hcju/proofwikiqa huggingface dataset:
Let $C$ be the set of points on the unit circle defined above. Let $P$ be the set: :$\set {z \in \C: \cmod z = 1}$ Let $z = x + i y \in C$. $z$ can be expressed in polar form as: :$z = \polar {r, \theta}$ where: :$x = r \cos \theta$ :$y = r \sin \theta$ By definition of unit circle, $z$ is $1$ unit away from $\tuple {0, 0}$. By Pythagoras's Theorem: :$\sqrt {\paren {r \cos \theta}^2 + \paren {r \sin \theta}^2} = 1$ from which: :$\sqrt {x^2 + y^2} = 1$ By definition of complex modulus: :$\cmod z = 1$ and so $z \in P$. Thus $C \subseteq P$. Let $z \in P$. Then by definition $\cmod z = 1$. By Modulus of Complex Number equals its Distance from Origin, $z$ is $1$ unit away from $\tuple {0, 0}$. By definition of $C$, it follows that $z \in C$. Thus $P \subseteq C$. The result follows by definition of set equality. {{Qed}}
From Equation of Circle, the equation of a circle with radius $R$ and center $\tuple {a, b}$ is: :$\paren {x - a}^2 + \paren {y - b}^2 = R^2$ Substituting $\tuple {0, 0}$ for $\tuple {a, b}$ and $1$ for $R$ gives the result. {{qed}}
We apply the Method of Truth Tables to the proposition. As can be seen by inspection, in each case, the truth values in the appropriate columns match for all boolean interpretations. $\begin{array}{|c|ccc||c|ccc|} \hline p & p & \implies & \top & \top \\ \hline F & F & T & T & T \\ T & T & T & T & T \\ \hline \end{array}$ {{qed}}
We apply the Method of Truth Tables to the proposition. As can be seen by inspection, in each case, the truth values in the appropriate columns match for all boolean interpretations. $\begin{array}{|c|ccc||c|ccc|} \hline p & \top & \implies & p \\ \hline F & T & F & F \\ T & T & T & T \\ \hline \end{array}$ {{qed}}
We apply the Method of Truth Tables to the proposition. As can be seen by inspection, in each case, the truth values in the appropriate columns match for all boolean interpretations. $\begin{array}{|c|ccc||cc|} \hline p & p & \implies & \bot & \neg & p \\ \hline F & F & T & F & T & F \\ T & T & F & F & F & T \\ \hline \end{array}$ {{qed}}
We apply the Method of Truth Tables to the proposition. As can be seen by inspection, in each case, the truth values in the appropriate columns match for all boolean interpretations. $\begin{array}{|c|ccc||c|} \hline p & \bot & \implies & p & \top \\ \hline F & F & T & F & T \\ T & F & T & T & T \\ \hline \end{array}$ {{qed}}
We apply the Method of Truth Tables to the proposition. As can be seen by inspection, in each case, the truth values in the appropriate columns match for all boolean interpretations. $\begin{array}{|c|ccc||c|ccc|} \hline p & p & \iff & \top & p \\ \hline F & F & F & T & F \\ T & T & T & T & T \\ \hline \end{array}$ {{qed}}
{{BeginTableau|p \oplus \top \vdash \neg p}} {{Premise|1|p \oplus \top}} {{SequentIntro|2|1|\left({p \lor \top} \right) \land \neg \left({p \land \top}\right)|1|Definition of Exclusive Or}} {{SequentIntro|3|1|\top \land \neg \left({p \land \top}\right)|1|Disjunction with Tautology}} {{SequentIntro|4|1|\neg \left({p \land \top}\right)|1|Conjunction with Tautology}} {{SequentIntro|5|1|\neg p|1|Conjunction with Tautology}}  {{EndTableau}} {{qed|lemma}} {{BeginTableau|\neg p \vdash p \oplus \top}} {{Assumption|1|\neg p}} {{TopIntro|2}} {{SequentIntro|3||p \lor \top|2|Disjunction with Tautology}} {{SequentIntro|4|1|\neg \left({p \land \top}\right)|1|Conjunction with Tautology}} {{Conjunction|5|1|\left({p \lor \top}\right) \land \neg \left({p \land \top}\right)|3|4}} {{SequentIntro|6|1|p \oplus \top|5|Definition of Exclusive Or}} {{EndTableau}} {{qed}}
