hcju/proofwikiqa huggingface dataset:
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. $\begin{array}{|ccccc||cccc|} \hline \neg & p & \land & \neg & q & \neg & (p & \lor & q) \\ \hline T & F & T & T & F & T & F & F & F \\ T & F & F & F & T & F & F & T & T \\ F & T & F & T & F & F & T & T & F \\ F & T & F & F & T & F & T & T & T \\ \hline \end{array}$ {{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. $\begin{array}{|ccc||cccccc|} \hline p & \land & q & \neg & (\neg & p & \lor & \neg & q)  \\ \hline F & F & F & F & T & F & T & T & F \\ F & F & T & F & T & F & T & F & T \\ T & F & F & F & F & T & T & T & F \\ T & T & T & T & F & T & F & F & T \\ \hline \end{array}$ {{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. $\begin{array}{|ccc||cccccc|} \hline p & \lor & q & \neg & (\neg & p & \land & \neg & q) \\ \hline F & F & F & F & T & F & T & T & F \\ F & T & T & T & T & F & F & F & T \\ T & T & F & T & F & T & F & T & F \\ T & T & T & T & F & T & F & F & T \\ \hline \end{array}$ {{qed}}
We only have to check condition $(2)$, as a simply connected set is path-connected by definition. === Necessary condition === Suppose that $X$ is simply connected. Let $\gamma_1, \gamma_2: \closedint 0 1 \to X$ be two paths with $\map {\gamma_1} 0 = \map {\gamma_2} 0$ and $\map {\gamma_1} 1 = \map {\gamma_2} 1$. Define the mapping $-\gamma_2: \closedint 0 1 \to X$ by $-\map {\gamma_2} t = \map {\gamma_2} {1 - t}$. From Continuity of Composite Mapping, it follows that $-\gamma_2$ is a path. Let $c: \closedint 0 1 \to X$ be the constant path defined by $\map c t = \map {\gamma_1} 0$. When $\equiv$ denotes equivalence of homotopy classes, we have: {{begin-eqn}} {{eqn | l = \gamma_1       | o = \equiv       | r = \gamma_1 * c       | c = here, $\gamma_1 * c$ denotes the concatenation of $\gamma_1$ and $c$ }} {{eqn | o = \equiv       | r = \gamma_1 * \paren {-\gamma_2 * \gamma_2 }       | c = by assumption, as $\paren {-\gamma_2 * \gamma_2 }$ and $c$ are both closed paths with initial point $\map {\gamma_1} 0$ }} {{eqn | o = \equiv       | r = \paren {\gamma_1 * \paren {-\gamma_2} } * \gamma_2  }} {{eqn | o = \equiv       | r = c * \gamma_2       | c = by assumption }} {{eqn | o = \equiv       | r = \gamma_2 }} {{end-eqn}} {{qed|lemma}} === Sufficient Condition === Let $x_0 \in X$. We show that the fundamental group $\map {\pi_1} {X, x_0}$ is trivial. Let $\gamma_1, \gamma_2: \closedint 0 1 \to X$ be two closed paths with initial point $x_0$. By assumption, $\gamma_1$ and $\gamma_2$ are freely homotopic and belong to the same homotopy class. Hence, $\map {\pi_1} {X, x_0}$ has only one element. {{qed}}
We apply the Method of Truth Tables to the proposition. As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations. $\begin{array}{|ccccc||ccccccc|} \hline p & \land & (q & \lor & r) & (p & \land & q) & \lor & (p & \land & r) \\ \hline F & F & F & F & F & F & F & F & F & F & F & F \\ F & F & F & T & T & F & F & F & F & F & F & T \\ F & F & T & T & F & F & F & T & F & F & F & F \\ F & F & T & T & T & F & F & T & F & F & F & T \\ T & F & F & F & F & T & F & F & F & T & F & F \\ T & T & F & T & T & T & F & F & T & T & T & T \\ T & T & T & T & F & T & T & T & T & T & F & F \\ T & T & T & T & T & T & T & T & T & T & T & T \\ \hline \end{array}$ {{qed}}
