hcju/proofwikiqa huggingface dataset:
{{begin-eqn}} {{eqn | ll= \forall z \in \left\{ {z \in \C: \ e^z - e^{-z} \ne 0}\right\}:       | o =        | r = \frac 2 {e^z - e^{-z} }       | c = Definition of Hyperbolic Cosecant: Definition 1 }} {{eqn | ll= \forall z \in \left\{ {z \in \C: \ \frac {e^z - e^{-z} } 2 \ne 0}\right\}:       | r = 1 / \frac {e^z - e^{-z} } 2       | c =  }} {{eqn | ll= \forall z \in \left\{ {z \in \C: \ \sinh z \ne 0}\right\}:       | r = 1 / \sinh z       | c = Definition of Hyperbolic Cosecant: Definition 2 }} {{end-eqn}} {{qed}} Category:Hyperbolic Cosecant Function 5v8tbqpe8tmw20rcp3g5063dh13pesx
{{begin-eqn}} {{eqn | ll= \forall z \in \left\{ {z \in \C: \ e^z + e^{-z} \ne 0}\right\}:       | o =        | r = \frac 2 {e^z + e^{-z} }       | c = Definition of Hyperbolic Secant: Definition 1 }} {{eqn | ll= \forall z \in \left\{ {z \in \C: \ \frac {e^z + e^{-z} } 2 \ne 0}\right\}:       | r = 1 / \frac {e^z + e^{-z} } 2       | c =  }} {{eqn | ll= \forall z \in \left\{ {z \in \C: \ \cosh z \ne 0}\right\}:       | r = 1 / \cosh z       | c = Definition of Hyperbolic Secant: Definition 2 }} {{end-eqn}} {{qed}} Category:Hyperbolic Secant Function h8gqzgxmo92a683mv5754wid0m2jd5o
{{begin-eqn}} {{eqn | ll= \forall z \in \left\{ {z \in \C: \ e^{2 z} - 1 \ne 0}\right\}:       | o =        | r = \frac {e^{2 z} + 1} {e^{2 z} - 1}       | c = Definition of Hyperbolic Cotangent: Definition 3 }} {{eqn | ll= \forall z \in \left\{ {z \in \C: \ e^z - e^{-z} \ne 0}\right\}:       | r = \frac {e^z \left({e^z + e^{-z} }\right)} {e^z \left({e^z - e^{-z} }\right)}       | c =  }} {{eqn | ll= \forall z \in \left\{ {z \in \C: \ e^z - e^{-z} \ne 0}\right\}:       | r = \frac {e^z + e^{-z} } {e^z - e^{-z} }       | c = Definition of Hyperbolic Cotangent: Definition 1 }} {{eqn | ll= \forall z \in \left\{ {z \in \C: \ \frac {e^z - e^{-z} } 2 \ne 0}\right\}:       | r = \frac {\left({\dfrac {e^z + e^{-z} } 2}\right) } {\left({\dfrac {e^z - e^{-z} } 2}\right) }       | c =  }} {{eqn | ll= \forall z \in \left\{ {z \in \C: \ \sinh z \ne 0}\right\}:       | r = \frac {\cosh z} {\sinh z}       | c = Definition of Hyperbolic Cotangent: Definition 2 }} {{end-eqn}} {{qed}} Category:Hyperbolic Cotangent Function rhocl5nkolp0ngc62ttnk9jfbl7wx7f
Let $z = r e^{i \theta}$ such that $z \ne 0$. Let $F = \set {\ln r + i \theta + 2 k \pi i: k \in \Z}$. Let $G = \set {w \in \C: e^w = z}$. We will demonstrate that $F = G$. === Definition 1 implies Definition 2 === Let $w  = x + i y$ such that $w \in F$. Then: :$x + i y = \ln r + i \theta + 2 k \pi i$ for some $k \in \Z$. Equating real and imaginary parts: :$x = \ln r$ :$y = \theta + 2 k \pi$ Then: {{begin-eqn}} {{eqn | l = e^w       | r = e^{x + i y}       | c =  }} {{eqn | r = e^x e^{i y}       | c = Exponential of Sum }} {{eqn | r = e^{\ln r} e^{i \paren {\theta + 2 k \pi} }       | c = Definition of $w$ }} {{eqn | r = e^{\ln r} e^{i \theta}       | c = Periodicity of Complex Exponential Function }} {{eqn | r = r e^{i \theta}       | c = Exponential of Natural Logarithm }} {{eqn | r = z       | c = Definition of $z$ }} {{end-eqn}} Thus $w \in G$ and so $f \subseteq G$. {{qed|lemma}} === Definition 2 implies Definition 1 === Let $w \in G$. By definition: :$\exists z \in \C_{\ne 0}: z = e^w = r e^{i \theta}$ Thus: {{begin-eqn}} {{eqn | l = r e^{i \theta}       | r = e^{\ln r} e^{i \theta}       | c = Exponential of Natural Logarithm }} {{eqn | r = e^{\ln r + i \theta}       | c = Exponential of Sum }} {{eqn | r = e^w       | c = by definition }} {{end-eqn}} Thus $w$ is of the form $\ln r + i \theta + k \pi i$, where $k = 0$. Therefore $w \in F$, and so $G \subseteq F$. {{qed|lemma}} So by definition of set equality: :$F = G$ {{qed}}
From Domain of Real Natural Logarithm: : $\operatorname{Dom} \left({\ln'}\right) = \R_{>0}$ From Image of Real Natural Logarithm: : $\operatorname{Im} \left({\ln'}\right) = \R$ Let $z \in \C$ such that $z = x + i y$. Let $z$ be expressed in exponential form as $z = r e^{i \theta}$. Let $x > 0$ and $y = 0$. Thus $z \in \R_{>0}$. Then: {{begin-eqn}} {{eqn | l = \ln z       | r = \left\{ {\ln' r + i \theta + 2 \pi k i}\right\}       | c =  }} {{eqn | r = \left\{ {\ln' x + 2 \pi k i}\right\}       | c = as $r = x$ and $\theta = 0$ }} {{end-eqn}} In order for $\ln z \in \R$ it is necessary that $k = 0$. Thus $\ln z = \ln' x$. Hence the result. {{qed}} Category:Natural Logarithms j4h0xmkgtlc49xsftaqfnqphpm2d3yj
Using a tableau proof for instance 1 of a Gentzen proof system: {| border="1" |+$\vdash \left({p \lor q}\right) \implies \left({q \lor p}\right)$ |- ! Line !! ! Pool ! Formula ! Rule ! Depends upon ! Notes |- | 1 || || | $\neg p, q, p$ | Axiom || || |- | 2 || || | $\neg q, q, p$ | Axiom || || |- | 3 || || | $\neg \left({p \lor q}\right), q, p$ | $\alpha$-Rule: $\alpha\lor$ | 1, 2 || |- | 4 || || | $\neg \left({p \lor q}\right), q \lor p$ | $\beta$-Rule: $\beta\lor$ | 3 || |- | 5 || || | $\left({p \lor q}\right) \implies \left({q \lor p}\right)$ | $\beta$-Rule: $\beta\implies$ | 4 || |} {{qed}}
This is a specific instance of Provable by Gentzen Proof System iff Negation has Closed Tableau: Set of Formulas. {{qed}}
By definition of natural logarithm: :$\ln^{-1} = \exp$ From Exponential Tends to Zero and Infinity: :$\Dom \exp = \openint {-\infty} {+\infty}$ :$\Img \exp = \openint 0 {+\infty}$ From Exponential is Strictly Increasing: :$\exp$ is strictly increasing. From Strictly Monotone Real Function is Bijective, $\exp: \R \to \R_{>0}$ is a bijection. Thus: :$\Dom \ln = \Img \exp$ and so $\Dom \ln = \R_{>0}$. {{qed}} Category:Natural Logarithms 14154zzvbik35lknflmgrxyz0xxq2th
By definition of natural logarithm: :$\ln^{-1} = \exp$ From Exponential Tends to Zero and Infinity: :$\Dom \exp = \openint {-\infty} {+\infty}$ :$\Img \exp = \openint 0 {+\infty}$ From Exponential is Strictly Increasing: :$\exp$ is strictly increasing. From Strictly Monotone Real Function is Bijective, $\exp: \R \to \R_{>0}$ is a bijection. Thus: :$\Img \ln = \Dom \exp$ and so $\Img \ln = \R$. {{qed}} Category:Natural Logarithms 4d0rbbqnorxwqsc2hkxxvh1ektnvkzw
