hcju/proofwikiqa huggingface dataset:
It is useful to first state the Cayley tables for the three logical operations $\lor$, $\land$ and $\neg$: :$\begin{array}{c|cc}  \lor & \bot & \top \\ \hline  \bot & \bot & \top \\  \top & \top & \top \end{array} \qquad \begin{array}{c|cc}  \land & \bot & \top \\ \hline  \bot  & \bot & \bot \\  \top  & \bot & \top \end{array} \qquad \begin{array}{c|cc}       & \bot & \top \\ \hline  \neg & \top & \bot \end{array}$ Let us now verify the axioms for a Boolean algebra in turn. === $(BA \ 0)$: Closure === It is immediate from the Cayley tables that $S$ is closed under $\lor$, $\land$ and $\neg$. {{qed|lemma}} === $(BA \ 1)$: Commutativity === Follows from the Rule of Commutation. {{qed|lemma}} === $(BA \ 2)$: Distributivity === Follows from the Rule of Distribution {{qed|lemma}} === $(BA \ 3)$: Identities === Follows from Conjunction with Tautology and Disjunction with Contradiction. {{qed|lemma}} === $(BA \ 4)$: Complements === Follows from Contradiction is Negation of Tautology and Tautology is Negation of Contradiction. {{qed|lemma}} Having verified all axioms, we conclude $\mathbf 2$ is a Boolean algebra. {{qed}}
Let us verify the axioms for a Boolean algebra in turn. === $(BA \ 0)$: Closure === Follows from Induced Operations Preserve Closure. {{qed|lemma}} === $(BA \ 1)$: Commutativity === Follows from Induced Operations Preserve Commutativity. {{qed|lemma}} === $(BA \ 2)$: Distributivity === Follows from Induced Operations Preserve Distributivity. {{qed|lemma}} === $(BA \ 3)$: Identities === Follows from Identity for Induced Operation. {{qed|lemma}} === $(BA \ 4)$: Complements === Follows from Induced Operations Preserve Identities. {{qed|lemma}} Having verified all the axioms, we conclude $\mathbf 2^X$ is a Boolean algebra. {{qed}}
From Symmetric Difference with Intersection forms Ring, $\left({S, *, \cap}\right)$ is a ring with unity. By Intersection is Idempotent, $\cap$ is idempotent. It follows that $\left({S, *, \cap}\right)$ is a Boolean ring. {{qed}}
From Support Operation Inverse to Characteristic Function Operation, $\chi_{\left({\cdot}\right)}$ is a bijection. It therefore suffices to establish it is a ring homomorphism. By Characteristic Function of Symmetric Difference: :$\chi_{A * B} = \chi_A + \chi_B - 2 \chi_A \chi_B$ Since $\mathbf 2^S$ is a Boolean ring, by Idempotent Ring has Characteristic Two, the right-hand side reduces to: :$\chi_{A * B} = \chi_A + \chi_B$ showing that $\chi_{\left({\cdot}\right)}$ preserves ring addition. By Characteristic Function of Intersection: Variant 1: :$\chi_{A \cap B} = \chi_A \cdot \chi_B$ showing that $\chi_{\left({\cdot}\right)}$ preserves the ring product. Hence $\chi_{\left({\cdot}\right)}$ is a ring homomorphism. The result follows. {{qed}}
{{BeginTableau|\vdash \paren {\paren {p \implies q} \implies p} \implies p}} {{Assumption|1|\paren {p \implies q} \implies p}} {{SequentIntro|2|1|p|1|Peirce's Law: Formulation 1: $\paren {p \implies q} \implies p \vdash p$}} {{Implication|3||\paren {\paren {p \implies q} \implies p} \implies p|1|2}} {{EndTableau}} {{qed}}
We apply the Method of Truth Tables to the proposition. As can be seen by inspection, the truth values under the main connective are $T$ for all boolean interpretations. $\begin{array}{|ccccc|c|c|}\hline ((p & \implies & q) & \implies & p) & \implies & p \\ \hline F & T & F & F & F & T & F \\ F & T & T & F & F & T & F \\ T & F & F & T & T & T & T \\ T & T & T & T & T & T & T \\ \hline \end{array}$ {{qed}}