We apply the Method of Truth Tables to the proposition. As can be seen by inspection, in each case, the truth values in the appropriate columns match for all boolean interpretations. $\begin{array}{|c|ccc||cc|} \hline p & p & \oplus & \top & \neg & p \\ \hline F & F & T & T & T & F \\ T & T & F & T & F & T \\ \hline \end{array}$ {{qed}}
{{BeginTableau|p \iff \bot \vdash \neg p}} {{Premise|1|p \iff \bot}} {{BiconditionalElimination|2|1|p \implies \bot|1|1}} {{SequentIntro|3|1|\neg p|2|Contradictory Consequent}} {{EndTableau}} {{qed|lemma}} {{BeginTableau|\neg p \vdash p \iff \bot}} {{Assumption|1|\neg p}} {{SequentIntro|2|1|p \implies \bot|1|Contradictory Consequent}} {{TopIntro|3}} {{SequentIntro|4||\bot \implies p|3|Contradictory Antecedent}} {{BiconditionalIntro|5|1|p \iff \bot|2|4}} {{EndTableau}} {{qed}}
We apply the Method of Truth Tables to the proposition. As can be seen by inspection, in each case, the truth values in the appropriate columns match for all boolean interpretations. $\begin{array}{|c|ccc||cc|} \hline p & p & \iff & \bot & \neg & p \\ \hline F & F & T & F & T & F \\ T & T & F & F & F & T \\ \hline \end{array}$ {{qed}}
{{BeginTableau|p \oplus \bot \vdash p}} {{Premise|1|p \oplus \bot}} {{SequentIntro|2|1|\left({p \lor \bot} \right) \land \neg \left({p \land \bot}\right)|1| {{Defof|Exclusive Or}} }} {{SequentIntro|3|1|p \land \neg \left({p \land \bot}\right)|2|Disjunction with Contradiction}} {{SequentIntro|4|1|p \land \neg \bot|3|Conjunction with Contradiction}} {{SequentIntro|5|1|p \land \top|4|Tautology is Negation of Contradiction}} {{SequentIntro|6|1|p|5|Conjunction with Tautology}} {{EndTableau}} {{qed|lemma}} {{BeginTableau|p \vdash p \oplus \bot}} {{Premise|1|p}} {{SequentIntro|2|1|p \land \top|1|Conjunction with Tautology}} {{SequentIntro|3|1|\left({p \lor \bot}\right) \land \top|2|Disjunction with Contradiction}} {{SequentIntro|4|1|\left({p \lor \bot}\right) \land \neg \bot|3|Tautology is Negation of Contradiction}} {{SequentIntro|5|1|\left({p \lor \bot}\right) \land \neg \left({p \land \bot}\right)|4|Conjunction with Contradiction}} {{SequentIntro|6|1|p \oplus \bot|5| {{Defof|Exclusive Or}} }} {{EndTableau}} {{qed}}
We apply the Method of Truth Tables to the proposition. As can be seen by inspection, in each case, the truth values in the appropriate columns match for all boolean interpretations. $\begin{array}{|c|ccc||c|ccc|} \hline p & p & \oplus & \bot & p \\ \hline F & F & F & F & F \\ T & T & T & F & T \\ \hline \end{array}$ {{qed}}
=== Definition 1 equivalent to Definition 3 === {{begin-eqn}} {{eqn | o =        | r = \map \Re {\overline {z_1} z_2}       | c = {{Defof|Dot Product|subdef = Complex|index = 3}} }} {{eqn | r = \map \Re {\paren {x_1 - i y_1} {x_2 + i y_2} }       | c = {{Defof|Complex Conjugate}} }} {{eqn | r = \map \Re {\paren {x_1 x_2 + y_1 y_2} + i \paren {x_1 y_2 - x_2 y_1} }       | c = {{Defof|Complex Multiplication}} }} {{eqn | r = x_1 x_2 + y_1 y_2       | c = {{Defof|Real Part}} }} {{end-eqn}} {{qed|lemma}} === Definition 2 equivalent to Definition 3 === {{begin-eqn}} {{eqn | o =        | r = \map \Re {\overline {z_1} z_2}       | c = {{Defof|Dot Product|subdef = Complex|index = 3}} }} {{eqn | r = r_1 r_2 \map \cos {\theta_2 - \theta_1}       | c = Complex Dot Product in Exponential Form }} {{eqn | r = \cmod {z_1} \, \cmod {z_2} \map \cos {\theta_2 - \theta_1}       | c = {{Defof|Polar Form of Complex Number}} }} {{eqn | r = \cmod {z_1} \, \cmod {z_2} \cos \theta       | c = where $\theta = \theta_2 - \theta_1$ is the angle between $z_1$ and $z_2$ }} {{end-eqn}} {{qed|lemma}} === Definition 1 equivalent to Definition 4 === {{begin-eqn}} {{eqn | o =        | r = \frac {\overline {z_1} z_2 + z_1 \overline {z_2} } 2       | c = {{Defof|Dot Product|subdef = Complex|index = 4}} }} {{eqn | r = \frac {\paren {x_1 - i y_1} \paren {x_2 + i y_2} + \paren {x_1 + i y_1} \paren {x_2 - i y_2} } 2       | c = {{Defof|Complex Conjugate}} }} {{eqn | r = \frac {\paren {\paren {x_1 x_2 + y_1 y_2} + i \paren {x_1 y_2 - x_2 y_1} } + \paren {\paren {x_1 x_2 + y_1 y_2} + i \paren {-x_1 y_2 + x_2 y_1} } } 2       | c = {{Defof|Complex Multiplication}} }} {{eqn | r = x_1 x_2 + y_1 y_2       | c = after algebra }} {{end-eqn}} {{qed}}
=== Definition 1 equivalent to Definition 3 === {{begin-eqn}} {{eqn | o =        | r = \map \Im {\overline {z_1} z_2}       | c = {{Defof|Vector Cross Product|subdef = Complex|index = 3}} }} {{eqn | r = \map \Im {\paren {x_1 - i y_1} \paren {x_2 + i y_2} }       | c = {{Defof|Complex Conjugate}} }} {{eqn | r = \map \Im {\paren {x_1 x_2 + y_1 y_2} + i \paren {x_1 y_2 - x_2 y_1} }       | c = {{Defof|Complex Multiplication}} }} {{eqn | r = x_1 y_2 - x_2 y_1       | c = {{Defof|Imaginary Part}} }} {{end-eqn}} {{qed|lemma}} === Definition 2 equivalent to Definition 3 === {{begin-eqn}} {{eqn | o =        | r = \map \Im {\overline {z_1} z_2}       | c = {{Defof|Vector Cross Product|subdef = Complex|index = 3}} }} {{eqn | r = r_1 r_2 \sin \paren {\theta_2 - \theta_1}       | c = Complex Cross Product in Exponential Form }} {{eqn | r = \cmod {z_1} \, \cmod {z_2} \map \sin {\theta_2 - \theta_1}       | c = {{Defof|Polar Form of Complex Number}} }} {{eqn | r = \cmod {z_1} \, \cmod {z_2} \sin \theta       | c = where $\theta = \theta_2 - \theta_1$ is the angle between $z_1$ and $z_2$ }} {{end-eqn}} {{qed|lemma}} === Definition 1 equivalent to Definition 4 === {{begin-eqn}} {{eqn | o =        | r = \frac {\overline {z_1} z_2 - z_1 \overline {z_2} } {2 i}       | c = Definition of Complex Dot Product: Definition 4 }} {{eqn | r = \frac {\paren {x_1 - i y_1} \paren {x_2 + i y_2} - \paren {x_1 + i y_1} \paren {x_2 - i y_2} } {2 i}       | c = {{Defof|Complex Conjugate}} }} {{eqn | r = \frac {\paren {\paren {x_1 x_2 + y_1 y_2} + i \paren {x_1 y_2 - x_2 y_1} } - \paren {\paren {x_1 x_2 + y_1 y_2} + i \paren {-x_1 y_2 + x_2 y_1} } } {2 i}       | c = {{Defof|Complex Multiplication}} }} {{eqn | r = x_1 y_2 - x_2 y_1       | c = after algebra }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l =\overline {z_1} z_2        | r = \paren {z_1 \circ z_2} + i \paren {z_1 \times z_2}       | c = Product of Complex Number with Conjugate by Dot and Cross Product }} {{eqn | r = \cmod {z_1} \, \cmod {z_2} \cos \theta + i \paren {z_1 \times z_2}       | c = {{Defof|Dot Product|subdef = Complex|index = 2}} }} {{eqn | r = \cmod {z_1} \, \cmod {z_2} \cos \theta + i \cmod {z_1} \, \cmod {z_2} \sin \theta       | c = {{Defof|Vector Cross Product|subdef = Complex|index = 2}} }} {{eqn | r = \cmod {z_1} \, \cmod {z_2} \paren {\cos \theta + i \sin \theta}       | c =  }} {{eqn | r = \cmod {z_1} \, \cmod {z_2} e^{i \theta}       | c = Euler's Formula }} {{end-eqn}} {{qed}}