This is an immediate consequence of: * Provable by Gentzen Proof System iff Negation has Closed Tableau * Soundness and Completeness of Semantic Tableaus {{qed}}
Using a tableau proof for instance 1 of a Hilbert proof system: {{BeginTableau|p \implies p|nohead=1}} {{TableauLine|n = 1   | f = \paren {p \implies \paren {\paren {p \implies p} \implies p} } \implies \paren {\paren {p \implies \paren {p \implies p} } \implies \paren {p \implies p} }   | rtxt = Axiom 2   | c = $\mathbf A = p, \mathbf B = p \implies p, \mathbf C = p$ }} {{TableauLine|n = 2   | f = p \implies \paren {\paren {p \implies p} \implies p}   | rtxt = Axiom 1   | c = $\mathbf A = p, \mathbf B = p \implies p$ }} {{ModusPonens|3||\paren {p \implies \paren {p \implies p} } \implies \paren {p \implies p}|1|2}} {{TableauLine|n = 4   | f = p \implies \paren {p \implies p}   | rtxt = Axiom 1   | c = $\mathbf A = p, \mathbf B = p$ }} {{ModusPonens|5||p \implies p|3|4}} {{EndTableau}} {{qed}}
For any proof of $U, \mathbf A \vdash \mathbf B$, we indicate how to transform it into a proof of $U \vdash \mathbf A \implies \mathbf B$ without using the deduction rule. This is done by applying the Second Principle of Mathematical Induction to the length $n$ of the proof of $U,\mathbf A \vdash \mathbf B$. If $n = 1$, then one of the following must occur: * $\mathbf B \in U$ * $\mathbf B = \mathbf A$ * $\mathbf B$ is an axiom or a theorem of $\mathscr H$ In the first case, obviously $U \vdash \mathbf B$. By '''Axiom 1''', $U \vdash \mathbf B \implies \left({\mathbf A \implies \mathbf B}\right)$. By Modus Ponens, $U \vdash \mathbf A \implies \mathbf B$. In the second case, $U \vdash \mathbf A \implies \mathbf A$ by the Law of Identity. Finally, in the third case, we have $U \vdash \mathbf B$. As in the first case, we conclude $U \vdash \mathbf A \implies \mathbf B$. If $n > 1$, the only other option for arriving at $U, \mathbf A \vdash \mathbf B$ is through Modus Ponens. That is to say, two earlier lines of the proof contain: :$U, \mathbf A \vdash \mathbf C$ :$U, \mathbf A \vdash \mathbf C \implies \mathbf B$ for some WFF $\mathbf C$. But then these sequents have shorter proofs. Hence, they satisfy the induction hypothesis. Thus, we may infer: :$U \vdash \mathbf A \implies \mathbf C$ :$U \vdash \mathbf A \implies \left({\mathbf C \implies \mathbf B}\right)$ This allows us to give the following proof of $U \vdash \mathbf A \implies \mathbf B$: {{BeginTableau|U \vdash \mathbf A \implies \mathbf B|nohead=1}} {{TableauLine|n = 1   |f = U \vdash \mathbf A \implies \mathbf C   |rtxt = Hypothesis }} {{TableauLine|n = 2   |f = U \vdash \mathbf A \implies \left({\mathbf C \implies \mathbf B}\right)   |rtxt = Hypothesis }} {{TableauLine|n = 3   |f = U \vdash \left({\mathbf A \implies \left({\mathbf C \implies \mathbf B}\right)}\right) \implies \left({\left({\mathbf A \implies \mathbf C}\right)\implies \left({\mathbf A \implies \mathbf B}\right)}\right)   |rtxt = Axiom 2 }} {{ModusPonens|4||U \vdash \left({\mathbf A \implies \mathbf C}\right)\implies \left({\mathbf A \implies \mathbf B}\right)|2|3}} {{ModusPonens|5||U \vdash \mathbf A \implies \mathbf B|1|4}} {{EndTableau}} The result follows by the Second Principle of Mathematical Induction. {{qed}}
