hcju/proofwikiqa huggingface dataset:
By the definition of a ring monomorphism it is sufficient to prove for all $r, r' \in R$ that: :$\quad \map \phi {r + r'} = \map \phi r +' \map \phi {r'}$ :$\quad \map \phi {r \circ r'} = \map \phi r \circ' \map \phi {r'}$ :$\quad r \ne r' \implies \map \phi r \ne \map \phi {r'}$ That is, for all $r, r' \in R$ it needs to be shown that: :$(1): \quad f_{r + r' } = f_r +' f_{r'}$ :$(2): \quad f_{r \circ r' } = f_r \circ' f_{r'}$ :$(3): \quad r \ne r' \implies f_r \ne f_{r'}$ === $(1): f_{r + r'} = f_r +' f_{r'}$ === For all $s \in S$: {{begin-eqn}} {{eqn | l = \map {f_{r + r'} } s       | r = r + r'       | c = Definition of $f_{r + r'}$ }} {{eqn | r = \map {f_r} s + \map {f_{r'} } s       | c = Definition of $f_r$ and $f_{r'}$ }} {{eqn | r = \map {\paren {f_r +' f_{r'} } } s       | c = {{Defof|Pointwise Operation}} $+'$ }} {{end-eqn}} The result follows. {{qed|lemma}} === $(2): f_{r \circ r'} = f_r \circ' f_{r'}$ === For all $s \in S$: {{begin-eqn}} {{eqn | l = \map {f_{r \circ r'} } s       | r = r \circ r'       | c = Definition of $f_{r \circ r'}$ }} {{eqn | r = \map {f_r} s \circ \map {f_{r'} } s       | c = Definition of $f_r$ and $f_{r'}$ }} {{eqn | r = \map {\paren {f_r \circ' f_{r' } } } s       | c = {{Defof|Pointwise Operation}} $\circ'$ }} {{end-eqn}} The result follows. {{qed|lemma}} === $(3): r \ne r' \implies f_r \ne f_{r'}$ === Let $s \in S$, then: :$\map {f_r} s = r \ne r' = \map {f_{r'} } s$ So: :$f_r \ne f_{r'}$ and the proof is complete. {{qed}} Category:Rings of Mappings Category:Ring Monomorphisms 3dcnd863674x5uee7e9j6paviat5gm9
By Cauchy Sequences form Ring with Unity, $\CC$ is a subring of the ring of sequences over $R$. Let $i: \CC \to  R^\N$ be the inclusion mapping of $\CC$ into the ring of sequences. By Embedding Ring into Ring Structure Induced by Ring Operations the composition $i \circ \phi: R \to R^\N$ is a ring monomorphism. Since for all $a \in R$, we have that: :$\map {\paren {i \circ \phi} } a = \map \phi a$ The result follows. {{qed}}
By the definitional abbreviation for the conditional: :$\mathbf A \implies \mathbf B =_{\text{def}} \neg \mathbf A \lor \mathbf B$ the Rule of Idempotence can be written as: : $\neg \left({p \lor p}\right) \lor p$ This evaluates as follows: :$\begin{array}{|cccc|c|c|} \hline \neg & (p & \lor & p) & \lor & p \\ \hline 2 & 0 & 0 & 0 & 0 & 0 \\ 1 & 1 & 1 & 1 & 1 & 1 \\ 0 & 2 & 2 & 2 & 0 & 2 \\ \hline \end{array}$ {{qed}} Category:Formal Semantics asmpzabvq5k0cfuq8tclysj6vp07fjk
By the definitional abbreviation for the conditional: :$\mathbf A \implies \mathbf B =_{\text{def}} \neg \mathbf A \lor \mathbf B$ the Rule of Commutation can be written as: :$\neg \left({p \lor q}\right) \lor \left({q \lor p}\right)$ This evaluates as follows: :$\begin{array}{|cccc|c|ccc|} \hline \neg & (p & \lor & q) & \lor & (q & \lor & p) \\ \hline 2 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 2 & 0 & 0 & 1 & 0 & 1 & 0 & 0 \\ 2 & 0 & 0 & 2 & 0 & 2 & 0 & 0 \\ 2 & 1 & 0 & 0 & 0 & 0 & 0 & 1 \\ 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\ 0 & 1 & 2 & 2 & 0 & 2 & 2 & 1 \\ 2 & 2 & 0 & 0 & 0 & 0 & 0 & 2 \\ 0 & 2 & 2 & 1 & 0 & 1 & 2 & 2 \\ 0 & 2 & 2 & 2 & 0 & 2 & 2 & 2 \\ \hline \end{array}$ {{qed}} Category:Formal Semantics 04e31w1poro9dfijnymy5oqr9occrh6
