hcju/proofwikiqa huggingface dataset:
{{BeginTableau|\neg \left({ p \downarrow \left({q \downarrow r}\right) \implies \left({p \downarrow q}\right) \downarrow r }\right)}} {{Assumption|1|\neg p \land r}} {{Simplification|2|1|\neg p|1|1}} {{Simplification|3|1|r|1|2}} {{Addition|4|1|q \lor r|3|2}} {{DoubleNegIntro|5|1|\neg \neg \left({q \lor r}\right)|4}} {{SequentIntro|6|1|\neg \left({q \downarrow r}\right)|5|Definition of Logical NOR}} {{Conjunction|7|1|\neg p \land \neg \left({q \downarrow r}\right)|2|6}} {{DeMorgan|8|1|\neg \left({p \lor \left({q \downarrow r}\right) }\right)|7|Conjunction of Negations}} {{SequentIntro|9|1|p \downarrow \left({q \downarrow r}\right)|8|Definition of Logical NOR}} {{Addition|10|1|\left({p \downarrow q}\right) \lor r|3|2}} {{DoubleNegIntro|11|1|\neg \neg \left({\left({p \downarrow q}\right) \lor r}\right)|10}} {{SequentIntro|12|1|\neg \left({\left({p \downarrow q}\right) \downarrow r}\right)|11|Definition of Logical NOR}} {{DoubleNegIntro|13|1|\neg \neg \left({p \downarrow \left({q \downarrow r}\right)}\right)|9}} {{Conjunction|14|1|\left({\neg \neg \left({p \downarrow \left({q \downarrow r}\right)}\right)}\right) \land \left({\neg \left({\left({p \downarrow q}\right) \downarrow r}\right)}\right)|13|12}} {{DeMorgan|15|1|\neg \left({\neg \left({p \downarrow \left({q \downarrow r}\right)}\right) \lor \left({\left({p \downarrow q}\right) \downarrow r}\right)}\right)|14|Conjunction of Negations}} {{SequentIntro|16|1   |\neg \left({ p \downarrow \left({q \downarrow r}\right) \implies \left({p \downarrow q}\right) \downarrow r }\right)   |15   |Rule of Material Implication }} {{EndTableau}} Taking $p = \bot$ and $r = \top$, we have $\vdash \neg p \land r$, discharging the last assumption. Hence the result. {{qed}}
Apply the Method of Truth Tables: :$\begin{array}{|ccccc||ccccc|} \hline p & \downarrow & (q & \downarrow & r) & (p & \downarrow & q) & \downarrow & r \\ \hline F & F & F & T & F & F & T & F & F & F \\ F & T & F & F & T & F & T & F & F & T \\ F & T & T & F & F & F & F & T & T & F \\ F & T & T & F & T & F & F & T & F & T \\ T & F & F & T & F & T & F & F & T & F \\ T & F & F & F & T & T & F & F & F & T \\ T & F & T & F & F & T & F & T & T & F \\ T & F & T & F & T & T & F & T & F & T \\ \hline \end{array}$ As can be seen by inspection, the truth values under the main connectives do not match for all boolean interpretations. {{qed}}
It is sufficient to show that two distinct representations represents two distinct numbers. Let $a$ and $b$ two real numbers representable as the form above. Their signs are easy to distinguish, so we consider $\size a$ and $\size b$. There is some $n$ such that: :$\size a, \size b < 3^n$ In that case, $\dfrac {\size a} {3^n}$ can be represented as: :$0.a_1 a_2 a_3 \ldots$ and $\dfrac {\size b} {3^n}$ can be represented as: :$0.b_1 b_2 b_3 \ldots$ where $a_i, b_i$ are either $0$ or $2$. Let $N$ be the smallest integer such that $a_N \ne b_N$. {{WLOG}} assume that $a_N = 2$ and $b_N = 0$. We have: {{begin-eqn}} {{eqn | l = \frac {\size a} {3^n}       | r = \sum_{j \mathop = 1}^\infty \frac {a_j} {3^j} }} {{eqn | r = \sum_{j \mathop = 1}^{N - 1} \frac {a_j} {3^j} + \sum_{j \mathop = N}^\infty \frac {a_j} {3^j} }} {{eqn | o = \ge       | r = \sum_{j \mathop = 1}^{N - 1} \frac {a_j} {3^j} + \frac 2 {3^N}       | c = because $a_N = 2$, $a_j \ge 0$ }} {{eqn | o = >       | r = \sum_{j \mathop = 1}^{N - 1} \frac {a_j} {3^j} + \frac 1 {3^N} }} {{eqn | r = \sum_{j \mathop = 1}^{N - 1} \frac {b_j} {3^j} + \sum_{j \mathop = N + 1}^\infty \frac 2 {3^j}       | c = Sum of Infinite Geometric Sequence }} {{eqn | o = \ge       | r = \sum_{j \mathop = 1}^{N - 1} \frac {b_j} {3^j} + \sum_{j \mathop = N}^\infty \frac {b_j} {3^j}       | c = because $b_N = 0$, $b_j \le 2$ }} {{eqn | r = \frac {\size b} {3^n} }} {{end-eqn}} and thus $\size a$ and $\size b$ are distinct. {{qed}}