Let us use the following abbreviations {{begin-eqn}} {{eqn | l = \phi       | o = \text{ for }       | r = p \implies q }} {{eqn | l = \psi       | o = \text{ for }       | r = q \implies r }} {{eqn | l = \chi       | o = \text{ for }       | r = p \implies r }} {{end-eqn}} {{BeginTableau|\left({p \implies q}\right) \implies \left({\left({q \implies r}\right) \implies \left({p \implies r}\right)}\right)}} {{TheoremIntro|1|\left({\phi \land \psi}\right) \implies \chi|Hypothetical Syllogism: Formulation 3}} {{SequentIntro|2||\phi \implies \left({\psi \implies \chi}\right)|1|Rule of Exportation}} {{EndTableau}} Expanding the abbreviations leads us back to: : $\vdash \left({p \implies q}\right) \implies \left({\left({q \implies r}\right) \implies \left({p \implies r}\right)}\right)$ {{qed}}
{{BeginTableau|\vdash \paren {p \implies q} \implies \paren {\paren {q \implies r} \implies \paren {p \implies r} }|instance 1 of a Hilbert proof system}} {{Assumption|1|p}} {{Assumption|2|p \implies q}} {{ModusPonens|3|1, 2|q|1|2}} {{Assumption|4|q \implies r}} {{ModusPonens|5|1, 2, 4|r|3|4}} {{TableauLine|n = 6   | pool = 2, 4   | f = p \implies r   | rlnk = Definition:Deduction Rule   | rtxt = Deduction Rule   | dep  = 5 }} {{TableauLine|n = 7   | pool = 2   | f = \paren {q \implies r} \implies \paren {p \implies r}   | rlnk = Definition:Deduction Rule   | rtxt = Deduction Rule   | dep  = 6 }} {{TableauLine|n = 8   | f = \paren {p \implies q} \implies \paren {\paren {q \implies r} \implies \paren {p \implies r} }   | rlnk = Definition:Deduction Rule   | rtxt = Deduction Rule   | dep  = 7 }} {{EndTableau}} {{qed}}
Using a tableau proof for instance 1 of a Hilbert proof system: {{BeginTableau|\vdash \paren {p \implies \paren {q \implies r} } \implies \paren {q \implies \paren {p \implies r} } |nohead = 1}} {{Assumption|1|p}} {{Assumption|2|p \implies \paren {q \implies r} }} {{ModusPonens|3|1, 2|q \implies r|1|2}} {{Assumption|4|q}} {{ModusPonens|5|1, 2, 4|r|3|4}} {{TableauLine|n = 6   | pool = 2, 4   | f = p \implies r   | rlnk = Definition:Deduction Rule   | rtxt = Deduction Rule   | dep  = 5 }} {{TableauLine|n = 7   | pool = 2   | f = q \implies \paren {p \implies r}   | rlnk = Definition:Deduction Rule   | rtxt = Deduction Rule   | dep  = 6 }} {{TableauLine|n = 8   | f = \paren {p \implies \paren {q \implies r} } \implies \paren {q \implies \paren {p \implies r} }   | rlnk = Definition:Deduction Rule   | rtxt = Deduction Rule   | dep  = 7 }} {{EndTableau}} {{qed}}
{{BeginTableau|\vdash q \implies \paren {p \implies q} }} {{Assumption|1|q}} {{SequentIntro|2|1|p \implies q|1|True Statement is implied by Every Statement: Formulation 1}} {{Implication|3||q \implies \paren {p \implies q}|1|2}} {{EndTableau}} {{qed}}
