hcju/proofwikiqa huggingface dataset:
=== Necessary Condition === Suppose $y \in K_{II}$. $x \ne 0$ by Cofinal to Zero iff Ordinal is Zero. If $x = z^+$ for some $z$, then $z = \bigcup x$ by Union of Successor Ordinal. Therefore, $z$ would be the least upper bound of $x$. Since $\operatorname{cof} \left({ x,y }\right)$, it follows that $\forall a \in x: \exists b \in y: f\left({ b }\right) \ge a$ by the definition of cofinal. So $f\left({b}\right) \ge z$ for some $b \in y$. But since $y$ is a limit ordinal, $b^+ \in y$ by Successor in Limit Ordinal. Therefore, $f\left({b^+}\right) \in x$ and $f\left({b^+}\right) > f\left({b}\right) \ge z$, which contradicts the fact that $z$ is an upper bound of $x$. Therefore, $x \ne z^+$ for any $z$ and $x \in K_{II}$. {{qed|lemma}} === Sufficient Condition === Suppose $x \in K_{II}$. $y \ne 0$ by Cofinal to Zero iff Ordinal is Zero. Moreover, if $y = z^+$ for some ordinal $z$, then $z = \bigcup y$ by Union of Successor Ordinal. Therefore, $z$ would be the least upper bound of $y$. Since $\operatorname{cof} \left({ x,y }\right)$, it follows that $\forall a \in x: \exists b \in y: f\left({ b }\right) \ge a$ by the definition of cofinal. But $z \in y$, so $f\left({z}\right) \in x$ by the definition of $f$. Therefore, $f\left({z}\right)^+ \in x$ by the fact that $x$ is a limit ordinal. This means that $\exists b \in y: f\left({b}\right) \ge f\left({z}\right)^+$ This would mean that $f\left({b}\right) > f\left({z}\right)$ by Ordinal is Less than Successor. But this means that $b \ge z$ since otherwise, $f\left({b}\right) < f\left({ z }\right)$ by the definition of strictly increasing. Furthermore, $b \ne z$, since otherwise, $f\left({b}\right) = f\left({z}\right)$ by Substitutivity of Equality. It follows that $b > z$. This contradicts the fact that $z$ is an upper bound of $y$. Therefore, $y \ne z^+$ for any $z$ and $y \in K_{II}$. {{qed}}
Let $D$ be the subset of $\Z_{>0}$ defined as: :$D = \set {a s + b t: s, t \in \Z, a s + b t > 0}$ Setting $s = 1$ and $t = 0$ it is clear that $a = \paren {a \times 1 + b \times 0} \in D$. So $D \ne \O$. By Set of Integers Bounded Below by Integer has Smallest Element, $D$ has a smallest element $d$, say. Thus $d = a s + b t$ for some $s, t \in \Z$. === Proof of $(1)$ === From the Division Theorem we can write $a = q d + r$ for some $q, r$ with $0 \le r < d$. If $r \ne 0$ we have: {{begin-eqn}} {{eqn | l = r       | r = a - q d       | c =  }} {{eqn | r = a - q \paren {a s + b t}       | c =  }} {{eqn | r = a \paren {1 - q s} + b \paren {-q t}       | c =  }} {{end-eqn}} So by definition of $D$, it is clear that $r \in D$. But $r < d$ which contradicts the stipulation that $d$ is the smallest element of $D$. Thus $r = 0$ and so $a = q d$. That is $d \divides a$. By the same argument, it follows that $d \divides b$ also. {{qed|lemma}} === Proof of $(2)$ === Let $c \divides a$ and $c \divides b$. Then: : $\exists u, v \in \Z: a = c u, b = c v$ Thus: {{begin-eqn}} {{eqn | l = d       | r = a s + b t       | c =  }} {{eqn | r = c u s + c v t       | c =  }} {{eqn | r = c \paren {u s + v t}       | c =  }} {{end-eqn}} demonstrating that $c \divides d$. {{qed}}