{{BeginTableau|\vdash \paren {\paren {q \lor r} \land p} \iff \paren {\paren {q \land p} \lor \paren {r \land p} } }} {{Assumption|1|\paren {q \lor r} \land p}} {{SequentIntro|2|1|\paren {q \land p} \lor \paren {r \land p}|1|Conjunction is Right Distributive over Disjunction: Formulation 1}} {{Implication|3||\paren {\paren {q \lor r} \land p} \implies \paren {\paren {q \land p} \lor \paren {r \land p} }|1|2}} {{Assumption|4|\paren {q \land p} \lor \paren {r \land p} }} {{SequentIntro|5|4|\paren {q \lor r} \land p|4|Conjunction is Right Distributive over Disjunction: Formulation 1}} {{Implication|6||\paren {\paren {q \land p} \lor \paren {r \land p} } \implies \paren {\paren {q \lor r} \land p}|4|5}} {{BiconditionalIntro|7||\paren {\paren {q \lor r} \land p} \iff \paren {\paren {q \land p} \lor \paren {r \land p} }|3|6}} {{EndTableau}} {{qed} Category:Rule of Distribution 7umzufsmkl5vcnhblyr3fhdma3gi018
We apply the Method of Truth Tables to the proposition. As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations. :$\begin{array}{|ccccc||ccccccc|} \hline (q & \lor & r) & \land & p & (q & \land & p) & \lor & (r & \land & p) \\ \hline F & F & F & F & F & F & F & F & F & F & F & F \\ F & F & F & F & T & F & F & T & F & F & F & T \\ F & T & T & F & F & F & F & F & F & T & F & F \\ F & T & T & T & T & F & F & T & T & T & T & T \\ T & T & F & F & F & T & F & F & F & F & F & F \\ T & T & F & T & T & T & T & T & T & F & F & T \\ T & T & T & F & F & T & F & F & F & T & F & F \\ T & T & T & T & T & T & T & T & T & T & T & T \\ \hline \end{array}$ {{qed}}
Consider the differential equation: :$D_z \map f z = i \cdot \map f z$ === Step 1 === We will prove that $z = \cos \theta + i \sin \theta$ is a solution. {{begin-eqn}} {{eqn | l = z       | r = \cos \theta + i \sin \theta       | c =  }} {{eqn | l = \frac {\d z} {\d \theta}       | r = -\sin \theta + i \cos \theta       | c = Derivative of Sine Function, Derivative of Cosine Function, Linear Combination of Derivatives }} {{eqn | r = i^2 \sin \theta + i\cos \theta       | c = $i^2 = -1$ }} {{eqn | r = i \paren {i \sin \theta + \cos \theta} }} {{eqn | r = i z }} {{end-eqn}} {{qed|lemma}} === Step 2 === We will prove that $y = e^{i\theta}$ is a solution. {{begin-eqn}} {{eqn | l = y       | r = e^{i\theta}       | c =  }} {{eqn | l = \frac {\d y} {\d \theta}       | r = i e^{i \theta}       | c = Derivative of Exponential Function, Chain Rule for Derivatives, Linear Combination of Derivatives }} {{eqn | r = i y }} {{end-eqn}} {{qed|lemma}} === Step 3 === Consider the initial condition $\map f 0 = 1$. {{begin-eqn}} {{eqn | l = \bigintlimits y {\theta \mathop = 0} {}       | r = e^{0 i} }} {{eqn | r = 1 }} {{eqn | l = \bigintlimits z {\theta \mathop = 0} {}       | r = \cos 0 + i \sin 0 }} {{eqn | r = 1 }} {{end-eqn}} So $y$ and $z$ are both particular solutions. But a particular solution to a differential equation is unique. {{MissingLinks|"Specific solution" and the proof that such a specific solution is unique.}} Therefore $y = z$, that is, $e^{i \theta} = \cos \theta + i \sin \theta$. {{qed}}
