_2$
Thus $\forall n > N$:
:$y_n < \dfrac {l + m} 2 < x_n$
It has been shown that:
:$l > m \implies \forall n \in \N: \exists m \ge n: x_n > y_n$
Taking the contrapositive:
:$\exists N \in \N: \forall n \ge N: x_n \le y_n \implies l \le m$
Hence the result.
{{qed}}
Theorem: Inequality Rule for Real Sequences

Let $\sequence {x_n}$ and $\sequence {y_n}$ be sequences in $\R$.
Let $\sequence {x_n}$ and $\sequence {y_n}$ be convergent to the following limits:
{{begin-eqn}}
{{eqn | l = \lim_{n \mathop \to \infty} x_n
      | r = l
}}
{{eqn | l = \lim_{n \mathop \to \infty} y_n
      | r = m
}}
{{end-eqn}}
Let there exist $N \in \N$ such that:
:$\forall n \ge N: x_n \le y_n$
Then:
:$l \le m$

Proof:
Consider the sequence $\sequence {z_n}$ defined by:
:$z_n := y_n - x_n$
From Sum Rule for Real Sequences:
:$z_n \to m - l$ as $n \to \infty$
Furthermore, the assumption that $x_n \le y_n$ for all $n \in \N$ means that:
:$\forall n \in \N: z_n \ge 0$
Applying the Lower and Upper Bounds for Sequences to the sequence $\sequence {z_n}$ leads to the conclusion that $m - l \ge 0$.
That is:
:$l \le m$
{{qed}}
Theorem: Left Regular Representation of Subset Product

Let $\struct {S, \circ}$ be a magma.
Let $T \subseteq S$ be a subset of $S$.
Let $\lambda_a: S \to S$ be the left regular representation of $S$ with respect to $a$.
Then:
:$\lambda_a \sqbrk T = \set a \circ T = a \circ T$
where $a \circ T$ denotes subset product with a singleton.

Proof:
{{begin-eqn}}
{{eqn | l = \lambda_a \sqbrk T
      | r = \set {s \in S: \exists t \in T: s = \map {\lambda_a} t}
      | c = {{Defof|Image of Subset under Mapping}}
}}
{{eqn | r = \set {s \in S: \exists t \in T: s = a \circ t}
      | c = {{Defof|Left Regular Representation}}
}}
{{eqn | r = \set {a \circ t: t \in T}
      | c = 
}}
{{eqn | r = a \circ T
      | c = {{Defof|Subset Product}}
}}
{{end-eqn}}
{{qed}}
Theorem: Right Regular Representation of Subset Product

Let $\struct {S, \circ}$ be a magma.
Let $T \subseteq S$ be a subset of $S$.
Let $\rho_a: S \to S$ be the right regular representation of $S$ with respect to $a$.
Then:
:$\rho_a \sqbrk T = T \circ \set a = T \circ a$
where $T \circ a$ denotes subset product with a singleton.

Proof:
{{begin-eqn}}
{{eqn | l = \rho_a \sqbrk T
      | r = \set {s \in S: \exists t \in T: s = \map {\rho_a} t}
      | c = {{Defof|Image of Subset under Mapping}}
}}
{{eqn | r = \set {s \in S: \exists t \in T: s = t \circ a}
      | c = {{Defof|Right Regular Representation}}
}}
{{eqn | r = \set {t \circ a: t \in T}
      | c = 
}}
{{eqn | r = T \circ a
      | c = {{Defof|Subset Product}}
}}
{{end-eqn}}
{{qed}}
Theorem: Order of Cycle is Length of Cycle

Let $S_n$ denote the symmetric group on $n$ letters.
Let $\pi \in S_n$ be a cyclic permutation of length $k$.
Then:
:$\order \pi = k$
where:
:$\order \pi$ denotes the order of $\pi$ in $S_n$.