From Bézout's Lemma we have: Let $a, b \in \Z$ such that $a$ and $b$ are not both zero. Let $\gcd \set {a, b}$ be the greatest common divisor of $a$ and $b$. Then: :$\exists x, y \in \Z: a x + b y = \gcd \set {a, b}$ Furthermore, $\gcd \set {a, b}$ is the smallest positive integer combination of $a$ and $b$. In this instance, $a, b \in \Z_{>0}$ and are therefore both non-zero. The result then follows by definition of greatest common divisor: $d = \gcd \set {a, b}$ {{iff}}: : $(1): \quad d \divides a \land d \divides b$ : $(2): \quad c \divides a \land c \divides b \implies c \divides d$ {{qed}}
=== Necessary Condition === {{:Matrix is Invertible iff Determinant has Multiplicative Inverse/Necessary Condition}} === Sufficient Condition === {{:Matrix is Invertible iff Determinant has Multiplicative Inverse/Sufficient Condition}}
Let $G_1$ and $G_2$ be prime groups, both of finite order $p$. From Prime Group is Cyclic, both $G_1$ and $G_2$ are cyclic. The result follows directly from Cyclic Groups of Same Order are Isomorphic. {{Qed}}
Let $G_1$ and $G_2$ be infinite cyclic groups. From Infinite Cyclic Group is Isomorphic to Integers we have: :$G_1 \cong \left({\Z, +}\right) \cong G_2$ where $\left({\Z, +}\right)$ is the additive group of integers. From Isomorphism is Equivalence Relation it follows that: :$G_1 \cong  G_2$ {{Qed}}
Let $H = \bigcap \mathbb S$. Let $H_k$ be any element of $\mathbb S$. Then: {{begin-eqn}} {{eqn | l = a, b       | o = \in       | r = H       | c =  }} {{eqn | ll= \leadsto       | lo= \forall k:       | l = a, b       | o = \in       | r = H_k       | c = {{Defof|Intersection of Set of Sets}} }} {{eqn | ll= \leadsto       | lo= \forall k:       | l = a \circ b^{-1}       | o = \in       | r = H_k       | c = Group properties }} {{eqn | ll= \leadsto       | l = a \circ b^{-1}       | o = \in       | r = H       | c = {{Defof|Intersection of Set of Sets}} }} {{eqn | ll= \leadsto       | l = H       | o = \le       | r = G       | c = One-Step Subgroup Test }} {{end-eqn}} {{qed|lemma}} Now to show that $\struct {H, \circ}$ is the largest such subgroup. Let $K$ be a subgroup of $\struct {G, \circ}$ such that: :$\forall S \in \mathbb S: K \subseteq S$ Then by definition $K \subseteq H$. Let $x, y \in K$. Then: :$x \circ y^{-1} \in K \implies x \circ y^{-1} \in H$ Thus any subgroup of all elements of $\mathbb S$ is also a subgroup of $H$ and so no larger than $H$. Thus $H = \bigcap \mathbb S$ is the largest subgroup of $S$ contained in each element of $\mathbb S$. {{qed}}
From the definition of join, $H \vee K$ is the smallest subgroup of $G$ containing $H \cup K$. The result follows from Union of Subgroups. {{qed}}
From Powers of Element form Subgroup, $H = \set {a^n: n \in \Z}$ is a subgroup of $G$. Let $K \le G: a \in K$. Then $\forall n \in \Z: a^n \in K$. Thus, $H \subseteq K$. {{qed}}
Let $T = \bigcap \mathbb S$. Let $T_k$ be any element of $\mathbb S$. Then: {{begin-eqn}} {{eqn | o =        | r = a, b \in T       | c =  }} {{eqn | o = \implies       | r = \forall k: a, b \in T_k       | c = {{Defof|Set Intersection}} }} {{eqn | o = \implies       | r = \forall k: a \circ b \in T_k       | c = Subsemigroups are closed }} {{eqn | o = \implies       | r = a \circ b \in T       | c = {{Defof|Set Intersection}} }} {{end-eqn}} So $\left({T, \circ}\right)$ is a subsemigroup of $\left({S, \circ}\right)$. Now to show that $\left({T, \circ}\right)$ is the largest such subsemigroup. Let $x, y \in T$. Then $\forall K \subseteq T: x \circ y \in K \implies x \circ y \in T$. Thus any $K \in \mathbb S: K \subseteq T$ and thus $T$ is the largest subsemigroup of $S$ contained in each member of $\mathbb S$. {{qed}}