Let $\CC_n$ be defined as in $(1)$. Let $x \in \closedint 0 1$. We need to show that: :$x$ can be written in base $3$ without using the digit $1$ {{iff}}: ::$\forall n \in \Z, n \ge 1: x \in C_n$ First we note that from Sum of Infinite Geometric Sequence: :$\displaystyle 1 = \sum_{n \mathop = 0}^\infty \frac 2 3 \paren {\frac 1 3}^n$ that is: :$1 = 0.2222 \ldots_3$ Thus any real number which, expressed in base $3$, ends in $\ldots 10000 \ldots$ can be expressed as one ending in $\ldots 02222 \ldots$ by dividing the above by an appropriate power of $3$. {{proof wanted|Still some work to do yet, obviously ...<br />New formulation demands new approach}}
Suppose $G = \struct {V, E}$ has a finite number of vertices. Let $V = \set {v_1, v_2, \ldots, v_n}$ where $n = \card V$ is the cardinality of $V$. As $G$ is locally finite, each element of $V$ has a finite number of incident edges. For each $v_k \in V$, let $r_k$ be the degree of $v_k$. Then: :$\displaystyle \card E \le \sum_{i \mathop = 1}^n r_k$ where $\card E$ is the cardinality of $E$, that is, the number of edges in $G$. As $\displaystyle \sum_{i \mathop = 1}^n r_k$ is the sum of a finite number of finite numbers, it is itself finite. So $\card E$ is itself finite, and so $G$ has: :a finite number of vertices :a finite number of edges and so is by definition a finite graph. By transposition, if $G$ is infinite, it must contain an infinite number of vertices. {{qed}}
By definition, the Cantor set is the complement of a union of open sets relative to the closed interval $\closedint 0 1$. By the definition of a topology, that union is itself open in $\R$. The closed interval $\closedint 0 1$ is itself the complement of a union of open sets $\openint \gets 0 \cup \openint 1 \to$. Hence the result. {{qed}}
Let $g, h \in G$. It is given that $S$ is a generating set for $G$. It follows that there exist $s_1, \ldots, s_n \in S$ such that $g^{-1} \circ h = s_1 \circ \cdots \circ s_n$. Therefore $d_S \left({g, h}\right) \le n$, establishing that $\R$ is a valid codomain for the mapping $d_S$ with domain $G \times G$. This is the form a mapping must have to be able to be a metric. Now checking the other defining properties for a metric in turn: === M1 === Clearly the empty sequence can be formed with elements from $S$. It also has length zero. Therefore, we have for any $g \in G$ that $d_S \left({g, g}\right) = 0$. === M2 === Let $g, h, k \in G$. Let $d_S \left({g, h}\right) = n, d_S \left({h, k}\right) = m$. Let $s_1, \ldots, s_n, r_1, \ldots, r_m \in S$ be such that: :$g^{-1} \circ h = s_1 \circ \cdots \circ s_n$ :$h^{-1} \circ k = r_1 \circ \cdots \circ r_m$ From these equations, obtain: {{begin-eqn}} {{eqn | l = g^{-1} \circ k       | r = \left({g^{-1} \circ h}\right) \circ \left({h^{-1} \circ k}\right) }} {{eqn | r = \left({s_1 \circ \cdots \circ s_n}\right) \circ \left({r_1 \circ \cdots \circ r_m}\right) }} {{eqn | ll= \implies       | l = g \circ s_1 \circ \cdots s_n \circ r_1 \circ \cdots r_n       | r = k }} {{end-eqn}} Therefore, $d_S \left({g, k}\right) \le m +n = d_S \left({g, h}\right) + d_S \left({h, k}\right)$. === M3 === Let $g, h \in G$. Let $d_S \left({g, h}\right) = n$. Furthermore, let $s_1, \ldots, s_n \in S$ be such that: :$g^{-1} \circ h = s_1 \circ \cdots \circ s_n$ From Inverse of Group Product, obtain: {{begin-eqn}} {{eqn | l = \left({g^{-1} \circ h}\right)^{-1}       | r = \left({s_1 \circ \cdots \circ s_n}\right)^{-1} }} {{eqn | l = h^{-1} \circ g       | r = s_n^{-1} \circ \cdots \circ s_1^{-1} }} {{end-eqn}} By assumption $S$ is closed under taking inverses, and hence the latter expression yields a valid sequence for the metric $d_S$. It follows that $d_S \left({h, g}\right) \le d_S \left({g, h}\right)$. Switching the roles of $g$ and $h$ in the above, we obtain the converse inequality, and hence equality. === M4 === There is only one word of length zero, namely the empty word. However, the empty word sends an element $g$ to itself. Hence $g, h \in G, g \ne h \implies d_S \left({g, h}\right) > 0$. Thus $d_S$ is a metric. {{qed}} Category:Group Theory Category:Word Metric 6a41qg2j6p5nbo9grxnhyuggwxoasmz
We have Cantor Set is Closed in Real Number Space. Taking, for example, $0 \in \CC$ and $1 \in \R$ it is clear that: :$\forall x \in \CC: \map d {0, x} \le 1$ and so $\CC$ is bounded. Hence the result from the Heine-Borel Theorem. {{qed}}
We have that the Cantor space is a metric subspace of the real number space $\R$, and hence a metric space. We also have Cantor Space is Compact. The result follows from Compact Metric Space is Complete. {{qed}}
We have that the Cantor space is a metric subspace of the real number space $\R$, and hence a metric space. The result follows from Metric Space fulfils all Separation Axioms. {{qed}}
We have that the Cantor space is a topological subspace of the real number space with the usual (Euclidean) topology $\struct {\R, \tau_d}$. We also have that the Real Number Line is Second-Countable. The result follows from Second-Countability is Hereditary. {{qed}}
Let $U \in \tau_d$ be open in $T$. Let $p \in U$. Then $\exists x \in U: \exists \epsilon \in \R: \map d {x, p} < \epsilon$. Hence the result. {{qed}}
By definition, $T$ is '''scattered''' {{iff}} it contains no non-empty subset which is dense-in-itself. We have that Cantor Space is Dense-in-itself. Hence the result by definition of a scattered space. {{qed}}
From Cantor Space is Dense-in-itself, $\CC$ contains no isolated points. We also have that the Cantor Set is Closed in Real Number Space. The result follows from the definition of perfect set. {{qed}}
From Cantor Set is Closed in Real Number Space, $\CC$ is closed. So from Closed Set equals its Closure: :$\CC^- = \CC$ where $\CC^-$ denotes the closure of $\CC$. Let $0 \le a < b \le 1$. Then $I = \openint a b$ is an open interval of $\closedint 0 1$. Let $\epsilon = b - a$. Clearly $\epsilon > 0$. Let $n \in \N$ such that $3^{-n} < \epsilon$. {{explain|Need to fill in the details to show that between $a$ and $b$ there is an interval $3^{-\paren {n + 1} }$ wide that contains numbers with $1$ in its $n + 1$th digit in its ternary expansion (or something like that) - intuitively obvious but fiddly to prove.}} So there exists an open interval of $\closedint 0 1$ which has been deleted from $\closedint 0 1$ during the process of creating $\CC$. Thus no open interval of $\closedint 0 1$ is disjoint from ''all'' the open intervals deleted from $\closedint 0 1$. So an open interval of $\closedint 0 1$ can not be a subset of $\CC = \CC^-$. Hence the result, by definition of nowhere dense. {{qed}}
