hcju/proofwikiqa huggingface dataset:
=== Necessary Condition === First, suppose that $\le$ is compatible with the ring structure of $R$. Hence: :$(OR1): \quad \le$ is compatible with $+$ :$(OR2): \quad \forall x, y \in R: 0_R \le x, 0_R \le y \implies 0_R \le x \circ y$. $(1)$: Let $x, y \in R: 0_R \le x, 0_R \le y$. Then $0_R + 0_R \le x + y$ by the fact that $\le$ is compatible with $+$. Thus $0_R \le x + y$ and thus $x + y \in P$. $(2)$: By Properties of Ordered Ring item 4, $-P = \set {x \in R: x \le 0_R}$. Let $x \in P \cap \paren {-P}$. Then $x \le 0_R$ and $0_R \le x$. So from the antisymmetric nature of $\le$, $x = 0_R$. $(3)$: This is equivalent to $0_R \le x, 0_R \le y \implies 0_R \le x \circ y$ which is one of the properties of being compatible with the ring structure of $R$. $(4)$: Now if $\le$ is a total ordering, then $\forall x \in R$, either $x \le 0_R$ or $0_R \le x$, and the result follows. === Sufficient Condition === Let $P \subseteq R$ such that $(1)$, $(2)$ and $(3)$ are satisfied. By item $(OR2)$ of Properties of Ordered Ring, we have: :$x \le y \iff 0 \le y + \paren {-x}$ so there is at most one ordering on $R$ compatible with the ring structure of $R$ such that $P = R_{\ge 0}$, namely, the one that satisfies: :$x \le y \iff y + \left({-x}\right) \in P$ Now we need to show that $\le$ thus defined has the required properties. ;Reflexivity: By $(2)$: :$\forall x \in R: x \le x$ because $x + \paren {-x} = 0_R \in P$ ;Antisymmetry: Let $x \le y$ and $y \le x$. Then $y + \paren {-x} \in P$ and $-\paren {y + \paren {-x} } = x + \paren {-y} \in P$. Thus by $(2)$, $y + \paren {-x} = 0_R$ and thus $y = x$. ;Transitivity: If $x \le y$ and $y \le z$, then $y + \paren {-x} \in P$ and $z + \paren {-y} \in P$. But as $z + \paren {-x} = z + \paren {-y} + y + \paren {-x}$, we have that $z + \paren {-x} \in P$ from $(1)$. Hence $x \le z$. Let $x \le y$. Then $z + x \le z + y$ since $\paren {z + y} + \paren {-\paren {x + x} } = y + \paren {-x} \in P$. Finally, Ordering Compatible with Ring holds because of $(3)$. If $(4)$ holds, then $\forall x, y \in R$, either $y + \paren {-x} \in P$ or $x + \paren {-y} = -\paren {y + \paren {-x} }  \in P$, that is, either $x \le y$ or $y \le x$. {{qed}}
From Symmetric Difference on Power Set forms Abelian Group, $\struct {\powerset S, *}$ is an abelian group, where $\O$ is the identity and each element is self-inverse. From Power Set with Intersection is Monoid, $\struct {\powerset S, \cap}$ is a commutative monoid whose identity is $S$. Also Intersection Distributes over Symmetric Difference. Thus $\struct {\powerset S, \cap}$ is a commutative ring with a unity which is $S$. From Intersection with Empty Set: :$\forall A \in \powerset S: A \cap \O = \O = \O \cap A$ Thus $\O$ is indeed the zero. However, from Set Intersection Not Cancellable, it follows that $\struct {\powerset S, *, \cap}$ is not an integral domain. {{qed}}
From Power Set is Closed under Symmetric Difference and Power Set is Closed under Intersection, we have that both $\struct {\powerset S, *}$ and $\struct {\powerset S, \cap}$ are closed. Hence $\powerset S$ is a ring of sets, and hence a commutative ring. From Intersection with Subset is Subset, we have $A \subseteq S \iff A \cap S = A$. Thus we see that $S$ is the unity. Also during the proof of Power Set with Intersection is Monoid, it was established that $S$ is the identity of $\struct {\powerset S, \cap}$. We also note that set intersection is not cancellable, so $\struct {\powerset S, *, \cap}$ is not an integral domain. The result follows. {{qed}}
