hcju/proofwikiqa huggingface dataset:
To prove $G$ is isomorphic to $\struct {\R, +}$, it is sufficient to find a bijective homorphism $\phi: \to \R \to G$: :$\forall x, y \in G: \map \phi {x + y} = \map \phi x \circ \map \phi y$ From Group Examples: $\dfrac {x + y} {1 + x y}$: :the identity element of $G$ is $0$ :the inverse of $x$ in $G$ is $-x$. This also holds for $\struct {\R, +}$. This hints at the structure of a possible such $\phi$: :$(1): \quad$ that it is an odd function :$(2): \quad$ that it passes through $0$ :$(3): \quad$ that its image is the open interval $\openint {-1} 1$. One such function is the hyperbolic tangent function $\tanh$, which indeed has the above properties on $\R$. Then we have that: {{begin-eqn}} {{eqn | l = \map {\tanh } {x + y}       | r = \dfrac {\tanh x + \tanh y} {1 + \tanh x \tanh y}       | c = Hyperbolic Tangent of Sum }} {{eqn | r = \tanh x \circ \tanh y       | c =  }} {{end-eqn}} This demonstrates the homorphism between $\struct {\R,  +}$ and $\struct {G, \circ}$. We have that Real Hyperbolic Tangent Function is Strictly Increasing over $\R$. Hence from Strictly Monotone Real Function is Bijective, $\tanh: \R \to \openint {-1} 1$ is a bijection. Hence the result that $\struct {\R,  +} \cong \struct {G, \circ}$. {{Qed}}
{{begin-eqn}} {{eqn | l = \cmod {z + w}^2       | r = \paren {z + w} \paren {\overline z + \overline w}       | c = Product of Complex Number with Conjugate }} {{eqn | r = z \overline z + w \overline w + w \overline z + z \overline w       | c =  }} {{eqn | r = \cmod z^2 + \cmod w^2 + w \overline z + z \overline w       | c = Product of Complex Number with Conjugate }} {{eqn | r = \cmod z^2 + \cmod w^2 + w \overline z + \overline {\paren {\overline z} } \overline w       | c = Complex Conjugation is Involution }} {{eqn | r = \cmod z^2 + \cmod w^2 + \overline z w + \overline {\paren {\overline z w} }       | c = Product of Complex Conjugates and Complex Multiplication is Commutative }} {{eqn | r = \cmod z^2 + \cmod w^2 + 2 \map \Re {z \overline w}       | c = Sum of Complex Number with Conjugate }} {{eqn | o = \le       | r = \cmod z^2 + \cmod w^2 + 2 \cmod {z \overline w}       | c = Modulus Larger than Real Part }} {{eqn | r = \cmod z^2 + \cmod w^2 + 2 \cmod z \cmod {\overline w}       | c = Complex Modulus of Product of Complex Numbers }} {{eqn | r = \cmod z^2 + \cmod w^2 + 2 \cmod z \cmod w       | c = Complex Modulus equals Complex Modulus of Conjugate }} {{eqn | r = \paren {\cmod z + \cmod w}^2       | c = Square of Sum }} {{eqn | ll= \leadsto       | l = \cmod {z + w}       | o = \le       | r = \cmod z + \cmod w       | c =  }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \cmod {z_1 + z_2}       | o = \ge       | r = \cmod {z_1} - \cmod {z_2}       | c = Triangle Inequality for Complex Numbers: Corollary 1 }} {{eqn | l = \cmod {z_1 + z_2}       | o = \ge       | r = \cmod {z_2} - \cmod {z_1}       | c = Triangle Inequality for Complex Numbers: Corollary 1 }} {{eqn | r = -\paren {\cmod {z_1} - \cmod {z_2} }       | c =  }} {{eqn | ll= \leadsto       | l = \cmod {z_1 + z_2}       | o = \ge       | r = \cmod {\cmod {z_1} - \cmod {z_2} }       | c = Negative of Absolute Value: Corollary 2 }} {{end-eqn}} {{qed}}