Let $S_n$ and $C_n$ be as in the definition of the Cantor set as a limit of a decreasing sequence. Then the length of every interval in $S_n$ is seen to be $\dfrac 1 {3^n} = 3^{-n}$. Let $0 \le a < b \le 1$. Then $\left({a \,.\,.\, b}\right) \subseteq \left[{0 \,.\,.\, 1}\right]$ is an open interval. Let $n \in \N$ such that $3^{-n} < b - a$, so that the length of every interval in $S_n$ is $3^{-n} < b - a$. Therefore, as the intervals in $S_n$ do not overlap, no interval of length $b - a$ is contained in $C_n = \displaystyle \bigcup S_n$.  Consequently, as $\mathcal C \subseteq C_n$, no interval of length $b - a$ is contained in $\mathcal C$. Since the interval $\left({a \,.\,.\, b}\right)$ was of arbitrary length, there do not exist any open intervals in $\mathcal C$. Hence the result, by definition of nowhere dense. {{qed}}
From Cantor Space is Nowhere Dense, $T$ is nowhere dense in $\closedint 0 1$. So, trivially, $\CC$ is the union of nowhere dense subsets of $\closedint 0 1$. Hence the result from definition of meager. {{qed}}
We have that the Cantor Space is Complete Metric Space. By Baire Category Theorem, a complete metric space is also a Baire space. The result then follows by Baire Space is Non-Meager. {{qed}}
From the definition as a ternary representation, $\CC$ consists of all the elements of $\closedint 0 1$ which can be written without using a $1$. So let $x \in \CC$. Then in base $3$ notation, we have (as $0 \le x \le 1$): :$\displaystyle x = \sum_{i \mathop = 1}^\infty r_j 3^{-j}$ From the definition of the Cantor set, we have $\forall j: r_j \in \set {0, 2}$. Furthermore, from Representation of Ternary Expansions, the $r_j$ are unique. Now define the following function: :$\displaystyle f: \CC \to \closedint 0 1,\quad \map f {\sum_{i \mathop = 1}^\infty r_j 3^{-j} } = \sum_{i \mathop = 1}^\infty \frac {r_j} 2 2^{-j}$ Observe that $\forall j: \dfrac {r_j} 2 \in \set {0, 1}$. That the {{RHS}} expression is in fact an element of $\closedint 0 1$ now follows from binary notation. Furthermore by Existence of Base-N Representation, any element $y$ of $\closedint 0 1$ may be written in the following form (where $\forall j: b_j \in \set {0, 1}$): :$\displaystyle y = \sum_{i \mathop = 1}^\infty b_j 2^{-j}$ Obviously, $y = \map f x$, where $x \in \CC$ is defined as follows: :$\displaystyle x = \sum_{i \mathop = 1}^\infty 2 b_j 3^{-j}$ It follows that $f$ is surjective. From Closed Interval in Reals is Uncountable, the closed interval $\closedint 0 1$ is uncountable. The result follows. {{explain|it is obvious, but most likely there is some result that could be referred to}} {{qed}}
It follows from Representation of Ternary Expansions that every string of the form $0.nnnnn \ldots$ where $n \in \left\{{0, 2}\right\}$ is an element of $\mathcal C$. We also have that every string of the form $0.nnnnn \ldots$ where $n \in \left\{{0, 1}\right\}$ is an element of $\left[{0 \,.\,.\, 1}\right] \subset \R$ expressed in binary notation. Let $f: \mathcal C \to \left[{0 \,.\,.\, 1}\right]$ be the function defined by: :$\forall x \in \mathcal C: f \left({x}\right) = \text{ the number obtained by replacing every } 2 \text { in } x \text { with a } 1$ where $x$ is expressed in base $3$ notation. It is clear from the above that $f$ is a surjection. {{qed}}