By definition of complex dot product: :$z_1 \circ z_2 = \cmod {z_1} \, \cmod {z_2} \cos \theta$ :$\cmod {z_1}$ denotes the complex modulus of $z_1$ :$\theta$ denotes the angle from $z_1$ to $z_2$, measured in the positive direction. === Necessary Condition === Let $z_1$ and $z_2$ be perpendicular. Then either $\theta = 90^\circ$ or $\theta = 270^\circ$. Either way, from Cosine of Right Angle or Cosine of Three Right Angles: :$\cos \theta = 0$ and so: :$\cmod {z_1} \, \cmod {z_2} \cos \theta = 0$ Hence by definition: :$z_1 \circ z_2 = 0$ {{qed|lemma}} === Sufficient Condition === Let $z_1 \circ z_2 = 0$. Then by definition: :$\cmod {z_1} \, \cmod {z_2} \cos \theta = 0$ As neither $z_1 = 0$ or $z_2 = 0$ it follows that $\cos \theta = 0$. From Cosine of Half-Integer Multiple of Pi it follows that either: :$\theta = 90^\circ$ :$\theta = 270^\circ$ or: :$\theta$ is either of the above plus a full circle. That is, $z_1$ and $z_2$ are perpendicular. {{qed}}
By definition of complex cross product: :$z_1 \times z_2 = \cmod {z_1} \, \cmod {z_2} \sin \theta$ :$\cmod {z_1}$ denotes the complex modulus of $z_1$ :$\theta$ denotes the angle from $z_1$ to $z_2$, measured in the positive direction. === Necessary Condition === Let $z_1$ and $z_2$ be parallel. Then either $\theta = 0^\circ$ or $\theta = 180^\circ$. Either way, from Sine of Zero is Zero or Sine of Straight Angle: :$\sin \theta = 0$ and so: :$\cmod {z_1} \, \cmod {z_2} \sin \theta = 0$ Hence by definition: :$z_1 \times z_2 = 0$ {{qed|lemma}} === Sufficient Condition === Let $z_1 \times z_2 = 0$. Then by definition: :$\cmod {z_1} \, \cmod {z_2} \sin \theta = 0$ As neither $z_1 = 0$ or $z_2 = 0$ it follows that $\sin \theta = 0$. From Sine of Multiple of Pi it follows that either: :$\theta = 0^\circ$ :$\theta = 180^\circ$ or: :$\theta$ is either of the above plus a full circle. That is, $z_1$ and $z_2$ are parallel. {{qed}}
=== Necessary Condition === Let $\mathbf A$ be a tautology. Let $v$ be a boolean interpretation of $\mathbf A$. Then $v \left({\mathbf A}\right) = T$. Hence, by definition of the boolean interpretation of negation: :$v \left({\neg \mathbf A}\right) = F$ Since $v$ was arbitrary, it follows that $\neg \mathbf A$ is unsatisfiable. {{qed|lemma}} === Sufficient Condition === Let $\neg \mathbf A$ be unsatisfiable. Let $v$ be a boolean interpretation of $\neg \mathbf A$. Then $v \left({\neg \mathbf A}\right) = F$. Hence, by definition of the boolean interpretation of negation: :$v \left({\mathbf A}\right) = T$ Since $v$ was arbitrary, it follows that $\mathbf A$ is a tautology. {{qed}}
=== Necessary Condition === Let $\mathbf A$ be satisfiable. Then there exists a boolean interpretation $v$ of $\mathbf A$ such that: :$v \left({\mathbf A}\right) = T$ Hence, by definition of the boolean interpretation of negation: :$v \left({\neg \mathbf A}\right) = F$ It follows that $\neg \mathbf A$ is falsifiable. {{qed|lemma}} === Sufficient Condition === Let $\neg \mathbf A$ be falsifiable. Then there exists a boolean interpretation $v$ of $\neg \mathbf A$ such that: :$v \left({\neg \mathbf A}\right) = F$ Hence, by definition of the boolean interpretation of negation: :$v \left({\mathbf A}\right) = T$ It follows that $\mathbf A$ is satisfiable. {{qed}}
This is an immediate consequence of Subset of Satisfiable Set is Satisfiable. {{qed}}