We apply the Method of Truth Tables. As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations. $\begin{array}{|ccccc||c|}\hline ((p & \implies & q) & \implies & p) & p \\ \hline F & T & F & F & F & F \\ F & T & T & F & F & F \\ T & F & F & T & T & T \\ T & T & T & T & T & T \\ \hline \end{array}$ {{qed}} {{Namedfor|Charles Sanders Peirce}} Category:Peirce's Law Category:Truth Table Proofs h176q3qbxl3f7z8ak724c46xngv1otc
{{BeginTableau|\neg \left({p \implies q}\right) \vdash p \land \neg q}} {{Premise|1|\neg \left({p \implies q}\right)}} {{Assumption|2|\neg \left({p \land \neg q}\right)}} {{SequentIntro|3|2|p \implies q|2|Implication Equivalent to Negation of Conjunction with Negative}} {{NonContradiction|4|1, 2|3|1}} {{Reductio|5|1|p \land \neg q|2|4}} {{EndTableau}} {{qed}} {{LEM|Reductio ad Absurdum}} Category:Conjunction with Negative Equivalent to Negation of Implication 74vn9lygaw57ac3s1upxd6d0jrjstff
We apply the Method of Truth Tables. As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations. $\begin{array}{|cccc||cccc|} \hline p & \land & \neg & q & \neg & (p & \implies & q) \\ \hline F & F & T & F & F & F & T & F \\ F & F & F & T & F & F & T & T \\ T & T & T & F & T & T & F & F \\ T & F & F & T & F & T & T & T \\ \hline \end{array}$ {{qed}}
To prove a relation is an equivalence, we need to prove it is reflexive, symmetric and transitive. So, checking in turn each of the criteria for equivalence: === Reflexive === By hypothesis $\mathcal R$ is reflexive. {{Qed|lemma}} === Symmetric === By reflexivity: :$\tuple {x, x} \in \mathcal R$ If $\tuple {x, y} \in \mathcal R$ then by the definition of circular relation $\tuple {y, x} \in \mathcal R$. Hence $\mathcal R$ is symmetric. {{Qed|lemma}} === Transitive === Let $\tuple {x, y}, \tuple {y, z} \in \mathcal R$. By definition of circular relation: :$\tuple {z, x} \in \mathcal R$ By $\mathcal R$ being symmetric: :$\tuple {x, z} \in \mathcal R$ Hence $\mathcal R$ is transitive. {{Qed|lemma}} Thus is $\mathcal R$ is reflexive, symmetric and transitive, and therefore by definition an equivalence. {{Qed}}
Let $u, v: \closedint a b \times \set 0 \to \R$ be defined as in the Cauchy-Riemann Equations: :$\map u {t, y} = \map \Re {\map F z}$ :$\map v {t, y} = \map \Im {\map F z}$ where: :$\map \Re {\map F z}$ denotes the real part of $\map F z$ :$\map \Im {\map F z}$ denotes the imaginary part of $\map F z$. Then: {{begin-eqn}} {{eqn | l = \int_a^b \map f t \rd t       | r = \int_a^b \map {F'} {t + 0 i} \rd t       | c = by assumption }} {{eqn | r = \int_a^b \paren {\map {\dfrac {\partial u} {\partial t} } {t, 0} + i \map {\dfrac {\partial v} {\partial t} } {t, 0} } \rd t       | c = Cauchy-Riemann Equations }} {{eqn | r = \int_a^b \map {\dfrac {\partial u} {\partial t} } {t, 0} \rd t + i \int_a^b \map {\dfrac {\partial v} {\partial t} } {t, 0} \rd t       | c = {{Defof|Complex Riemann Integral}} }} {{eqn | r = \map u {b, 0} - \map u {a, 0} + i \paren {\map v {b, 0} - \map v {a, 0} }       | c = Fundamental Theorem of Calculus  }} {{eqn | r = \map F b - \map F a }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \int_C \map f z       | r = \sum_{k \mathop = 1}^n \int_{a_k}^{b_k} \map f {\map {\gamma_k} t} \map {\gamma_k'} t \rd t       | c = {{Defof|Complex Contour Integral}} }} {{eqn | r = \sum_{k \mathop = 1}^n \int_{a_k}^{b_k} \paren {\dfrac \rd {\rd t} \map F {\map {\gamma_k} t} } \rd t       | c = Derivative of