{{begin-eqn}} {{eqn | l = a       | r = b       | c =  }} {{eqn | ll= \vdash       | l = \map P a       | o = \iff       | r = \map P b       | c = Leibniz's Law }} {{eqn | ll= \vdash       | l = \map P b       | o = \iff       | r = \map P a       | c = Biconditional is Commutative }} {{eqn | ll= \vdash       | l = b       | r = a       | c = Leibniz's Law }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \map {\frac \d {\d x} } {\sinh u}       | r = \map {\frac \d {\d u} } {\sinh u} \frac {\d u} {\d x}       | c = Chain Rule for Derivatives }} {{eqn | r = \cosh u \frac {\d u} {\d x}       | c = Derivative of Hyperbolic Sine }} {{end-eqn}} {{qed}}
{{BeginTableau|p \iff q \vdash q \iff p}} {{Premise|1|p \iff q}} {{SequentIntro|2|1|\left({p \implies q}\right) \land \left({q \implies p}\right)|1|Rule of Material Equivalence}} {{Commutation|3|1|\left({q \implies p}\right) \land \left({p \implies q}\right)|2|Conjunction}} {{SequentIntro|4|1|q \iff p|3|Rule of Material Equivalence}} {{EndTableau}} {{qed|lemma}} {{BeginTableau|q \iff p \vdash p \iff q}} {{Premise|1|q \iff p}} {{SequentIntro|2|1|\left({q \implies p}\right) \land \left({p \implies q}\right)|1|Rule of Material Equivalence}} {{Commutation|3|1|\left({p \implies q}\right) \land \left({q \implies p}\right)|2|Conjunction}} {{SequentIntro|4|1|p \iff q|3|Rule of Material Equivalence}} {{EndTableau}} {{qed}}
{{begin-eqn}} {{eqn | l = \map {\frac \d {\d x} } {\cosh u}       | r = \map {\frac \d {\d u} } {\cosh u} \frac {\d u} {\d x}       | c = Chain Rule for Derivatives }} {{eqn | r = \sinh u \frac {\d u} {\d x}       | c = Derivative of Hyperbolic Cosine }} {{end-eqn}} {{qed}}
Let $a, b \in \CC$ such that $a < b$. Then $b - a = \epsilon$. Consider $n \in \N$ such that $3^{-n} < \epsilon$. {{wtd|To be proved that between $a + 3^{-\paren {n + 1} }$ and $a + 2 \times 3^{-\paren {n + 1} }$ there exists an interval $I$ which is excluded from $\CC$. Or something like that.<br/>Then we pick some $r \in I$ and continue as follows.}} So $\exists r \in \R: a < r < b, r \notin \CC$. Let $A = \CC \cap \hointr 0 r$ and $B = \CC \cap \hointl r 1$. Thus $A \mid B$ is a separation of $\CC$ such that $a \in A, b \in B$. Such a separation can be found for any two distinct $a, b \in \CC$. Hence the result by definition of totally separated. {{qed}}
From Cantor Space satisfies all Separation Axioms we have that $T$ is a $T_2$ (Hausdorff) space. Consider the real number $\dfrac 1 4 = 0.020202 \ldots_3$. We have that: :$C_1 := \CC \cap \hointr 0 {\dfrac 1 4}$ :$C_2 := \CC \cap \hointl {\dfrac 1 4} 1$ are disjoint sets both of which are open sets of $T$. However, as $\dfrac 1 4 \in \CC$ we have that: :$\dfrac 1 4 \in C_1^-$ and :$\dfrac 1 4 \in C_2^-$ where $C_1^-$ and $C_2^-$ are the closures of $C_1$ and $C_2$ in $T$. Thus $C_1^- \cap C_2^- \ne \O$. The result follows from Extremally Disconnected by Disjoint Open Sets. {{qed}}
In $\CC$, basis elements are sets of the form $\set {y: \size {x - y} < \epsilon}$ for $x \in \CC$ and some $\epsilon \in \R_{>0}$. In $\displaystyle \prod_{n \mathop = 1}^\infty A_n$, sets of the form $\set {\sequence {a_i} \in \prod A_n: a_i \text { is fixed for } 1 \le i \le n}$ forms a basis for the Tychonoff topology. Consider the function $f$ taking each point from $\tuple {a_1, a_2, \ldots, a_n, \ldots}$ in $\prod A_n$ to the point $0 \cdotp a_1 a_2 \ldots a_n \ldots_3$ in $\CC$. As $f$ takes basis elements to basis elements, both $f$ and $f^{-1}$ are seen to be continuous. {{qed}}
Let $\BB$ be a basis of $T$. Let $A \in \BB$. By definition of $\BB$, $A$ is an open set of $T$. But the Cantor Space is Totally Separated. Therefore $A$ is not a connected set. Hence the result from definition of a locally connected space. {{qed}}
