hcju/proofwikiqa huggingface dataset:
{{MissingLinks|among general reference to results, maybe also some links to details on this "method of variations" approach}} Put $y = g \left({z}\right)$. Let $\delta z \in \C \setminus \left\{ {0} \right\}$. Put $\delta y = g \left({z + \delta z}\right) - y$, so: :$g \left({z + \delta z}\right) = y + \delta y$ As $\delta z \to 0$, we have: :$(1): \quad \delta y \to 0$ by definition of continuity, as $g$ is continuous. :$(2): \quad \dfrac {\delta y} {\delta z} \to g'\left({z}\right)$ by definition of complex-differentiability. There are two cases to consider: === Case 1 === Suppose $g' \left({z}\right) \ne 0$. When $\delta z \to 0$, we have $\delta y \ne 0$ from $(2)$, if $\left\vert{\delta z}\right\vert$ is sufficiently small. Then: {{begin-eqn}} {{eqn | l = \lim_{\delta z \mathop \to 0} \frac {h \left({z + \delta z}\right) - h \left({z}\right)} {\delta z}       | r = \lim_{\delta z \mathop \to 0} \frac {f \left({g \left({z + \delta z}\right)}\right) - f \left({g \left({z}\right)}\right)} {g \left({z + \delta z}\right) - g \left({z}\right)} \frac {g \left({z + \delta z}\right) - g \left({z}\right)} {\delta z}       | c =  }} {{eqn | r = \lim_{\delta z \mathop \to 0} \frac {f \left({y + \delta y}\right) - f \left({y}\right)} {\delta y} \frac {\delta y} {\delta z}       | c =  }} {{eqn | r = f' \left({y}\right) g^{\prime} \left({z}\right)       | c = As $\delta y \to 0$, then $\dfrac {f \left({y + \delta y}\right) - f \left({y}\right)} {\delta y} \to f^{\prime} \left({y}\right)$ }} {{end-eqn}} hence the result. {{qed|lemma}} === Case 2 === Now suppose $g' \left({z}\right) = 0$. When $\delta z \to 0$, there are two possibilities: === Case 2a === If $\delta y = 0$, then: :$\dfrac {h \left({z + \delta z}\right) - h \left({z}\right)} {\delta z} = 0 = f' \left({y}\right) g' \left({z}\right)$ Hence the result. {{qed|lemma}} === Case 2b === If $\delta y \ne 0$, then: :$\dfrac {h \left({z + \delta z}\right) - h \left({z}\right)} {\delta z} = \dfrac {f \left({y + \delta y}\right) - f \left({y}\right)} {\delta y} \dfrac {\delta y} {\delta z}$ As $\delta y \to 0$: : $(1): \quad \dfrac {f \left({y + \delta y}\right) - f \left({y}\right)} {\delta y} \to f' \left({y}\right)$ : $(2): \quad \dfrac {\delta y} {\delta z} \to 0$ Thus: : $\displaystyle \lim_{\delta z \mathop \to 0} \frac {h \left({z + \delta z}\right) - h \left({z}\right)} {\delta z} \to 0 = f' \left({y}\right) g' \left({z}\right)$ Again, hence the result. {{qed|lemma}} All cases have been covered, so by Proof by Cases, the result is complete. {{qed}}
$\phi$ is either bijective or strictly increasing. From Continuous Real Function on Closed Interval is Bijective iff Strictly Monotone, it follows that in both cases $\phi$ is both bijective and strictly monotone. As $\map \phi c < \map \phi d$, $\phi$ must be strictly increasing. As $\phi$ is strictly increasing, we have $c = \map {\phi^{-1} } {a_0} < \map {\phi^{-1} } {a_1} < \cdots < \map {\phi^{-1} } {a_n} = d$. Hence, $\map {\phi^{-1} } {a_0}, \map {\phi^{-1} } {a_1}, \dotsc, \map {\phi^{-1} } {a_n}$ form a subdivision of $\closedint a b$. Define $J_i := \closedint {\map {\phi^{-1} } {a_{i - 1} } } {\map {\phi^{-1} } {a_i} }$ as a closed real interval. From Derivative of Complex Composite Function, it follows that $\sigma$ restricted to $J_i$ is complex differentiable. Its derivative is given by: :$\map {\sigma' \restriction_{J_i} } t = \map {\gamma' \restriction_{I_i} } {\map \phi t} \map {\phi'} t$ From Derivative of Monotone Function, it follows that for all $t \in \closedint c d$, we have $\map {\phi'} t > 0$. As $\map {\gamma' \restriction_{I_i} } {\map \phi t} \ne 0$ by definition of contour, we have $\map {\sigma' \restriction_{J_i} } t \ne 0$. Then $\sigma$ fulfils both requirement $(1)$ and $(2)$ from the definition of contour, so $\sigma$ is a contour. {{qed}}
By definition of parameterization, $\gamma_i = \sigma_i \circ \phi_i$ for all $i \in \left\{ {1, \ldots, n}\right\}$. Here, $\phi_i: \left[{c_i \,.\,.\, d_i}\right] \to \left[{a_i \,.\,.\, b_i}\right]$ is a bijective differentiable strictly increasing function. Then: {{begin-eqn}} {{eqn | l = \int_C f \left({z}\right) \rd z        | r = \sum_{i \mathop = 1}^n \int_{a_i}^{b_i} f \left({\gamma_i \left({t}\right) }\right) \gamma_i' \left({t}\right) \rd t       | c = {{Defof|Complex Contour Integral}} }} {{eqn | r = \sum_{i \mathop = 1}^n \int_{\phi_i^{-1} \left({a_i}\right) }^{\phi_i^{-1} \left({b_i}\right)} f \left({\gamma_i \left({\phi_i \left({u}\right) }\right) }\right) \gamma_i' \left({\phi_i \left({u}\right) }\right) \phi_i' \left({u}\right) \rd u       | c = substitution with $t = \phi \left({u}\right)$ }} {{eqn | r = \sum_{i \mathop = 1}^n \int_{\phi_i^{-1} \left({a_i}\right) }^{\phi_i^{-1} \left({b_i}\right) } f \left({ \sigma_i \left({u}\right) }\right) \sigma_i' \left({u}\right) \rd u       | c = Derivative of Complex Composite Function }} {{eqn | r = \sum_{i \mathop = 1}^n \int_{c_i}^{d_i} f \left({ \sigma_i \left({u}\right) }\right) \sigma_i' \left({u}\right) \rd u       | c = Reparameterization of Directed Smooth Curve Maps Endpoints To Endpoints }} {{end-eqn}} {{qed}}
From Unity of Ring is Idempotent, $1_R$ is an idempotent element of $R$. Hence $1_R \in A$. Recall that the ring product of $A$ is a restriction from that of $R$. Hence, for each $x \in A$: :$x \circ 1_R = x = 1_R \circ x$ so that $1_R$ is a unity for $A$, as desired. {{qed}} Category:Ring Theory nspjhifbhflpwipy9lij3cnpga97hy1
From Ring of Idempotents is Idempotent Ring, $\struct {A, \oplus, \circ}$ is an idempotent ring. By Unity is Unity in Ring of Idempotents, $\struct {A, \oplus, \circ}$ is also a unitary ring. Hence, by definition, $\struct {A, \oplus, \circ}$ is a Boolean ring. {{qed}}
Since $R$ has more than two elements, there exist distinct non-zero elements $x, y \in R$. Note that $x + y \ne 0_R$ since $x$ and $y$ are distinct (by Idempotent Ring has Characteristic Two). If $x \circ y = 0$, $x$ is a proper zero divisor. If $x \circ y \ne 0$, then: {{begin-eqn}} {{eqn|l = \left({x \circ y}\right) \circ \left({x + y}\right)      |r = x \circ y \circ x + x \circ y \circ y }} {{eqn|r = x \circ y + x \circ y      |c = $R$ is an idempotent ring, Idempotent Ring is Commutative }} {{eqn|r = 0_R      |c = Idempotent Ring has Characteristic Two }} {{end-eqn}} Hence $x \circ y$ is a proper zero divisor. The result follows, from Proof by Cases. {{qed}}
