hcju/proofwikiqa huggingface dataset:
We will prove by induction of the cardinality of finite subset of $H$. === Base case === :$\forall A \subseteq S: \left\vert{A}\right\vert = 1 \implies \exists \sup A$ where $\left\vert{A}\right\vert$ denotes the cardinality of $A$. Let $A \subseteq S$ such that :$\left\vert{A}\right\vert = 1$ By Cardinality of Singleton: :$\exists a: A = \left\{ {a}\right\}$ By definition of subset: :$a \in S$ Thus by Supremum of Singleton: :the supremum of $A$ exists in $\left({S, \preceq}\right)$. === Induction Hypothesis  === :$n \ge 1 \land \forall A \subseteq S: \left\vert{A}\right\vert = n \implies \exists \sup A$ === Induction Step === :$\forall A \subseteq S: \left\vert{A}\right\vert = n+1 \implies \exists \sup A$ Let $A \subseteq S$ such that :$\left\vert{A}\right\vert = n+1$ By definition of cardinality of finite set: :$A \sim \N_{< n+1}$ where $\sim$ denotes set equivalence. By definition of set equivalence: :there exists a bijection $f: \N_{< n+1} \to A$ By Restriction of Injection is Injection: :$f \restriction_{\N_{< n}}: \N_{< n} \to f^\to\left({\N_{< n}}\right)$ is an injection. By definition :$f \restriction_{\N_{< n}}: \N_{< n} \to f^\to\left({\N_{< n}}\right)$ is a surjection. By definition :$f \restriction_{\N_{< n}}: \N_{< n} \to f^\to\left({\N_{< n}}\right)$ is a bijection. By definition of set equivalence: :$\N_{< n} \sim f^\to\left({\N_{< n}}\right)$ By definition of cardinality of finite set: :$\left\vert{f^\to\left({\N_{< n}}\right)}\right\vert = n$ By definitions of image of set and subset: :$f^\to\left({\N_{< n}}\right) \subseteq A$ By Subset Relation is Transitive: :$f^\to\left({\N_{< n}}\right) \subseteq S$ By Induction Hypothesis: :$\exists \sup \left({ f^\to\left({\N_{< n}}\right)}\right)$ By definition $\N_{< n+1}$ :$n \in \N_{< n+1}$ By definition of mapping: :$f \left({n}\right) \in A$ By definition of subset: :$f \left({n}\right) \in S$ {{begin-eqn}} {{eqn |l = A       |r = f^\to\left({\N_{< n+1} }\right) | c = definition of surjection }} {{eqn |r =  f^\to\left({\N_{< n}\cup \left\{ {n}\right\} }\right) | c = definition of initial segment of natural numbers }} {{eqn |r = f^\to\left({\N_{< n} }\right) \cup f^\to\left({\left\{ {n}\right\} }\right) | c =Image of Union under Mapping }} {{eqn |r = f^\to\left({\N_{< n} }\right) \cup \left\{ {f\left({n }\right)}\right\} | c =Image of Singleton under Mapping }} {{end-eqn}} Thus: {{begin-eqn}} {{eqn | l =\sup A       | r = \sup\left\{ {\sup \left({f^\to\left({\N_{< n} }\right)}\right), \sup \left\{ {f\left(n\right)}\right\} }\right\}       | c = Supremum of Suprema }} {{eqn | r = \sup\left\{ {\sup \left({f^\to\left({\N_{< n} }\right)}\right), f\left(n\right)}\right\}       | c = Supremum of Singleton }} {{eqn | r = \sup \left({f^\to\left({\N_{< n} }\right)}\right) \vee f\left(n\right)       | c = definition of join operation }} {{end-eqn}} Thus $A$ admits a supremum in $\left({S, preceq}\right)$. {{qed}}
From Variance as Expectation of Square minus Square of Expectation: :$\var X = \expect {X^2} - \paren {\expect X}^2$ From Expectation of Exponential Distribution: :$\expect X = \beta$ The expectation of $X^2$ is: {{begin-eqn}} {{eqn | l = \expect {X^2}       | r = \int_{x \mathop \in \Omega_X} x^2 \, \map {f_X} x \rd x       | c = {{Defof|Expectation of Continuous Random Variable}} }} {{eqn | r = \int_0^\infty x^2 \frac 1 \beta \map \exp {-\frac x \beta} \rd x       | c = PDF of Exponential Distribution }} {{eqn | r = \intlimits {-x^2 \map \exp {-\frac x \beta} } 0 \infty + \int_0^\infty 2 x \map \exp {-\frac x \beta} \rd x       | c = Integration by Parts }} {{eqn | r = 0 + 2 \beta \int_0^\infty x \frac 1 \beta \, \map \exp {-\frac x \beta} \rd x        | c = algebraic manipulation }} {{eqn | r = 2 \beta \, \expect X       | c = Expectation of Exponential Distribution }} {{eqn | r = 2 \beta^2 }} {{end-eqn}} Thus the variance of $X$ is: {{begin-eqn}} {{eqn | l = \var X       | r = \expect {X^2} - \paren {\expect X}^2 }} {{eqn | r = 2 \beta^2 - \beta^2  }} {{eqn | r = \beta^2 }} {{end-eqn}} {{qed}}
By Moment Generating Function of Exponential Distribution, the moment generating function $M_X$ of $X$ is given by:  :$\map {M_X} t = \dfrac 1 {1 - \beta t}$  From Variance as Expectation of Square minus Square of Expectation: :$\var X = \expect {X^2} - \paren {\expect X}^2$  From Moment in terms of Moment Generating Function, we also have:  :$\expect {X^2} = \map {M_X''} 0$ In Expectation of Exponential Distribution: Proof 2, it is shown that:  :$\map {M_X'} t = \dfrac \beta {\paren {1 - \beta t}^2}$  We have:  {{begin-eqn}} {{eqn	| l = \map {M_X''} t 	| r = \frac \d {\d t} \paren {\frac \beta {\paren {1 - \beta t}^2} } }} {{eqn	| r = \frac {2 \beta^2} {\paren {1 - \beta t}^3} 	| c = Chain Rule for Derivatives, Derivative of Power  }} {{end-eqn}}  Setting $t = 0$ gives:  {{begin-eqn}} {{eqn	| l = \expect {X^2} 	| r = \frac {2 \beta^2} {\paren {1 - 0 \beta}^3} }} {{eqn	| r = 2 \beta^2 }} {{end-eqn}} By Expectation of Exponential Distribution, we have:  :$\expect X = \beta$ So:  {{begin-eqn}} {{eqn	| l = \var X 	| r = 2 \beta^2 - \beta^2 }} {{eqn	| r = \beta^2 }} {{end-eqn}} {{qed}}
=== Sufficient Condition === Assume that :$\left\vert{A}\right\vert = 1$ By definition of cardinality of finite set: :$A \sim \N_{< 1} = \left\{ {0}\right\}$ where $\sim$ denotes set equivalence. By Set Equivalence is Equivalence Relation: :$\left\{ {0}\right\} \sim A$ By definition of set equivalence there exists a bijection: :$f: \left\{ {0}\right\} \to A$ By definition of bijection: :$f$ is s surjection. Thus {{begin-eqn}} {{eqn | l = A       | r = f^\to\left({\left\{ {0}\right\} }\right)       | c = {{Defof|Surjection}} }} {{eqn | r = \left\{ {f\left({0}\right)}\right\}       | c = Image of Singleton under Mapping }} {{end-eqn}} {{qed|lemma}} === Necessary Condition === Assume that: :$\exists a: A = \left\{ {a}\right\}$ Define a mapping $f: A \to \left\{ {0}\right\}$: :$F\left({a}\right) = 0$ It is easy to see by definition that :$f$ is an injection and a surjection. By definition :$f$ is bijection. By definition of set equivalence: :$A \sim \left\{ {0}\right\} = \N_{< 1}$ Thus by definition of cardinality of finite set: :$\left\vert{A}\right\vert = 1$ {{qed}}
This follows by mutatis mutandis of the proof of Existence of Non-Empty Finite Suprema in Join Semilattice. {{qed}}
:400px Let $AB$ be a diameter of a circle whose center is at $O$. By definition of diameter, $AB$ passes through $O$. $\angle AOB\cong\angle BOA$ because they are both straight angles. Thus, the arcs are congruent by Equal Angles in Equal Circles: :$\stackrel{\frown}{AB}\cong\stackrel{\frown}{BA}$ Hence, a circle is split into two equal arcs by a diameter. {{qed}}