Recall the axioms of $\mathscr H$: {{begin-axiom}} {{axiom | lc = '''Axiom 1:'''         | m  = \mathbf A \implies \paren {\mathbf B \implies \mathbf A} }} {{axiom | lc = '''Axiom 2:'''         | m  = \paren {\mathbf A \implies \paren {\mathbf B \implies \mathbf C} } \implies \paren {\paren {\mathbf A \implies \mathbf B} \implies \paren {\mathbf A \implies \mathbf C} } }} {{axiom | lc = '''Axiom 3:'''         | m  = \paren {\neg \mathbf B \implies \neg \mathbf A} \implies \paren {\mathbf A \implies \mathbf B} }} {{end-axiom}} That these are tautologies is shown on, respectively: :True Statement is implied by Every Statement :Self-Distributive Law for Conditional :Rule of Transposition That Modus Ponens infers tautologies from tautologies is shown on: :Modus Ponendo Ponens Since: :All axioms of $\mathscr H$ are tautologies; :All rules of inference of $\mathscr H$ preserve tautologies it is guaranteed that every formal proof in $\mathscr H$ results in a tautology. That is, all $\mathscr H$-theorems are tautologies. {{qed}}
=== Definition 1 implies Definition 2 === Let $f: S \to T$ be an injection. By Injection to Image is Bijection, $f$ is a bijection from $S$ to the image of $f$. {{qed|lemma}} === Definition 2 implies Definition 1 === Let $T' \subseteq T$ such that there exists a bijection $f: S \to T'$. Let $i: T' \to T$ be the inclusion of $T'$ in $T$. Then by Composite of Injections is Injection, $i \circ f: S \to T$ is an injection. {{qed}}
Let $b, b' \in S$ be immediate successors of $a$. We have that $\preceq$ is a total ordering. {{WLOG}}: :$b \preceq b'$ By virtue of $b'$ being a immediate successor of $a$: :$\neg \exists c \in S: a \prec c \prec b'$ However, since $b$ is also an immediate successor: :$a \prec b$ Hence, it cannot be the case that $b \prec b'$. Since $b \preceq b'$, it follows that $b = b'$. Hence the result. {{qed}}
Let $b, b' \in S$ be immediate predecessors of $a$. We have that $\preceq$ is a total ordering. {{WLOG}}: :$b \preceq b'$ By virtue of $b$ being a immediate predecessor of $a$: :$\neg \exists c \in S: b \prec c \prec a$ However, since $b'$ is also an immediate predecessor: :$b' \prec a$ Hence, it cannot be the case that $b \prec b'$. Since $b \preceq b'$, it follows that $b = b'$. Hence the result. {{qed}}
By definition of the natural numbers: :$\N = \displaystyle \bigcap \II$ where $\II$ is the collection of all inductive sets. Suppose that $n \in \N$. Then by definition of intersection: :$\forall I \in \II: n \in I$ Because all these $I$ are inductive: :$\forall I \in \II: n + 1 \in I$ Again by definition of intersection: :$n + 1 \in \N$ {{qed}}
By the given definition of the natural numbers: :$\N = \bigcap \II$ where $\II$ is the collection of all inductive sets. The result is a direct application of Intersection of Inductive Set as Subset of Real Numbers is Inductive Set. {{qed}}