=== Necessary condition === Let $f$ be a bijection. From Continuous Injection of Interval is Strictly Monotone, it follows that $f$ is strictly monotone. {{qed|lemma}} === Sufficient condition === Let $f$ be strictly monotone. From Strictly Monotone Real Function is Bijective, it follows that $f$ is a bijection on its image. From Image of Interval by Continuous Function is Interval, it follows that the image of $f$ is a real interval. Suppose that $f$ is strictly increasing. Then by definition $\map f c < \map f d$. It follows that: {{begin-eqn}} {{eqn | l = \map f c       | r = a }} {{eqn | l = \map f d       | r = b }} {{end-eqn}} It follows that for all $x \in \closedint c d$: {{begin-eqn}} {{eqn | l = \map f x       | o = \ge       | r = \map f c }} {{eqn | r = a }} {{eqn | l = \map f x       | o = \le       | r = \map f d }} {{eqn | r = b }} {{end-eqn}} Thus, the image of $f$ is $\closedint a b$. {{qed|lemma}} Suppose instead that $f$ is strictly decreasing. Then by definition $\map f c > \map f d$. It follows that: {{begin-eqn}} {{eqn | l = \map f c       | r = b }} {{eqn | l = \map f d       | r = a }} {{end-eqn}} It follows that for all $x \in \closedint c d$: {{begin-eqn}} {{eqn | l = \map f x       | o = \le       | r = \map f c }} {{eqn | r = b }} {{eqn | l = \map f x       | o = \ge       | r = \map f d }} {{eqn | r = a }} {{end-eqn}} Thus the image of $f$ is again $\closedint a b$. {{qed}} Category:Continuous Functions Category:Real Functions Category:Monotone Mappings Category:Bijections 9qldbiiz1uw3zwfxzb3epczq8pqb78p
By definition of contour, each $C_i$ and $D_j$ is a directed smooth curve for all $i \in \left\{ {1, \ldots, n}\right\}, j \in \left\{ {1, \ldots, m}\right\}$. By definition of contour, $\gamma_i \left({b_i}\right) = \gamma_{i+1} \left({a_{i+1} }\right)$ and $\sigma_j \left({d_j}\right) = \sigma_{j+1} \left({c_{j+1} }\right)$ for all $i \in \left\{ {1, \ldots, n-1}\right\}, j \in \left\{ {1, \ldots, m-1}\right\}$. By assumption, $\gamma_n \left({b_n}\right) = \sigma_1 \left({c_1}\right)$. Hence, $C_1, \ldots, C_n, D_1, \ldots, D_m$ defines a contour. {{qed}}
By definition of contour, $C$ is a finite sequence $C_1, \ldots, C_n$ of directed smooth curves. Let $C_i$ be parameterized by the smooth path $\gamma_i: \left[{a_i \,.\,.\, b_i}\right] \to \C$ for all $i \in \left\{ {1, \ldots, n}\right\}$. Similarly, $D$ is a finite sequence $D_1, \ldots, D_m$ of directed smooth curves. Let $D_j$ be parameterized by the smooth path $\sigma_j: \left[{c_j \,.\,.