Let $f$ be continuous at $x$. Then by definition: :For every neighborhood $N$ of $\map f x$ in $\R$, there exists a neighborhood $M$ of $x$ in $S$ such that $\map {f^\to} M \subseteq N$. From: :Open Ball in Real Number Line is Open Interval :Open Ball of Metric Space is Open Set :Set is Open iff Neighborhood of all its Points  it follows that: :For every $\epsilon \in \R_{>0}$, there exists a neighborhood $M$ of $x$ in $S$ such that $\map {f^\to} M \subseteq \openint {\map f x - \epsilon} {\map f x + \epsilon}$ By definition of a neighborhood in $\struct {S, \tau}$: :for all neighborhoods $M$ there exists $U \in \tau$ such that $z \in U \subseteq M$ Thus: :For every $\epsilon \in \R_{>0}$, there exists $U \in \tau$ with $x \in U$ such that $\map {f^\to} U \subseteq \openint {\map f x - \epsilon} {\map f x + \epsilon}$
Let $f$ satisfy: :$\forall \epsilon \in \R_{>0} : \exists U \in \tau : x \in U : \map {f^\to} U \subseteq \openint {\map f x - \epsilon} {\map f x + \epsilon}$ Let $N$ be a neighborhood of $\map f x$. By definition of a neighborhood of $\map f x$: :there exists $U$ such that $z \in U \subseteq N$ By definition of the topology induced by the Euclidean Metric on Real Number Line: :there exists an open $\epsilon$-ball $\map {B_\epsilon} {\map f x}$ such that $x \in \map {B_\epsilon} {\map f x} \subseteq U$ From Open Ball in Real Number Line is Open Interval: :$\map {B_\epsilon} {\map f x} = \openint {\map f x - \epsilon} {\map f x + \epsilon}$ Thus: :$\exists U \in \tau : x \in U : \map {f^\to} U \subseteq \openint {\map f x - \epsilon} {\map f x + \epsilon}$ From Set is Open iff Neighborhood of all its Points, $U$ is a neighborhood of $x$. Since $N$ was an arbitrary neighborhood of $\map f x$ it follows that $f$ is continuous at $x$.
{{begin-eqn}} {{eqn | l = \max \set {a, b}       | r = a + b - \min \set {a, b}       | c = Sum Less Minimum is Maximum }} {{eqn | r = a + b - \frac 1 2 \paren {a + b - \size {a - b} }       | c = Min is Half of Sum Less Absolute Difference }} {{eqn | r = \frac 1 2 \paren {2 a + 2 b - \paren {a + b - \size {a - b} } } }} {{eqn | r = \frac 1 2 \paren {2 a + 2 b - a - b + \size {a - b} }  }} {{eqn | r = \frac 1 2 \paren {a + b + \size {a - b} }  }} {{end-eqn}} {{qed}}
From Sum Less Minimum is Maximum: :$\forall x \in S : \max \set {\map f x, \map g x} = \map f x + \map g x  - \min \set {\map f x, \map g x}$ Thus:  :$\max \set {f, g} = f + g - \min \set{f, g}$ From Minimum Rule for Continuous Functions: :$\min \set {f, g}$ is continuous From Multiple Rule for Continuous Mappings into Topological Ring: :$-\min \set {f, g}$ is continuous  From Sum Rule for Continuous Mappings into Topological Ring: :$f + g -\min \set {f, g}$ is continuous  Thus: :$\max \set {f, g}$ is continuous  {{qed}}
=== Definition 1 implies Definition 2 === Let $\struct {S, \tau}$ satisfy: :$\forall x, y \in S, x \ne y: \exists U, V \in \tau: x \in U, y \in V: U^- \cap V^- = \O$  Let $x, y \subseteq S , x \ne y $ be arbitrary. Then: :$\exists U, V \in \tau: x \in U, y \in V: U^- \cap V^- = \O$  Let $N_x = U$ and $N_y = V$. From Set is Subset of Itself then: :$\exists N_x, N_y \subseteq S: \exists U, V \in \tau: x \in U \subseteq N_x, y \in V \subseteq N_y: N_x^- \cap N_y^- = \O$ Because $x, y \in S$ were arbitrary: :$\forall x, y \in S, x \ne y : \exists N_x, N_y \subseteq S: \exists U, V \in \tau: x \subseteq U \subseteq N_x, y \subseteq V \subseteq N_y: N_x^- \cap N_y^- = \O$ {{qed|lemma}} === Definition 2 implies Definition 