{{handwaving}} By the definitional abbreviation for the conditional: :$\mathbf A \implies \mathbf B =_{\text{def}} \neg \mathbf A \lor \mathbf B$ the Factor Principle can be written as: :$\neg \left({\neg p \lor q}\right) \lor \left({\neg \left({r \lor p}\right) \lor \left ({r \lor q}\right)}\right)$ This evaluates as follows: :$\begin{array}{|ccccc|c|cccccccc|} \hline \neg & (\neg & p & \lor & q) & \lor & (\neg & (r & \lor & p) & \lor & (r & \lor & q)) \\ \hline 0 & 2 & 0 & 0 & 0 & 0 & 2 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 2 & 1 & 0 & 0 & 0 & 2 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 2 & 2 & 0 & 0 & 0 & 2 & 0 & 0 & 2 & 0 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 & 0 & 0 & 2 & 1 & 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 2 & 1 & 0 & 0 & 0 & 1 & 1 & 1 & 1 & 0 & 1 & 0 & 0 \\ 0 & 2 & 2 & 0 & 0 & 0 & 0 & 1 & 2 & 2 & 0 & 1 & 0 & 0 \\ 0 & 2 & 0 & 0 & 0 & 0 & 2 & 2 & 0 & 0 & 0 & 2 & 0 & 0 \\ 0 & 2 & 1 & 0 & 0 & 0 & 0 & 2 & 2 & 1 & 0 & 2 & 0 & 0 \\ 0 & 2 & 2 & 0 & 0 & 0 & 0 & 2 & 2 & 2 & 0 & 2 & 0 & 0 \\ 0 & 2 & 0 & 0 & 1 & 0 & 2 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ 1 & 1 & 1 & 1 & 1 & 0 & 2 & 0 & 0 & 1 & 0 & 0 & 0 & 1 \\ 2 & 0 & 2 & 2 & 1 & 0 & 2 & 0 & 0 & 2 & 0 & 0 & 0 & 1 \\ 0 & 2 & 0 & 0 & 1 & 0 & 2 & 1 & 0 & 0 & 2 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\ 2 & 0 & 2 & 2 & 1 & 0 & 0 & 1 & 2 & 2 & 0 & 1 & 1 & 1 \\ 0 & 2 & 0 & 0 & 1 & 0 & 2 & 2 & 0 & 0 & 2 & 2 & 2 & 1 \\ 1 & 1 & 1 & 1 & 1 & 0 & 0 & 2 & 2 & 1 & 0 & 2 & 2 & 1 \\ 2 & 0 & 2 & 2 & 1 & 0 & 0 & 2 & 2 & 2 & 0 & 2 & 2 & 1 \\ 0 & 2 & 0 & 2 & 2 & 0 & 2 & 0 & 0 & 0 & 0 & 0 & 0 & 2 \\ 0 & 1 & 1 & 2 & 2 & 0 & 2 & 0 & 0 & 1 & 0 & 0 & 0 & 2 \\ 2 & 0 & 2 & 0 & 2 & 0 & 2 & 0 & 0 & 2 & 0 & 0 & 0 & 2 \\ 0 & 2 & 0 & 2 & 2 & 0 & 2 & 1 & 0 & 0 & 2 & 1 & 2 & 2 \\ 0 & 1 & 1 & 2 & 2 & 0 & 1 & 1 & 1 & 1 & 2 & 1 & 2 & 2 \\ 2 & 0 & 2 & 0 & 2 & 0 & 0 & 1 & 2 & 2 & 0 & 1 & 2 & 2 \\ 0 & 2 & 0 & 2 & 2 & 0 & 2 & 2 & 0 & 0 & 2 & 2 & 2 & 2 \\ 0 & 1 & 1 & 2 & 2 & 0 & 0 & 2 & 2 & 1 & 0 & 2 & 2 & 2 \\ 2 & 0 & 2 & 0 & 2 & 0 & 0 & 2 & 2 & 2 & 0 & 2 & 2 & 2 \\ \hline \end{array}$ {{qed}} {{proofread}} Category:Formal Semantics 2il6dm5w31g149leqiwrt09afiaiw9k
By the definitional abbreviation for the conditional: :$\mathbf A \implies \mathbf B =_{\text{def}} \neg \mathbf A \lor \mathbf B$ the Rule of Idempotence can be written as: : $\neg \left({p \lor p}\right) \lor p$ This evaluates as follows: :$\begin{array}{|cccc|c|c|} \hline \neg & (p & \lor & p) & \lor & p \\ \hline 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 1 & 1 & 0 & 1 \\ 0 & 2 & 2 & 2 & 0 & 2 \\ 2 & 3 & 3 & 3 & 0 & 3 \\ \hline \end{array}$ {{qed}} Category:Formal Semantics 01yq3v78ftnoxwrey60d9xk091x9ahx