Complex Composite Function }} {{eqn | r = \sum_{k \mathop = 1}^n \paren {\map F {\map {\gamma_k} {b_k} } - \map F {\map {\gamma_k} {a_k} } }       | c = Fundamental Theorem of Calculus for Complex Riemann Integrals }} {{eqn | r = \map F {\map {\gamma_n} {b_n} } - \map F {\map {\gamma_1} {a_1} }       | c = the sum is telescoping }} {{eqn | r = \map F w - \map F z       | c = {{Defof|Endpoints of Contour (Complex Plane)}} }} {{end-eqn}} If $C$ is a closed contour, then $z = w$. It follows that: :$\map F w - \map F z = 0$ {{qed}}
First, we prove by induction over $k \in \N_{\ge 1}$ that: :$\displaystyle f^{\left({k}\right) } \left({z}\right) = \sum_{n \mathop = k}^\infty a_n \left({z - \xi}\right)^{n - k} n^{\underline k}$ where $n^{\underline k}$ denotes the falling factorial. === Basis for the Induction === For $k = 1$, it follows from Derivative of Complex Power Series that  :$\displaystyle f^{\left({k}\right) } \left({z}\right) = \sum_{n \mathop = 1}^\infty n a_n \left({z - \xi}\right)^{n - 1}$ As $n = n^{\underline 1}$, this proves the hypothesis. From Radius of Convergence of Derivative of Complex Power Series, it follows that the equation holds for all $z \in \C$ with $\left\vert{z - \xi}\right\vert < R$. === Induction Hypothesis === For fixed $k \in \N_{\ge 1}$, the hypothesis is that: :$\displaystyle f^{\left({k}\right) } \left({z}\right) = \sum_{n \mathop = k}^\infty a_n \left({z - \xi}\right)^{n-k} n^{\underline k}$ and $f^{\left({k}\right) }$ has radius of convergence $R$. === Induction Step === Note that $f^{\left({k+1}\right) }$ is the derivative of $f^{\left({k}\right) }$. From Radius of Convergence of Derivative of Complex Power Series, it follows that $f^{\left({k+1}\right) }$ has radius of convergence $R$. For $z \in \C$ with $\left\vert{z - \xi}\right\vert < R$, we have: {{begin-eqn}} {{eqn | l = f^{\left({k+1}\right) }       | r = \dfrac{\mathrm d}{\mathrm dz} \left({\sum_{n \mathop = k}^\infty a_n \left({z - \xi}\right)^{n-k} n^{\underline k} }\right)       | c = Induction Hypothesis }} {{eqn | r = \dfrac{\mathrm d}{\mathrm dz} \left({\sum_{n \mathop = 0}^\infty a_{n+k} \left({z - \xi}\right)^{n} \left({n+k}\right)^{\underline k} }\right)       | c = subtracting $k$ from $n$ }} {{eqn | r = \sum_{n \mathop = 1}^\infty n a_{n+k} \left({z - \xi}\right)^{n-1} \left({n+k}\right)^{\underline k}       | c = Derivative of Complex Power Series }} {{eqn | r = \sum_{n \mathop = 1}^\infty a_{n+k} \left({z - \xi}\right)^{n-1} \left({n+k}\right)^{\underline{k+1} }       | c = {{Defof|Falling Factorial}} }} {{eqn | r = \sum_{n \mathop = k+1}^\infty a_{n} \left({z - \xi}\right)^{n - \left({k+1}\right) } n^{\underline{k+1} }       | c = adding $k$ to $n$ }} {{end-eqn}} {{qed|lemma}} With $k \in \N$, we have: {{begin-eqn}} {{eqn | l = f^{\left({k}\right)} \left({\xi}\right)       | r = \sum_{n \mathop = k}^\infty a_n \left({\xi - \xi}\right)^{n - k} n^{\underline k} }} {{eqn | r = a_k 0^0 k^{\underline k} }} {{eqn | r = a_k k! }} {{end-eqn}} Hence, it follows that $a_n = \dfrac{f^{\left({n}\right)} \left({\xi}\right)} {n!}$. By definition of Taylor series, it follows that $f$ is equal to its Taylor series expansion about $\xi$. {{qed}}