1 === Let $\struct {S, \tau}$ satisfy: :$\forall x, y \in S, x \ne y : \exists N_x, N_y \subseteq S: \exists U, V \in \tau: x \subseteq U \subseteq N_x, y \subseteq V \subseteq N_y: N_x^- \cap N_y^- = \O$ Let $x, y \subseteq S , x \ne y $ be arbitrary. Then: :$\exists N_x, N_y \subseteq S: \exists U, V \in \tau: x \in U \subseteq N_x, y \in V \subseteq N_y: N_x^- \cap N_y^- = \O$ From Topological Closure of Subset is Subset of Topological Closure: :$U^- \subseteq N_x^-, V^- \subseteq N_y^-$ From Subsets of Disjoint Sets are Disjoint then: :$U^- \cap V^- = \O$ Thus: :$\exists U, V \in \tau: x \in U, y \in V: U^- \cap V^- = \O$ Because $x, y \in S$ were arbitrary: :$\forall x, y \in S, x \ne y: \exists U, V \in \tau: x \in U, y \in V: U^- \cap V^- = \O$  {{qed}} Category:Separation Axioms 94sbbc25rb3r6p73c0tk31ct6hzpnri
Suppose $T$ is a completely Hausdorff space. As $S_\alpha \ne \O$ we also have $S \ne \O$ by the axiom of choice. Let $\alpha \in I$. From Subspace of Product Space Homeomorphic to Factor Space, $\struct {S_\alpha, \tau_\alpha}$ is homeomorphic to a subspace $T_\alpha$ of $T$. From completely Hausdorff property is hereditary, $T_\alpha$ is completely Hausdorff. From Completely Hausdorff Space is Preserved under Homeomorphism, $\struct {S_\alpha, \tau_\alpha}$ is completely Hausdorff. As $\alpha \in I$ was arbitrary, the result follows.
We have that a metric space is fully $T_4$. Then we have that a fully $T_4$ space is $T_4$. {{qed}} Category:Metric Spaces Category:T4 Spaces 91f9trk2m9kj7okqupxtfry047n4g3s
Let $x, y \in S : x \ne y$ be arbitrary. Then $x_\alpha \ne y_\alpha$ for some $\alpha \in I$. Since $\struct {S_\alpha, \tau_\alpha}$ is completely Hausdorff spaces then: :$\exists U, V \in \tau_\alpha: x_\alpha \in U, y_\alpha \in V : U^- \cap V^- = \O$ Let $\pr_\alpha: S \to S_\alpha$ be the projection of $S$ to $S_\alpha$. Then: {{begin-eqn}} {{eqn | l = \map {\pr_\alpha^\gets} {U^-} \cap \map {\pr_\alpha^\gets} {V^-}       | r = \map {\pr_\alpha^\gets} {U^- \cap V^-}       | c = Preimage of Intersection under Mapping }} {{eqn | r = \map {\pr_\alpha^\gets} \O }} {{eqn | r = \O }} {{end-eqn}}  From Preimage of Subset is Subset of Preimage: :$\map {\pr_\alpha^\gets} U \subseteq \map {\pr_\alpha^\gets} {U^-}$ and :$\map {\pr_\alpha^\gets} V \subseteq \map {\pr_\alpha^\gets} {V^-}$ From Projection from Product Topology is Open and Continuous:General Result :$\pr_\alpha: S \to S_\alpha$ is continuous. From Continuity Defined from Closed Sets: :$\map {\pr_\alpha^\gets} {U^-}, \map {\pr_\alpha^\gets} {V^-}$ are closed in $T$   From Set Closure is Smallest Closed Set in Topological Space: :$\paren{\map {\pr_\alpha^\gets} U}^- \subseteq \map {\pr_\alpha^\gets} {U^-}$ and :$\paren{\map {\pr_\alpha^\gets} V}^- \subseteq \map {\pr_\alpha^\gets} {V^-}$ From Subsets of Disjoint Sets are Disjoint: :$\paren{\map {\pr_\alpha^\gets} U}^- \cap \paren{\map {\pr_\alpha^\gets} V}^- = \O$ By definition of the projection $\pr_\alpha$: :$\map {\pr_\alpha} x = x_\alpha \in U$ By definition of the preimage under $\pr_\alpha$: :$x \in \map {\pr_\alpha^\gets} U$ Similarly: :$y \in \map {\pr_\alpha^\gets} V$ By definition of the Tychonoff topology $\tau$: :$\map {\pr_\alpha^\gets} U, \map {\pr_\alpha^\gets} V \in \tau$ As $x, y \in S$ were arbitrary, it follows that $T$ is a completely Hausdorff space by definition.