\, d_j}\right] \to \C$ for all $j \in \left\{ {1, \ldots, m}\right\}$. Put $\gamma_i = \sigma_{i - n}, a_i = c_{i - n}$ and $b_i = d_{i - n}$ for all $i \in \left\{ {n + 1, \ldots, n + m}\right\}$. Then $C \cup D$ is a sequence of $n + m$ directed smooth curves which are parameterized by $\gamma_1, \ldots, \gamma_{n + m}$. Then: {{begin-eqn}} {{eqn | l = \displaystyle \int_{C \cup D} f \left({z}\right) \rd z       | r = \sum_{i \mathop = 1}^{n + m} \int_{a_i}^{b_i} f \left({\gamma_i \left({t}\right) }\right) \gamma_i' \left({t}\right) \rd t       | c = {{Defof|Complex Contour Integral}} }} {{eqn | r = \sum_{i \mathop = 1}^n \int_{a_i}^{b_i} f \left({\gamma_i \left({t}\right) }\right) \gamma_i' \left({t}\right) \rd t + \sum_{j \mathop = 1}^m \int_{c_j}^{d_j} f \left({\sigma_j \left({t}\right) }\right) \sigma_j' \left({t}\right) \rd t }} {{eqn | r = \int_C f \left({z}\right) \rd z + \int_D f \left({z}\right) \rd z       | c = {{Defof|Complex Contour Integral}} }} {{end-eqn}} {{qed}}
Checking in turn each of the criteria for an equivalence: === Reflexive === Let $\gamma: \left[{ a \,.\,.\, b }\right] \to \C$ be a smooth path. Define $\phi: \left[{ a \,.\,.\, b }\right] \to \left[{ a \,.\,.\, b }\right]$ as the identity function, so : $\phi \left({t}\right) = t $ for all $t \in \left[{ a \,.\,.\, b }\right]$ From Identity Mapping is Bijection, it follows that $\phi$ is bijective. From Derivative of Identity Function, it follows that: $\phi' \left({t}\right) = 1 > 0$ for all $t \in \left[{ a \,.\,.\, b }\right]$. Then Derivative of Monotone Function shows that $\phi$ is strictly increasing. Hence, $\gamma \sim \gamma$, so $\sim$ is reflexive. {{qed|lemma}} === Symmetric === Let $\gamma : \left[{ a \,.\,.\, b }\right] \to \C$ and $\sigma : \left[{ c \,.\,.\, d }\right] \to \C$ be smooth paths such that $\gamma \sim \sigma$. That is, $\sigma = \gamma \circ \phi$ where $\phi: \left[{c \,.\,.\, d}\right] \to \left[{a \,.\,.\, b}\right]$ is bijective, differentiable and strictly increasing. From Bijection iff Inverse is Bijection, it follows that $\phi$ has an inverse function $\phi^{-1}: \left[{a \,.\,.\, b}\right] \to \left[{c \,.\,.\, d}\right]$. Then $\gamma = \sigma \circ \phi^{-1}$. From Derivative of Inverse Function: : $D \phi^{-1} \left({t}\right) = \dfrac 1 {\phi' \left({\phi^{-1} \left({t}\right) }\right)}$ for all $t \in \left[{a \,.\,.\, b}\right]$. From Derivative of Monotone Function: :$\phi' \left({\phi^{-1} \left({t}\right) }\right) > 0$ Then: :$D \phi^{-1} \left({t}\right) > 0$ so $\phi^{-1}$ is strictly increasing. Hence, $\sigma \sim \gamma$, so $\sim$ is symmetric. {{qed|lemma}} === Transitive === Let $\gamma : \left[{ a \,.\,.\, b }\right] \to \C$, $\sigma : \left[{ c \,.\,.\, d }\right] \to \C$ and $\rho : \left[{ g \,.\,.\, h }\right] \to \C$ be smooth paths such that $\gamma \sim \sigma$ and $\sigma \sim \rho$. That is: : $\sigma = \gamma \circ \phi$ and $\rho = \sigma \circ \psi$ where $\phi: \left[{c \,.\,.\, d}\right] \to \left[{a \,.\,.\, b}\right]$ and $\psi: \left[{g \,.\,.\, h}\right] \to \left[{c \,.\,.