hcju/proofwikiqa huggingface dataset:
By the definition of the usual ordering on the extended real numbers: :${\le} = {\le_\R} \cup \set {\tuple {x, +\infty}: x \in \overline \R} \cup \set {\tuple {-\infty, x}: x \in \overline \R}$ Suppose $x \in \overline \R$ and $+\infty \le x$. That is: : $\tuple {+\infty, x} \in {\le}$ By the definition of union, $\tuple {+\infty, x}$ must lie in one of the three sets whose union forms $\le$. Since ${\le_\R} \subseteq \R \times \R$ and $+\infty \notin \R$: :$\tuple {+\infty, x} \notin {\le_\R}$ Since $+\infty \ne -\infty$ by the definition of the extended real numbers: :$\tuple {+\infty, x} \notin \set {\tuple {-\infty, x}: x \in \overline \R}$ Therefore: :$\tuple {+\infty, x} \in \set {\tuple {x, +\infty}: x \in \overline \R}$ and we conclude that $x = +\infty$. That is, $+\infty$ is a maximal element of $\overline \R$. {{qed}}
By the definition of the usual ordering on the extended real numbers: :${\le} = {\le_\R} \cup \set {\tuple {x, +\infty}: x \in \overline \R} \cup \set {\tuple {-\infty, x}: x \in \overline \R}$ Suppose $x \in \overline \R$ and $x \le -\infty$. That is: : $\tuple {x, -\infty} \in {\le}$ By the definition of union, $\tuple {x, -\infty}$ must lie in one of the three sets whose union forms $\le$. Since ${\le_\R} \subseteq \R \times \R$ and $-\infty \notin \R$: :$\tuple {x, -\infty} \notin {\le_\R}$ Since $+\infty \ne -\infty$ by the definition of the extended real numbers: :$\tuple {x, -\infty} \notin \set {\tuple {x, +\infty}: x \in \overline \R}$ Therefore: :$\tuple {x, -\infty} \in \set {\tuple {-\infty, x}: x \in \overline \R}$ and we conclude that $x = -\infty$. That is, $-\infty$ is a minimal element of $\overline \R$. {{qed}}
By the definition of an ordered group, $\preceq$ is a relation compatible with $\circ$. Thus by Properties of Relation Compatible with Group Operation/CRG3, we obtain the first result: :$x \preceq y \iff y^{-1} \preceq x^{-1}$ By Reflexive Reduction of Relation Compatible with Group Operation is Compatible, $\prec$ is also compatible with $\circ$. Thus by again Properties of Relation Compatible with Group Operation/CRG3, we obtain the second result: :$x \prec y \iff y^{-1} \prec x^{-1}$ {{qed}} Category:Ordered Groups al7v6y3lv3z4b45o3yb0xk4d6bdtkjx
We will prove $(1)$ and $(2)$. $(3)$ can be proven using precisely the same argument as $(2)$ and $(4)$ follows from $(1)$. Suppose that $x \mathrel{\mathcal R} y$ and $z \mathrel{\mathcal R} w$. By the definition of compatibility: :$x \mathrel{\mathcal R} y \implies x \circ z \mathrel{\mathcal R} y \circ z$ :$z \mathrel{\mathcal R} w \implies y \circ z \mathrel{\mathcal R} y \circ w$ By transitivity: :$(1)\quad \left({x \mathrel{\mathcal R} y}\right) \land \left({z \mathrel{\mathcal R} w}\right) \implies x \circ z \mathrel{\mathcal R} y \circ w$ {{qed|lemma}} Suppose that $x \mathrel{\mathcal R} y$ and $z \mathrel{\mathcal R^=} w$. By Reflexive Closure of Relation Compatible with Operation is Compatible, $\mathcal R^=$ is compatible with $\circ$. By the definition of compatibility: :$x \mathrel{\mathcal R} y \implies x \circ z \mathrel{\mathcal R} y \circ z$ :$z \mathrel{\mathcal R^=} w \implies y \circ z \mathrel{\mathcal R^=} y \circ w$ By Extended Transitivity: :$(2)\quad \left({x \mathrel{\mathcal R} y}\right) \land \left({z \mathrel{\mathcal R} w}\right) \implies x \circ z \mathrel{\mathcal R} y \circ w$ {{qed|lemma}} $(3)$ can be proven using precisely the same argument as $(2)$. By Reflexive Closure of Relation Compatible with Operation is Compatible, $\mathcal R^=$ is compatible with $\circ$. By Reflexive Closure of Transitive Relation is Transitive, $\mathcal R^=$ is transitive. Thus $(4)$ follows from $(1)$. {{qed}} Category:Compatible Relations Category:Transitive Relations rpasew4agbzabt7qlr2cck5nchsk06h