From Radius of Convergence of Power Series over Factorial: Complex Case, it follows that the power series $\displaystyle \sum_{n \mathop = 0}^\infty \dfrac {z^n} {n!}$ is absolutely convergent over the entirety of $\C$. Hence, the definition of $\exp z$ as a sum of a series is valid. It remains to demonstrate the logical equivalence of all the definitions. === Sum of Series equivalent to Solution of Differential Equation === {{:Equivalence of Definitions of Complex Exponential Function/Sum of Series equivalent to Differential Equation}}{{qed|lemma}} === Sum of Series equivalent to Definition by Real Functions === {{:Equivalence of Definitions of Complex Exponential Function/Sum of Series equivalent to Definition by Real Functions}}{{qed|lemma}} === Sum of Series equivalent to Limit of Sequence === {{:Equivalence of Definitions of Complex Exponential Function/Sum of Series equivalent to Limit of Sequence}}{{qed}} Category:Exponential Function qjpkvc9f8dx1a4q0nwy5x8wan73qli9
Let $z \in \C \setminus \left\{ {0}\right\}$. Let $r = \cmod z$ be the modulus of $z$, and let $\theta = \arg \left({z}\right)$ be the argument of $z$. Then $r>0$. Let $\ln$ denote the real natural logarithm, and let $e$ denote the real exponential function. Then: {{begin-eqn}} {{eqn | l = \exp \left({\ln r + i \theta}\right)       | r = e^{ \ln r } \left({\cos \theta + i \sin \theta }\right)       | c = {{Defof|Exponential Function/Complex|subdef = Real Functions|Complex Exponential Function}} }} {{eqn | r = r \left({\cos \theta + i \sin \theta }\right)       | c = Exponential of Natural Logarithm }} {{eqn | r = z       | c = {{Defof|Polar Form of Complex Number}} }} {{end-eqn}} Hence, $z \in \operatorname{Im} \left({\exp}\right)$. Suppose instead that $z=0$. Let $z_0 = r_0 \left({\cos \theta_0 + i \sin \theta_0 }\right) \in \C$. From Exponential Tends to Zero and Infinity, it follows that $e^{ r_0 } \ne 0$. As $\cmod {\cos \theta_0 + i \sin \theta_0} = 1$, it follows that $\cos \theta_0 + i \sin \theta_0 \ne 0$. Then this equation has no solutions: :$0 = \exp z_0 = e^{ r_0 } \left({\cos \theta_0 + i \sin \theta_0 }\right)$ Hence, $\operatorname{Im} \left({\exp}\right) = \C \setminus \left\{ {0}\right\}$- {{qed}}
{{begin-eqn}} {{eqn | l = \map \exp {-z}       | r = \dfrac {\map \exp {-z} } {\map \exp 0}       | c = as $\map \exp 0 = 1$ by Exponential of Zero }} {{eqn | r = \dfrac {\map \exp {-z} } {\map \exp {z - z} } }} {{eqn | r = \dfrac {\map \exp {-z} } {\map \exp z \, \map \exp {-z} }       | c = Exponential of Sum: Complex Numbers }} {{eqn | r = \dfrac 1 {\map \exp z} }} {{end-eqn}} {{qed}}
By the definition of an ordered group, $\preceq$ is compatible with $\circ$. By the definition of an ordering, $\preceq$ is transitive. By Reflexive Reduction of Relation Compatible with Group Operation is Compatible, $\prec$ is also compatible with $\circ$. By Reflexive Reduction of Transitive Antisymmetric Relation is Strict Ordering, $\prec$ is also transitive. By the definition of an ordered group, $\left({S,\circ}\right)$ is a group, and therefore a semigroup. Thus the theorem holds by Transitive Relation Compatible with Semigroup Operation Relates Powers of Related Elements. {{qed}} Category:Ordered Semigroups hce7aaxz91tgmtbcgbfykv7efadcsbd
{{begin-eqn}} {{eqn | l = \map \exp {z + 2 k \pi i}       | r = \map \exp z \, \map \exp {2 k \pi i}       | c = Exponential of Sum: Complex Numbers }} {{eqn | r = \map \exp z \times 1       | c = Euler's Formula Example: $e^{2 k i \pi}$ }} {{eqn | r = \map \exp z }} {{end-eqn}} {{qed}}