This: :$e^{i \theta} = \cos \theta + i \sin \theta$ is logically equivalent to this: :$\dfrac{\cos \theta + i \sin \theta} {e^{i \theta}} = 1$ for every $\theta$. Note that the left expression is nowhere undefined. Taking the derivative of this: {{begin-eqn}} {{eqn | l = \dfrac {\mathrm d}{\mathrm d \theta} e^{-i \theta} \left({\cos \theta + i \sin \theta}\right)       | r = e^{-i\theta} \left({-\sin \theta + i \cos \theta}\right) + \left({-i e^{-i\theta} }\right) \left({\cos \theta + i \sin \theta}\right)       | c = Product Rule and Derivative of Exponential Function }} {{eqn | r = e^{-i \theta} \left({-\sin \theta + i \cos \theta - i \cos \theta - i^2 \sin \theta}\right)       | c = }} {{eqn | r = e^{-i \theta} \left({-\sin \theta + i \cos \theta - i \cos \theta + \sin \theta}\right)       | c = }} {{eqn | r = e^{-i \theta} \left({0}\right)       | c = }} {{eqn | r = 0       | c = }} {{end-eqn}} Thus the expression, as a function of $\theta$, is constant and so yields the same value for every $\theta$. We know the value at at least one point, that is, when $\theta = 0$: :$\dfrac {\cos 0 + i \sin 0} {e^{0 i}} = \dfrac {1 + 0} 1 = 1$ Thus it is $1$ for ''every'' $\theta$, which verifies the above. Hence the result. {{qed}}
We apply the Method of Truth Tables to the proposition. As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations. $\begin{array}{|ccccc||ccccccc|} \hline p & \lor & (q & \land & r) & (p & \lor & q) & \land & (p & \lor & r) \\ \hline F & F & F & F & F & F & F & F & F & F & F & F \\ F & F & F & F & T & F & F & F & F & F & T & T \\ F & F & T & F & F & F & T & T & F & F & F & F \\ F & T & T & T & T & F & T & T & T & F & T & T \\ T & T & F & F & F & T & T & F & T & T & T & F \\ T & T & F & F & T & T & T & F & T & T & T & T \\ T & T & T & F & F & T & T & T & T & T & T & F \\ T & T & T & T & T & T & T & T & T & T & T & T \\ \hline \end{array}$ {{qed}}
{{BeginTableau|\vdash \paren {\paren {q \land r} \lor p} \iff \paren {\paren {q \lor p} \land \paren {r \lor p} } }} {{Assumption|1|\paren {q \land r} \lor p}} {{SequentIntro|2|1|\paren {q \lor p} \land \paren {r \lor p}|1|Conjunction is Right Distributive over Disjunction: Formulation 1}} {{Implication|3||\paren {\paren {q \land r} \lor p} \implies \paren {\paren {q \lor p} \land \paren {r \lor p} }|1|2}} {{Assumption|4|\paren {q \lor p} \land \paren {r \lor p} }} {{SequentIntro|5|4|\paren {q \land r} \lor p|4|Conjunction is Right Distributive over Disjunction: Formulation 1}} {{Implication|6||\paren {\paren {q \lor p} \land \paren {r \lor p} } \implies \paren {\paren {q \land r} \lor p}|4|5}} {{BiconditionalIntro|7||\paren {\paren {q \land r} \lor p} \iff \paren {\paren {q \lor p} \land \paren {r \lor p} }|3|6}} {{EndTableau|qed}} Category:Rule of Distribution 7z7uuzjoo8r26oav447crim4y4t7oxm