Because $\left({G, \circ, \preceq}\right)$ is a group and $\preceq$ is compatible with $\circ$, $\prec$ is compatible with $\circ$ by Reflexive Reduction of Relation Compatible with Group Operation is Compatible. By the definition of an ordering, $\preceq$ is transitive and antisymmetric. Therefore by Reflexive Reduction of Transitive Antisymmetric Relation is Strict Ordering, $\prec$ is transitive. Thus the theorem holds by Operating on Transitive Relationships Compatible with Operation. {{qed}}
=== Proof of Forward Implication === {{:Rule of Transposition/Formulation 2/Forward Implication/Proof}} === Proof of Reverse Implication === {{:Rule of Transposition/Formulation 2/Reverse Implication/Proof}}
We prove the norm axioms. === Positive Definiteness === This is demonstrated in: : Complex Modulus equals Zero iff Zero : Complex Modulus is Non-Negative {{qed|lemma}} === Multiplicativity === Follows from Modulus of Product. {{qed|lemma}} === Triangle Inequality === Follows from Triangle Inequality for Complex Numbers. {{qed}} Category:Examples of Norms Category:Complex Modulus 5a2ojdgq19ezkpwv2hj8vxlwkl81k1r
By Complement of Relation Compatible with Group is Compatible, $\relcomp {G \times G} \QQ$ is compatible with $\circ$. Thus by Intersection of Relations Compatible with Operation is Compatible, $\RR \cap \relcomp {G \times G} \QQ$ is compatible with $\circ$. But: :$\RR \cap \relcomp {G \times G} \QQ = \RR \setminus \QQ$ so the theorem holds. {{qed}} Category:Compatible Relations jyg5vn2asa1vqciut2pcdlutk31nlto
We proceed by mathematical induction. By definition of power: : $x^1 = x$ : $y^1 = y$ Hence, by assumption: : $x^1 \mathrel{\mathcal R} y^1$ Suppose now that for $n \ge 1$: :$x^n \mathrel{\mathcal R} y^n$ Recall the assumption that $x \mathrel{\mathcal R} y$. Applying Operating on Transitive Relationships Compatible with Operation to these relations yields: :$x^n \circ x \mathrel{\mathcal R} y^n \circ y$ By definition of power: : $x^{n+1} = x^n \circ x$ and: : $y^{n+1} = y^n \circ y$ The result follows by the Principle of Mathematical Induction. {{qed}} Category:Compatible Relations Category:Semigroups oe270hq5prl06s9p8hnlb1f8xq4878z
=== Reflexive === Follows from Reflexive Closure is Reflexive. {{qed|lemma}} === Transitive === Follows from Reflexive Closure of Transitive Relation is Transitive. {{qed|lemma}} === Antisymmetric === Let $a, b \in S$. Suppose that $a \ne b$. Then by the definition of the diagonal relation, :$\left({(a,b)}\right) \notin \Delta_S$ and  :$\left({(b,a)}\right) \notin \Delta_S$. Since $\prec$ is antisymmetric, $\left({(a,b)}\right)$ and $\left({(b,a)}\right)$ are not both in $\prec$. Thus $a \not\preceq b$ or $b \not\preceq a$. {{qed}} Category:Order Theory Category:Reflexive Closures Category:Transitive Relations Category:Symmetric Relations k54ogs7jzmwk2efbsg8nhelsh36zc79