Let $AB$ be a diameter of a circle $ADBE$ whose center is at $C$. By definition of diameter, $AB$ passes through $C$. {{AimForCont}} that $AB$ does not bisect $ADBE$, but that $ADBC$ is larger than $AEBC$. :400px Thus it will be possible to find a diameter $DE$ passing through $C$ such that $DC \ne CE$. Both $DC$ and $CE$ are radii of $ADBE$. By Euclid's definition of the circle: :''{{:Definition:Euclid's Definitions - Book I/15 - Circle}}'' :''{{:Definition:Euclid's Definitions - Book I/16 - Center of Circle}}'' That is, all radii of $ADBE$ are equal. But $DC \ne CE$. From this contradiction it follows that $AB$ bisects the circle. {{qed}}
Let $\triangle ABC$ be a triangle. From Sum of Angles of Triangle equals Two Right Angles: Proof 2: :$\paren 1: \angle ABC + \angle BCA + \angle CAB = 180^\circ$ Extend $AB$ to $D$. By Two Angles on Straight Line make Two Right Angles: :$\paren 1: \angle ABC + \angle CBD = 180^\circ$ Combining $\paren 1$ and $\paren 2$ and using Equality is Transitive: :$\angle ABC + \angle BCA + \angle CAB = \angle ABC + \angle CBD$ By using commom notion 3: :$\angle BCA + \angle CAB = \angle CBD$ By using Equality is Symmetric: :$\angle CBD = \angle BCA + \angle CAB$ {{qed}}
:400px From Parallelograms with Same Base and Same Height have Equal Area: :$ACDE = ACHR = ATUJ$ and: :$BCFG = BCHS = BIUT$ Hence the result. {{qed}}
: 800px Consider the cone in which $\alpha$ is half the opening angle. Let the tilting angle of the slicing plane to the horizontal be $\beta$. The conic section thus defined is given here as an ellipse, but the argument holds for the other cases also. Inside the cone, let a sphere be inscribed which is tangent to the slicing plane at the point $F$. Let the sphere be tangent to the cone along the circle $C$. Let $d$ be the line in which the slicing plane intersects the plane of the circle $C$. It is to be demonstrated that the conic section induced by the slicing plane has $F$ as its focus and $d$ as its directrix. Let $P$ be a point on the conic section. Let $PQ$ be constructed parallel to the axis of the cone such that $Q$ lies on the plane of the circle $C$. Let $R$ be the point where the generatrix of the cone through $P$ intersects $C$. Let a perpendicular be constructed from $P$ to $d$, intersecting $d$ at $D$. Hence: :$\beta = \angle PDQ$ We have that $PR$ and $PF$ are both tangents to the sphere from the same point $P$. Thus $PF$ and $PR$ are the same length: :$(1): \quad PR = PF$ From the right triangle $\triangle PQR$: :$PQ = PR \cos \alpha$ From the right triangle $\triangle PQD$: :$PQ = PD \sin \beta$ Thus: {{begin-eqn}} {{eqn | l = PR \cos \alpha       | r = PD \sin \beta       | c =  }} {{eqn | ll= \leadsto       | l = \frac {PR} {PD}       | r = \frac {\sin \beta} {\cos \alpha}       | c =  }} {{eqn | ll= \leadsto       | l = \frac {PF} {PD}       | r = \frac {\sin \beta} {\cos \alpha}       | c = from $(1)$ }} {{eqn | ll= \leadsto       | l = \frac {PF} {PD}       | r = \frac {\cos \gamma} {\cos \alpha}       | c = where $\gamma = \angle DPQ$ }} {{end-eqn}} Let us define: :$e = \dfrac {\cos \gamma} {\cos \alpha}$ For a given slicing plane and cone this is constant. It is noted that: :$(1): \quad$ when $\gamma < \alpha$, that is, the tilting angle is less than half the opening angle of the cone, the conic section is an ellipse :$(2): \quad$ when $\gamma = \alpha$, that is, the tilting angle equals half the opening angle of the cone, the conic section is a parabola :$(3): \quad$ when $\gamma > \alpha$, that is, the tilting angle is greater than half the opening angle of the cone, the conic section is a hyperbola :$(4): \quad$ when $\gamma = 0$, that is, the slicing plane is parallel to the plane of the circle $C$, the conic section is a circle and there is no directrix The result follows. {{qed}}