Let $a + i b \in z^{1/2}$. Then: {{begin-eqn}} {{eqn | l = \paren {a + i b}^2       | r = x + i y       | c = {{Defof|Square Root|subdef = Complex Number|index = 4|Square Root of Complex Number}} }} {{eqn | n = 1       | ll= \leadsto       | l = a^2 + 2 i a b - b^2       | r = x + i y       | c = Square of Sum and $i^2 = -1$ }} {{end-eqn}} Equating imaginary parts in $(1)$: {{begin-eqn}} {{eqn | l = 2 a b       | r = y       | c =  }} {{eqn | n = 2       | ll= \leadsto       | l = b       | r = \frac y {2 a}       | c = rearranging }} {{end-eqn}} Equating real parts in $(1)$: {{begin-eqn}} {{eqn | l = a^2 - b^2       | r = x       | c =  }} {{eqn | ll= \leadsto       | l = a^2 - \paren {\frac y {2 a} }^2       | r = x       | c = substituting for $b$ from $(2)$ }} {{eqn | ll= \leadsto       | l = 4 a^4 - 4 a^2 x - y^2       | r = 0       | c = multiplying by $4 a^2$ and rearranging }} {{eqn | ll= \leadsto       | l = a^2       | r = \frac {4 x \pm \sqrt {16 x^2 + 16 y^2} } 8       | c = Quadratic Formula }} {{eqn | ll= \leadsto       | l = a^2       | r = \frac {x \pm \sqrt {x^2 + y^2} } 2       | c = dividing top and bottom by $4$ }} {{eqn | n = 3       | ll= \leadsto       | l = a       | r = \pm \sqrt {\frac {x + \sqrt {x^2 + y^2} } 2}       | c = taking the square root }} {{end-eqn}} Note that in $(3)$, only the positive square root of the discriminant $x^2 + y^2$ is used. This is because the negative square root of $x^2 + y^2$ would yield $\dfrac {x - \sqrt {x^2 + y^2} } 2 < 0$. As $a \in \R$, it is necessary that $\dfrac {x + \sqrt {x^2 + y^2} } 2 > 0$. Hence $\sqrt {x^2 + y^2} > 0$. Then: {{begin-eqn}} {{eqn | l = b       | r = \frac y {2 a}       | c = from $(2)$ }} {{eqn | r = \frac y {2 \paren {\pm \sqrt {\dfrac {x + \sqrt {x^2 + y^2} } 2} } }       | c =  }} {{eqn | ll= \leadsto       | l = b^2       | r = \frac {y^2} {2 \paren {x + \sqrt {x^2 + y^2} } }       | c =  }} {{eqn | ll= \leadsto       | l = b^2       | r = \frac {y^2 \paren {x - \sqrt {x^2 + y^2} } } {2 \paren {x + \sqrt {x^2 + y^2} } \paren {x - \sqrt {x^2 + y^2} } }       | c = multiplying top and bottom by $x - \sqrt {x^2 + y^2}$ }} {{eqn | ll= \leadsto       | l = b^2       | r = \frac {y^2 \paren {x - \sqrt {x^2 + y^2} } } {2 \paren {x^2 - \paren {x^2 + y^2} } }       | c = Difference of Two Squares }} {{eqn | ll= \leadsto       | l = b^2       | r = \frac {y^2 \paren {x - \sqrt {x^2 + y^2} } } {- 2 y^2}       | c =  }} {{eqn | ll= \leadsto       | l = b^2       | r = \frac {-x + \sqrt {x^2 + y^2} } 2       | c =  }} {{end-eqn}} But from $(2)$ we have: :$b = \dfrac y {2 a}$ and so having picked either the positive square root or negative square root of either $a^2$ or $b^2$, the root of the other is forced. So: :if $y > 0$, then $a$ and $b$ are both of the same sign. Thus: :$b = 1 \times \sqrt {\dfrac {-x + \sqrt {x^2 + y^2} } 2}$ :if $y < 0$, then $a$ and $b$ are of opposite sign. Thus: :$b = \paren {-1} \times \sqrt {\dfrac {-x + \sqrt {x^2 + y^2} } 2}$ Hence: :$b = \dfrac y {\cmod y} \sqrt {\dfrac {-x + \sqrt {x^2 + y^2} } 2}$ {{qed}}