We apply the Method of Truth Tables to the proposition. As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations. $\begin{array}{|ccccc||ccccccc|} \hline (q & \land & r) & \lor & p & (q & \lor & p) & \land & (r & \lor & p) \\ \hline F & F & F & F & F & F & F & F & F & F & F & F \\ F & F & F & T & T & F & T & T & T & F & T & T \\ F & F & T & F & F & F & F & F & F & T & T & F \\ F & F & T & T & T & F & T & T & T & T & T & T \\ T & F & F & F & F & T & T & F & F & F & F & F \\ T & F & F & T & T & T & T & T & T & F & T & T \\ T & T & T & T & F & T & T & F & T & T & T & F \\ T & T & T & T & T & T & T & T & T & T & T & T \\ \hline \end{array}$ {{qed}}
Let $x \in \displaystyle \bigcup \mathbb S$. By the definition of union, there exists an $S \in \mathbb S$ such that $x \in S$. By premise, $S \subseteq T$. By the definition of subset, $x \in T$. Since this result holds for each $x \in \displaystyle \bigcup \mathbb S$: :$\displaystyle \bigcup \mathbb S \subseteq T$ {{qed}} Category:Union of Subsets is Subset is1e5gkt5f9egiiuv1vwh4v2k7b6qr3
By Union of Subsets is Subset: Set of Sets: :$\displaystyle \paren {\forall X \in \mathbb S: X \subseteq T} \implies \bigcup \mathbb S \subseteq T$ {{qed|lemma}} For the converse implication, suppose that $\displaystyle \bigcup \mathbb S \subseteq T$. Consider any $X \in \mathbb S$ and take any $x \in X$. From Set is Subset of Union: Set of Sets we have that $X \subseteq \bigcup \mathbb S$. Thus $\displaystyle x \in \bigcup \mathbb S$. But $\displaystyle \bigcup \mathbb S \subseteq T$. So it follows that $X \subseteq T$. So: :$\displaystyle \bigcup \mathbb S \subseteq T \implies \paren {\forall X \in \mathbb S: X \subseteq T}$ {{qed|lemma}} Hence: :$\displaystyle \paren {\forall X \in \mathbb S: X \subseteq T} \iff \bigcup \mathbb S \subseteq T$ {{Qed}}
By the definition of a relation from $S$ to $T$, each element of $\mathcal F$ is a subset of $S \times T$. By Intersection of Subsets is Subset: Set of Sets: : $\mathcal R \subseteq S \times T$ Therefore, by the definition of a relation from $S$ to $T$, $\mathcal R$ is a relation from $S$ to $T$. {{qed}} Category:Relation Theory tdt3qiqa479f55i5kwsfnrx83qyh4ag
Let $x \in \bigcap \mathbb S$. Then by the definition of intersection: :$\forall S \in \mathbb S: x \in S$. Since $\mathbb S$ is non-empty by the premise, it has some element $S$. Then $x \in S$. Since $S \in \mathbb S$, the premise shows that $S \subseteq T$. By the definition of subset, $x \in T$. Since this holds for each $x \in \bigcap \mathbb S$: :$\bigcap \mathbb S \subseteq T$ {{qed}} Category:Set Intersection Category:Subsets nvi46jila9jjopv6mlvzhcmdtcs3ngx
=== Definition 1 Implies Definition 2 === {{:Equivalence of Definitions of Initial Topology/Definition 1 Implies Definition 2}}{{qed|lemma}} === Definition 2 Implies Definition 1 === {{:Equivalence of Definitions of Initial Topology/Definition 2 Implies Definition 1}}{{qed}} Category:Initial Topology rmwc5xdissitxao55f9pfktz766vumr