Let $T = \struct {\CC, \tau_d}$ be the Cantor space. Let $A_n = \struct {\set {0, 2}, \tau_n}$ be the discrete space of the two points $0$ and $2$. Let $\displaystyle A = \prod_{n \mathop = 1}^\infty A_n$. Let $T' = \struct {A, \tau}$ be the product space where $\tau$ is the Tychonoff topology on $A$. From Cantor Space as Countably Infinite Product, $T'$ is homeomorphic to $T$. From Totally Disconnected and Locally Connected Space is Discrete, we have that $A_n$ is locally connected. But we also have that the Cantor Space is not Locally Connected. Hence the result. {{qed}}
{{begin-eqn}} {{eqn | l = \map {\frac \d {\d x} } {\tanh u}       | r = \map {\frac \d {\d u} } {\tanh u} \frac {\d u} {\d x}       | c = Chain Rule for Derivatives }} {{eqn | r = \sech^2 u \frac {\d u} {\d x}       | c = Derivative of Hyperbolic Tangent }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \cosh^2 x - \sinh^2 x       | r = \paren {\frac {e^x + e^{-x} } 2}^2 - \paren {\frac {e^x - e^{-x} } 2}^2         | c = {{Defof|Hyperbolic Cosine}} and {{Defof|Hyperbolic Sine}} }} {{eqn | r = \paren {\frac {\paren {e^x}^2 + 2 \paren {e^x} \paren {e^{-x} } + \paren {e^{-x} }^2} 4} - \paren {\frac {\paren {e^x}^2 - 2 \paren {e^x} \paren {e^{-x} } + \paren {e^{-x} }^2} 4}         | c = Square of Sum }} {{eqn | r = \paren {\frac {e^{2 x} + 2 + e^{-2 x} } 4} - \paren {\frac {e^{2 x} - 2 + e^{-2 x} } 4}         | c = Exponential of Sum }} {{eqn | r = \frac {e^{2 x} - e^{2 x} + e^{-2 x} - e^{-2 x} + 2 + 2} 4 }} {{eqn | r = 1 }} {{end-eqn}} {{qed}}
{{:Graph of Arccosecant Function}} Let $y = \arccsc x$ where  $x < -1$ or $x > 1$. Then: {{begin-eqn}} {{eqn | l = y       | r = \arccsc x       | c =  }} {{eqn | ll= \leadsto       | l = x       | r = \csc y       | c = where $y \in \closedint 0 \pi \land y \ne \dfrac pi 2$ }} {{eqn | ll= \leadsto       | l = \frac {\d x} {\d y}       | r = -\csc y \cot y       | c = Derivative of Cosecant Function }} {{eqn | ll= \leadsto       | l = \frac {\d y} {\d x}       | r = \frac {-1} {\csc y \cot y}       | c = Derivative of Inverse Function }} {{eqn | ll= \leadsto       | l = \paren {\frac {\d y} {\d x} }^2       | r = \frac 1 {\csc^2 y \cot^2 y}       | c = squaring both sides }} {{eqn | r = \frac 1 {\csc^2 y \paren {\csc^2 y - 1} }       | c = Difference of Squares of Cosecant and Cotangent }} {{eqn | r = \frac 1 {x^2 \paren {x^2 - 1} }       | c = Definition of $x$ }} {{eqn | ll= \leadsto       | l = \size {\dfrac {\d y} {\d x} }       | r = \dfrac 1 {\size x \sqrt {x^2 - 1} }       | c = squaring both sides }} {{end-eqn}} Since $\dfrac {\d y} {\d x} = \dfrac {-1} {\csc y \cot y}$, the sign of $\dfrac {\d y} {\d x}$ is opposite to the sign of $\csc y \cot y$. Writing $\csc y \cot y$ as $\dfrac {\cos y} {\sin^2 y}$, it is evident that the sign of $\dfrac {\d y} {\d x}$ is opposite to the sign of $\cos y$. From Sine and Cosine are Periodic on Reals, $\cos y$ is never negative on its domain ($y \in \closedint {-\dfrac \pi 2} {\dfrac \pi 2} \land y \ne 0$). However, by definition of the arccosecant of $x$: :$0 < \arccsc x < \dfrac \pi 2 \implies x > 1$ :$-\dfrac \pi 2 < \arccsc x < 0 \implies x < -1$ Thus: :$\dfrac {\map \d {\arccsc x} } {\d x} = \dfrac {-1} {\size x \sqrt {x^2 - 1} } = \begin {cases} \dfrac {-1} {x \sqrt {x^2 - 1} } & : 0 < \arccsc x < \dfrac \pi 2 \ (\text {that is: $x > 1$}) \\ \dfrac {+1} {x \sqrt {x^2 - 1} } & : -\dfrac \pi 2 < \arccsc x < 0 \ (\text {that is: $x < -1$}) \\ \end{cases}$ {{qed}}