Let $a, b, c \in S$. Suppose that $a \mathrel{\mathcal R^=} b$ and $b \mathrel{\mathcal R^=} c$. If $a = b$, then since $b \mathrel{\mathcal R^=} c$, also $a \mathrel{\mathcal R^=} c$. If $b = c$, then since $a \mathrel{\mathcal R^=} b$, also $a \mathrel{\mathcal R^=} c$. The only case that remains is that $a \ne b$ and $b \ne c$. Then by the definition of $\mathcal R^=$, $a \mathrel{\mathcal R} b$ and $b \mathrel{\mathcal R} c$. Since $\mathcal R$ is transitive, it follows that: :$a \mathrel{\mathcal R} c$ and hence also $a \mathrel{\mathcal R^=} c$. Thus $\mathcal R^=$ is transitive. {{qed}}
Let $a, b \in S$. Suppose that $a \mathrel{\mathcal R^=} b$ and $b \mathrel{\mathcal R^=} a$. By definition of $\mathcal R^=$, this means: :$a \mathrel{\mathcal R} b$ or $a = b$ :$b \mathrel{\mathcal R} a$ or $b = a$ If $a = b$ or $b = a$ we are done, by definition of an antisymmetric relation. So suppose $a \mathrel{\mathcal R} b$ and $b \mathrel{\mathcal R} a$. Since $\mathcal R$ is antisymmetric, it follows that $a = b$ in this case as well. Hence $\mathcal R^=$ is also antisymmetric. {{qed}} Category:Reflexive Closures 0a3vbanfzlhyi6dvj8t7lzegrqzusl5
We have, by definition of the usual ordering on $\overline \R$: :$\forall x \in \overline \R: x \le +\infty$ That is, $+\infty$ is the greatest element of $\overline \R$. {{qed}}
We have, by definition of the usual ordering on $\overline \R$: :$\forall x \in \overline \R: -\infty \le x$ That is, $-\infty$ is the smallest element of $\overline \R$. {{qed}}
By Relation Compatible with Group Operation is Strongly Compatible, $\mathcal R$ is strongly compatible with $\circ$. Thus by the definition of strong compatibility, the theorem holds. {{qed}} Category:Compatible Relations Category:Group Theory phn74ihn9gsu7nic8mwcdnr01wk6iyr
We can apply CRG1 to $x$, $y$, and $x^{-1}$ to obtain the following equivalences: :$x \mathrel {\mathcal R} y \iff x \circ x^{-1} \mathrel {\mathcal R} y \circ x^{-1}$ :$x \mathrel {\mathcal R} y \iff x^{-1} \circ x \mathrel {\mathcal R} x^{-1} \circ y$ Applying CRG1 to $x$, $y$, and $y^{-1}$, on the other hand, yields: :$x \mathrel {\mathcal R} y \iff x \circ y^{-1} \mathrel {\mathcal R} y \circ y^{-1}$ :$x \mathrel {\mathcal R} y \iff y^{-1} \circ x \mathrel {\mathcal R} y^{-1} \circ y$ By the definition of inverses: :$x \circ x^{-1} = x^{-1} \circ x = y \circ y^{-1} = y^{-1} \circ y = e$ Making these substitutions proves the theorem. {{qed}} Category:Compatible Relations Category:Group Theory 6qzoin644p2bfszzgxuo3hyjxsolcde
Applying User:Dfeuer/CRG2$(1)$ to $x$ and $e$ gives: :$x \mathrel{\mathcal R} e \iff e \mathrel{\mathcal R} e \circ x^{-1}$ Applying User:Dfeuer/CRG2$(3)$ to $e$ and $x$ gives: :$e \mathrel{\mathcal R} x \iff e \circ x^{-1} \mathrel{\mathcal R} e$ Since $e \circ x^{-1} = x^{-1} \circ e = x^{-1}$ for all $x \in G$, the theorem holds. {{qed}} Category:Compatible Relations Category:Group Theory 2s2daaweuwdvg32qu0gzsa8fkb12fkj
Let $\succeq$ be the dual ordering of $\preceq$. By definition of dual statement: :$b \preceq a$ is dual to: :$b \succeq a$ and by definition of dual ordering, this is equivalent to: :$a \preceq b$ By virtue of Dual of Dual Statement (Order Theory), the converse follows. The other claims are proved on the following pages, in order: :Succeed is Dual to Precede :Strictly Succeed is Dual to Strictly Precede :Upper Bound is Dual to Lower Bound :Supremum is Dual to Infimum :Maximal Element is Dual to Minimal Element :Greatest Element is Dual to Smallest Element :Weak Lower Closure is Dual to Weak Upper Closure :Strict Lower Closure is Dual to Strict Upper Closure :Join is Dual to Meet :Lower Set is Dual to Upper Set {{qed}}