By definition of an ordering: :$0 \preceq 0$ Thus from axiom $(\text {NO} 3)$: :$\exists p \in S: 0 \circ p = 0$ By the definition of zero: :$0 \preceq 0 \circ 0$ and $0 \preceq p$ Thus since $\preceq$ is compatible with $\circ$: :$0 \circ 0 \preceq 0 \circ p = 0$ Thus: :$0 \circ 0 \preceq 0$ and $0 \preceq 0 \circ 0$  Hence, as $\preceq$ is antisymmetric, it follows that: :$0 \circ 0 = 0$ Because $\struct {S, \circ, \preceq}$ is a semigroup, $\circ$ is associative.  So: :$\forall n \in S: \paren {n \circ 0} \circ 0 = n \circ \paren {0 \circ 0} = n \circ 0$ Thus from axiom $(\text {NO} 2)$: :$\forall n \in S: n \circ 0 = n$ Similarly: :$\forall n \in S: 0 \circ \paren {0 \circ n} = \paren {0 \circ 0} \circ n = 0 \circ n$ meaning: :$\forall n \in S: 0 \circ n = n$ Thus: :$\forall n \in S: n \circ 0 = n = 0 \circ n$ and so $0$ is the identity for $\circ$. {{Qed}}
From Shape of Sine Function: :$\forall x \in \left[{0 \,.\,.\, \dfrac \pi 2}\right]: 0 \le \sin x \le 1$ and so on the same interval: :$0 \le \sin^{n + 1} x \le \sin^n x$ therefore: :$\forall n \in \N: 0 < I_{n + 1} < I_n$ From Reduction Formula for Integral of Power of Sine: :$\displaystyle \int \sin^n x \rd x = \dfrac {n - 1} n \int \sin^{n - 2} x \rd x - \dfrac {\sin^{n - 1} x \cos x} n$ Thus: {{begin-eqn}} {{eqn | l = I_n       | r = \int_0^{\frac \pi 2} \sin^n x \rd x       | c =  }} {{eqn | r = \dfrac {n - 1} n \int_0^{\frac \pi 2} \sin^{n - 2} x \rd x - \left[{\dfrac {\sin^{n - 1} x \cos x} n}\right]_0^{\pi / 2}       | c =  }} {{eqn | r = \dfrac {n - 1} n I_{n - 2} - \frac {\left({\sin^{n - 1} \frac \pi 2 \cos \frac \pi 2 - \sin^{n - 1} 0 \cos 0}\right)} n       | c =  }} {{eqn | r = \dfrac {n - 1} n I_{n - 2} - \frac {\left({0 - \sin^{n - 1} 0 \cos 0}\right)} n       | c = Cosine of Right Angle }} {{eqn | r = \dfrac {n - 1} n I_{n - 2} - \frac {\left({0 - 0}\right)} n       | c = Sine of Zero is Zero }} {{eqn | ll= \implies       | l = n I_n       | r = \left({n - 1}\right) I_{n - 2}       | c = multiplying both sides by $n$ }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = n I_n       | r = \left({n - 1}\right) I_{n - 2}       | cc= Reduction Formula for Definite Integral of Power of Sine }} {{eqn | ll= \implies       | l = 1       | o = <       | r = \frac {I_{ 2n} } {I_{2 n + 1} }       | cc=  }} {{eqn | o = <       | r = \frac {I_{2 n - 1} } {I_{2 n + 1} }       | cc= Reduction Formula for Definite Integral of Power of Sine }} {{eqn | r = \frac {2 n + 1}  {2 n}       | cc= Reduction Formula for Definite Integral of Power of Sine }} {{eqn | r = \to 0       | c = as $n \to \infty$ }} {{end-eqn}} {{qed}}
{{improve|This would be more rigorous if implemented as a formal induction proof.}} Let $I_n = \displaystyle \int_0^{\frac \pi 2} \sin^n x \rd x$. Then: {{begin-eqn}} {{eqn | l = I_{2 n}       | r = \frac {2 n - 1} {2 n} I_{2 n - 2}       | c = Reduction Formula for Definite Integral of Power of Sine }} {{eqn | r = \frac {\paren {2 n - 1} \paren {2 n - 3} } {2 n \paren {2 n - 2} } I_{2 n - 4}       | c = Reduction Formula for Definite Integral of Power of Sine again }} {{eqn | r = \frac {\paren {2 n - 1} \paren {2 n - 3} \cdots 1} {2 n \paren {2 n - 2} \cdots 2} I_0       | c = Reduction Formula for Definite Integral of Power of Sine until the end }} {{eqn | r = \frac {\paren {2 n - 1} \paren {2 n - 3} \cdots 1} {2 n \paren {2 n - 2} \cdots 2} \int_0^{\pi / 2} \rd x       | c = Definition