From Connected Domain is Connected by Staircase Contours, it follows that there exists a staircase contour $C_w$ in $D$ with start point $z_0$ and end point $w$. If $C_w'$ is another staircase contour with the same endpoints as $C_w$, then $C_w' \cup \paren {-C_w}$ is a closed staircase contour. Then the definition of $F$ is independent of the choice of contour, as: {{begin-eqn}} {{eqn | l = \int_{C_w} \map f z \rd z       | r = \int_{C_w} \map f z \rd z + \int_{C_w' \cup \paren {-C_w} } \map f z \rd z       | c = by assumption }} {{eqn | r = \int_{C_w} \map f z \rd z + \int_{C_w'} \map f z \rd z + \int_{-C_w} \map f z \rd z       | c = Contour Integral of Concatenation of Contours }} {{eqn | r = \int_{C_w} \map f z \rd z + \int_{C_w'} \map f z \rd z - \int_{C_w} \map f z \rd z       | c = Contour Integral along Reversed Contour }} {{eqn | r = \int_{C_w'} \map f z \rd z }} {{end-eqn}} We now show that $F$ is the primitive of $f$. Let $\epsilon \in \R_{>0}$. By definition of continuity, there exists $r \in \R_{>0}$ such that the open ball $\map {B_r} w \subseteq D$, and for all $z \in \map {B_r} w$: :$\size {\map f z - \map f w} < \dfrac \epsilon 2$ Let $h = x+iy \in \C \setminus \set 0$ with $x, y \in \R$ such that $\size h < r$. Let $\LL$ be the staircase contour that goes in a horizontal line from $w$ to $w + x$, and continues in a vertical line from $w + x$ to $w + h$. As $w + x, w + h \in \map {B_r} w$, it follows from Open Ball is Convex Set that $\LL$ is a contour in $\map {B_r} w$. Then $C_w \cup \LL$ is a staircase contour from $z_0$ to $w + h$, so: {{begin-eqn}} {{eqn | l = \map F {w + h} - \map F w       | r = \int_{C_w \cup \LL} \map f z \rd z - \int_{C_w} \map f z \rd z }} {{eqn | r = \int_\LL \map f z \rd z       | c = Contour Integral of Concatenation of Contours }} {{end-eqn}} From Derivative of Complex Polynomial, it follows that $\dfrac \rd {\rd z} \map f w z = \map f w$, so: {{begin-eqn}} {{eqn | l = \int_\LL \map f w \rd z       | r = \map f w \paren {w + h} - \map f w w       | c = Fundamental Theorem of Calculus for Contour Integrals }} {{eqn | r = h \map f w }} {{end-eqn}} We can now show that $\map {F'} w = \map f w$, as: {{begin-eqn}} {{eqn | l = \size {\dfrac {\map F {w + h} - \map F w} h - \map f w}       | r = \size {\dfrac 1 h \int_\LL \map f z \rd z - \dfrac 1 h h \map f w}       | c = by the above calculations }} {{eqn | r = \size {\dfrac 1 h} \size {\int_\LL \paren {\map f z - \map f w} \rd z}       | c = Linear Combination of Contour Integrals }} {{eqn | o = <        | r = \size {\dfrac 1 h} \dfrac \epsilon 2 \map L \LL       | c = Estimation Lemma, as $z \in \map {B_r} w$ }} {{eqn | r = \dfrac {\size x + \size y} {\size h} \dfrac \epsilon 2        | c = the lengths of the line segments are $\size x$ and $\size y$ }} {{eqn | o = \le       | r = \epsilon       | c = Modulus Larger than Real Part and Imaginary Part }} {{end-eqn}} When $h$ tends to $0$, we have $\map {F'} w = \map f w$ by definition of differentiability. {{qed}}
For all $n > N$, put $a_n = 0$. Then: :$\displaystyle f \left({z}\right) = \sum_{n \mathop = 0}^\infty a_n z^n$ The result now follows from Derivative of Complex Power Series. {{qed}} Category:Complex Differential Calculus t58tbo04g3a75np4x3mq3x7asipi2is