By definition, the set of words in $H \cup K$ is: :$S = \map W {H \cup K} := \set {s_1 \circ s_2 \circ \cdots \circ s_n: n \in \N_{>0}: s_i \in H \cup K 1 \le i \le n}$ Let $h \in H$. Then setting $n = 1$ in the above definition and letting $s_1 = h$ it follows that $H \subseteq S$. Similarly it is seen that $K \subseteq S$ So from Union is Smallest Superset it follows that $H \cup K \subseteq S$. From Set of Words Generates Group it follows that $S$ is the subgroup of $G$ generated by $H \cup K$. {{qed}}
From Order of Subgroup Product: : $(1): \quad \size {H K} = \dfrac {\order H \order K} {\order {H \cap K} }$ From Subset Product is Subset of Generator, we have that: : $H K \subseteq H \vee K$ where $H K$ is the subset product of $H$ and $K$. Thus: : $(2): \quad \size {H \vee K} \ge \size {H K}$ The result follows by substituting for $\size {H K}$ from $(1)$ into $(2)$. {{qed}}
By Abelian Group Factored by Prime, the subgroup $H_2$ defined as: :$H_2 := \set {g \in G: g^2 = e}$ has precisely two elements. One of them has to be $e$, since $e^2 = e$. The result follows. {{qed}}
There are two possibilities for any $h \in H$. :$(1): \quad \operatorname{Im}^{-1} \left({h}\right) = \varnothing$ :$(2): \quad \operatorname{Im}^{-1} \left({h}\right) \ne \varnothing$ If $(1)$, then the first disjunct of the result is satisfied. Now suppose $(2)$ holds. Let $e_G$ and $e_H$ be the identity elements of $G$ and $H$ respectively. Let $K = \ker \left({\phi}\right)$. Let $x, y \in G$ such that $\phi \left({x}\right) = \phi \left({y}\right)$. Then: {{begin-eqn}} {{eqn | l = e_H       | r = \phi \left({x}\right) * \left({\phi \left({y}\right)}\right)^{-1}       | c = Definition of Inverse Element }} {{eqn | r = \phi \left({x}\right) * \phi \left({y^{-1} }\right)       | c = Homomorphism to Group Preserves Inverses }} {{eqn | r = \phi \left({x \circ y^{-1} }\right)       | c = $\phi$ is a homomorphism }} {{eqn | ll=\implies       | l = x \circ y^{-1}       | o = \in       | r = K       | c = Definition of Kernel }} {{eqn | ll= \implies       | l = x       | o = \in       | r = y K       | c = Element in Left Coset iff Product with Inverse in Subgroup }} {{end-eqn}} Thus: :$\left\{{x: \phi \left({x}\right) = \phi \left({y}\right)}\right\}$ is a subset of $y K$. From Kernel is Normal Subgroup of Domain we have that $y K = K y$. Now suppose $x \in K y$. Then, by definition, $x = k y$ for some $k \in K$. Thus: {{begin-eqn}} {{eqn | l = x       | r = k y       | c =  }} {{eqn | ll= \implies       | l = \phi \left({x}\right)       | r = \phi \left({k}\right) * \phi \left({y}\right)       | c = $\phi$ is a homomorphism }} {{eqn | r = e_H * \phi \left({y}\right)       | c = Definition of Kernel: $k \in K$ }} {{eqn | r = \phi \left({y}\right)       | c = Definition of Identity Element }} {{end-eqn}} A similar process gives that $x \in y K \implies \phi \left({x}\right) = \phi \left({y}\right)$. Hence the result. {{qed}}