Let $\sequence{\alpha_n}$ be the sequence of partial sums: :$\forall n \in \N: \alpha _n = \sum_{i \mathop = 0}^n d_{n + m} p^{n + m}$ Let $\sequence{\beta_n}$ be the sequence of partial sums: :$\forall n \in \N: \beta _n = \sum_{i \mathop = 0}^n d_{n + l} p^{n + l}$ Then: {{begin-eqn}} {{eqn | l = \beta_n       | r = \sum_{i \mathop = 0}^n d_{n + l} p^{n + l} }} {{eqn | r = \sum_{i \mathop = l}^{n + l} d_n p^n }} {{eqn | r = \sum_{i \mathop = m}^{l - 1} d_n p^n  +  \sum_{i \mathop = l}^{n + l} d_n p^n }} {{eqn | r = \sum_{i \mathop = m}^{n + l} d_n p^n }} {{eqn | r = \sum_{i \mathop = 0}^{n + l - m} d_{n + m} p^{n + m} }} {{eqn | r = \alpha_{n + l - m} }} {{end-eqn}} By definition of $l$: :$m \le l$ So: :$\forall n \in \N : n + l - m \ge n$ Thus $\sequence{\beta_n}$ is a subsequence of $\sequence{\alpha_n}$ by definition. From Subsequence of Cauchy Sequence in Normed Division Ring is Cauchy Sequence: :$\displaystyle \sum_{i \mathop = l}^\infty d_i p^i$ is a Cauchy Sequence in $\Q$. From Subsequence is Equivalent to Cauchy Sequence: :$\displaystyle \lim_{n \mathop \to \infty} {\alpha_n - \beta_n} = 0$ That is, the sequence $\sequence {\alpha_n - \beta_n}$ is a null sequence. By definition of $p$-adic numbers as a quotient of Cauchy sequences: :$\sequence {\alpha_n}$ and $\sequence{\beta_n}$ represent the same $p$-adic number Since $\sequence {\alpha_n}$ represents $\mathbf a$, it follows that $\sequence {\beta_n}$ represents $\mathbf a$ {{qed}} Category:P-adic Number Theory cge4m7jficc932hj75o7c357grzf3ek
From Subsequence of Cauchy Sequence in Normed Division Ring is Cauchy Sequence: :$\sequence {x_{m_n} }$ is a Cauchy sequence Let $\epsilon > 0$. By definition of a Cauchy sequence: :$\exists N: \forall n, m > N: \norm {x_n - x_m } < \epsilon$ From Index of Subsequence not Less than its Index: $\forall n \in \N : m_n \ge n$ Thus: :$\exists N: \forall n > N: \norm {x_n - x_{m_n} } < \epsilon$ By definition of convergence: :$\displaystyle \lim_{n \mathop \to \infty} {x_n - x_{m_n} } = 0$ {{qed}} Category:Cauchy Sequences Category:Normed Division Rings 9kmhm554fykctwjkx1wpwshgnupjzs3
Suppose $\forall i: 0 \le i \le r - 2: a_i = p - 1$.  Then: {{begin-eqn}} {{eqn | l = n + 1       | r = 1 + \sum_{i \mathop = 0}^r a_ip^i       | c = from $\displaystyle n = \sum_{k \mathop = 0}^r a_k p^k$ }} {{eqn | r = 1 + a_r p^r + a_{r - 1} p^{r - 1} + \sum_{i \mathop = 0}^{r - 2} (p - 1) p^i       | c = from $a_i = p - 1$ for all $0 \le i \le r - 2$ }} {{eqn | r = 1 + a_r p^r + a_{r - 1} p^{r - 1} + \paren {p^{r - 1} - 1}       | c = by Difference of Two Powers }} {{eqn | r = p^{r - 1} \paren {p a_r + a_{r - 1} + 1}       | c = factorisation }} {{end-eqn}} Since $p a_r + a_{r - 1} + 1$ is an integer, we have $p^{r - 1} \divides \paren {n + 1}$. Now we prove the contrapositive of our theorem. Suppose $r > 2$ and $p^{r - 1} \nmid \paren {n + 1}$. By the contrapositive of the result above: :$\exists i: 0 \le i \le r - 2, a_i \neq p - 1$ Let $I$ be the smallest integer for which $a_I \neq p - 1$. Then for every $0 \le i < I$, we have: :$a_i = p - 1$ Let $m = p^r - 1$. Then $m = \paren{p - 1} \displaystyle \sum_{i \mathop = 0}^r p^i$ is a representation of $m$ to the base $p$. We