Let $S_n = B^-_{\rho_n} \left({x_n}\right)$ be the closed ball of radius $\rho_n$ about the point $x_n$. That is, let $S_n = \left\{{x \in A: d \left({x_n, x}\right) \le \rho_n}\right\}$. Then the sequence $\left\langle{x_n}\right\rangle$ forms a Cauchy sequence: :$d \left({x_n, x_{n+p}}\right) < \rho_n$ for any $p \ge 0$ since $S_{n+p} \subseteq S_n$. However, $\rho_n \to 0$ as $n \to \infty$ and therefore $d \left({x_n, x_{n+p}}\right) \to 0$ as $n \to \infty$ for any $p \ge 0$. Since the space $M$ is complete, there exists $x \in X$ such that $x_n \to x$ as $n \to \infty$. Then since the subsequence $\left\langle{x_k}\right\rangle_{k = n}^\infty$ is contained entirely in $S_n$ and converges to $x$, $x$ is a member of the closure of $S_n$ by Closure of Subset of Metric Space by Convergent Sequence. Since $S_n$ is closed, by Closed Set Equals its Closure, we have: :$\forall n \in \N: x \in S_n$ Hence: :$\displaystyle x \in \bigcap_{n \mathop = 1}^\infty S_n$ Suppose now that $y \ne x$. Then it follows that $d \left({x, y}\right) > 0$, and hence that for some $n$: :$d \left({x, y}\right) > 2 \rho_n$ Since $x, x_n \in S_n$, it then follows that: :$d \left({x, x_n}\right) \le \rho_n$ Now: {{begin-eqn}} {{eqn|l = 2 \rho_n      |o = <      |r = d \left({x, y}\right) }} {{eqn|o = \le      |r = d \left({x, x_n}\right) + d \left({x_n, y}\right)       |c = Axiom $(M2)$ for metrics }} {{eqn|o = \le      |r = \rho_n + d \left({x_n, y}\right) }} {{end-eqn}} From the above, it follows that: :$d \left({x_n, y}\right) > \rho_n$ so that $y \notin S_n$, and consequently: :$\displaystyle y \notin \bigcap_{i \mathop = 1}^\infty S_n$ Hence: :$\displaystyle \bigcap_{n \mathop = 1}^\infty S_n = \left\{{x}\right\}$ {{qed}} Category:Complete Metric Spaces Category:Named Theorems pzhppcz0hslzecdycn1u21bshgjp17h
{{BeginTableau|\left({p \implies q}\right) \land \left({\neg p \implies q}\right) \vdash q}} {{Premise|1|\left({p \implies q}\right) \land \left({\neg p \implies q}\right)}} {{Simplification|2|1|p \implies q|1|1}} {{Simplification|3|1|\neg p \implies q|1|2}} {{SequentIntro|4|1|p \lor \neg p \implies q \lor q|2, 3|Constructive Dilemma}} {{ExcludedMiddle|5|p \lor \neg p}} {{ModusPonens|6|1|q \lor q|4|5}} {{Idempotence|7|1|q|6|Disjunction}} {{EndTableau}} {{Qed}}
From the Constructive Dilemma we have: : $p \implies q, r \implies s \vdash p \lor r \implies q \lor s$ from which, changing the names of letters strategically: : $p \implies q, \neg p \implies q \vdash p \lor \neg p \implies q \lor q$ From Law of Excluded Middle, we have: : $\vdash p \lor \neg p$ From the Rule of Idempotence we have: : $q \lor q \vdash q$ and the result follows by Hypothetical Syllogism. {{qed}}
We apply the Method of Truth Tables. As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations. $\begin{array}{|cccccccc||c|} \hline (p & \implies & q) & \land & (\neg & p & \implies & q) & q \\ \hline F & T & F & F & T & F & F & F & F \\ F & T & T & T & T & F & T & T & T \\ T & F & F & F & F & T & T & F & F \\ T & T & T & T & F & T & T & T & T \\ \hline \end{array}$ {{qed}}
{{BeginTableau|q \vdash \left({p \implies q}\right) \land \left({\neg p \implies q}\right)}} {{Premise|1|q}} {{SequentIntro|2|1|p \implies q|1|True Statement is implied by Every Statement}} {{SequentIntro|3|1|\neg p \implies q|1|True Statement is implied by Every Statement}} {{Conjunction|4|1|\left({p \implies q}\right) \land \left({\neg p \implies q}\right)|2|3}} {{EndTableau}} {{qed}} Category:Principle of Dilemma 9yblx1tnx4w60wc2mrbf3rgkgc15rnc
{{BeginTableau|\vdash q \implies \paren {\paren {p \implies q} \land \paren {\neg p \implies q} } }} {{Assumption|1|q|}} {{SequentIntro|2|1|\paren {p \implies q} \land \paren {\neg p \implies q}|1|Principle of Dilemma: Formulation 1: Reverse Implication}} {{Implication|3||q \implies \paren {\paren {p \implies q} \land \paren {\neg p \implies q} }|1|2}} {{EndTableau}} {{qed}} Category:Principle of Dilemma d9q4uvsexr57d799yi2aupfipflfpf2