{{handwaving}} By the definitional abbreviation for the conditional: :$\mathbf A \implies \mathbf B =_{\text{def}} \neg \mathbf A \lor \mathbf B$ the Rule of Addition can be written as: : $\neg q \lor \left({p \lor q}\right)$ This evaluates as follows: :$\begin{array}{|cc|c|ccc|} \hline \neg & q & \lor & (p & \lor & q) \\ \hline 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 1 \\ 0 & 2 & 0 & 0 & 0 & 2 \\ 2 & 3 & 0 & 0 & 0 & 3 \\ 1 & 0 & 0 & 1 & 0 & 0 \\ 0 & 1 & 0 & 1 & 1 & 1 \\ 0 & 2 & 0 & 1 & 2 & 2 \\ 2 & 3 & 0 & 1 & 3 & 3 \\ 1 & 0 & 0 & 2 & 0 & 0 \\ 0 & 1 & 0 & 2 & 2 & 1 \\ 0 & 2 & 0 & 2 & 2 & 2 \\ 2 & 3 & 0 & 2 & 0 & 3 \\ 1 & 0 & 0 & 3 & 0 & 0 \\ 0 & 1 & 0 & 3 & 3 & 1 \\ 0 & 2 & 0 & 3 & 3 & 2 \\ 2 & 3 & 0 & 3 & 3 & 3 \\ \hline \end{array}$ {{qed}} Category:Formal Semantics 5hf2kbaogy667algmc2jorctor673g1
Let $T$ be the codomain of a surjection $f$ from $S$ to $T$. By the Quotient Theorem for Surjections, $f$ induces an equivalence $\mathcal R_f$ on $T$. :$f = r \circ q_{\mathcal R_f}$ where: :$\mathcal R_f$ is the equivalence induced by $f$ on $T$ :$r: S / \mathcal R_f \to T$ is a bijection from the quotient set $S / \mathcal R_f$ to $T$ :$q_{\mathcal R_f}: S \to S / \mathcal R_f$ is the quotient mapping induced by $\mathcal R_f$. From the Fundamental Theorem on Equivalence Relations, $\mathcal R_f$ induces a partition on $S$. From Cardinality of Set of Induced Equivalence Classes of Surjection, $\mathcal R_f$ has $m$ components. From Number of Set Partitions by Number of Components, there are $\displaystyle {m \brace n}$ different ways of performing such a partitioning. From Cardinality of Set of Injections, there are $n!$ different bijections from $S / \mathcal R_f \to T$ The total number of surjections is then the product of these: :$C = n! \displaystyle {m \brace n}$ {{qed}}
As $T$ is a surjection, $T = f \sqbrk S$. Thus Image of Set Difference under Mapping can be applied: :$f \sqbrk S \setminus f \sqbrk A \subseteq f \sqbrk {S \setminus A}$ {{qed}}
By the definitional abbreviation for the conditional: :$\mathbf A \implies \mathbf B =_{\text{def}} \neg \mathbf A \lor \mathbf B$ the Rule of Idempotence can be written as: : $\neg \left({p \lor p}\right) \lor p$ This evaluates as follows: :$\begin{array}{|cccc|c|c|} \hline \neg & (p & \lor & p) & \lor & p \\ \hline 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 1 & 1 & 0 & 1 \\ 3 & 2 & 2 & 2 & 0 & 2 \\ 0 & 3 & 3 & 3 & 0 & 3 \\ \hline \end{array}$ {{qed}} Category:Formal Semantics 5wfsgwvosz5m5ujv2gjtoogcqm6423d
Trivially, we have: :$\forall x \in \R: \map f {1 \cdot x} = 1 \map f x$ Next, suppose that: :$\map f {n x} = n \map f x$ By additivity of $f$, we have: {{begin-eqn}} {{eqn | l = \map f {\paren {n + 1} x}       | r = \map f {n x + x}       | c =  }} {{eqn | r = \map f {n x} + \map f x = n \map f x + \map f x       | c =  }} {{eqn | r = \paren {n + 1} \map f x       | c =  }} {{end-eqn}} Hence by the Principle of Mathematical Induction: :$\forall n \in \N, x \in \R: \map f {n x} = n \map f x$ As Additive Function is Odd Function and Odd Function of Zero is Zero, we conclude: :$\forall p \in \Z, x \in \R: \map f {p