\, d}\right]$ are bijective, differentiable and strictly increasing. Then: :$\rho = \gamma \circ \left({\phi \circ \psi}\right)$ From Composite of Bijections is Bijection, it follows that $\phi \circ \psi$ is bijective. From Derivative of Composite Function, it follows that: :$D \left({\phi \circ \psi}\right) \left({t}\right) = \phi' \left({\psi \left({t}\right) }\right) \psi' \left({t}\right)$ for all $t \in \left[{ g \,.\,.\, h }\right]$. As $\phi'\left({\psi \left({t}\right) }\right) > 0$ and $\psi' \left({t}\right) > 0$, this implies that: :$D \left({\phi \circ \psi}\right) \left({t}\right) > 0$ so $\phi \circ \psi$ is strictly increasing. Hence, $\gamma \sim \rho$, so $\sim$ is transitive. {{qed|lemma}} Therefore, $\sim$ is an equivalence relation. {{qed}}
From the Trichotomy Law for Real Numbers exactly one of the following holds: :* $x < y$ and so $\max \left\{{x, y}\right\} = y$ :* $x = y$ and so $\max \left\{{x, y}\right\} = x = y$ :* $y < x$ and so $\max \left\{{x, y}\right\} = x$ By the definition of the absolute value function for each case respectively we have: :* $\left\vert{x - y}\right\vert = y - x$ :* $\left\vert{x - y}\right\vert = 0$ :* $\left\vert{x - y}\right\vert = x - y$ Thus the equation holds by $+$ being commutative and associative as for each case: :* $\frac 1 2 \left({x + y + y - x} \right) = y$ :* $\frac 1 2 \left({x + y + 0} \right) = x = y$ :* $\frac 1 2 \left({x + y + x - y} \right) = x$ {{qed}} Category:Analysis Category:Max and Min Operations noh4h8i0dl1dakz0153irzt311b6pr9
By definition of directed smooth curve, there exists a bijective differentiable strictly increasing function $\phi: \left[{c \,.\,.\, d}\right] \to \left[{a \,.\,.\, b}\right]$ such that $\sigma = \gamma \circ \phi$. From Surjection by Restriction of Codomain, it follows that there exists a function $\tilde{\gamma}: \left[{a \,.\,.\, b}\right] \to \operatorname{Im} \left({\gamma}\right)$ that is the surjective restriction of $\gamma$ to $\operatorname{Im} \left({\gamma}\right)$. Put $\tilde{\sigma} := \tilde{\gamma} \circ \phi : \left[{c \,.\,.\, d}\right] \to \operatorname{Im} \left({\gamma}\right)$. From Composite of Surjections is Surjection, it follows that $\tilde{\sigma}$ is surjective. Then: :$\operatorname{Im} \left({\tilde{\sigma} }\right) = \operatorname{Im} \left({\gamma}\right)$ Now: : $\sigma \left({t}\right) = \tilde{\sigma} \left({t}\right)$ for all $t \in \left[{c \,.\,.\, d}\right] = \operatorname{Dom} \left({\sigma}\right) = \operatorname{Dom} \left({\tilde{\sigma} }\right)$ where $\operatorname{Dom} \left({\sigma}\right)$ denotes the domain of $\sigma$. It follows that: :$\operatorname{Im} \left({\sigma}\right) = \operatorname{Im} \left({\tilde{\sigma} }\right)$ Hence: :$\operatorname{Im} \left({\tilde{\gamma} }\right) = \operatorname{Im} \left({\sigma}\right)$ {{qed}}