Let the sequence $\sequence {\cmod {\dfrac {a_{n+1} } {a_n} } }_{n \mathop \in \N}$ converge. Let $\epsilon \in \R_{>0}$, and let $z \in \C$. Let $\cmod {z - \xi} = R - \epsilon$. By definition of radius of convergence, it follows that $S \paren z$ is absolutely convergent. Then from the Ratio Test: :$\lim_{n \mathop \to \infty} \cmod {\dfrac {a_{n+1} \paren {z - \xi}^{n + 1} } {a_n \paren {z - \xi}^n} } \le 1$ By Multiple Rule for Complex Sequences, this inequality can be rearranged to obtain: {{begin-eqn}} {{eqn | l = \lim_{n \mathop \to \infty} \cmod {\dfrac {a_{n + 1} } {a_n} }        | o = \le       | r = \cmod {\dfrac {\paren {z - \xi}^{n + 1} } {\paren {z - \xi}^n } }^{-1} }} {{eqn | r = \dfrac 1 {\cmod {x - \xi} }} {{eqn | r = \dfrac 1 {R - \epsilon} }} {{end-eqn}} {{qed|lemma}} Let $\cmod {z - \xi} = R + \epsilon$. Then $S \paren z$ is divergent, so the Ratio Test shows that: :$\displaystyle \lim_{n \mathop \to \infty} \cmod {\dfrac {a_{n + 1} \paren {z - \xi}^{n + 1} } {a_n \paren {z - \xi}^n} } \ge 1$ Similarly, this inequality can be rearranged as: :$\displaystyle \lim_{n \mathop \to \infty} \cmod {\dfrac {a_{n + 1} } {a_n} } \ge \dfrac 1 {R + \epsilon}$ {{qed|lemma}} As $\epsilon > 0$ was arbitrary, it follows that: :$\displaystyle \lim_{n \mathop \to \infty} \cmod {\dfrac {a_{n + 1}} {a_n} } = \dfrac 1 R$ Hence the result. {{qed}}
From the ratio test, $S \paren x$ is convergent if: :$\displaystyle \lim_{n \mathop \to \infty} \size {\frac {a_{n + 1} \paren {x - \xi}^{n + 1} } {a_n \paren {x - \xi}^n} } < 1$ Thus: {{begin-eqn}} {{eqn | l = \size {\frac {a_{n + 1} \paren {x - \xi}^{n + 1} } {a_n \paren {x - \xi}^n} }       | o = <       | r = 1       | c =  }} {{eqn | ll= \iff       | l = \size {\frac {a_{n + 1} } {a_n} } \size {x - \xi}       | o = <       | r = 1       | c =  }} {{eqn | ll= \iff       | l = \size {\frac {a_{n + 1} } {a_n} }       | o = <       | r = \frac 1 {\size {x - \xi} }       | c =  }} {{end-eqn}} The result follows from the definition of radius of convergence. {{qed}}
This proof assumes that $V$ is a semi-inner product space over $\R$. Then for all $x, y \in V$, we have $\innerprod x y = \innerprod y x$ by property $(1')$ of semi-inner products. Define $f_{x, y}: \R \to \R_{\ge 0}$ by: :$\map {f_{x, y} } \lambda = \innerprod {x - \lambda y} {x - \lambda y}$ Then by property $(4)$ of semi-inner product: :$\forall \lambda \in \R: \map {f_{x, y} } \lambda \ge 0$ For all $\lambda \in \R$, it follows that: {{begin-eqn}} {{eqn | l = \map {f_{x, y} } \lambda       | r = \innerprod {x - \lambda y} {x - \lambda y} }} {{eqn | r = \innerprod x x + \innerprod x {-\lambda y} + \innerprod {-\lambda y} x + \innerprod {-\lambda y} {-\lambda y}       | c = Property $(3)$ of Semi-Inner Product }} {{eqn | r = \innerprod x x - 2 \lambda \innerprod x y + \lambda^2 \innerprod y y       | c = Properties $(1')$ and $(2)$ of Semi-Inner Product }} {{eqn | r = a \lambda^2 + b \lambda + c }} {{end-eqn}} where we have put $a = \innerprod y y$, $b = -2 \innerprod x y$, and $c = \innerprod x x$. Then $f_{x, y}$ is a quadratic polynomial which satisfies $\map {f_{x, y} } \lambda \ge 0$. Hence $f_{x, y}$ has at most one distinct real root. From Solution to Quadratic_Equation, it follows that the discriminant $\Delta$ satisfies: :$\Delta = b^2 - 4 a c \le 0$ Therefore: :$4 \innerprod x y^2 - 4 \innerprod x x \innerprod y y \le 0$ which we rearrange as: :$\size {\innerprod x y}^2 = \innerprod x y^2 \le \innerprod x x \innerprod y y$ {{qed}}