=== Proof for Morphisms-Only Category Theory === {{finish}} === Proof for Object Category Theory === The seven axioms are: {{begin-eqn}} {{eqn|l = \operatorname{dom} \operatorname{id}_A = A      |o = \qquad      |r = \operatorname{cod} \operatorname{id}_A = A }} {{eqn|l = f \circ 1_{\operatorname{dom} f} = f      |o =       |r = 1_{\operatorname{cod} f} \circ f = f }} {{eqn|l = \operatorname{dom} \left({g \circ f}\right) = \operatorname{dom} f      |o =       |r = \operatorname{cod} \left({g \circ f}\right) = \operatorname{cod} g }} {{eqn|l = h \circ \left({g \circ f}\right)      |r = \left({h \circ g}\right) \circ f }} {{end-eqn}} Their duals are: {{begin-eqn}} {{eqn|l = \operatorname{cod} \operatorname{id}_A = A      |o = \qquad      |r = \operatorname{dom} \operatorname{id}_A = A }} {{eqn|l = 1_{\operatorname{cod} f} \circ f = f      |o =       |r = f \circ 1_{\operatorname{dom} f} = f }} {{eqn|l = \operatorname{cod} \left({f \circ g}\right) = \operatorname{cod} f      |o =       |r = \operatorname{dom} \left({f \circ g}\right) = \operatorname{dom} g }} {{eqn|l = \left({f \circ g}\right) \circ h      |r = f \circ \left({g \circ h}\right) }} {{end-eqn}} It is seen that only names of the bound variables $f,g$ and $h$ have been changed at some places. Therefore, we conclude: :$\mathrm{CT}^* = \mathrm{CT}$ {{qed}}
The first statement is Kernel is Normal Subgroup of Domain: :The kernel of $\phi$ is a normal subgroup of the domain of $\phi$: ::$\map \ker \phi \lhd \Dom \phi$ The second statement is Quotient Group Epimorphism is Epimorphism: :The mapping $\phi: G \to G / N$, defined as: ::$\phi: G \to G / N: \map \phi x = x N$ :is a group epimorphism, and its kernel is $N$. {{qed}}
By definition, $\kappa_x: G \to G$ is a mapping defined as: :$\forall g \in G: \map {\kappa_x} g = x g x^{-1}$ Let $n \in N$. Then: {{begin-eqn}} {{eqn | l = \map {\kappa_x} n       | r = x n x^{-1}       | c =  }} {{eqn | o = \in       | r = N       | c = {{Defof|Normal Subgroup}} }} {{end-eqn}} {{Qed}}
Let $N' := \phi \left({N}\right)$. From Group Homomorphism Preserves Subgroups, $N'$ is a subgroup of $H$. It remains to show that $N'$ is normal in $H$. Let $h \in H$ be any arbitrary element of $H$. Let $n' \in N'$ be any arbitrary element of $N'$. Because $\phi$ is an epimorphism, it is by definition surjective. Therefore: : $\exists n \in N: \phi \left({n}\right) = n'$ : $\exists g \in G: \phi \left({g}\right) = h$ Then: {{begin-eqn}} {{eqn | l = g \circ n \circ g^{-1}       | o = \in       | r = N       | c = {{Defof|Normal Subgroup}} }} {{eqn | ll= \implies       | l = \phi \left({g \circ n \circ g^{-1} }\right)       | o = \in       | r = N'       | c = as $N'$ is the image of $N$ }} {{eqn | ll= \implies       | l = \phi \left({g}\right) * \phi \left({n}\right) * \phi \left({g^{-1} }\right)       | o = \in       | r = N'       | c = as $\phi$ has the morphism property }} {{eqn | ll= \implies       | l = \phi \left({g}\right) * \phi \left({n}\right) * \phi \left({g}\right)^{-1}       | o = \in       | r = N'       | c = Group Homomorphism Preserves Inverses }} {{eqn | ll= \implies       | l = h * n' * h^{-1}       | o = \in       | r = N'       | c = Definition of $g$ and $n$ }} {{end-eqn}} As $h$ and $n'$ were arbitrary, it follows that: :$\forall h \in H, n' \in N': h * n' * h^{-1} \in N'$ and so $N'$ is a normal subgroup of $H$. {{qed}}
Let $H$ be a subgroup of $G$ such that $K \subseteq H$. Consider the restriction of $\phi$ to $H$. By Group Homomorphism Preserves Subgroups, $\phi_{\restriction_H} \sqbrk H$ is a group. From Group Epimorphism Induces Bijection between Subgroups it follows that the First Isomorphism Theorem can be applied. Hence the result. {{qed}}