=== Necessary Condition === Let $\struct {K, +, \circ}$ be a subfield of $\struct {F, +, \circ}$. Then the conditions $(1)$ to $(4)$ all hold by virtue of the field axioms. {{qed|lemma}} === Sufficient Condition === Suppose the conditions $(1)$ to $(4)$ hold. From the Division Subring Test, it follows that $\struct {K, +, \circ}$ is a division ring. As $\struct {F, +, \circ}$ is a field, then $\circ$ is commutative on all of $F$. Therefore $\circ$ is commutative also on $K$ by Restriction of Commutative Operation is Commutative. Thus $\struct {K, +, \circ}$ is a field. {{qed}}
$1_D = 1_F$ by Subdomain Test. The sum and product of two elements of $K$ are also in $K$ by Addition of Division Products and Product of Division Products. The additive and product inverses of $K$ are also in $K$ by Negative of Division Product and Inverse of Division Product. Thus by Subfield Test, $\struct {K, +, \circ}$ is a subfield of $\struct {F, +, \circ}$ which clearly contains $\struct {D, +, \circ}$. Hence the result. {{qed}}
By definition, $\struct {K, \circ_R}$ and $\struct {L, \circ_S}$ are inverse completions of $\struct {R, \circ_R}$ and $\struct {S, \circ_S}$ respectively. {{Questionable|this is not the definition}} So by the Extension Theorem for Homomorphisms, there is one and only one monomorphism $\psi: \struct {K, \circ_R} \to \struct {L, \circ_S$ extending $\phi$. Thus: :$\forall x \in R, y \in R^*: \map \psi {\dfrac x y} = \dfrac {\map \phi x} {\map \phi y}$ By the Extension Theorem for Isomorphisms, $\psi$ is an isomorphism if $\phi$ is. Thus, $\forall x, y \in R, z, w \in R^*$: {{begin-eqn}} {{eqn | l = \map \psi {\frac x z +_R \frac y w}       | r = \map \psi {\frac {\paren {x \circ_R w} +_R \paren {y \circ_R z} } {z \circ_R w} }       | c = Addition of Division Products }} {{eqn | r = \frac {\map \phi {\paren {x \circ_R w} +_R \paren {y \circ_R z} } } {\map \phi {z \circ_R w} }       | c = Definition of $\psi$ }} {{eqn | r = \frac {\paren {\map \phi x \circ_S \map \phi w} +_S \paren {\map \phi y \circ_S \map \phi z} } {\map \phi z \circ_S \map \phi w}       | c = Morphism Property }} {{eqn | r = \frac {\map \phi x} {\map \phi z} +_S \frac {\map \phi y} {\map \phi w}       | c = Addition of Division Products }} {{eqn | r = \map \psi {\frac x z} +_S \map \psi {\frac y w}       | c = Definition of $\psi$ }} {{end-eqn}} Thus $\psi: K \to L$ is a monomorphism. {{qed}}
By definition of field of quotients: :$\forall z \in K: \exists x, y \in D: z = \dfrac x y, y \in D_{\ne 0}$ Suppose $z = x' / y'$ such that $y' \notin D_{>0}$. Then $y' < 0$ as $D$ is totally ordered. Then: {{begin-eqn}} {{eqn | l = x' / y'       | r = x' \circ \paren {y'}^{-1}       | c = {{Defof|Division}} }} {{eqn | r = \paren {-x'} \circ \paren {-y'}^{-1}       | c = Product of Ring Negatives }} {{eqn | r = \paren {-x'} / \paren {-y'}       | c = {Defof|Division}} }} {{end-eqn}} {{explain|first to second step above assumes more than is stated: $-\paren {y'}^{-1} {{=}} \paren {-y'}^{-1}$ - add that step and link to that result}} If $y' < 0$, then $\paren {-y'} > 0$ from Properties of Ordered Ring $(4)$. So all we need to do is set $x = -x', y = -y'$ and the result follows. {{qed}} Category:Fields of Quotients Category:Integral Domains 1i02g1dcijv3u3rs5199dhvjn1utptf