By definition of reparameterization, there exists a bijective differentiable strictly increasing real function $\phi: \left[{c \,.\,.\, d}\right] \to \left[{a \,.\,.\, b}\right]$ such that $\sigma = \gamma \circ \phi$. As $\phi^{-1} \left({a}\right) \in \left[{c \,.\,.\, d}\right]$: : $c \le \phi^{-1} \left({a}\right)$ As $\phi$ is strictly increasing: : $\phi \left({c}\right) \le \phi \left({\phi^{-1} \left({a}\right) }\right) = a$ As $\phi \left({c}\right) \in \left[{a \,.\,.\, b}\right]$: : $\phi \left({c}\right) = a$ Hence: :$\sigma \left({c}\right) = \gamma \circ \phi \left({c}\right) = \gamma \left({a}\right)$ As $\phi^{-1} \left({b}\right) \in \left[{c \,.\,.\, d}\right]$: : $d \ge \phi^{-1} \left({b}\right)$ As $\phi$ is strictly increasing: : $\phi \left({d}\right) \ge \phi \left({\phi^{-1} \left({b}\right) }\right) = b$ As $\phi \left({d}\right) \in \left[{a \,.\,.\, b}\right]$: : $\phi \left({d}\right) = b$ Hence: :$\sigma \left({d}\right) = \gamma \circ \phi \left({d}\right) = \gamma \left({b}\right)$ {{qed}}
We have: {{begin-eqn}} {{eqn | l = \paren {-x} \circ \paren {-y}       | r = -\paren {x \circ \paren {-y} }       | c = Product with Ring Negative }} {{eqn | r = -\paren {-\paren {x \circ y} }       | c = Product with Ring Negative }} {{eqn | r = x \circ y       | c = Negative of Ring Negative }} {{end-eqn}} {{qed}}
Let $\mathbf U = \sqbrk a_{m n}$ be an upper triangular matrix. By definition: :$\forall a_{i j} \in \mathbf U: i > j \implies a_{i j} = 0$ Let $\mathbf U^\intercal = \sqbrk b_{n m}$ be the transpose of $\mathbf U$. That is: :$\mathbf U^\intercal = \sqbrk b_{n m}: \forall i \in \closedint 1 n, j \in \closedint 1 n: b_{i j} = a_{j i}$ Thus: :$\forall b_{j i} \in \mathbf U^\intercal: i > j \implies b_{j i} = 0$ By exchanging $i$ and $j$ in the notation of the above: :$\forall b_{i j} \in \mathbf U^\intercal: i < j \implies b_{i j} = 0$ Thus by definition it is seen that $\mathbf U^\intercal$ is a lower triangular matrix. {{qed}} Category:Triangular Matrices Category:Transposes of Matrices 8brtoc9pbbp6vnhh6u25d3syuohyc91
Derivatives are defined in terms of limits. The result follows from Limit iff Limits from Left and Right. {{qed}} Category:Analysis mp510lwkzpr8y4j9e78aa5dcv5wyvvk
So as to alleviate possible confusion over notation, let the composite of $\phi$ and $\psi$ be denoted $\psi \bullet \phi$ instead of the more usual $\psi \circ \phi$. Then what we are trying to prove is denoted: :$\paren {\psi \bullet \phi}: \struct {R_1, +_1, \circ_1} \to \struct {R_3, +_3, \circ_3}$ is a homomorphism. To prove the above is the case, we need to demonstrate that the morphism property is held by $+_1$ and $\circ_1$ under $\psi \bullet \phi$. We take two elements $x, y \in R_1$, and put them through the following wringer with respect to $+_1$: {{begin-eqn}} {{eqn | l = \map {\paren {\psi \bullet \phi} } {x +_1 y}       | r = \map \psi {\map \phi {x +_1 y} }       | c = {{Defof|Composition of Mappings}} }} {{eqn | r = \map \psi {\map \phi x +_2 \map \phi y}       | c = Morphism Property applied to $+_1$ under $\phi$ }} {{eqn | r = \map \psi {\map \phi x} +_3 \map \psi {\map \phi y}       | c = Morphism Property applied to $+_2$ under $\psi$ }} {{eqn | r = \map {\paren {\psi \bullet \phi} } x +_3 \map {\paren {\psi \bullet \phi} } y       | c = {{Defof|Composition of Mappings}} }} {{end-eqn}} The same applies to $\circ_1$: {{begin-eqn}} {{eqn | l = \map {\paren {\psi \bullet \phi} } {x \circ_1 y}       | r = \map \psi {\map \phi {x \circ_1 y} }       | c = {{Defof|Composition of Mappings}} }} {{eqn | r = \map \psi {\map \phi x \circ_2 \map \phi y}       | c = Morphism Property applied to $\circ_1$ under $\phi$ }} {{eqn | r = \map \psi {\map \phi x} \circ_3 \map \psi {\map \phi y}       | c = Morphism Property applied to $\circ_2$ under $\psi$ }} {{eqn | r = \map {\paren {\psi \bullet \phi} } x \circ_3 \map {\paren {\psi \bullet \phi} } y       | c = {{Defof|Composition of Mappings}} }} {{end-eqn}} Disentangling the confusing and tortuous expressions above, we (eventually) see that this shows that the morphism property is indeed held by both $+_1$ and $\circ_1$ under $\psi \bullet \phi$. Thus $\paren {\psi \bullet \phi}: \struct {R_1, +_1, \circ_1} \to \struct {R_3, +_3, \circ_3}$ is a homomorphism. {{qed}}
A ring epimorphism is a ring homomorphism which is also a surection. From Composition of Ring Homomorphisms is Ring Homomorphism, $\psi \circ \phi$ is a ring homomorphism. From Composite of Surjections is Surjection, $\psi \circ \phi$ is a surection. {{qed}}
A ring monomorphism is a ring homomorphism which is also an injection. From Composition of Ring Homomorphisms is Ring Homomorphism, $\psi \circ \phi$ is a ring homomorphism. From Composite of Injections is Injection, $\psi \circ \phi$ is an injection. {{qed}}
A ring isomorphism is a ring homomorphism which is also a bijection. From Composition of Ring Homomorphisms is Ring Homomorphism, $\psi \circ \phi$ is a ring homomorphism. From Composite of Bijections is Bijection, $\psi \circ \phi$ is a bijection. {{qed}}
A ring endomorphism is a ring homomorphism $f$ from a set to itself. That is: :$\operatorname{Dom} \left({\phi}\right) = \operatorname{Cdm} \left({\phi}\right)$ :$\operatorname{Dom} \left({\psi}\right) = \operatorname{Cdm} \left({\psi}\right)$ From Composition of Ring Homomorphisms is Ring Homomorphism, $\psi \circ \phi$ is a ring homomorphism. By definition of composition of mappings: :$\operatorname{Cdm} \left({\phi}\right) = \operatorname{Dom} \left({\psi}\right)$ Thus: :$\operatorname{Dom} \left({\phi}\right) = \operatorname{Cdm} \left({\psi}\right) = R$ and so $\psi \circ \phi$ is a ring endomorphism. {{qed}}
A ring automorphism is a ring isomorphism $f$ from a set to itself. That is: :$\operatorname{Dom} \left({\phi}\right) = \operatorname{Cdm} \left({\phi}\right)$ :$\operatorname{Dom} \left({\psi}\right) = \operatorname{Cdm} \left({\psi}\right)$ From Composition of Ring Isomorphisms is Ring Isomorphism, $\psi \circ \phi$ is a ring isomorphism. By definition of composition of mappings: :$\operatorname{Cdm} \left({\phi}\right) = \operatorname{Dom} \left({\psi}\right)$ Thus: :$\operatorname{Dom} \left({\phi}\right) = \operatorname{Cdm} \left({\psi}\right) = R$ and so $\psi \circ \phi$ is a ring automorphism. {{qed}}