=== Necessary Condition === Complex Plane is Metric Space shows that $\C$ is topologically equivalent to the Euclidean space $\R^2$. The result follows from Connected Open Subset of Euclidean Space is Path-Connected. {{explain|It still needs to be established that connectedness is preserved by a homeomorphism.}} {{qed|lemma}} === Sufficient Condition === The result follows from Path-Connected Space is Connected. {{qed}} Category:Complex Analysis t5htvab26mdou1b95gqf3n80posp7i4
{{BeginTableau  |\vdash \paren {\paren {p \implies q} \land \paren {p \implies \neg q} } \implies \neg p}} {{Assumption    |1|\paren {p \implies q} \land \paren {p \implies \neg q} }} {{Simplification|2|1|p \implies q|1|1}} {{Simplification|3|1|p \implies \neg q|1|2}} {{SequentIntro  |4|1|\neg p|2, 3|Proof by Contradiction: Formulation 1}} {{Implication   |5| |\paren {\paren {p \implies q} \land \paren {p \implies \neg q} } \implies \neg p|1|4}} {{EndTableau}} {{Qed}}
=== $(1)$ implies $(2)$ === If $F$ is a primitive of $f$, we have: {{begin-eqn}} {{eqn | l = \int_{C_1} \map f z \rd z        | r = \map F {z_2} - \map F {z_1}       | c = Fundamental Theorem of Calculus for Contour Integrals }} {{eqn | r = \int_{C_2} \map f z \rd z  }} {{end-eqn}} {{qed|lemma}} === $(2)$ implies $(3)$ === Let $C$ be a closed contour in $D$ with endpoints $z_0$. Let the constant function $\gamma: \closedint 0 1 \to D$ with $\map \gamma t = z_0$ be the parameterization of a contour $C_0$. Then: {{begin-eqn}} {{eqn | l = \oint_C \map f z \rd z       | r = \oint_{C_0} \map f z \rd z       | c = by assumption }} {{eqn | r = \int_0^1 \map f {\map \gamma t} \map {\gamma'} t \rd t       | c = {{Defof|Complex Contour Integral}} }} {{eqn | r = 0       | c = by Derivative of Complex Polynomial, as $\gamma$ is constant }} {{end-eqn}} {{qed|lemma}} === $(3)$ implies $(1)$ === This follows from Zero Staircase Integral Condition for Primitive. {{qed|lemma}} === Construction of a Primitive === If the conditions hold, we choose $z_0 \in D$ and define a function $F: D \to \C$ of $f$ by: :$\displaystyle \map F w = \int_{C_w} \map f z \rd z$ where $C_w$ is any contour in $D$ with start point $z_0$ and end point $w$. From Connected Domain is Connected by Staircase Contours, it follows that we can choose $C_w$ to be a staircase contour. If $C_w'$ is another contour in $D$ with the same endpoints as $C_w$, then: :$\displaystyle \int_{C_w'} \map f z \rd z = \int_{C_w} \map f z \rd z$ by condition $(2)$, so $F$ is well-defined. From Zero Staircase Integral Condition for Primitive, it follows that $F$ is a primitive of $f$. {{qed}}
Let: :$p \in a^\succ$ By the definition of strict upper closure: : $a \prec p$ By the definition of total ordering, $p \prec b$ or $p \succcurlyeq b$. But if $p \prec b$ then $a \prec p \prec b$, contradicting the premise. Thus $p \succcurlyeq b$, so $p \in b^\succcurlyeq$. By definition of subset: :$a^\succ \subseteq b^\succcurlyeq$ Let: : $p \in b^\succcurlyeq$ By the definition of weak upper closure: :$b \preccurlyeq p$ Since $a \prec b$, Extended Transitivity shows that $a \prec p$. Thus: :$p \in a^\succ$ By definition of subset: :$b^\succcurlyeq \subseteq a^\succ$ Therefore by definition of set equality: :$a^\succ = b^\succcurlyeq$ {{qed|lemma}} A similar argument shows that $b^\prec = a^\preccurlyeq$. {{qed}} Category:Total Orderings icypkz4hxwvkb4jujqksn97vn3531ez