Let the surface of the sphere of radius $r$ be divided into many small areas. If they are made small enough, they can be approximated to plane figures. Let the areas of these plane figures be denoted: :$a_1, a_2, a_3, \ldots$ Let the sphere of radius $r$ be divided into as many pyramids whose apices are at the center and whose bases are these areas. From Volume of Pyramid, their volumes are: :$\dfrac {r a_1} 3, \dfrac {r a_2} 3, \dfrac {r a_3} 3, \ldots$ The volume $\mathcal V$ of the sphere is given by the sum of the volumes of each of these pyramids: {{begin-eqn}} {{eqn | l = \mathcal V       | r = \dfrac {r a_1} 3 + \dfrac {r a_2} 3 + \dfrac {r a_3} 3 + \cdots       | c =  }} {{eqn | r = \dfrac r 3 \left({a_1 + a_2 + a_3 + \cdots}\right)       | c =  }} {{end-eqn}} But $a_1 + a_2 + a_3 + \cdots$ is the surface area of the sphere. Hence: {{begin-eqn}} {{eqn | l = \mathcal V       | r = \dfrac r 3 \left({a_1 + a_2 + a_3 + \cdots}\right)       | c =  }} {{eqn | r = \dfrac r 3 A       | c =  }} {{end-eqn}} It needs to be noted that this proof is intuitive and non-rigorous. {{qed}}
Let $L$ be a lower bound for $A$. By definition of lower bound: :$\forall x \in A: L \preceq x$ By definition of subset: :$\forall x \in B: x \in A$ Hence: :$\forall x \in B: L \preceq x$ Thus by definition: :$L$ is a lower bound for $B$. {{qed}}
Assume that: : $U$ is upper bound for $A$. By definition of upper bound: :$\forall x \in A: x \preceq U$ By definition of subset: :$\forall x \in B: x \in A$ Hence: :$\forall x \in B: x \preceq U$ Thus by definition : $U$ is sn upper bound for $B$. {{qed}}
Consider an arbitrary element $R$ of $\SS$. Let $B$ be half the perimeter of $R$. Let $x$ be the length of one side of $R$. Then the length of an adjacent side is $B - x$. The area $\AA$ of $R$ is then given by: :$\AA = x \paren {B - x}$ Let $\AA$ be expressed in functional notation as: :$\map f x = x \paren {B - x}$ We have: {{begin-eqn}} {{eqn | l = \map f {x + E} - \map f x       | r = \paren {x + E} \paren {B - \paren {x + E} } - x \paren {B - x}       | c =  }} {{eqn | r = E B - 2 E x - E^2       | c =  }} {{eqn | ll= \leadsto       | l = \frac {\map f {x + E} - \map f x} E       | r = B - 2 x - E       | c =  }} {{eqn | ll= \leadsto       | l = \intlimits {\frac {\map f {x + E} - \map f x} E} {E \mathop = 0} {}       | r = B - 2 x       | c =  }} {{end-eqn}} Thus from Derivative at Maximum or Minimum, when $B - 2 x = 0$, the area of $R$ is at a maximum or a minimum. That is: :$x = \dfrac B 2 = \dfrac L 4$ Geometrical analysis shows that for this length of side the area is not a minimum because that happens when $x = 0$ or $x = B$. Hence the result. {{qed}}
This follows by mutatis mutandis of the proof of Mapping Preserves Finite and Filtered Infima. {{qed}}
First some lemmata: === Lemma 1 === {{:Integer as Sum of Polygonal Numbers/Lemma 1}}{{qed|lemma}} === Lemma 2 === {{:Integer as Sum of Polygonal Numbers/Lemma 2}}{{qed|lemma}} === Lemma 3 === {{:Integer as Sum of Polygonal Numbers/Lemma 3}}{{qed|lemma}} First note that Cauchy's Lemma (Number Theory) gives us: {{:Cauchy's Lemma (Number Theory)}}{{qed|lemma}} Now we consider the numbered statements as asserted. For $(1)$, we have Integer is Sum of Three Triangular Numbers. For $(2)$, we have Lagrange's Four Square Theorem. For each $m \ge 3$, we prove that $n$ is the sum of at most $m + 2$ polygonal numbers of order $m + 2$. We choose to consider