We apply the Method of Truth Tables to the proposition: :$\paren {p \implies q} \implies r \vdash \paren {p \implies r} \implies \paren {q \implies r}$ As can be seen for all boolean interpretations by inspection, where the truth value under the main connective on the {{LHS}} is $T$, that under the one on the {{RHS}} is also $T$: $\begin{array}{|ccccc||ccccccc|} \hline (p & \implies & q) & \implies & r & (p & \implies & r) & \implies & (q & \implies & r) \\ \hline F & T & F & F & F & F & T & F & T & F & T & F \\ F & T & F & T & T & F & T & T & T & F & T & T \\ F & T & T & F & F & F & T & F & F & T & F & F \\ F & T & T & T & T & F & T & T & T & T & T & T \\ T & F & F & T & F & T & F & F & T & F & F & F \\ T & F & F & T & T & T & T & T & T & F & T & T \\ T & T & T & F & F & T & F & F & T & T & F & F \\ T & T & T & T & T & T & T & T & T & T & T & T \\ \hline \end{array}$ {{qed|lemma}} The two formulas are not equivalent, as the relevant columns do not match exactly. For example, when $p =  q = r = F$ we have that: :$\paren {p \implies q} \implies r = F$ but: :$\paren {p \implies r} \implies \paren {q \implies r} = T$ Hence the result: :$\paren {p \implies q} \implies r \vdash \paren {p \implies r} \implies \paren {q \implies r}$ but: :$\paren {p \implies r} \implies \paren {q \implies r} \not \vdash \paren {p \implies q} \implies r$ {{qed}} Category:Truth Table Proofs Category:Conditional is not Left Self-Distributive jghkmpapahabeztjswp0cgg1xrdszb5
{{ProofWanted|Can be seen to be logically equivalent to Conditional is not Left Self-Distributive/Formulation 1 by application of the Rule of Implication and Modus Ponendo Ponens.}}
=== Necessary Condition === {{:Cauchy-Riemann Equations/Necessary Condition}} === Sufficient Condition === {{:Cauchy-Riemann Equations/Sufficient Condition}} === Expression of Derivative === {{:Cauchy-Riemann Equations/Expression of Derivative}} {{Namedfor|Augustin Louis Cauchy|name2 = Georg Friedrich Bernhard Riemann|cat = Cauchy|cat2 = Riemann}}
=== Proof of Forward Implication === {{:Self-Distributive Law for Conditional/Forward Implication/Formulation 2/Proof 1}} === Proof of Reverse Implication === {{:Self-Distributive Law for Conditional/Reverse Implication/Formulation 2/Proof}} {{BeginTableau|\vdash \paren {p \implies \paren {q \implies r} } \iff \paren {\paren {p \implies q} \implies \paren {p \implies r} } }} {{TheoremIntro|1|\paren {p \implies \paren {q \implies r} } \implies \paren {\paren {p \implies q} \implies \paren {p \implies r} }|Self-Distributive Law for Conditional: Forward Implication: Formulation 2}} {{TheoremIntro|2|\paren {\paren {p \implies q} \implies \paren {p \implies r} } \implies \paren {p \implies \paren {q \implies r} }|Self-Distributive Law for Conditional: Reverse Implication: Formulation 2}} {{BiconditionalIntro|3||\paren {p \implies \paren {q \implies r} } \iff \paren {\paren {p \implies q} \implies \paren {p \implies r} }|1|2}} {{EndTableau}} {{qed}}
=== Proof of Forward Implication === {{:Self-Distributive Law for Conditional/Forward Implication/Formulation 1/Proof}} === Proof of Reverse Implication === {{:Self-Distributive Law for Conditional/Reverse Implication/Formulation 1/Proof}}