also have: {{begin-eqn}} {{eqn | l = n - m       | r = \sum_{i \mathop = 0}^r a_ip^i - \paren {p^r - 1}       | c = from $\displaystyle n = \sum_{k \mathop = 0}^t a_k p^k$ }} {{eqn | r = 1 - p^r + a_r p^r + \sum_{i = I}^{r - 1} a_i p^i + \sum_{i = 0}^{I - 1} (p - 1) p^i       | c = from $a_i = p - 1$ for all $0 \le i < I$ }} {{eqn | r = 1 + \paren {a_r - 1} p^r + \sum_{i = I}^{r - 1} a_i p^i + \paren {p^I - 1}       | c = by Difference of Two Powers }} {{eqn | r = \paren {a_r - 1} p^r + \sum_{i = I + 1}^{r - 1} a_i p^i + \paren {a_I + 1} p^I        | c =  }} {{end-eqn}} Notice that: :$0 \le a_r - 1 < p - 1$ :$0 \le a_i < p$ for $I + 1 \le i \le r - 1$ :$1 \le a_I + 1 < p - 1 + 1 = p$ So $\paren {a_r - 1} p^r + \displaystyle \sum_{i = I + 1}^{r - 1} a_i p^i + \paren {a_I + 1} p^I$ is a representation of $n - m$ to the base $p$ (possibly with leading zeroes). Now we add $n-m$ and $m$ in base $p$:  place value   p^r   p^{r-1}     p^{I+1}   p^I  p^{I-1} p^{I-2}       1                               n - m =  (a_r-1) a_{r-1} ... a_{I+1} (a_I+1)   0       0    ...   0    +)     m =           (p-1)       (p-1)   (p-1)  (p-1)   (p-1)      (p-1)                   -------------------------------------------------------------------------     n     =     *       *        a_{I+1}   a_I   (p-1)   (p-1)      (p-1)                                    carry   carry                                            So there are at least two carries in this addition, at place values $p^I$ and $p^{I + 1}$. By Kummer's Theorem, we have: :$p^2 \divides \dbinom {n - m + m} m = \dbinom n m$ Thus the contrapositive is proved. {{qed}}
{{begin-eqn}} {{eqn | l = \dfrac {\d^m} {\d z^m} \map G z        | r = \sum_{k \mathop \ge 0} \dfrac {\paren {k + m}!} {k!} a_{k + m} z^k       | c = Derivative of Generating Function: General Result }} {{eqn | r = \dfrac {m!} {0!} a_m + \sum_{k \mathop \ge 1} \dfrac {\paren {k + m}!} {k!} a_{k + m} z^k       | c =  }} {{eqn | r = m! a_m + \sum_{k \mathop \ge 1} \dfrac {\paren {k + m}!} {k!} a_{k + m} z^k       | c = Factorial of Zero }} {{eqn | ll= \leadsto       | l = \map {G^{\paren m} } 0       | r = m! a_m       | c =  }} {{eqn | ll= \leadsto       | l = \dfrac 1 {m!} \map {G^{\paren m} } 0       | r = a_m       | c =  }} {{end-eqn}} {{qed}} Category:Generating Functions olu9mfz3kbo57heaq3f4n6sybnxawtz
Proof by induction: === Basis for the Induction === When $n = 0$, we have: :$\dfrac {\d^0} {\d z^0} \map G z = \map G z$ Also: :$\displaystyle \sum_{k \mathop \ge 0} \dfrac {\paren {k + 0}!} {k!} a_{k + 0} z^k = \sum_{k \mathop \ge 0} a_k z^k$ This is our base case. === Induction Hypothesis === :$\forall n \in \N: n \ge 0: \dfrac {\d^n} {\d z^n} \map G z = \displaystyle \sum_{k \mathop \ge 0} \dfrac {\paren {k + n}!} {k!} a_{k + n} z^k$ This is our induction hypothesis. It is to be demonstrated that: :$\dfrac {\d^{n + 1} } {\d z^{n + 1} } \map G z = \displaystyle \sum_{k \mathop \ge 0} \dfrac {\paren {k + n + 1}!} {k!} a_{k + n + 1} z^k$ === Induction Step === This is our induction step: Consider $m = n + 1$. {{begin-eqn}} {{eqn | l = \dfrac {\d^{n + 1} } {\d z^{n + 1} } \map G z       | r = \dfrac \d {\d z} \paren {\frac {\d^n} {\d z^n} \map G z }       | c =  }} {{eqn | r = \dfrac \d {\d z} \paren {\sum_{k \mathop \ge 0} \dfrac {\paren {k + n}!