Since $\mathcal F$ is $\mathscr M$-satisfiable, there exists some model $\mathcal M$ of $\mathcal F$: :$\mathcal M \models_{\mathscr M} \mathcal F$ Thus for every $\psi \in \mathcal F$: :$\mathcal M \models_{\mathscr M} \psi$ Now, for every $\psi$ in $\mathcal F'$: :$\psi \in \mathcal F$ by definition of subset. Hence: :$\forall \psi \in \mathcal F': \mathcal M \models_{\mathscr M} \psi$ that is, $\mathcal M$ is a model of $\mathcal F'$. Hence $\mathcal F'$ is $\mathscr M$-satisfiable. {{qed}} Category:Formal Semantics 5tyf4z8g4z7byror23id2vjcjr5wld2
Since $\mathcal F$ is $\mathscr M$-satisfiable, there exists some model $\mathcal M$ of $\mathcal F$: :$\mathcal M \models_{\mathscr M} \mathcal F$ Since $\psi$ is a tautology, also: :$\mathcal M \models_{\mathscr M} \psi$ Therefore, we conclude that: :$\mathcal M \models_{\mathscr M} \mathcal F \cup \left\{{\phi}\right\}$ i.e., $\mathcal F \cup \left\{{\phi}\right\}$ is satisfiable.
This is an immediate consequence of Superset of Unsatisfiable Set is Unsatisfiable. {{qed}}
By assumption, $\FF$ is unsatisfiable. Suppose now $\FF'$ were satisfiable. Then it would follow from Subset of Satisfiable Set is Satisfiable that $\FF$ were also satisfiable. We conclude that $\FF'$ must be unsatisfiable. {{qed}} Category:Formal Semantics fca2oi6sx6lh7lv4yg1ci7rqrwdwlqq
Suppose $\mathcal F \setminus \set {\phi}$ were satisfiable. Then by Satisfiable Set Union Tautology is Satisfiable, so would $\mathcal F$ be, for: :$\mathcal F = \paren {\mathcal F \setminus \set {\phi} } \cup \set {\phi}$ by Set Difference Union Intersection and Intersection with Subset is Subset. Therefore, $\mathcal F \setminus \set {\phi}$ must be unsatisfiable. {{qed}}
=== Necessary Condition === Assume that $\mathcal F \models_{\mathrm{BI}} \mathbf A$. Suppose $\displaystyle \bigwedge \mathcal F \implies \mathbf A$ were not a tautology. Then there exists a boolean interpretation $v$ such that: :$v \left({\displaystyle \bigwedge \mathcal F}\right) = T$ :$v \left({\mathbf A}\right) = F$ by definition of the boolean interpretation of $\implies$. It now follows from the boolean interpretation of conjunction that: :$v \left({\mathbf B}\right) = T$ for every $\mathbf B \in \mathcal F$. Hence, by definition of model: :$v \models_{\mathrm{BI}} \mathcal F$. It now follows from our assumption that $v \left({\mathbf A}\right) = T$. This is a contradiction, hence $\displaystyle \bigwedge \mathcal F \implies \mathbf A$ is a tautology. {{qed|lemma}} === Sufficient Condition === Assume that $\displaystyle \bigwedge \mathcal F \implies \mathbf A$ is a tautology. Let $v$ be an arbitrary model of $\mathcal F$. Then: :$v \left({\mathbf B}\right) = T$ for every $\mathbf B \in \mathcal F$, whence by the boolean interpretation of conjunction: :$v \left({\displaystyle \mathcal F}\right) = T$ Since $\displaystyle \bigwedge \mathcal F \implies \mathbf A$ is a tautology, it must be that: :$v \left({\mathbf A}\right) = T$ Hence, since $v$ was arbitrary: :$\mathcal F \models_{\mathrm{BI}} \mathbf A$ {{qed}}
Any model of $\FF \cup \FF'$ is a fortiori also a model of $\FF$. By definition of semantic consequence all models of $\FF$ are models of $\phi$. Therefore all models of $\FF \cup \FF'$ are also models of $\phi$. Hence: :$\FF \cup \FF' \models_{\mathscr M} \phi$ as desired. {{qed}} Category:Formal Semantics cowaaykdur0gx37z85obu9qhreoc7ck
Let $\mathcal M$ be a model of $\mathcal F \setminus \left\{{\psi}\right\}$. Since $\psi$ is a tautology, it follows that: :$\mathcal M \models_{\mathscr M} \psi$ Hence: :$\mathcal M \models \mathcal F$ which, by hypothesis, entails: :$\mathcal M \models \phi$ Since $\mathcal M$ was arbitrary, it follows by definition of semantic consequence that: :$\mathcal F \setminus \left\{{\psi}\right\} \models_{\mathscr M} \phi$ {{qed}}