From Group Epimorphism Preserves Normal Subgroups, $\phi \left({N}\right)$ is normal in $G_2$. Let $N' := \phi \left({N}\right)$. From Quotient Group Epimorphism is Epimorphism, we construct the (group) epimorphism $q: G_2 \to \dfrac {G_2}{N'}$. Now consider the composite mapping $q \circ \phi: G \to \dfrac {G_2}{N'}$ defined as: :$\forall q \in G: q \circ \phi \left({g}\right) = q \left({\phi \left({g}\right)}\right)$ From Composite of Group Epimorphisms is Epimorphism, $q \circ \phi$ is an epimorphism. Let $g \in G$ such that: : $q \circ \phi \left({g}\right) = e_{G_2/N'}$ where $e_{G_2/N'}$ is the identity of $\dfrac {G_2}{N'}$. Then $\phi \left({g}\right) \in \ker \left({q}\right)$. But then by Group Epimorphism Induces Bijection between Subgroups it must be that $g \in N$. Hence from the First Isomorphism Theorem, $\dfrac {G_2}{N'} \cong \dfrac {G_1} N$. {{qed}}
From Kernel is Normal Subgroup of Domain we have that $K$ is a normal subgroup of $G$. Thus the Third Isomorphism Theorem can be applied directly. {{qed}}
Let $\order G$ be a prime number. Then from Prime Group is Cyclic $G$ has a generator $\gen g$ such that $\order g = p$. Now suppose $e \ne g \in G$. Let $\order g = n$. Let $p \divides n$. Then by Subgroup of Finite Cyclic Group is Determined by Order the cyclic group $\gen g$ has an element $g$ of order $p$. Suppose $p \nmid n$. From Subgroup of Abelian Group is Normal, $\gen g$ is normal in $G$. Consider the quotient group $G' = \dfrac G {\gen g}$. As $p \nmid n$, $\gen e \subsetneq \gen g \subsetneq G$. Thus $\order {G'} < \order G$. But we have that $p \divides \order G$. It follows by induction that $G'$ has an element $h'$ of order $p$. Let $h$ be a preimage of $h'$ under the quotient epimorphism $\phi: G \to G'$. Then: :$\paren {h'}^p = e'$ where $e'$ is the identity of $G'$. Thus $h^p \in \gen g$ so $\paren {h^p}^n = \paren {h^n}^p = e$. Thus either $h^n$ has order $p$ or $h^n = e$. If $h^n = e$ then $\paren {h'}^n = e'$. But since $p$ is the order of $h'$ it would follow that $p \divides n$, contrary to assumption. {{qed}}
This result follows as a special case of Group has Subgroups of All Prime Power Factors. {{qed}}
It just remains to verify that these definitions satisfy the premises for applying Characterization of Metacategory via Equations. The first of these is the assertion: :$\operatorname{dom} \operatorname{id}_A = A$ Here, $A \in \mathbf C'_0$, so by above definition there is some $x \in \mathbf C_1$ with $A = \operatorname{dom} x$. Then also $\operatorname{id}_A = \operatorname{dom} x$; hence it suffices to verify: :$\operatorname{dom} \operatorname{dom} x = \operatorname{dom} x$ From axiom $(MOCT4)$, we have $R_\circ \left({x, \operatorname{dom} x, x}\right)$. From axiom $(MOCT2)$, this implies $\operatorname{dom} x = \operatorname{dom} \operatorname{dom} x$. {{qed|lemma}} Now, to verify: :$\operatorname{cod} \operatorname{id}_A = A$ Reasoning as above, this comes down to proving: :$\operatorname{cod} \operatorname{dom} x = \operatorname{dom} x$ Axiom $(MOCT4)$ gives again that $R_\circ \left({x, \operatorname{dom} x, x}\right)$ holds. Axiom $(MOCT1)$ thus yields that $\operatorname{cod} \operatorname{dom} x = \operatorname{dom} x$, as desired. {{qed|lemma}} For the next two conditions, we have to verify: :$f \circ \operatorname{id}_{\operatorname{dom} f} = f$ and $\operatorname{id}_{\operatorname{cod} f} \circ f = f$ Thus, rephrasing in the language of the morphisms-only metacategory $\mathbf C$, this means: :$R_\circ \left({f, \operatorname{id}_{\operatorname{dom} f}, f}\right)$ and $R_\circ \left({\operatorname{id}_{\operatorname{cod} f}, f, f}\right)$ By definition of $\operatorname{id}$, the first of these is contained in axiom $(MOCT4)$. For the second one, it is first to be shown that $\operatorname{cod} f$ is a legitimate object of $\mathbf C'_0$. Axiom $(MOCT4)$ again yields us that $R_\circ \left({\operatorname{cod} f, f, f}\right)$, from which by axiom $(MOCT2)$ it follows that: :$\operatorname{cod} f = \operatorname{dom} \operatorname{cod} f$ Now by definition of $\operatorname{id}$ the second statement can be reduced to $R_\circ \left({\operatorname{cod} f, f, f}\right)$ which we already saw to be an axiom. {{qed|lemma}} The conditions: :$\operatorname{dom} \left({g \circ f}\right) = \operatorname{dom} f$ and $\operatorname{cod} \left({g \circ f}\right) = \operatorname{cod} g$ follow literally from axiom $(MOCT2)$ and the definition of $g \circ f$. {{qed|lemma}} The final associativity axiom: :$h \circ \left({g \circ f}\right) = \left({h \circ g}\right) \circ f$ is phrased by $(MOCT3)$. {{qed|lemma}} Having verified all seven conditions, we conclude that $\mathbf C'$ with the defined operations constitutes a metacategory. {{qed}}
In order to check that $\mathbf C_1$ is a morphisms-only metacategory, we need to interpret the symbols $\operatorname{dom}$, $\operatorname{cod}$ and $R_\circ$. Set $\operatorname{dom} x := \operatorname{id}_{\operatorname{dom} x}$, and $\operatorname{cod} x := \operatorname{id}_{\operatorname{cod} x}$ with the symbols on the right as defined in $\mathbf C$. Further, put $R_\circ \left({x, y, z}\right) \iff x \circ y = z$. Now we can proceed to verify the axioms for a morphisms-only metacategory. For $(MOCT0)$, i.e. (omitting universal quantifiers): :$\left({R_\circ \left({x, y, z}\right) \land R_\circ \left({x, y, z'}\right)}\right) \implies z = z'$ we conclude that the antecedent holds iff $z = x \circ y = z'$, hence $(MOCT0)$ holds. {{qed|lemma}} The content of $(MOCT1)$ is: :$\operatorname{dom} x = \operatorname{cod} y \iff \exists z: R_\circ \left({x, y, z}\right)$ This is precisely saying that a composite morphism $x \circ y$ is defined iff $x$ and $y$ are composable morphisms. {{qed|lemma}} Axiom $(MOCT2)$ is the statement: :$R_\circ \left({x, y, z}\right) \implies \left({\operatorname{dom} z = \operatorname{dom} y \land \operatorname{cod} z = \operatorname{cod} x}\right)$ which prescribes the domain and codomain of $x \circ y$. That this is satisfied by $\mathbf C_1$ follows directly from axiom $(C1)$ for metacategories. {{qed|lemma}} Axiom $(MOCT3)$ is: :$R_\circ \left({x, y, a}\right) \land R_\circ \left({y, z, b}\right) \implies \left({\exists w: R_\circ \left({x, b, w}\right) \land R_\circ \left({a, z, w}\right)}\right)$ which expresses that $\circ$ is an associative operation, exactly as axiom $(C3)$ for metacategories specifies. {{qed|lemma}} Finally, axiom $(MOCT4)$ is the statement: :$R_\circ \left({x, \operatorname{dom} x, x}\right) \land R_\circ \left({\operatorname{cod} x, x, x}\right)$ which will hold iff $x \circ \operatorname{id}_{\operatorname{dom} x} = x$ and $\operatorname{id}_{\operatorname{cod} x} \circ x = x$. This is the content of axiom $(C2)$ for metacategories. {{qed|lemma}} Having verified the axioms, we conclude $\mathbf C_1$ to be a morphisms-only metacategory. {{qed}}