First, note that from Divided by Positive Element of Field of Quotients: :$\forall z \in K: \exists x, y \in R: z = \dfrac x y, y \in R_+^*$ Now we show that $P$ satistfies conditions $(1)$ to $(4)$ of Positive Elements of Ordered Ring. From Addition of Division Products and Product of Division Products, it is clear that $P$ satisfies $(1)$ and $(3)$. {{qed|lemma}} Next we establish $(2)$. Let $z \in P \cap \paren {-P}$. Then $z \in P$ and $z \in \paren {-P}$. Thus: :$\exists x_1, x_2 \in D_+, y_1, y_2 \in D_+^*: z = x_1 / y_1, -z = x_2 / y_2$ So: :$x_1 / y_1 = -\paren {x_2 / y_2} \implies x_1 \circ y_2 = -\paren {x_2 \circ y_1}$ But $0 \le x_1 \circ y_2$ and $-\paren {x_2 \circ y_1} \le 0$. So $x_1 \circ y_2 = 0$. As $0 < y_2$, this means $x = 0$ and therefore $z = 0$. Thus $(2)$ has been established. {{qed|lemma}} Now to show that $(4)$ holds. Let $z = x / y$ where $x \in D, y \in D_+$. Suppose $0 \le x$. Then $z \in P$. However, suppose $x < 0$. Then $0 < \paren {-x}$ so $-z = \paren {-x} / y \in P$. Thus: :$z = -\paren {-z} \in -P$ So: :$P \cup \paren {-P} = K$ So by Positive Elements of Ordered Ring, the relation $\le'$ on $K$ defined by $P$ is a total ordering on $K$ compatible with its ring structure. {{qed|lemma}} Now we need to show that the ordering induced on $D$ by $\le'$ is indeed $\le$. Let $z \in D_+$. Then $z = z / 1_D \in P$, as $1_D \in D_+^*$. Thus $D_+ \subseteq P$ and $D_+ \subseteq D$ so $D_+ \subseteq D \cap P$ from Intersection is Largest Subset. Conversely, let $z \in D \cap P$. Then: :$\exists x \in D_+, y \in D_+^*: z = x / y$ If $x = 0$ then $z = 0$, and if $0 < x$ then as $z \circ y = x$ and $0 < y$, it follows that $0 < z$ by item 1 of Properties of a Totally Ordered Ring. So $\forall z \in D: 0 \le z \iff z \in P$. Thus it follows that $z \in D \cap P \implies z \in D_+$, i.e. $D \cap P \subseteq D_+$. Thus $D_+ = D \cap P$. By item $(2)$ of Properties of Ordered Ring, we have: :$x \le y \iff 0 \le y + \left({-x}\right)$ and thus it follows that the ordering induced on $D$ by $\le'$ is $\le$. {{qed|lemma}} Now we need to show uniqueness. Let $\preceq$ be a total ordering on $K$ which is compatible with its ring structure and induces on $D$ the ordering $\le$. Let $Q = \set {z \in K: 0 \preceq z}$. We now show that $Q = P$. Let $x \in D_+, y \in D_+^*$. Then $0 \preceq x$ and $0 \prec 1 / y$ from item 4 of Properties of a Totally Ordered Ring. Thus by compatibility with ring structure, $0 \preceq x / y$. Hence $P \subseteq Q$. Conversely, let $z \in Q$. Let $z = x / y$ where $x \in D, y \in D_+^*$. Then $x = z \circ y$ and by compatibility with ring structure, $0 \preceq x$. Thus $0 \le x$ and hence $z \in P$, and so $Q \subseteq P$. So $Q = P$. Therefore, by item $(2)$ of Properties of Ordered Ring, it follows that $\preceq$ is the same as $\le$. {{Qed}} {{improve|Put this in the context of Ordered Integral Domain.}}