{{begin-eqn}} {{eqn | l = \map {\frac \d {\d x} } {\sech u}       | r = \map {\frac \d {\d u} } {\sech u} \frac {\d u} {\d x}       | c = Chain Rule for Derivatives }} {{eqn | r = -\sech u \tanh u \frac {\d u} {\d x}       | c = Derivative of Hyperbolic Secant }} {{end-eqn}} {{qed}}
From the Binomial Theorem: : $\displaystyle \paren {a + b}^p = \sum_{k \mathop = 0}^p \binom p k a^k b^{p - k}$ Also note that: : $\displaystyle \sum_{k \mathop = 0}^p \binom p k a^k b^{p-k} = a^p + \sum_{k \mathop = 1}^{p - 1} \binom p k a^k b^{p - k} + b^p$ So: {{begin-eqn}} {{eqn | l = \forall k: 0 < k < p: \binom p k       | o = \equiv       | r = 0       | rr= \pmod p       | c = Binomial Coefficient of Prime }} {{eqn | ll= \leadsto       | l = \binom p k a^k b^{p - k}       | o = \equiv       | r = 0       | rr= \pmod p       | c = {{Defof|Modulo Multiplication}} }} {{eqn | ll= \leadsto       | l = \sum_{k \mathop = 1}^{p - 1} \binom p k a^k b^{p - k}       | o = \equiv       | r = 0       | rr= \pmod p       | c = {{Defof|Modulo Addition}} }} {{eqn | ll= \leadsto       | l = \sum_{k \mathop = 0}^p \binom p k a^k b^{p - k}       | o = \equiv       | r = a^p + b^p       | rr= \pmod p       | c = from above }} {{end-eqn}} {{qed}}
This proof utilizes the Method of Disks and thus is dependent on Volume of Cylinder. From the Method of Disks, the volume of the cone can be found by the definite integral: :$\displaystyle (1): \quad V = \pi \int_0^{AC} \paren {\map R x}^2 \rd x$ where $\map R x$ is the function describing the line which is to be rotated about the $x$-axis in order to create the required solid of revolution. In this example, $\map R x$ describes the line segment $\overline {AB}$, and so: :$\map R x = \dfrac r h x$ We have also defined: :$\overline {AC}$ as the axis of the cone, whose length is $h$ :$A$ as the origin. So the equation $(1)$ is interpreted as: {{begin-eqn}} {{eqn | l = V       | r = \pi \int_0^h \paren {\frac r h x}^2 \rd x       | c =  }} {{eqn | r = \intlimits {\pi \paren {\frac r h}^2 \frac {x^3} 3} {x \mathop = 0} {x \mathop = h}       | c = Constant Multiple Rule, Power Rule }} {{eqn | r = \frac 1 3 \pi r^2 h       | c =  }} {{end-eqn}} {{qed}}
Each of the open set axioms are examined in turn: === $(O1)$: Union of Open Sets === Let $\left({U_j}\right)_{j \in J}$ be an arbitrary indexed set, such that: :$\forall j \in J: U_j \in \tau$ Thus for all $i \in I$, we have by definition of set intersection that: :$\forall j \in J: U_j \in \tau_i$ Since $\tau_i$ is a topology for every $i \in I$, by definition we have: : $\displaystyle \forall i \in I: \bigcup_{j \mathop \in J} {U_j} \in \tau_i$ Therefore we have: : $\displaystyle \bigcup_{j \mathop \in J} {U_j} \in \bigcap_{i \mathop \in I} {\tau_i} = \tau$ {{qed|lemma}} === $(O2)$: Intersection of Open Sets === Let $U_1, U_2 \in \tau$. Then by definition of set intersection: : $\forall i \in I: U_1, U_2 \in \tau_i$ Since $\tau_i$ is a topology for each $i \in I$, we obtain that: : $\forall i \in I: U_1 \cap U_2 \in \tau_i$ Therefore we have: : $\displaystyle U_1 \cap U_2 \in \bigcap_{i \mathop \in I} {\tau_i} = \tau$ {{qed|lemma}} === $(O3)$: Set Itself === By the definition of a topology: : $\forall i \in I: S \in \tau_i$ Thus by definition of set intersection we have that: : $\displaystyle S \in \bigcap_{i \mathop \in I} {\tau_i} = \tau$ {{qed|lemma}} Thus, by definition, $\tau = \displaystyle \bigcap_{i \mathop \in I} {\tau_i}$ is a topology. {{qed}}