Let $S$ be a set. Let $f: S \to \R$ be a real-valued function. === Definition 1 implies Definition 2 === Let $f$ be bounded according to definition 1: :bounded above in $S$ by $H \in \R$ :bounded below in $S$ by $L \in \R$ Thus by definition: :$\forall x \in S: L \le \map f x$ :$\forall x \in S: \map f x \le H$ By Equivalence of Definitions of Bounded Subset of Real Numbers: :$\exists K \in \R_{\le 0}: \forall x \in S: \size {\map f x} \le K$ So $f$ is bounded according to definition 2. {{qed|lemma}} === Definition 2 implies Definition 1 === Let $f$ be bounded according to definition 2: :$\exists K \in \R_{\le 0}: \forall x \in S: \size {\map f x} \le K$ Then by Equivalence of Definitions of Bounded Subset of Real Numbers: :$\exists K \in \R_{\le 0}: \forall x \in S: -K \le \map f x$ and so $f$ is bounded below in $S$ by $-K \in \R$ and :$\exists K \in \R_{\le 0}: \forall x \in S: \map f x \le K$ and so $f$ is bounded above in $S$ by $K \in \R$. So $f$ is bounded according to definition 2. {{qed|lemma}} === Definition 1 implies Definition 3 === Let $f$ be bounded according to definition 1: :bounded above in $S$ by $H \in \R$ :bounded below in $S$ by $L \in \R$ Thus by definition: :$\forall x \in S: L \le \map f x$ :$\forall x \in S: \map f x \le H$ That is: :$\map f x \in \closedint L H$ Thus $f$ is bounded according to definition 3. {{qed|lemma}} === Definition 3 implies Definition 1 === Let $f$ be bounded according to definition 3: :$\exists a, b \in \R_{\le 0}: \forall x \in S: \map f x \in \closedint a b$ Thus by definition: :$\forall x \in S: a \le \map f x$ and so $f$ is bounded below in $S$ by $a \in \R$ :$\forall x \in S: \map f x \le b$ and so $f$ is bounded above in $S$ by $b \in \R$. So $f$ is bounded according to definition 1. {{qed}} Category:Boundedness 773q2gghv21uzjiclkxh4gx6m4wwws4
Let $H'$ denote the set of limit points of $H$. === Definition 1 implies Definition 2 === Let $H$ be closed in $M$ by definition 1. Then by definition $A \setminus H$ is open in $M$. Let $b \in M: b \in A \setminus H$. Then by definition of open set: :$\exists \delta \in \R_{>0}: B_\delta \left({b; d}\right) \subseteq A \setminus H$ where $B_\delta \left({b; d}\right)$ denotes the open $\delta$-ball of $b$. From Intersection with Complement is Empty iff Subset: :$B_\delta \left({b; d}\right) \cap H = \varnothing$ So by definition of limit point: :$b \notin H'$ Hence: :$A \setminus H \cap H' = \varnothing$ By Intersection with Complement is Empty iff Subset: :$H' \subseteq H$ Thus $H$ contains all its limit points. That is, $H$ be closed in $M$ by definition 2. {{qed|lemma}} === Definition 2 implies Definition 1 === Let $H$ be closed in $M$ by definition 2. That is: :$H' \subseteq H$ Then by Set Complement inverts Subsets: :$A \setminus H \subseteq A \setminus H'$ Let $b \in A \setminus H$. Then by definition of relative complement: :$b \notin H'$ Then by definition of limit point: :$\exists \delta \in \R_{>0}: B_\delta \left({b; d}\right) \cap H = \varnothing$ and so: :$B_\delta \left({b; d}\right) \subseteq A \setminus H$ As $b$ is arbitrary, it follows that $A \setminus H$ is open in $M$. Hence by definition, $H$ is closed in $M$ by definition 1. {{qed}}