} {k!} a_{k + n} z^k}       | c = by induction hypothesis }} {{eqn | r = \sum_{k \mathop \ge 0} \paren {\dfrac \d {\d z} \dfrac {\paren {k + n}!} {k!} a_{k + n} z^k}       | c =  }} {{eqn | r = \sum_{k \mathop \ge 0} \dfrac {k \paren {k + n}!} {k!} a_{k + n} z^{k - 1}       | c = Power Rule for Derivatives }} {{eqn | r = \sum_{k \mathop \ge 1} \dfrac {k \paren {k + n}!} {k!} a_{k + n} z^{k - 1}       | c = as the zeroth term vanishes when $k = 0$ }} {{eqn | r = \sum_{k \mathop \ge 1} \dfrac {\paren {k + n}!} {\paren {k - 1}!} a_{k + n} z^{k - 1}       | c = {{Defof|Factorial}} }} {{eqn | r = \sum_{k \mathop \ge 0} \dfrac {\paren {k + n + 1}!} {k!} a_{k + n + 1} z^k       | c = Translation of Index Variable of Summation }} {{end-eqn}} Hence the result by induction. {{qed}} Category:Generating Functions 66edb3o96gii1knsc3212pvvokcurm5
The general solution to the 1-D wave equation: :$u_{tt} = c^2 u_{xx} \quad \text{for } - \infty < x < \infty$ is given by :$\map u {x,t} = \map f {x + c t} + \map g {x - c t}$ where $f,g$ are arbitrary twice-differentiable functions. From initial conditions we have: {{begin-eqn}} {{eqn | l = \map \phi x       | m = \map u {x, 0}       | mo = =        | r = \map f x + \map g x }} {{eqn | l = \map \psi x       | m = \map {u_t} {x, 0}       | mo = =       | r = c \map {f'} x - c \map {g'} x       | c = Chain Rule for Partial Derivatives }} {{end-eqn}} So we have: {{begin-eqn}} {{eqn | l = \map {\phi'} x       | r = \map {f'} x + \map {g'} x       | c = Sum Rule for Derivatives }} {{eqn | l = \dfrac {\map \psi x} c       | r = \map {f'} x - \map {g'} x }} {{end-eqn}} Solving the equations give: {{begin-eqn}} {{eqn | l = \map {f'} x       | r = \dfrac 1 2 \paren {\map {\phi'} x + \dfrac {\map \psi x} c} }} {{eqn | l = \map {g'} x       | r = \dfrac 1 2 \paren {\map {\phi'} x - \dfrac {\map \psi x} c} }} {{end-eqn}} Integrating both equations and using Fundamental Theorem of Calculus: {{begin-eqn}} {{eqn | l = \map f x       | r = \dfrac 1 2 \map \phi x + \dfrac 1 {2c} \int_0^x \map \psi s \rd s + A }} {{eqn | l = \map g x       | r = \dfrac 1 2 \map \phi x - \dfrac 1 {2c} \int_0^x \map \psi s \rd s + B }} {{end-eqn}} for some constants $A,B$. From $\map \phi x = \map f x + \map g x$, we have $A + B = 0$.  Therefore: {{begin-eqn}} {{eqn | l = \map u {x,t}       | r = \map f {x + c t} + \map g {x - c t} }} {{eqn | r = \dfrac 1 2 \map \phi {x + c t} + \dfrac 1 {2c} \int_0^{x + c t} \map \psi s \rd s + A + \dfrac 1 2 \map \phi {x - c t} - \dfrac 1 {2c} \int_0^{x - c t} \map \psi s \rd s + B       | c = substitution }} {{eqn | r = \dfrac 1 2 \paren {\map \phi {x + c t} + \map \phi {x - c t} } + \dfrac 1 {2 c} \int_{x - c t}^{x + c t} \map \psi s \rd s       | c = simplification }} {{end-eqn}} {{qed}} {{Namedfor|Jean le Rond d'Alembert|cat = D'Alembert}}
Let $S$ be the set defined as: :$S = \set {s: \exists t \in A: t = \map g s} \cup \set \O$ By definition, we have that: :$\O \in S$ :$s \in S \implies \map g s \in S$ So, by definition, $S$ is inductive under $g$. Because $A$ is inductive under $g$: :$\O \in A$ and so by definition of subset: :$S \subseteq A$ It remains to be shown that $S$ is minimally inductive under $g$. {{ProofWanted}}