By Field of Quotients is Unique, all we need to show is: :$\forall x_1, x_2 \in R, y_1, y_2 \in R_{> 0}: \dfrac {x_1} {y_1} \le \dfrac {x_2} {y_2} \iff \dfrac {\map \phi {x_1}} {\map \phi {y_1} } \le \dfrac {\map \phi {x_2} } {\map \phi {y_2} }$ Let $x_1 / y_1 \le x_2 / y_2$, where $y_1, y_2 \in R_{> 0}$. As $y_1, y_2 \in R_{> 0}$, it follows that $0 < y_1 \circ_1 y_2$ and $0 < 1 / \paren {y_1 \circ_1 y_2}$. We also have: :$0 < \map \phi {y_1} \circ_2 \map \phi {y_2} = \map \phi {y_1 \circ_1 y_2}$ Therefore: :$x_1 \circ_1 y_2 = \frac {x_1} {y_1} \circ_1 \paren {y_1 \circ y_2} \le \dfrac {x_2} {y_2} \circ_1 \paren {y_1 \circ_1 y_2} = x_2 \circ_1 y_1$ Conversely, let $x_1 \circ_1 y_2 \le x_2 \circ_1 y_1$. Then: :$\dfrac {x_1} {y_1} = x_1 \circ_1 y_2 \circ_1 \paren {\dfrac 1 {y_1 \circ_1 y_2} } \le x_2 \circ_1 y_1 \circ_1 \paren {\dfrac 1 {y_1 \circ_1 y_2} } = \dfrac {x_2} {y_2}$ That is, we have: :$\dfrac {x_1} {y_1} \le \dfrac {x_2} {y_2} \iff x_1 \circ_1 y_2 \le x_2 \circ_1 y_1$ Similarly: :$\dfrac {\map \phi {x_1} } {\map \phi {y_1} } \le \dfrac {\map \phi {x_2} } {\map \phi {y_2} } \iff \map \phi {x_1} \circ_2 \map \phi {y_2} \le \map \phi {x_2} \circ_2 \map \phi {y_1}$ Now $\phi: R \to S$ is an order embedding. Therefore: :$x_1 \circ_1 y_2 \le x_2 \circ_1 y_1 \iff \map \phi {x_1 \circ_1 y_2} \le \map \phi {x_2 \circ_1 y_1}$ The result follows. {{qed}}
Let $z$ be cancellable and $x \prec y$. Then by the definition of ordered semigroup: : $x \circ z \preceq y \circ z$ From the fact that $z$ is cancellable: : $x \circ z = y \circ z \iff x = y$ Thus as $x \circ z \ne y \circ z$ it follows from Strictly Precedes is Strict Ordering that: : $x \circ z \prec y \circ z$ Similarly, $z \circ x \prec z \circ y$ follows from $z \circ x \preceq z \circ y$. {{qed}}
=== Necessary Condition === {{begin-eqn}} {{eqn | l = x       | o = \prec       | r = y       | c =  }} {{eqn | ll= \implies       | l = e       | r = x^{-1} \circ x \prec x^{-1} \circ y       | c = Strict Ordering Preserved under Product with Cancellable Element }} {{eqn | ll= \implies       | l = y^{-1}       | r = e \circ y^{-1} \prec x^{-1} \circ y \circ y^{-1} = x^{-1}       | c =  }} {{eqn | ll= \implies       | l = y^{-1}       | o = \prec       | r = x^{-1}       | c =  }} {{end-eqn}} {{qed|lemma}} === Sufficient Condition === {{begin-eqn}} {{eqn | l = y^{-1}       | o = \prec       | r = x^{-1}       | c =  }} {{eqn | ll= \implies       | l = x       | r = \left({x^{-1} }\right)^{-1} \prec \left({y^{-1} }\right)^{-1} = y       | c =  }} {{eqn | ll= \implies       | l = x       | o = \prec       | r = y       | c =  }} {{end-eqn}} {{qed}}
Since $\left({S, \circ, \preceq}\right)$ is an ordered group, we have, for all $x,y,z \in S$: :$(R): \quad x \prec y \implies x \circ z \prec y \circ z$ :$(L): \quad x \prec y \implies z \circ x \prec z \circ y$ Using $(R)$ and $(L)$, let us demonstrate the equivalences. === $(1) \iff (2)$ and $(1) \iff (5)$ === We have by applying $(R)$ twice: {{begin-eqn}} {{eqn|l = x      |o = \prec      |r = y }} {{eqn|ll= \implies      |l = x \circ z      |o = \prec      |r = y \circ z }} {{eqn|ll= \implies      |l = x \circ z \circ z^{-1}      |o = \prec      |r = y \circ z \circ z^{-1} }} {{end-eqn}} This last identity comes down to $x \prec y$ by definition of inverse. Selecting $z = x^{-1}$ proves $(1) \iff (5)$ as well. {{qed|lemma}} === $(1) \iff (3)$ and $(1) \iff (6)$ === We have by applying $(L)$ twice: {{begin-eqn}} {{eqn|l = x      |o = \prec      |r = y }} {{eqn|ll= \implies      |l = z \circ x      |o = \prec      |r = z \circ y }} {{eqn|ll= \implies      |l = z^{-1} \circ z \circ x      |o = \prec      |r = z^{-1} \circ z \circ x }} {{end-eqn}} This last identity comes down to $x \prec y$ by definition of inverse. Selecting $z = x^{-1}$ proves $(1) \iff (6)$ as well. {{qed|lemma}} === $(4) \iff (5)$ === Applying that we know $(1) \iff (2)$, we obtain: {{begin-eqn}} {{eqn|l = y^{-1}      |o = \prec      |r = x^{-1} }} {{eqn|ll= \iff      |l = e      |o = \prec      |r = y \circ x^{-1} }} {{end-eqn}} since $y \circ y^{-1} = e$. {{qed|lemma}} The remaining equivalences follow from Material Equivalence is Equivalence. {{qed}}
This follows: : $(1): \quad$ As a direct consequence of Mapping from Totally Ordered Set is Order Embedding iff Strictly Increasing : $(2): \quad$ From the definition of monomorphism as a homomorphism which is an injection. {{Qed}}
First note that as $\left({S, \circ, \preceq}\right)$ is well-ordered and hence totally ordered, the Trichotomy Law applies. Thus: {{begin-eqn}} {{eqn | l = \forall m \in S:       | o =        | r = m \notin n^\prec       | c =  }} {{eqn | o = \iff       | r = \neg \ m \prec n       | c = Definition of Strict Lower Closure }} {{eqn | o = \iff       | r = m = n \lor n \prec m       | c = Trichotomy Law }} {{eqn | o = \iff       | r = n \preceq m       | c = Definition of Strictly Precede }} {{end-eqn}} So: {{begin-eqn}} {{eqn | l = \forall p \in S:       | o =        | r = p \notin \left({n \circ 1}\right)^\prec       | c =  }} {{eqn | o = \iff       | r = n \circ 1 \preceq p       | c = from the above }} {{eqn | o = \iff       | r = n \prec p       | c = Sum with One is Immediate Successor in Naturally Ordered Semigroup }} {{end-eqn}} Similarly: {{begin-eqn}} {{eqn | l = \forall p \in S:       | o =        | r = p \notin n^\prec \cup \left\{ {n}\right\}       | c =  }} {{eqn | o = \iff       | r = n \preceq p \land n \ne p       | c = from the above }} {{eqn | o = \iff       | r = n \prec p       | c = Definition of Strictly Precede }} {{end-eqn}} So: :$p \notin n^\prec \cup \left\{{n}\right\} \iff p \notin \left({n \circ 1}\right)^\prec$ Thus: :$\complement_S \left({\left({n \circ 1}\right)^\prec}\right) = \complement_S \left({n^\prec \cup \left\{{n}\right\}}\right)$ from the definition of relative complement. So: {{begin-eqn}} {{eqn | l = \left({n \circ 1}\right)^\prec       | r = \complement_S \left({\complement_S \left({\left({n \circ 1}\right)^\prec}\right)}\right)       | c = Relative Complement of Relative Complement }} {{eqn | r = \complement_S \left({\complement_S \left({n^\prec \cup \left\{ {n}\right\} }\right)}\right)       | c = from above }} {{eqn | r = n^\prec \cup \left\{ {n}\right\}       | c = Relative Complement of Relative Complement }} {{end-eqn}} {{qed}}
Let $m \preceq n$. Then: {{begin-eqn}} {{eqn | o =        | r = x \in \left[{m \,.\,.\, n \circ 1}\right]       | c =  }} {{eqn | o = \iff       | r = m \preceq x \land x \preceq \left({n \circ 1}\right)       | c = Definition of Closed Interval }} {{eqn | o = \iff       | r = m \preceq x \land \left({x \prec n \circ 1 \lor x = n \circ 1}\right)       | c = Definition of Strictly Precedes }} {{end-eqn}} {{begin-eqn}} {{eqn | o =        | r = x \in \left[{m \,.\,.\, n}\right] \cup \left\{ {n \circ 1}\right\}       | c =  }} {{eqn | o = \iff       | r = m \preceq x \land \left({x \preceq n \lor x = n \circ 1}\right)       | c = Definitions of Closed Interval and Union }} {{eqn | o = \iff       | r = m \preceq x \land \left({x \prec n \lor x = n \lor x = n \circ 1}\right)       | c = Definition of Strictly Precedes }} {{eqn | o = \iff       | r = m \preceq x \land \left({x \prec n \circ 1 \lor x = n \circ 1}\right)       | c = Definition of Strictly Precedes }} {{end-eqn}} Thus: :$\left[{m \,.\,.\, n \circ 1}\right] = \left[{m \,.\,.\, n}\right] \cup \left\{{n \circ 1}\right\}$ {{qed}}