{{NotZFC}} Let $G$ be the smallest transitive set containing $S$ as a subset. By Set Contained in Smallest Transitive Set, $G$ must exist. By Transitive Set Contained in Von Neumann Hierarchy Level, $G \subseteq V_i$ for some ordinal $i$. Therefore: :$G \in V_{i+1}$ {{explain|Proper Well-Ordering determines Smallest Elements only applies once it's been shown that initial segments are sets}} We have that the ordinals are well-ordered. From Proper Well-Ordering determines Smallest Elements, there exists a smallest ordinal $k$ such that $G \in V_{k+1}$. Hence, by definition, $G$ has rank $k$. {{qed}} Category:Von Neumann Hierarchy m3gvls68ke0ptcnnudc49tjecr6nq8v
The proof follows from Quotient Mapping on Structure is Canonical Epimorphism. When $N \lhd G$, we have: {{begin-eqn}} {{eqn | ll= \forall x, y \in G:       | l = \map {q_N} {x y}       | r = x y N       | c = {{Defof|Quotient Group Epimorphism}} }} {{eqn | r = \paren {x N} \paren {y N}       | c = {{Defof|Quotient Group}} }} {{eqn | r = \map {q_N} x \map {q_N} y       | c = {{Defof|Quotient Group Epimorphism}} }} {{end-eqn}} Therefore $q$ is a homomorphism. We have that: :$\forall x \in G: x N \in G / N = \map {q_N} x$ so $q$ is surjective. Therefore $q$ is an epimorphism. Let $x \in G$. {{begin-eqn}} {{eqn | l = x       | o = \in       | r = \map \ker {q_N}       | c =  }} {{eqn | ll= \leadstoandfrom       | l = \map {q_N} x       | r = e_{G/N}       | c = {{Defof|Kernel of Group Homomorphism}} }} {{eqn | ll= \leadstoandfrom       | l = x N       | r = e_{G/N} N       | c = {{Defof|Quotient Group Epimorphism}} }} {{eqn | ll= \leadstoandfrom       | l = x N       | r = N       | c = Coset by Identity }} {{eqn | ll= \leadstoandfrom       | l = x       | o = \in       | r = N       | c = Left Coset Equals Subgroup iff Element in Subgroup }} {{end-eqn}} thus proving that $\map \ker {q_N} = N$ from definition of subset. {{qed}}
{{NotZFC}} By Non-Maximal Element of Well-Ordered Class has Immediate Successor, $\operatorname{succ} \left({i}\right)$ is guaranteed to exist. Let $i \prec j$. Let $S_i := \left\{{q: i \prec q}\right\}$. {{explain|Important to specify what the domain of $q$ actually is -- presumably $S$.}} Then $\operatorname{succ} \left({i}\right)$ is the minimal element of $S_i$. By supposition, $j \in S_i$. Thus: :$j \not \prec \operatorname{succ} \left({i}\right)$ Since Well-Ordering is Total Ordering: :$\operatorname{succ} \left({i}\right) \preceq j$ Thus by supposition: :$f \left({\operatorname{succ} \left({i}\right)}\right) \le f \left({j}\right)$ Since $f \left({i}\right) < \left({\operatorname{succ} \left({i}\right)}\right)$: :$f \left({i}\right) < f \left({j}\right)$ by transitivity. {{qed}} Category:Class Mappings Category:Increasing Mappings p00orr6p1txznme27vyt9qae4lud9ij
Let $x$ be a non-maximal element of $C$. Let $S$ be the class of successors of $x$ in $C$. $S$ is non-empty because $x$ is not maximal. By Proper Well-Ordering Determines Smallest Elements, $S$ has a minimal element, $y$. Then $x < y$ by the definition of $S$. Suppose that for some $z \in C$, $x < z < y$. Then by the definition of $S$, $z \in S$. This contradicts the minimality of $y$. Thus $y$ is the immediate successor element of $x$. {{qed}} Category:Well-Orderings ciuv64qtc1uddocrcxqcqjjcgfadvyu