By definition of Bell numbers: :$B_{n + 1}$ is the number of partitions of a (finite) set whose cardinality is $n + 1$. Let $k \in \set {k \in \Z: 0 \le k \le n}$. Let us form a partition of a (finite) set $S$ with cardinality $n + 1$ such that one component has $n + 1 - k > 0$ elements. We can do this by first choosing $1$ element from $S$. We put this element into that single component. Then choose $k$ more elements from $S$, and let the remaining $n - k$ elements be put into the same component as the first element. From Cardinality of Set of Subsets and the definition of binomial coefficient, there are $\dbinom n k$ ways to do this. For the chosen $k$ elements, there are $B_k$ ways to partition them. Thus there are $\dbinom n k B_k$ possible partitions for $S$: :$\dbinom n k$ of selecting $n - k$ elements to form one component with the one singled-out element :for each of these, $B_k$ ways to partition the remaining $k$ elements. Summing the number of ways to do this over all possible $k$: :$\displaystyle B_{n + 1} = \sum_{k \mathop = 0}^n \dbinom n k B_k$ {{qed}} Category:Bell Numbers 58cj8hqf8fx22rjop7d9p1u0i2xwsvz
Let $\N$ be considered as the naturally ordered semigroup: :$\struct {\N, +, \le}$ The result follows from Sum with One is Immediate Successor in Naturally Ordered Semigroup.
Let $\N$ be defined as the von Neumann construction $\omega$. By definition of the ordering on von Neumann construction: :$m \le n \iff m \subseteq n$ From Von Neumann Construction of Natural Numbers is Minimally Inductive, $\omega$ is minimally inductive class under the successor mapping. The result is then a direct application of Characteristics of Minimally Inductive Class under Progressing Mapping: Image of Proper Subset is Subset: :$m \subset n \implies m^+ \subseteq n$ {{qed}}
Let $\N$ be defined as the von Neumann construction $\omega$. By definition of the ordering on von Neumann construction: :$m \le n \iff m \subseteq n$ From Von Neumann Construction of Natural Numbers is Minimally Inductive, $\omega$ is minimally inductive class under the successor mapping. From Characteristics of Minimally Inductive Class under Progressing Mapping: Mapping Preserves Subsets: :$\forall m, n \in \omega: m \subseteq n \implies m^+ \subseteq n^+$ {{qed}}
From Von Neumann Construction of Natural Numbers is Minimally Inductive, $\omega$ is minimally inductive class under the successor mapping. The result is a direct application of Non-Empty Bounded Subset of Minimally Inductive Class under Progressing Mapping has Greatest Element. {{qed}}
By definition of the ordering on von Neumann construction: :$m \le n \iff m \subseteq n$ === Necessary Condition === Let $m \in n$. Then from Natural Number is Transitive Set: :$m \subseteq n$ But if $m = n$, we would have: :$m \in m$ which contradicts Natural Number is Ordinary Set. So we have: :$m \le n$ and $m \ne n$ that is: :$m < n$ {{qed|lemma}} === Sufficient Condition === Let $m < n$. Then by Natural Number m is Less than n implies n is not Greater than Successor of n: :$m^+ \le n$ But we have $m \in m^+$ by construction. Hence: :$m \in n$ {{qed}}