x} = p \map f x$ Let $p \ne 0$. By substituting $y = p x$, the above gives: :$\forall p \in \Z \setminus \set 0, y \in \R: \map f y = p \map f {\dfrac y p}$ In other words: :$\forall p \in \Z \setminus \set 0, y \in \R: \map f {\dfrac y p} = \dfrac 1 p \map f y$ Given $p, q \in \Z, q \ne 0$, we have: {{begin-eqn}} {{eqn | l = \map f {\dfrac p q} x       | r = \map f {p \dfrac x q}       | c =  }} {{eqn | r = p \map f {\dfrac x q}       | c =  }} {{eqn | r = p \dfrac {\map  f x } q       | c =  }} {{eqn | r = \dfrac p q \map f x       | c =  }} {{end-eqn}} Therefore we conclude: :$\forall r \in \Q, x \in \R: \map f {r x} = r \map f x$ {{qed}} Category:Additive Functions sod129ok9yo83743zxodu3g3r9dnleq
{{handwaving}} By the definitional abbreviation for the conditional: :$\mathbf A \implies \mathbf B =_{\text{def}} \neg \mathbf A \lor \mathbf B$ the Rule of Addition can be written as: : $\neg q \lor \left({p \lor q}\right)$ This evaluates as follows: :$\begin{array}{|cc|c|ccc|} \hline \neg & q & \lor & (p & \lor & q) \\ \hline 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 1 \\ 3 & 2 & 0 & 0 & 0 & 2 \\ 0 & 3 & 0 & 0 & 0 & 3 \\ 1 & 0 & 0 & 1 & 0 & 0 \\ 0 & 1 & 0 & 1 & 1 & 1 \\ 3 & 2 & 0 & 1 & 2 & 2 \\ 0 & 3 & 0 & 1 & 3 & 3 \\ 1 & 0 & 0 & 2 & 0 & 0 \\ 0 & 1 & 0 & 2 & 2 & 1 \\ 3 & 2 & 0 & 2 & 2 & 2 \\ 0 & 3 & 0 & 2 & 0 & 3 \\ 1 & 0 & 0 & 3 & 0 & 0 \\ 0 & 1 & 0 & 3 & 3 & 1 \\ 3 & 2 & 0 & 3 & 0 & 2 \\ 0 & 3 & 0 & 3 & 3 & 3 \\ \hline \end{array}$ {{qed}} Category:Formal Semantics 9qn6oodis4tald5nfqxy6512l4cavu8
By the definitional abbreviation for the conditional: :$\mathbf A \implies \mathbf B =_{\text{def}} \neg \mathbf A \lor \mathbf B$ the Rule of Commutation can be written as: :$\neg \left({p \lor q}\right) \lor \left({q \lor p}\right)$ This evaluates as follows: :$\begin{array}{|cccc|c|ccc|} \hline \neg & (p & \lor & q) & \lor & (q & \lor & p) \\ \hline 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 1 & 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 2 & 0 & 2 & 0 & 0 \\ 1 & 0 & 0 & 3 & 0 & 3 & 0 & 0 \\ 1 & 1 & 0 & 0 & 0 & 0 & 0 & 1 \\ 0 & 1 & 1 & 1 & 0 & 1 & 1 & 1 \\ 3 & 1 & 2 & 2 & 0 & 2 & 2 & 1 \\ 0 & 1 & 3 & 3 & 0 & 3 & 3 & 1 \\ 1 & 2 & 0 & 0 & 0 & 0 & 0 & 2 \\ 3 & 2 & 2 & 1 & 0 & 1 & 2 & 2 \\ 3 & 2 & 2 & 2 & 0 & 2 & 2 & 2 \\ 1 & 2 & 0 & 3 & 0 & 3 & 0 & 2 \\ 1 & 3 & 0 & 0 & 0 & 0 & 0 & 3 \\ 0 & 3 & 3 & 1 & 0 & 1 & 3 & 3 \\ 1 & 3 & 0 & 2 & 0 & 2 & 0 & 3 \\ 0 & 3 & 3 & 3 & 0 & 3 & 3 & 3 \\ \hline \end{array}$ {{qed}} Category:Formal Semantics emlja2kbsygf1qleayueqncwq850mjn
As $f$ is additive, we have: {{begin-eqn}} {{eqn | l = f \paren 1       | r = f \paren {0 + 1}       | c =  }} {{eqn | r = f \paren 0 + f \paren 1       | c =  }} {{end-eqn}} that is: :$f \paren 0 = 0$ {{qed}} Category:Additive Functions b7tdfjld57tam4s0nxbe9x7k62jlq6j
From Additive Function of Zero is Zero: :$\map f 0 = 0$ Thus, for all $x \in \R$, we have: {{begin-eqn}} {{eqn | l = 0       | r = \map f 0       | c =  }} {{eqn | r = \map f {x + \paren {-x} }       | c =  }} {{eqn | r = \map f x + \map f {-x}       | c =  }} {{end-eqn}} It follows that the function $f$ is odd: :$\forall x \in \R: \map f {-x} = -\map f x$. {{qed}} Category:Additive Functions Category:Odd Functions kktj3q1hz1r50dop4kgcdw1dz9fa7hv