Let $x, y \in V$. Let $\lambda \in \mathbb K$. Then: {{begin-eqn}} {{eqn | l = 0       | o = \le       | r = \innerprod {x - \lambda y} {x - \lambda y}       | c = Property $(4)$ of Semi-Inner Product }} {{eqn | r = \innerprod x x + \innerprod x {-\lambda y} + \innerprod {-\lambda y} x + \innerprod {-\lambda y} {-\lambda y}       | c = Property $(3)$ of Semi-Inner Product }} {{eqn | r = \innerprod x x - \lambda^* \innerprod x y - \lambda \innerprod x y^* + \lambda \lambda^* \innerprod y y       | c = Property $(1)$ and $(2)$ of Semi-Inner Product }} {{end-eqn}} where $\lambda^*$ denotes the complex conjugate of $\lambda$. (If $\mathbb K$ is a subfield of $\R$ , then $\lambda^* = \lambda$.) First, suppose $\innerprod y y \ne 0$. Insert $\lambda = \innerprod x y \innerprod y y^{-1}$ in the inequality: {{begin-eqn}} {{eqn | l = 0       | o = \le       | r = \innerprod x x - \innerprod x y^* \paren {\innerprod y y^{-1} }^* \innerprod x y - \innerprod x y \innerprod y y^{-1} \innerprod x y^* + \innerprod x y \innerprod y y^{-1} \innerprod x y^* \paren {\innerprod y y^{-1} }^* \innerprod y, y       | c = Product of Complex Conjugates }} {{eqn | r = \innerprod x x - \size {\innerprod x y}^2 \paren {\innerprod y y^{-1} }^* - \size {\innerprod x y}^2 \innerprod y y^{-1} + \size {\innerprod x y}^2 \paren {\innerprod y y^{-1} }^*        | c = Modulus in Terms of Conjugate }} {{eqn | r = \innerprod x x - \size {\innerprod x y}^2 \innerprod y y^{-1}  }} {{end-eqn}} Reorder the inequality to get: :$\size {\innerprod x y}^2 \le \innerprod x x \innerprod y y$ {{qed|lemma}} Next, suppose $\innerprod y y = 0$. Let $n \in \N \subset \mathbb K$ be a natural number. Insert $\lambda = n \innerprod x y$ in the inequality: {{begin-eqn}} {{eqn | l = 0       | o = \le       | r = \innerprod x x - n^* \innerprod x y^* \innerprod x y - n \innerprod x y \innerprod x y^* + 0       | c = Product of Complex Conjugates }} {{eqn | r = \innerprod x x - n \innerprod x y^* \innerprod x y - n \innerprod x y \innerprod x y^*       | c = Complex Number equals Conjugate iff Wholly Real }} {{eqn | r = \innerprod x x - 2 n \size {\innerprod x y}^2       | c = Modulus in Terms of Conjugate }} {{end-eqn}} Rearrange the inequality to get: :$\innerprod x x \ge 2 n \size {\innerprod x y}^2$ As this inequality holds for all $n \in \N$, it follows that: :$\size {\innerprod x y}^2 = 0$. Then: :$\size {\innerprod x y}^2 \le 0 = \innerprod x x \innerprod y y$ {{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. $\begin{array}{|ccc||ccccc|} \hline p & \land & q & \neg & (p & \implies & \neg & q) \\ \hline F & F & F & F & F & T & T & F \\ F & F & T & F & F & T & F & T \\ T & F & F & F & T & T & T & F \\ T & T & T & T & T & F & F & T \\ \hline \end{array}$ {{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. $\begin{array}{|ccc||cccc|} \hline p & \implies & q & \neg & p & \lor & q \\ \hline F & T & F & T & F & T & F \\ F & T & T & T & F & T & T \\ T & F & F & F & T & F & F \\ T & T & T & F & T & T & T \\ \hline \end{array}$ {{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. $\begin{array}{|ccc||cccc|} \hline p & \lor & q & \neg & p & \implies & q \\ \hline F & F & F & T & F & F & F \\ F & T & T & T & F & T & T \\ T & T & F & F & T & T & F \\ T & T & T & F & T & T & T \\ \hline \end{array}$ {{qed}}