polygonal numbers of order $m + 2$, rather than $m$-gonal numbers, because the former have a neater closed form: :$\map P {m + 2, k} = \dfrac m 2 \paren {k^2 - k} + k$ For $n < 116 m$ the result is shown in Lemma 1. It remains to investigate the integers $n \ge 116 m$. Define: :$I = \openint {\dfrac 2 3 + \sqrt {8 \paren {\dfrac n m} - 8} } {\dfrac 1 2 + \sqrt {6 \paren {\dfrac n m} - 3} }$ By Lemma 2: :For $\dfrac n m \ge 116$, the length of $I$ is greater than $4$. Thus $I$ must contain at least $2$ consecutive odd integers. Let $b_1, b_2$ be those consecutive odd integers. Because $b_1 + m - 1 = b_2 + m - 3$: :the set $\set {b + r: b \in \set {b_1, b_2}, r \in \set {0, 1, \dots, m - 2} }$ contains a complete residue system modulo $m$. Hence one can choose some $b, r$ and write $n \equiv b + r \pmod m$. Define: :$a = 2 \paren {\dfrac {n - b - r} m} + b = \paren {1 - \dfrac 2 m} b + 2 \paren {\dfrac {n - r} m}$ Since $m \divides \paren {n - b - r}$ and $b$ is odd: :$a$ is an odd integer. Since $b \in I$, from Lemma 3 we have: :$b^2 < 4 a$ :$3 a < b^2 + 2 b + 4$ Then by Cauchy's Lemma (Number Theory), we can write: :$a = s^2 + t^2 + u^2 + v^2$ :$b = s + t + u + v$ for nonnegative integers $s, t, u, v$. Thus: {{begin-eqn}} {{eqn | l = n       | r = \frac m 2 \paren {a - b} + b + r       | c = From $a = 2 \paren {\dfrac {n - b - r} m} + b$ }} {{eqn | r = \frac m 2 \paren {s^2 + t^2 + u^2 + v^2 - s - t - u - v} + s + t + u + v + r }} {{eqn | r = \paren {\frac m 2 \paren {s^2 - s} + s} + \paren {\frac m 2 \paren {t^2 - t} + t} + \paren {\frac m 2 \paren {u^2 - u} + u} + \paren {\frac m 2 \paren {v^2 - v} + v} + r \times 1 }} {{eqn | r = \map P {m + 2, s} + \map P {m + 2, t} + \map P {m + 2, u} + \map P {m + 2, v} + r \map P {m + 2, 1} }} {{end-eqn}} Since $r \le m - 2$, the above expression comprises at most $m + 2$ polygonal numbers of order $m + 2$. {{qed}}
Let $x, y \in S$ such that: :$x \preceq y$ By Infimum of Singleton: :$\set x$ and $\set y$ admit infima in $\struct {S, \preceq}$ By Infimum of Upper Closure of Set: :$\set x^\succeq$ and $\set y^\succeq$ admit infima in $\struct {S, \preceq}$ where $\set x^\succeq$ denotes the upper closure of $\set x$. By Upper Closure of Singleton :$x^\succeq$ and $y^\succeq$ admit infima in $\struct {S, \preceq}$ By Upper Closure of Element is Filter: :$x^\succeq$ and $y^\succeq$ are filter in $\struct {S, \preceq}$ By assumption and definition of mapping preserves the infimum on subset: :$\map {f^\to} {x^\succeq}$ and $\map {f^\to} {y^\succeq}$ admit infima in $\struct {T, \precsim}$ and :$\inf \set {\map {f^\to} {x^\succeq} } = \map f {\inf \set {x^\succeq} }$ and $\inf \set {\map {f^\to} {y^\succeq} } = \map f {\inf \set {y^\succeq} }$ By Infimum of Upper Closure of Element: :$\inf \set {x^\succeq} = x$ and $\inf \set {y^\succeq} = y$ By Upper Closure is Decreasing: :$y^\succeq \subseteq x^\succeq$ By Image of Subset under Relation is Subset of Image/Corollary 2: :$\map {f^\to} {y^\succeq} \subseteq \map {f^\to} {x^\succeq}$ Thus by Infimum of Subset: :$\map f x \precsim \map f y$ Thus by definition: :$f$ is increasing. {{qed}}
This follows by mutatis mutandis of the proof of Supremum of Lower Closure of Set. {{qed}}
Let $V_K$ be the volume of $K$. Let $V_C$ be the volume of a cylinder of base $A$ and of height $h$. From Volume of Cylinder: :$V_C = A h$ From Volume of Cone is Third of Cylinder on Same Base and of Same Height: {{begin-eqn}} {{eqn | l = V_K       | r = \dfrac {V_C} 3       | c =  }} {{eqn | r = \dfrac {A h} 3       | c =  }} {{end-eqn}} {{qed}} Category:Cones Category:Volume Formulas l2jfjl03yymwq4tnmi2q8qaskbw69s9