of $I_n$ }} {{eqn | r = \frac {\paren {2 n - 1} \paren {2 n - 3} \cdots 1} {2 n \paren {2 n - 2} \cdots 2} \bigintlimits x 0 {\pi / 2}       | c = Integral of Constant }} {{eqn | r = \frac {\paren {2 n - 1} \paren {2 n - 3} \cdots 1} {2 n \paren {2 n - 2} \cdots 2} \frac \pi 2       | c = Integral of Constant }} {{eqn | r = \frac {2n \paren {2 n - 1} \paren {2 n - 2} \paren {2 n - 3} \cdots 2 \cdot 1} {\paren {2 n}^2 \paren {2 n - 2}^2 \cdots 2^2} \frac \pi 2       | c = multiplying top and bottom by bottom }} {{eqn | r = \frac {2n \paren {2 n - 1} \paren {2 n - 2} \paren {2 n - 3} \cdots 2 \cdot 1} {\paren {2^n}^2 n^2 \paren {n - 1}^2 \cdots 1^2} \frac \pi 2       | c = extracting factor of $\paren {2^n}^2$ from the bottom }} {{eqn | r = \frac {\paren {2 n}!} {\paren {2^n}^2 \paren {n!}^2} \frac \pi 2       | c = {{Defof|Factorial}} }} {{eqn | r = \frac {\paren {2 n}!} {\paren {2^n n!}^2} \frac \pi 2       | c = rearranging  }} {{end-eqn}} {{qed}}
{{improve|This would be more rigorous if implemented as a formal induction proof.}} Let $I_n = \displaystyle \int_0^{\frac \pi 2} \sin^n x \rd x$. Then: {{begin-eqn}} {{eqn | l = I_{2 n + 1}       | r = \frac {2 n} {2 n + 1} I_{2 n - 1}       | c = Reduction Formula for Definite Integral of Power of Sine }} {{eqn | r = \frac {2 n \paren {2 n - 2} } {\paren {2 n + 1} \paren {2 n - 1} } I_{2 n - 3}       | c = Reduction Formula for Definite Integral of Power of Sine again }} {{eqn | r = \frac {2 n \paren {2 n - 2} \dotsm 2} {\paren {2 n + 1} \paren {2 n - 1} \dotsm 3} I_1       | c = Reduction Formula for Definite Integral of Power of Sine until the end }} {{eqn | r = \frac {2 n \paren {2 n - 2} \dotsm 2} {\paren {2 n + 1} \paren {2 n - 1} \dotsm 3} \int_0^{\pi / 2} \sin x \rd x       | c = Definition of $I_n$ }} {{eqn | r = \frac {2 n \paren {2 n - 2} \dotsm 2} {\paren {2 n + 1} \paren {2 n - 1} \dotsm 3} \intlimits {-\cos x} 0 {\pi / 2}       | c = Primitive of Sine Function }} {{eqn | r = \frac {2 n \paren {2 n - 2} \dotsm 2} {\paren {2 n + 1} \paren {2 n - 1} \dotsm 3} \paren {0 - \paren {-1} }       | c = Cosine of Right Angle and Cosine of Zero is One }} {{eqn | r = \frac {2 n \paren {2 n - 2} \dotsm 2} {\paren {2 n + 1} \paren {2 n - 1} \dotsm 3}       | c =  }} {{eqn | r = \frac {\paren {2 n}^2 \paren {2 n - 2}^2 \dotsm 2^2} {\paren {2 n + 1} \paren {2 n} \paren {2 n - 1} \paren {2 n - 2} \paren {2 n - 3} \dotsm 3 \cdot 2}       | c = multiplying top and bottom by top }} {{eqn | r = \frac {\paren {2 n}^2 n^2 \paren {n - 1}^2 \dotsm 1^2} {\paren {2 n + 1} \paren {2 n} \paren {2 n - 1} \paren {2 n - 2} \paren {2 n - 3} \dotsm 3 \cdot 2}       | c = extracting factor of $\left({2^n}\right)^2$ from the top }} {{eqn | r = \frac {\paren {2^n n!}^2} {\paren {2 n + 1}!}       | c = {{Defof|Factorial}} }} {{end-eqn}} {{qed}}
From axiom $(\text {NO} 4)$, we have: :$\exists m, n \in S: m \ne n$ That is, there are at least two distinct elements in $S$. Therefore, there must be at least one element in $S^* = S \setminus \set 0$. So: :$S^* = S \setminus \set 0 \ne \O$ {{qed}}