=== Necessary Condition === Suppose $S$ contained the complementary pair $\left\{{p, \neg p}\right\}$. By the Principle of Non-Contradiction, $\left\{{p, \neg p}\right\}$ is unsatisfiable. By Superset of Unsatisfiable Set is Unsatisfiable, so is $S$. Hence if $S$ is satisfiable, it cannot contain any complementary pairs. {{qed|lemma}} === Sufficient Condition === Suppose $S$ contains no complementary pair. Define the boolean interpretation $v: \mathcal P_0 \to \left\{{T, F}\right\}$ by: :$v \left({p}\right) = \begin{cases} T & \text{if $p \in S$}\\ F & \text{otherwise} \end{cases}$ Let $\mathbf A \in S$ be an arbitary literal. If $\mathbf A = p$ is a positive literal, then $v \left({\mathbf A}\right) = T$ is true by the first clause of the definition. Now suppose $\mathbf A = \neg p$ is a negative literal. Then $p \notin S$, for $S$ contains no complementary pairs. Therefore $v \left({p}\right) = F$, and hence: :$v \left({\neg p}\right) = v \left({\mathbf A}\right) = T$ Since $v$ has been demonstrated to be a model of $S$, $S$ is satisfiable. {{qed}}
Let $S \subseteq \C$. === Definition 1 implies Definition 2 === Let $z_0$ be an exterior point of $S$ by definition 1. Let $N_\epsilon \left({z_0}\right)$ be an $\epsilon$-neighborhood of $z_0$ such that $N_\epsilon \left({z_0}\right) \cap S = \varnothing$. That is, $N_\epsilon \left({z_0}\right)$ has no points which are also in $S$. By definition, it follows that $z_0$ is not a boundary point of $S$. Aiming for a contradiction, suppose $z_0$ is an interior point of $S$. Let $N_{\epsilon'} \left({z_0}\right)$ be an $\epsilon'$-neighborhood such that $N_{\epsilon'} \left({z_0}\right) \subseteq S$. By Empty Intersection iff Subset of Complement: : $N_\epsilon \left({z_0}\right) \subseteq \complement_\C \left({s}\right)$ {{explain|Need a result to show that neighborhoods of a given point cannot be disjoint}} Thus $z_0$ is an exterior point of $S$ by definition 2. {{qed|lemma}} === Definition 2 implies Definition 1 === Let $z_0$ be an exterior point of $S$ by definition 2. As $z_0$ is not an interior point of $S$ there exists no $\epsilon$-neighborhood of $z_0$ which is disjoint from $\complement_\C \left({S}\right)$. That is, every $\epsilon$-neighborhood of $z_0$ contains points which are not in $S$. As $z_0$ is not a boundary point of $S$, there exists at least one $\epsilon$-neighborhood of $z_0$ which does not contain both points in $S$ and points not in $S$. But as every $\epsilon$-neighborhood of $z_0$ contains points which are not in $S$, it follows there must be at least one $\epsilon$-neighborhood of $z_0$ disjoint from $S$. Thus $z_0$ is an exterior point of $S$ by definition 1. {{qed}} Category:Complex Analysis pl9wkbxw2v52gx6jo64t60r52h3rebz
Let $S \subseteq \C$. === Definition 1 implies Definition 2 === Let $S$ be an open set in $\C$ by definition 1. That is: : $\forall z_0 \in S: \exists \epsilon \in \R_{>0}: N_{\epsilon} \left({z_0}\right) \subseteq S$ Thus every point in $S$ has an $\epsilon$-neighborhood $N_{\epsilon} \left({z_0}\right)$ such that $N_{\epsilon} \left({z_0}\right) \subseteq S$. That is, by definition, every point of $S$ is an interior point. Thus $S$ is an open set in $\C$ by definition 2. {{qed|lemma}} === Definition 2 implies Definition 1 === Let $S$ be an open set in $\C$ by definition 2. That is, every point of $S$ is an interior point. This means: : $\forall z_0 \in S: \exists \epsilon \in \R_{>0}: N_{\epsilon} \left({z_0}\right) \subseteq S$ Thus $S$ is an open set in $\C$ by definition 1. {{qed}} Category:Complex Analysis 9n6tna5wwh2g0jpnkb5payht4ksuwln