:600px Consider a cylinder $C$ whose base is a circle of radius $r$ and whose height is $h$. Consider a cuboid $K$ whose height is $h$ and whose base has the same area as the base of $C$. Let the area of those bases be $A$. Let the cylinder $C$ be positioned with its base in the same plane as the base of $K$. By Cavalieri's Principle $C$ and $K$ have the same volume. The bases of $C$ are circles of radius $r$. From Area of Circle, the area of each base therefore gives: :$A = \pi r^2$ From Volume of Cuboid, $K$ has volume given by: :$V_K = A h = \pi r^2 h$ Hence the result. {{qed}} Category:Cylinders Category:Volume Formulas oy3b69d3hb9fkft0wxnc17jp4r6u9ea
By definition of Derivative of Curve at Point, the tangent to $C$ at the point $\tuple {x, y}$ is the derivative of its equation at that point. Thus: {{begin-eqn}} {{eqn | l = \frac {\d x} {\d \theta}       | r = a \paren {1 - \cos \theta}       | c =  }} {{eqn | l = \frac {\d y} {\d \theta}       | r = a \sin \theta       | c =  }} {{eqn | ll= \leadsto       | l = \frac {\d y} {\d x}       | r = \frac {\d y} {\d \theta} / \frac {\d x} {\d \theta}       | c =  }} {{eqn | r = \frac {a \sin \theta} {a \paren {1 - \cos \theta} }       | c =  }} {{eqn | r = \frac {\sin \theta} {1 - \cos \theta}       | c =  }} {{eqn | r = \frac 1 {\map \tan {\theta / 2} }       | c = Half Angle Formula for Tangent: Corollary 2 }} {{eqn | r = \cot \frac \theta 2       | c =  }} {{end-eqn}} {{qed}}
Consider a polygon $ABCD$ being rolled along a straight line in the same way as the generating circle of $C$. Let $A', B', C', D'$ be the points around which the $ABCD$ rotates while rolling. :600px The point $A$ traces out in succession several arcs of circles with centers $B', C', D'$. The tangent to each of these arcs is perpendicular to the line joining the point of tangency to the corresponding point of rotation. Consider a circle as being a polygon with an infinite number of sides. It follows that in the limit the tangent to $C$ at any given point is perpendicular to the line joining the point of tangency to the corresponding point where the generating circle touches the straight line along which the generating circle rolls. :600px Let $O$ be the center of the generating circle. Let $P$ be the point of tangency. Let $Q$ be the point where the generating circle touches the straight line along which the generating circle rolls. Let $S$ be the point where the tangent to $C$ meets the straight line along which the generating circle rolls. As $OP$ and $OQ$ are both radii of the generating circle: :$OP = OQ = a$ and so $\triangle POQ$ is isosceles. Let $R$ be dropped perpendicular to $PQ$ at $R$. As $\triangle POQ$ is isosceles, $\triangle OPR = \triangle OQR$ and so $PR = RQ$. Thus $\angle POR = \angle QOR = \dfrac \theta 2$. As $OR \perp PR$ and $PR \perp PS$, we have that $\triangle OPR$ is similar to $\triangle QPS$. Thus $\angle PQS = \angle POR$ and so $\angle PSQ$ is $\dfrac \pi 2 - \angle POR = \dfrac \pi 2 - \dfrac \theta 2$. By definition, $\tan \angle PSQ$ is the slope of the tangent $\dfrac {\d y} {\d x}$ to $C$. Thus: {{begin-eqn}} {{eqn | l = \dfrac {\d y} {\d x}       | r = \tan \angle PSQ       | c =  }} {{eqn | r = \map \tan {\dfrac \pi 2 - \dfrac \theta 2}       | c =  }} {{eqn | r = \cot \frac \theta 2       | c = Tangent of Complement equals Cotangent }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \left\{ {s}\right\}^\succeq       | r = \bigcup \left\{ {t^\succeq: t \in \left\{ {s}\right\} }\right\}       | c = definition of upper closure of set }} {{eqn | r = \bigcup \left\{ {s^\succeq}\right\}       | c = definition of singleton }} {{eqn | r = s^\succeq       | c = Union of Singleton }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \left\{ {s}\right\}^\preceq       | r = \bigcup \left\{ {t^\preceq: t \in \left\{ {s}\right\} }\right\}       | c = Definition of lower closure of set }} {{eqn | r = \bigcup \left\{ {s^\preceq}\right\}       | c = Definition of Singleton }} {{eqn | r = s^\preceq       | c = Union of Singleton }} {{end-eqn}} {{qed}}
By Singleton is Directed and Filtered Subset :$\set s$ is a filtered subset of $S$ By Filtered iff Upper Closure Filtered: :$\set s^\succeq$ is a filtered subset of $S$ By Upper Closure is Upper Set: :$\set s^\succeq$ is a upper set in $S$ By Upper Closure of Singleton :$\set s^\succeq = s^\succeq$ By definition of reflexivity: :$s \preceq s$ By definition of upper closure of element: :$s \in s^\succeq$ Thus by definition: :$s^\succeq$ is non-empty, filtered and upper. Thus by definition: :$s^\succeq$ is a filter in $\struct {S, \preceq}$ {{qed}}
Consider the logarithmic spiral $S$ expressed as: :$r = a e^{b \theta}$ :400px Let $\psi$ be the angle between the tangent to $S$ and the radius vector. The  derivative of $r$ {{WRT}} $\theta$ is: :$\dfrac {\mathrm d r} {\mathrm d \theta} = a b e^{b \theta} = b r$ and thus: {{begin-eqn}} {{eqn | l = \tan \psi       | r = \frac r {\frac {\mathrm d r} {\mathrm d \theta} }       | c =  }} {{eqn | r = \frac r {b r}       | c =  }} {{eqn | r = \frac 1 b       | c =  }} {{eqn | ll= \leadsto       | l = \cot \psi       | r = b       | c =  }} {{end-eqn}} Thus for a given logarithmic spiral, $\psi$ is constant and equal to $\operatorname{arccot} b$. {{qed}}
{{begin-eqn}} {{eqn | l = \map \inf {s^\succeq}       | r = \map \inf {\set s^\succeq}       | c = Upper Closure of Singleton }} {{eqn | r = \map \inf {\set s}       | c = Infimum of Upper Closure of Set }} {{eqn | r = s       | c = Infimum of Singleton }} {{end-eqn}} {{qed}}
From Volume of Solid of Revolution: {{begin-eqn}} {{eqn | l = V       | r = \pi \int_1^a \frac 1 {x^2} \, \mathrm d x       | c =  }} {{eqn | r = \pi \left[{-\dfrac 1 x}\right]_1^a       | c = Primitive of Power }} {{eqn | r = \pi \left[{\dfrac 1 x}\right]_a^1       | c =  }} {{eqn | r = \pi \left({1 - \dfrac 1 a}\right)       | c =  }} {{end-eqn}} {{qed}}
By Singleton is Directed and Filtered Subset :$\left\{ {s}\right\}$ is a directed subset of $S$ By Directed iff Lower Closure Directed: :$\left\{ {s}\right\}^\preceq$ is a directed subset of $S$ By Lower Closure is Lower Set: :$\left\{ {s}\right\}^\preceq$ is a lower set in $S$ By Lower Closure of Singleton :$\left\{ {s}\right\}^\preceq = s^\preceq$ By definition of reflexivity: :$s \preceq s$ By definition of lower closure of element: :$s \in s^\preceq$ Thus by definition: :$s^\succeq$ is non-empty directed and lower. Thus by definition: :$s^\preceq$ is a ideal in $\left({S, \preceq}\right)$ {{qed}}
{{begin-eqn}} {{eqn | l = \sup \left({s^\preceq}\right)       | r = \sup \left({\left\{ {s}\right\}^\preceq}\right)       | c = Lower Closure of Singleton }} {{eqn | r = \sup \left({\left\{ {s}\right\} }\right)       | c = Supremum of Lower Closure of Set }} {{eqn | r = s       | c = Supremum of Singleton }} {{end-eqn}} {{qed}}