Note that from Image of Set Difference under Mapping: :$f \sqbrk {S_1} \setminus f \sqbrk {S_2} \subseteq f \sqbrk {S_1 \setminus S_2}$ By Proof by Counterexample it is demonstrated that the inclusion does not necessarily apply in the other direction. Let: :$S_1 = \set {x \in \Z: x \le 0}$ :$S_2 = \set {x \in \Z: x \ge 0}$ :$f: \Z \to \Z: \forall x \in \Z: \map f x = x^2$ We have: :$S_1 \setminus S_2 = \set {-1, -2, -3, \ldots}$ :$f \sqbrk {S_1} = \set {0, 1, 4, 9, 16, \ldots} = f \sqbrk {S_2}$ Then from Set Difference with Self is Empty Set: :$f \sqbrk {S_1} \setminus f \sqbrk {S_2} = \O$ but: :$f \sqbrk {S_1 \setminus S_2} = f \sqbrk {\set {x \in \Z: x > 0} } = \set {1, 4, 9, 16, \ldots}$
$\alpha$ is an instance of a cross-relation. We also have that Natural Number Addition is Commutative. The result therefore follows from Cross-Relation is Equivalence Relation. {{qed}}
We have from Cross-Relation on Real Numbers is Equivalence Relation that $\alpha$ is an equivalence relation. Thus: {{begin-eqn}} {{eqn | l = \tuple {x_1, y_1}       | o = \alpha       | r = \tuple {x_2, y_2}       | c =  }} {{eqn | ll= \iff       | l = x_1 + y_2       | r = x_2 + y_1       | c =  }} {{eqn | ll= \iff       | l = x_1 - y_1       | r = x_2 - y_2       | c =  }} {{end-eqn}} Thus each equivalence classes consists of sets of points such that: :$x - y = c$ That is: :$y = x + c$ Thus from Equation of Straight Line in Plane, this is the equation of a straight line whose slope is $1$. {{qed}}
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}}
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}}
=== Closure === Integer Addition is Closed. {{qed|lemma}} === Associativity === Integer Addition is Associative. {{qed|lemma}} The criteria for $\struct {\Z, +}$ to be a semigroup are seen to be satisfied.  {{qed}}
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}}
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 $M >= K: \norm {x_M - l} >= \epsilon$ 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 = {{NormAxiom|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 p5b1gjcn4xk1dx8b3mtgnv2vjlyqab6
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. === $(1): \quad \phi$ is a homomorphism === By definition, $\phi$ is the composition of two mappings: :$\phi = q \circ \phi'$ where: :$\text{(a)}: \quad \phi': R \to \CC$, defined by: $\forall a \in R, \map {\phi'} a = \sequence {a, a, a, \dotsc}$ :$\text{(b)}: \quad q$ is the quotient mapping $q: \CC \to \CC \, \big / \NN$ defined by: $\map q {\sequence {x_n} } = \sequence {x_n} + \NN$ By Embedding Normed Division Ring into Ring of Cauchy Sequences, $\phi'$ is a ring monomorphism. By Quotient Ring Epimorphism is Epimorphism, then $q$ is a ring epimorphism. By Composition of Ring Homomorphisms is Ring Homomorphism then the composition $\phi = q \circ \phi'$ is a ring