We apply the Method of Truth Tables to the proposition: $p \implies \left({q \implies r}\right) \dashv \vdash \left({p \implies q}\right) \implies \left({p \implies r}\right)$ As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations. $\begin{array}{|ccccc||ccccccc|} \hline p & \implies & (q & \implies & r) & (p & \implies & q) & \implies & (p & \implies & r) \\ \hline F & T & F & T & F & F & T & F & T & F & T & F \\ F & T & F & T & T & F & T & F & T & F & T & T \\ F & T & T & F & F & F & T & T & T & F & T & F \\ F & T & T & T & T & F & T & T & T & F & T & T \\ T & T & F & T & F & T & F & F & T & T & F & F \\ T & T & F & T & T & T & F & F & T & T & T & T \\ T & F & T & F & F & T & T & T & F & T & F & F \\ T & T & T & T & T & T & T & T & T & T & T & T \\ \hline \end{array}$ {{qed}}
Let $z = x + i y$. Then:   {{begin-eqn}} {{eqn | l = \dfrac {\partial f} {\partial x} \left({z}\right)       | r = \dfrac {\partial u} {\partial x} \left({x, y}\right) + i \dfrac {\partial v} {\partial x} \left({x, y}\right) }} {{eqn | r = \operatorname{Re} \left({f' \left({z}\right) }\right) + i \operatorname{Im} \left({f' \left({z}\right) }\right)       | c = from the last part of the proof for sufficient condition }} {{eqn | r = f' \left({z}\right) }} {{end-eqn}} Similarly: {{begin-eqn}} {{eqn | l = -i \dfrac {\partial f} {\partial y} \left({z}\right)       | r = -i \left({\dfrac {\partial u} {\partial y} \left({x, y}\right) + i \dfrac {\partial v} {\partial y} \left({x, y}\right) }\right) }} {{eqn | r = -i \left({ -\operatorname{Im} \left({f' \left({z}\right) }\right) + i \operatorname{Re} \left({f' \left({z}\right) }\right) }\right)       | c = from the last part of the proof for sufficient condition }} {{eqn | r = f' \left({z}\right) }} {{end-eqn}} {{qed}}
{{BeginTableau|\paren {p \implies q} \implies \paren {p \implies r} \vdash p \implies \paren {q \implies r} }} {{Premise|1|\paren {p \implies q} \implies \paren {p \implies r} }} {{Assumption|2|p}} {{Assumption|3|q}} {{SequentIntro|4|3|p \implies q|3|True Statement is implied by Every Statement}} {{ModusPonens|5|1, 3|p \implies r|1|4}} {{ModusPonens|6|1, 2, 3|r|5|2}} {{Implication|7|1, 2|q \implies r|3|6}} {{Implication|8|1|p \implies \paren {q \implies r}|2|7}} {{EndTableau}} {{qed}}
{{BeginTableau|\paren {p \implies \paren {q \implies r} } \implies \paren {\paren {p \implies q} \implies \paren {p \implies r} } }} {{Assumption  |1|p \implies \paren {q \implies r} }} {{SequentIntro|2|1|\paren {p \implies q} \implies \paren {p \implies r}|1|Self-Distributive Law for Conditional: Formulation 1}} {{Implication |3||\paren {p \implies \paren {q \implies r} } \implies \paren {\paren {p \implies q} \implies \paren {p \implies r} }|1|2}} {{EndTableau}} {{Qed}}
{{BeginTableau |\vdash \paren {\paren {p \implies q} \implies \paren {p \implies r} } \implies \paren {p \implies \paren {q \implies r} } }} {{Assumption   |1|\paren {p \implies q} \implies \paren {p \implies r} }} {{SequentIntro |2|1|p \implies \paren {q \implies r}|1|Self-Distributive Law for Conditional: Formulation 1}} {{Implication  |3||\paren {\paren {p \implies q} \implies \paren {p \implies r} } \implies \paren {p \implies \paren {q \implies r} }|1|2}} {{EndTableau}} {{Qed}}
By definition, $a$ succeeds $b$ {{iff}}: :$b \preceq a$ The dual of this statement is: :$a \preceq b$ by Dual Pairs (Order Theory). By definition, this means $a$ precedes $b$. The converse follows from Dual of Dual Statement (Order Theory). {{qed}}
By definition, $a$ strictly succeeds $b$ {{iff}}: :$b \preceq a$ and $b \ne a$ The dual of this statement is: :$a \succeq b$ and $b \ne a$ by Dual Pairs (Order Theory). By definition, this means $a$ strictly precedes $b$. The converse follows from Dual of Dual Statement (Order Theory). {{qed}}