Proof:
Let $\pi = \tuple {a_0, a_1, \ldots, a_{k - 1} }$.
Observe that:
{{begin-eqn}}
{{eqn | l = \paren {\paren {j + n} \pmod k} + 1
      | r = \paren {j + n + 1} \pmod k
      | c = 
}}
{{eqn | ll=\leadsto
      | l = \map \pi {a_{\paren {j + n} \pmod k} }
      | r = a_{\paren {j + n + 1} \pmod k}
      | c = 
}}
{{eqn | ll=\leadsto
      | q = \forall n \in \N, \forall j \in \Z / k \Z
      | l = \map {\pi^n} {a_j}
      | r = a_{\paren {j + n} \pmod k}
      | c = 
}}
{{eqn | ll=\leadsto
      | q = \forall j < k
      | l = \pi^k
      | r = I_{S_n}
      | c = 
}}
{{eqn | lo= \land
      | l = \pi^j
      | o = \ne
      | r = I_{S_n}
      | c = where $I_{S_n}$ denotes the identity mapping on $S_n$
}}
{{end-eqn}}
Hence the result, by definition of order of group element.
{{qed}}
Theorem: Norm Sequence of Cauchy Sequence has Limit

Let $\struct {R, \norm {\, \cdot \,} }$ be a normed division ring.
Let $\sequence {x_n}$ be a Cauchy sequence in $R$.
Then $\sequence {\norm {x_n} }$ has a limit in $\R$.
That is,
:$\exists l \in \R: \ds \lim_{n \mathop \to \infty} \norm {x_n} = l$

Proof:
It is first shown that $\sequence {\norm {x_n} }$ is a real Cauchy sequence in $\R$.
Let $\epsilon \in \R_{>0}$ be given.
By the definition of Cauchy sequence then:
:$\exists N \in \N: \forall n, m > N, \norm {x_n - x_m} < \epsilon$
By Reverse Triangle Inequality on Normed Division Ring, then:
:$\forall n, m > N: \cmod {\norm {x_n} - \norm {x_m} } \le \norm {x_n - x_m} < \epsilon$
From the definition of a real Cauchy sequence it follows that $\sequence {\norm {x_n} }$ is a real Cauchy sequence in $\R$.
By Cauchy's Convergence Criterion on Real Numbers $\sequence {\norm {x_n} }$ has a limit in $\R$.
{{qed}}
Theorem: Equivalent Cauchy Sequences have Equal Limits of Norm Sequences

Let $\struct {R, \norm {\, \cdot \,} }$ be a normed division ring.
Let $\sequence {x_n}$ and $\sequence {y_n}$ be Cauchy sequences in $R$.
Let $\ds \lim_{n \mathop \to \infty} {x_n - y_n} = 0$.
Then:
:$\ds \lim_{n \mathop \to \infty} \norm {x_n} = \lim_{n \mathop \to \infty} \norm {y_n}$

Proof:
Let:
:$l = \ds \lim_{n \mathop \to \infty} \norm {x_n}$
and:
:$m = \ds \lim_{n \mathop \to \infty} \norm {y_n}$
By Norm Sequence of Cauchy Sequence has Limit, both of these limits exist.
Then:
{{begin-eqn}}
{{eqn | l = \lim_{n \mathop \to \infty } \paren {\norm {x_n} - \norm {y_n} } 
      | r = l - m
      | c = Difference Rule for Real Sequences
}}
{{eqn | ll= \leadsto
      | l = \lim_{n \mathop \to \infty} \size {\norm {x_n} - \norm {y_n} }
      | r = \size {l - m}
      | c = Modulus of Limit
}}
{{end-eqn}}
By Reverse Triangle Inequality on Normed Division Ring, for $n \in \N$:
:$l = \size {\norm {x_n} - \norm {y_n} } \le \norm {x_n - y_n} = \to 0$
as $n \to \infty$.
By the Squeeze Theorem:
:$\ds \lim_{n \mathop \to \infty} \paren {\size {\norm {x_n} - \norm {y_n} } } = 0$
So:
:$\size {l - m} = 0$
and therefore:
$l = m$
{{qed}}
Category:Cauchy Sequences
Category:Normed Division Rings
bsbraz9abhwmqec6acxo35hneqdgyb9
Theorem: Rule of Conjunction/Sequent Form/Formulation 1

{{begin-eqn}}
{{eqn | l = p
      | o = 
}}
{{eqn | l = q
      | o = 
}}
{{eqn | ll= \vdash
      | l = p \land q
      | o = 
}}
{{end-eqn}}

Proof:
{{BeginTableau|p, q \vdash p \land q}}
{{Premise|1|p}}
{{Premise|2|q}}
{{Conjunction|3|1, 2|p \land q|1|2}}
{{EndTableau|qed}}
Theorem: Rule of Conjunction/Sequent Form/Formulation 1

{{begin-eqn}}
{{eqn | l = p
      | o = 
}}
{{eqn | l = q
      | o = 
}}
{{eqn | ll= \vdash
      | l = p \land q
      | o = 
}}
{{end-eqn}}