Let $T' = T \cup \left\{{x \in S: q \prec x}\right\}$. Then $T'$ satisfies the conditions of the Principle of Mathematical Induction. Therefore: :$T' = \left\{{x \in S: p \preceq x}\right\}$ Therefore: :$T = \left[{p \,.\,.\, q}\right]$ {{Qed}}
=== Proof that Mapping is Isomorphism === Let $T' = \Cdm g$, that is, the codomain of $g$. By Zero is Identity in Naturally Ordered Semigroup, $0'$ is the identity for $\circ'$. Thus: :$\map g 0 = \circ'^0 1' = 0'$ and so $0' \in T'$ Suppose $x' \in T'$. Then: :$\map g n = x'$ and so: {{begin-eqn}} {{eqn | l = x' \circ' 1'       | r = \circ'^n 1' \circ' 1'       | c =  }} {{eqn | r = \paren {\circ'^n 1'} \circ' \paren {\circ'^1 1'}       | c =  }} {{eqn | r = \circ'^{\paren {n \circ 1} } 1'       | c =  }} {{eqn | r = \map g {n \circ 1}       | c =  }} {{end-eqn}} So: :$x' \in T' \implies x' \circ' 1' \in T'$ Thus, by the Principle of Mathematical Induction applied to $S'$: :$T' = S'$ So: :$\forall a' \in S': \exists a \in S: \map g a = a'$ and so $g$ is surjective by definition. From Index Laws for Semigroup: Sum of Indices: :$\map g {a \circ b} = \map g a \circ' \map g b$ and therefore $g$ is a homomorphism from $\struct {S, \circ}$ to $\struct {S', \circ'}$. Now: {{begin-eqn}} {{eqn | ll= \forall p \in S:       | l = \circ'^p 1'       | o = \prec'       | r = \circ'^p 1' \circ' 1'       | c = Sum with One is Immediate Successor in Naturally Ordered Semigroup }} {{eqn | r = \paren {\circ'^p 1'} \circ' \paren {\circ'^1 1'}       | c =  }} {{eqn | r = \circ'^{\paren {p \circ 1} } 1'       | c = Index Laws for Semigroup: Sum of Indices }} {{eqn | ll= \leadsto       | l = \circ'^p 1'       | o = \prec'       | r = \circ'^{\paren {p \circ 1} } 1'       | c =  }} {{end-eqn}} For $p \in S$, let $S_p$ be the initial segment of $S$: :$S_p = \set {x \in S: x \prec p}$ Let: :$T = \set {p \in S: \forall a \in S_p: \circ'^a 1' \prec' \circ'^p 1'}$ Now $S_0 = \O \implies 0 \in T$. Suppose $p \in T$. Then: :$a \prec p \circ 1 \implies a \preceq p$ By Strict Lower Closure of Sum with One, either of these is the case: :$(1): \quad a \prec p: p \in T \implies \circ'^a 1' \prec' \circ'^p 1' \prec' \circ'^{\paren {p \circ 1} } 1'$ :$(2): \quad a = p: \circ'^a 1' = \circ'^p 1' \prec' \circ'^{\paren {p \circ 1} } 1'$ In either case, we have: :$p \in T \implies p \circ 1 \in T$, and by thePrinciple of Mathematical Induction: :$T = S$ So $n \prec p \implies \circ'^n 1' \prec' \circ'^p 1'$. Thus $g$ is a surjective monomorphism and therefore is an isomorphism from $\struct {S, \circ, \preceq}$ to $\struct {S', \circ', \preceq'}$. {{Qed}} === Proof that Isomorphism is Unique === Now we need to show that the isomorphism $g$ is unique. Let $f: S \to S'$ be another isomorphism different from $g$. {{AimForCont}} $\map f 1 \ne 1'$. We show by induction that $1' \notin \Cdm f$. {{finish}} ... Thus $1' \notin \Cdm f$ which is a contradiction. Thus $\map f 1 = 1$ and it follows {{finish}} that $f = g$. Thus the isomorphism $g$ is unique. {{Qed}}