By definition, $a$ is an upper bound for $T$ iff: :$\forall t \in T: t \preceq a$ The dual of this statement is: :$\forall t \in T: a \preceq t$ by Dual Pairs (Order Theory). By definition, this means $a$ is a lower bound for $T$. The converse follows from Dual of Dual Statement (Order Theory). {{qed}}
By definition, $a$ is a supremum for $T$ iff: :$a$ is an upper bound for $T$ :$a \preceq b$ for all upper bounds $b$ of $T$ The dual of this statement is: :$a$ is a lower bound for $T$ :$b \preceq a$ for all lower bounds $b$ of $T$ by Dual Pairs (Order Theory). By definition, this means $a$ is an infimum for $T$. The converse follows from Dual of Dual Statement (Order Theory). {{qed}}
By definition, $a$ is a maximal element of $T$ {{iff}}: :$\forall t \in T: a \preceq t$ implies $a = t$ The dual of this statement is: :$\forall t \in T: t \preceq a$ implies $a = t$ by Dual Pairs (Order Theory). By definition, this means $a$ is a minimal element of $T$. The converse follows from Dual of Dual Statement (Order Theory). {{qed}}
By definition, $a$ is the greatest element of $S$ {{iff}}: :$\forall b \in S: b \preceq a$ The dual of this statement is: :$\forall b \in S: a \preceq b$ by Dual Pairs (Order Theory). By definition, this means $a$ is the smallest element of $S$. The converse follows from Dual of Dual Statement (Order Theory). {{qed}}
=== Elements === By definition of lower closure, $b \in a^\preccurlyeq$ {{iff}}: :$b \preccurlyeq a$ The dual of this statement is: :$a \preccurlyeq b$ by Dual Pairs (Order Theory). By definition of upper closure, this means $b \in a^\succcurlyeq$. The converse follows from Dual of Dual Statement (Order Theory). {{qed|lemma}} === Sets === By the definition of lower closure, $b \in T^\preccurlyeq$ {{iff}}: :$\exists a \in T: b \preccurlyeq a$ The dual of this statement is: :$\exists a \in T: a \preccurlyeq b$ by Dual Pairs (Order Theory). By the definition of upper closure, this means $b \in T^\succcurlyeq$. {{qed}}
By definition of strict lower closure: :$b \in a^\prec$ {{iff}} :$b$ strictly precedes $a$ The dual of this statement is: :$b$ strictly succeeds $a$ by Dual Pairs (Order Theory). By definition of strict upper closure, this means: : $b \in a^\succ$ The converse follows from Dual of Dual Statement (Order Theory). {{qed}}
By definition of join, $c = a \vee b$ {{iff}}: :$c = \sup \left\{{a, b}\right\}$ where $\sup$ denotes supremum. The dual of this statement is: :$c = \inf \left\{{a, b}\right\}$ where $\inf$ denotes infimum, by Dual Pairs (Order Theory). By definition of meet, this means $c = a \wedge b$. {{qed}}
=== Proof of Forward Implication === {{:Rule of Exportation/Forward Implication/Formulation 1/Proof}} === Proof of Reverse Implication === {{:Rule of Exportation/Reverse Implication/Formulation 1/Proof}}
We apply the Method of Truth Tables to the proposition. As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations. $\begin{array}{|ccccc||ccccc|} \hline (p & \land & q) & \implies & r & p & \implies & (q & \implies & r) \\ \hline F & F & F & T & F & F & T & F & T & F \\ F & F & F & T & T & F & T & F & T & T \\ F & F & T & T & F & F & T & T & F & F \\ F & F & T & T & T & F & T & T & T & T \\ T & F & F & T & F & T & T & F & T & F \\ T & F & F & T & T & T & T & F & T & T \\ T & T & T & F & F & T & F & T & F & F \\ T & T & T & T & T & T & T & T & T & T \\ \hline \end{array}$ {{qed}}