This is a power series in the form $\displaystyle \sum_{n \mathop = 0}^\infty a_n \paren {z - \xi}^n$ where $\sequence {a_n} = \sequence {\dfrac 1 {n!} }$. Applying Radius of Convergence from Limit of Sequence: Complex Case, we find that: {{begin-eqn}} {{eqn | l = \lim_{n \mathop \to \infty} \cmod {\dfrac {a_{n + 1} } {a_n} }       | r = \lim_{n \mathop \to \infty} \cmod {\dfrac {\frac 1 {\paren {n + 1}!} } {\frac 1 {n!} } }       | c =  }} {{eqn | r = \lim_{n \mathop \to \infty} \cmod {\dfrac {n!} {\paren {n + 1}!} }       | c =  }} {{eqn | r = \lim_{n \mathop \to \infty} \cmod {\dfrac 1 {n + 1} }       | c =  }} {{eqn | r = 0       | c = Sequence of Powers of Reciprocals is Null Sequence }} {{end-eqn}} Hence the result. {{qed}}
This is a power series in the form $\displaystyle \sum_{n \mathop= 0}^\infty a_n \left({x - \xi}\right)^n$ where $\left \langle {a_n} \right \rangle = \left \langle {\dfrac 1 {n!}} \right \rangle$. Applying Radius of Convergence from Limit of Sequence, we find that: {{begin-eqn}} {{eqn | l = \lim_{n \mathop \to \infty} \left\vert{\dfrac {a_{n+1} } {a_n} }\right\vert       | r = \lim_{n \mathop \to \infty} \left\vert{\dfrac {\dfrac 1 {\left({n + 1}\right)!} } {\dfrac 1 {n!} } }\right\vert       | c =  }} {{eqn | r = \lim_{n \mathop \to \infty} \left\vert{\dfrac {n!} {\left({n + 1}\right)!} }\right\vert       | c =  }} {{eqn | r = \lim_{n \mathop \to \infty} \left\vert{\dfrac 1 {n+1} }\right\vert       | c =  }} {{eqn | r = 0       | c = Sequence of Powers of Reciprocals is Null Sequence }} {{end-eqn}} Hence the result. {{qed}}
{{begin-eqn}} {{eqn | l = \log_b x + \log_b y       | r = \map {\log_b} {b^{\log_b x + \log_b y} }       | c = {{Defof|General Logarithm}} }} {{eqn | r = \map {\log_b} {\paren {b^{\log_b x} } \paren {b^{\log_b y} } }       | c = Exponent Combination Laws: Product of Powers }} {{eqn | r = \map {\log_b} {x y}       | c = {{Defof|General Logarithm}} }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \log_b x + \log_b y       | r = \frac {\ln x} {\ln b} + \frac {\ln y} {\ln b}       | c = Change of Base of Logarithm }} {{eqn | r = \frac {\ln x + \ln y} {\ln b}       | c =  }} {{eqn | r = \frac {\map \ln {x y} } {\ln b}       | c = Sum of Logarithms: Proof for Natural Logarithm }} {{eqn | r = \map {\log_b} {x y}       | c = Change of Base of Logarithm }} {{end-eqn}} {{qed}}
We apply the Method of Truth Tables. $\begin{array}{|ccc||c|c|} \hline p & \land & q & p & q \\ \hline F & F & F & F & F \\ F & F & T & F & T \\ T & F & F & T & F \\ T & T & T & T & T \\ \hline \end{array}$ As can be seen, when $p \land q$ is true so are both $p$ and $q$. {{qed}}
We apply the Method of Truth Tables. $\begin{array}{|c|c||ccc|} \hline p & q & p & \land & q\\ \hline F & F & F & F & F \\ F & T & F & F & T \\ T & F & T & F & F \\ T & T & T & T & T \\ \hline \end{array}$ As can be seen, only when both $p$ and $q$ are true, then so is $p \land q$. {{qed}}