The result follows as a direct application of Strict Lower Closure of Sum with One. {{qed}} Category:Natural Numbers ebnlj9h3zjmkiqhic6nj6grj7lm6oxj
Consider the natural numbers $\N$ defined as the naturally ordered semigroup $\struct {S, \circ, \preceq}$. From its definition, $\struct {S, \circ, \preceq}$ is well-ordered by $\preceq$. The result follows. As $\N_{\ne 0} = \N \setminus \set 0$, by Set Difference is Subset $\N_{\ne 0} \subseteq \N$. As $\N$ is well-ordered, by definition, every subset of $\N$ has a smallest element. {{qed}}
Let $S$ be a non-empty subset of the set of natural numbers $\N$. We take as axiomatic that $\N$ is itself a subset of the set of real numbers $\R$. Thus $S \subseteq \R$. By definition: :$\forall n \in \N: n \ge 0$ and so: :$\forall n \in S: n \ge 0$ Hence $0$ is a lower bound of $S$. This establishes the fact that $S$ is bounded below. By the Continuum Property, we have that $S$ admits an infimum. Hence let $b = \inf S$. Because $b$ is the infimum of $S$, it follows that $b + 1$ is not a lower bound of $S$. So, for some $n \in S$: :$n < b + 1$ {{AimForCont}} $n$ is not the smallest element of $S$. Then there exists $m \in S$ such that: :$b \le m < n < b + 1$ from which it follows that: :$0 < n - m < 1$ But there exist no natural numbers $k$ such that $0 < k < 1$. Hence $n = b$ is the smallest element of $S$. {{qed}}
By Set Equivalence is Equivalence Relation, we have that: : $m = n \implies \N_m \sim \N_n$ It remains to show that: : $m \ne n \implies \N_m \nsim \N_n$. Since the naturals are totally ordered, it will be sufficient to show that: : $m \in \N_n \implies \N_m \nsim \N_n$ Let $S = \left\{{n \in \N: \forall m \in \N_n: \N_m \nsim \N_n}\right\}$. That is, $S$ is the set of all the natural numbers $n$ such that $\N_m \nsim \N_n$ for all $m \in \N_n$. We use mathematical induction to prove that $S = \N$. From Initial Segment of Natural Numbers determined by Zero is Empty: :$\N_0 = \varnothing$ Thus $0 \in S$. This is the basis for the induction. Now, assume the induction hypothesis that $n \in S$. We now complete the induction step, that is, to show that $n + 1 \in S$. Let $m \in \N_{n+1}$. If $m = 0$, then $\N_m \nsim \N_{n+1}$ because $\N_0 = \varnothing$ and $\N_{n+1} \ne \varnothing$. Aiming for a contradiction, suppose that $m \ge 1$ and $\N_m \sim \N_{n+1}$. Then, by Set Equivalence Less One Element, that means $\N_{m-1} \sim \N_n$. But then $m - 1 \in \N_n$ which contradicts the induction hypothesis that $n \in S$. Thus $n + 1 \in S$. The result follows from the fact that Set Equivalence is Equivalence Relation, in particular the symmetry clause. {{qed}}
The result follows directly from Equality of Natural Numbers and the fact that set equivalence is transitive. {{qed}}
Zero is defined as the cardinal of the empty set. The result follows from Finite Cardinals and Ordinals are Equivalent.
Let $A \subseteq B$. There are two cases: $(1): \quad A \ne B$ In this case: :$A \subsetneqq B$ and from Cardinality of Proper Subset of Finite Set: :$\card A < n$ $(2): \quad A = B$ In this case: :$\card A = \card B$ and so: :$\card A = n$ In both cases: :$\card A \le n$ Hence the result. {{qed}}