Let $\mathbf A$ be a square matrix over $R$ with two identical columns. Let $\mathbf A^\intercal$ denote the transpose of $\mathbf A$. Then $\mathbf A^\intercal$ has two identical rows. Then: {{begin-eqn}} {{eqn | l = \map \det {\mathbf A}       | r = \map \det {\mathbf A^\intercal}       | c = Determinant of Transpose }} {{eqn | r = 0       | c = Square Matrix with Duplicate Rows has Zero Determinant }} {{end-eqn}} {{qed}}
Let $e$ be the group identity of $G$. By User:Dfeuer/CRG2 $(1)$: :$(1): \quad x \mathrel{\mathcal R} y \iff e \mathrel{\mathcal R} y \circ x^{-1}$ By User:Dfeuer/CRG2 $(2)$, also: :$(2): \quad y^{-1} \mathrel{\mathcal R} x^{-1} \iff e \mathrel{\mathcal R} \left({y^{-1}}\right)^{-1} \circ x^{-1}$ By Inverse of Group Inverse $\left({y^{-1}}\right)^{-1} = y$. Thus, we can rewrite $(2)$ as: :$(3): \quad y^{-1} \mathrel{\mathcal R} x^{-1} \iff e \mathrel {\mathcal R} y \circ x^{-1}$ Now note that the {{RHS}} of $(3)$ is the same as the {{RHS}} in $(1)$. We conclude that: :$x \mathrel{\mathcal R} y \iff y^{-1} \mathrel{\mathcal R} x$ {{qed}} Category:Compatible Relations Category:Group Theory bxgnpz4l16tutays15dlp8v9poi7o7m
It is to be shown that, for all $a, b \in S$: :$a \preceq b$ {{iff}} $b = \sup \set {a, b}$ by definition of join. Here $\sup$ denotes supremum. Since any upper bound $c$ of $\set {a, b}$ must satisfy: :$b \preceq c$ it suffices to verify that: :$a \preceq b$ {{iff}} $b$ is an upper bound for $\set {a, b}$ Since $\preceq$ is reflexive, we know that: :$b \preceq b$ and therefore said equivalence is established. We conclude that $\preceq'$ and $\preceq$ coincide. {{qed}} {{expand|We need to start using a construct for duality. (Global) Dual is Ordering Induced by Meet Semilattice}} Category:Order Theory Category:Semilattices fgs0nl1upacwtukww0y2jip2lnphxmi
Denote the open ball of $0$ with radius $r \in \R_{>0}$ as $B_r \left({0}\right)$. Let $z \in D$. By the Alternative Differentiability Condition, it follows that there exists $r \in \R_{>0}$ such that for all $h \in B_r \left({0}\right) \setminus \left\{ {0}\right\}$: :$f\left({z + h}\right) = f \left({z}\right) + h \left({f' \left({z}\right) + \epsilon_f \left({h}\right) }\right)$ :$g\left({z + h}\right) = g \left({z}\right) + h \left({g' \left({z}\right) + \epsilon_g \left({h}\right) }\right)$ where $\epsilon_f, \epsilon_g: B_r \left({0}\right) \setminus \left\{ {0}\right\} \to \C$ are continuous functions that converge to $0$ as $h$ tends to $0$. Then: {{begin-eqn}} {{eqn | l = \left({f + g}\right) \left({z + h}\right)       | r = f \left({z}\right) + h \left({f' \left({z}\right) + \epsilon_f \left({h}\right) }\right) + g \left({z}\right) + h \left({g' \left({z}\right) + \epsilon_g \left({h}\right) }\right) }} {{eqn | r = \left({f + g}\right) \left({z}\right) + h \left({f' \left({z}\right) + g' \left({z}\right) + \left({ \epsilon_f + \epsilon_g }\right) \left({h}\right) }\right) }} {{end-eqn}} From Sum Rule for Continuous Functions, it follows that $\epsilon_f + \epsilon_g$ is a continuous function. From Sum Rule for Limits of Functions, it follows that $\displaystyle \lim_{h \to 0} \left({ \epsilon_f + \epsilon_g }\right) \left({h}\right) = 0$. Then the Alternative Differentiability Condition shows that: :$\left({f + g}\right)' \left({z}\right) = f' \left({z}\right) + g' \left({z}\right)$ {{qed}}