homomorphism {{qed|lemma}} === $(2): \quad \phi$ is an injection === Let $a, b \in R$. Suppose $\map \phi a = \map \phi b$. Then: {{begin-eqn}} {{eqn | l = \sequence {a, a, a, \dotsc} + \NN       | r = \sequence {b, b, b, \dotsc} + \NN       | c = Definition of $\phi$ }} {{eqn | ll= \leadsto       | l = \sequence {a, a, a, \dotsc} - \sequence {b, b, b, \dotsc}       | o = \in       | r = \NN       | c = Left Cosets are Equal iff Product with Inverse in Subgroup }} {{eqn | ll= \leadsto       | l = \sequence {a - b, a - b, a - b, \dotsc}       | o = \in       | r = \NN       | c = Ring operations on Ring of Cauchy Sequences }} {{eqn | ll= \leadsto       | l = \lim_{n \mathop \to \infty} {a - b}       | r = 0       | c = Definition of $\NN$ }} {{end-eqn}} By Constant Sequence Converges to Constant in Normed Division Ring then: :$\displaystyle \lim_{n \mathop \to \infty} {a - b} = a - b$ Hence $a-b = 0$. The result follows. {{qed|lemma}} === $(3): \quad \phi$ is distance-preserving === Let $a, b \in R$. Then: {{begin-eqn}} {{eqn | l = \norm {\phi \paren {a} - \phi \paren {b} }       | r = \norm {\phi \paren {a - b} }       | c = $\phi$ is a homomorphism }} {{eqn | r = \norm {\sequence {a - b, a - b, a - b, \dots} + \NN}       | c = Definition of $\phi$ }} {{eqn | r = \lim_{n \mathop \to \infty } \norm {a - b}       | c = Definition of $\norm {\, \cdot \,}$ }} {{eqn | r = \norm {a - b}       | c = Constant Sequence Converges to Constant in Normed Division Ring }} {{end-eqn}} {{qed}}
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 $\struct {T, \times}$. 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}}
<onlyinclude> Taking the group axioms in turn: === $\text G 0$: Closure === From Non-Zero Real Numbers Closed under Multiplication: Proof 2, $\R_{\ne 0}$ is closed under multiplication. Note that proof 2 needs to be used specifically here, as proof 1 rests on this result. {{qed|lemma}} === $\text G 1$: Associativity === Real Multiplication is Associative. {{qed|lemma}} === $\text G 2$: Identity === {{:Real Multiplication Identity is One}}{{qed|lemma}} === $\text G 3$: Inverses === {{:Inverses for Real Multiplication}}{{qed}}
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}}
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}}
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 =  }} {{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 =  }} {{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 =  }} {{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 =  }} {{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}}
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. === Lemma 1 === {{:Quotient Ring of Cauchy Sequences is Normed Division Ring/Lemma 1}}{{qed|lemma}} === Lemma 2 === {{:Quotient Ring of Cauchy Sequences is Normed Division Ring/Lemma 2}}{{qed|lemma}} === Lemma 3 === {{:Quotient Ring of Cauchy Sequences is Normed Division Ring/Lemma 3}}{{qed|lemma}} === Lemma 4 === {{:Quotient Ring of Cauchy Sequences is Normed Division Ring/Lemma 4}}{{qed|lemma}} The result follows. {{qed}}