Proof:
We apply the Method of Truth Tables.
$\begin{array}{|c|c||ccc|} \hline
p & q & p & \land & q\\
\hline
\F & \F & \F & \F & \F \\
\F & \T & \F & \F & \T \\
\T & \F & \T & \F & \F \\
\T & \T & \T & \T & \T \\
\hline
\end{array}$
As can be seen, only when both $p$ and $q$ are true, then so is $p \land q$.
{{qed}}
Theorem: Rule of Conjunction/Sequent Form/Formulation 2

:$\vdash p \implies \paren {q \implies \paren {p \land q} }$

Proof:
{{BeginTableau|\vdash p \implies \paren {q \implies \paren {p \land q} }|Instance 2 of the Hilbert-style systems}}
{{TableauLine
 | n = 1
 | f = \neg p \lor p
 | rlnk = Law of Excluded Middle/Sequent Form/Proof 2
 | rtxt = Law of Excluded Middle
}}
{{TableauLine
 | n = 2
 | f = \paren {\neg p \lor p} \implies \paren {p \lor \neg p}
 | rlnk = Definition:Hilbert Proof System/Instance 2
 | rtxt = Axiom $\text A 3$
 | c = $\neg p \,/\, p, p \,/\, q$
}}
{{TableauLine
 | n = 3
 | f = p \lor \neg p
 | rlnk = Definition:Hilbert Proof System/Instance 2
 | rtxt = Rule $\text {RST} 3$
 | dep  = 1, 2
}}
{{TableauLine
 | n = 4
 | f = \paren {\neg p \lor \neg q} \lor \neg \paren {\neg p \lor \neg q}
 | rlnk = Definition:Hilbert Proof System/Instance 2
 | rtxt = Rule $\text {RST} 1$
 | dep  = 3
 | c = $\paren {\neg p \lor \neg q} \,/\, p$
}}
{{TableauLine
 | n = 5
 | f = \paren {\neg p \lor \neg q} \lor \paren {p \land q}
 | rlnk = Definition:Hilbert Proof System/Instance 2
 | rtxt = Rule $\text {RST} 2 (1)$
}}
{{TableauLine
 | n = 6
 | f = \paren {\paren {\neg p \lor \neg q} \lor \paren {p \land q} } \implies \paren {\neg p \lor \paren {\neg q \lor \paren {p \land q} } }
 | rlnk = Rule of Association/Disjunction/Formulation 2/Reverse Implication
 | rtxt = Rule of Association
 | c = $\neg p \,/\, p, \neg q \,/\, q, \paren {p \land q} \,/\, r$
}}
{{TableauLine
 | n = 7
 | f = \neg p \lor \paren {\neg q \lor \paren {p \land q} }
 | rlnk = Definition:Hilbert Proof System/Instance 2
 | rtxt = Rule $\text {RST} 3$
 | dep  = 5, 6
}}
{{TableauLine
 | n = 8
 | f = p \implies \paren {q \implies \paren {p \land q} }
 | rlnk = Definition:Hilbert Proof System/Instance 2
 | rtxt = Rule $\text {RST} 2 (2)$
 | dep = 7
}}
{{EndTableau}}
{{qed}}
Theorem: Hilbert Proof System Instance 2 Independence Results/RST4 is Derivable

Rule of inference $RST \, 4$ is derivable from $RST \, 1, RST \, 2, RST \, 3$ and the axioms $(A1)$ through $(A4)$.

Proof:
Recall the statement of $RST \, 4$:
:If $\mathbf A$ and $\mathbf B$ are theorems of $\mathscr H_2$, then so is $\mathbf A \land \mathbf B$.
Suppose that $\mathbf A$ and $\mathbf B$ are theorems of $\mathscr H_2$.
From Rule of Conjunction/Sequent Form/Formulation 2, we have as a theorem:
:$p \implies \paren{ q \implies \paren{ p \land q } }$
which can be recursively checked to not have used $RST \, 4$ anywhere in a proof.
Applying $RST \, 1$, this becomes:
:$\mathbf A \implies \paren{ \mathbf B \implies \paren{ \mathbf A \land \mathbf B } }$
Using $RST \, 3$ twice, the result follows.
{{qed}}
Theorem: Subgroup Generated by One Element is Cyclic

Let $G$ be a group.
Let $a \in G$.
Then $\gen a$, the subgroup generated by $a$, is cyclic:

Proof:
By Subgroup Generated by One Element is Set of Powers:
:$\gen a = \set {a^n : n \in \Z}$
The result follows by definition of cyclic group.
{{qed}}