=== Necessary Conditions === Let $Y$ be connected in $\struct {S, \tau}$. {{AimForCont}} $Y$ is not convex in $S$. Then there exist $a, b, c \in S$ such that: :$a \prec b \prec c$ :$a, c \in Y$ but $b \notin Y$ Recall that: :$b^\prec$ denotes the (strict) lower closure of $b$: $b^\prec = \set {u \in S: u \prec b}$ :$b^\succ$ denotes the (strict) upper closure of $b$: $b^\succ = \set {u \in S: b \prec u}$ We have that $Y \cap b^\prec$ and $Y \cap b^\succ$ are separated in $Y$. {{explain|We probably need a link to a result proving the above statement.}} Hence is disconnected. Hence by Proof by Contradiction it follows that $Y$ is convex in $S$. {{qed|lemma}} {{AimForCont}} $Y$ is convex in $S$, but not a linear continuum. Then by Order Topology on Convex Subset is Subspace Topology, the subspace topology on $Y$ is the same as the order topology on $Y$. Thus by Linearly Ordered Space is Connected iff Linear Continuum, $Y$ is disconnected. Hence by Proof by Contradiction it follows that $Y$ is a linear continuum. {{qed|lemma}} === Sufficient Condition === Suppose that $Y$ is convex in $S$ and a linear continuum. Then the result follows from: : Order Topology on Convex Subset is Subspace Topology : Linearly Ordered Space is Connected iff Linear Continuum. {{qed}}
=== Proof for Morphisms-Only Category Theory === From a formal perspective, if one would have derived (for some collection of statements $\Delta$): :$\Delta \vdash \Sigma$ without using the axioms, then because $R_\circ$, $\operatorname{dom}$ and $\operatorname{cod}$ would still be undefined, it follows that necessarily also: :$\Delta^* \vdash \Sigma^*$ In this correspondence, taking $\Delta$ to be $\mathrm{CT}$, it follows that: :$\mathrm{CT} \vdash \Sigma$ implies: :$\mathrm{CT}^* \vdash \Sigma^*$ As the Category Axioms are Self-Dual, in that $\mathrm{CT}^* = \mathrm{CT}$, we obtain the result. {{qed}} === Proof for Object Category Theory === {{ProofWanted}}
From Dual Category of Dual Category, any metacategory $\mathbf C$ may be regarded as a dual category. Thus it is sufficient to verify $\Sigma^*$ holds in all dual categories. In any dual category $\mathbf C^{\text{op}}$, we see that: :$\operatorname{dom} f^{\text{op}} = \left({\operatorname{cod} f}\right)^{\text{op}}$ :$\operatorname{cod} f^{\text{op}} = \left({\operatorname{dom} f}\right)^{\text{op}}$ :$f^\text{op} \circ g^\text{op} = \left({g \circ f}\right)^{\text{op}}$ Thus, the interpretation of $\Sigma^*$ in $\mathbf C^{\text{op}}$ comes down to the interpretation of $\Sigma$ in $\mathbf C$. That is, except for amending every symbol $\star$ of $\mathbf C$ with a superscript "$\text{op}$", yielding $\star^\text{op}$. This operation does not change the validity of $\Sigma$ in any possible way, and $\Sigma$ was assumed to hold in $\mathbf C$. Hence we conclude that $\Sigma^*$ must hold in $\mathbf C^{\text{op}}$ as well. The result follows. {{qed}}
By the definition of a complex number, let $z = a + i b$ where $a$ and $b$ are real numbers. Then: {{begin-eqn}} {{eqn | l = z \overline z       | r = \paren {a + i b} \paren {a - i b}       | c = {{Defof|Complex Conjugate}} }} {{eqn | r = a^2 + a \cdot i b + a \cdot \paren {-i b} + i \cdot \paren {-i} \cdot b^2       | c = {{Defof|Complex Multiplication}} }} {{eqn | r = a^2 + i a b - i a b + b^2       | c =  }} {{eqn | r = a^2 + b^2       | c =  }} {{eqn | r = \paren {\sqrt {a^2 + b^2} }^2       | c =  }} {{eqn | r = \cmod z^2       | c = {{Defof|Complex Modulus}} }} {{end-eqn}} As $a^2 + b^2$ is wholly real, the result follows. {{qed}}