hcju/proofwikiqa huggingface dataset:
{{begin-eqn}} {{eqn | l = \map {\frac \d {\d x} } {\arccot u}       | r = \map {\frac \d {\d u} } {\arccot u} \frac {\d u} {\d x}       | c = Chain Rule for Derivatives }} {{eqn | r = -\dfrac 1 {1 + u^2} \frac {\d u} {\d x}       | c = Derivative of Arccotangent Function }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \map {\frac \d {\d x} } {\arcsec u}       | r = \map {\frac \d {\d u} } {\arcsec u} \frac {\d u} {\d x}       | c = Chain Rule for Derivatives }} {{eqn | r = \dfrac 1 {\size u \sqrt {u^2 - 1} } \frac {\d u} {\d x}       | c = Derivative of Arcsecant Function }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \map {\frac \d {\d x} } {\arccsc u}       | r = \map {\frac \d {\d u} } {\arccsc u} \frac {\d u} {\d x}       | c = Chain Rule for Derivatives }} {{eqn | r = -\dfrac 1 {\size u \sqrt {u^2 - 1} } \frac {\d u} {\d x}       | c = Derivative of Arccosecant Function }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \map f x       | r = \map f {-x}       | c = {{Defof|Even Function}} }} {{eqn | ll= \leadsto       | l = \frac \d {\d x} \map f x       | r = \frac \d {\d x} \map f {-x}       | c = differentiating both sides {{WRT|Differentiation}} $x$ }} {{eqn | ll= \leadsto       | l = \map {f'} x       | r = \map {f'} {-x} \times \paren {-1}       | c = Chain Rule for Derivatives }} {{eqn | r = -\map {f'} {-x}       | c =  }} {{end-eqn}} Hence the result by definition of odd function. {{qed}} Category:Even Functions Category:Differential Calculus q12f604kwqnh0uz516rkh6s0hud65ih
{{begin-eqn}} {{eqn | l = \map f x       | r = -\map f {-x}       | c = {{Defof|Odd Function}} }} {{eqn | ll= \leadsto       | l = \frac \d {\d x} \map f x       | r = -\frac \d {\d x} \map f {-x}       | c = differentiating both sides {{WRT|Differentiation}} $x$ }} {{eqn | ll= \leadsto       | l = \map {f'} x       | r = -\map {f'} {-x} \times \paren {-1}       | c = Chain Rule for Derivatives }} {{eqn | r = \map {f'} {-x}       | c =  }} {{end-eqn}} Hence the result by definition of even function. {{qed}} Category:Odd Functions Category:Differential Calculus efatzudp5edp8236s4w6wm3n5kyic8k
{{AimForCont}} $n$ has an odd divisor $d$. By Sum of Two Odd Powers: :$\paren {n^{n/d} + 1} \divides \paren {\paren {n^{n/d}}^d + 1^d} = S_n$ thus $S_n$ is composite, which is a contradiction. Hence $n$ has no odd divisors. That is, $n$ is a power of $2$. Write $n = 2^m$. {{AimForCont}} that $m$ has an odd divisor $f$. By Sum of Two Odd Powers: :$\paren {2^{n m/f} + 1} \divides \paren {\paren {2^{n m/f}}^f + 1^f} = 2^{m n} + 1 = S_n$ thus $S_n$ is composite, which is a contradiction. Hence $m$ has no odd divisors. That is, $m$ is a power of $2$. Therefore we must have: :$n = 2^{2^k}$ for some integer $k$. {{qed}}
From Newton's Second Law of Motion, the force on $B$ equals its mass multiplied by its acceleration. Thus the force $F$ on $B$ is given by: :$F = m \dfrac {\d^2 y} {\d t^2}$ where it is assumed that the acceleration is in a downward direction. The force on $B$ due to gravity is $m g$. The force on $B$ due to the air it is passing through is $k$ multiplied by the speed of $B$, in the opposite direction to its travel. That is:: :$k \dfrac {d y} {d t}$ Hence the required differential equation: :$m \dfrac {\d^2 y} {\d t^2} = m g - k \dfrac {d y} {d t}$ {{qed}}
From Motion of Cart attached to Wall by Spring under Damping: :$\dfrac {\d^2 \mathbf x} {\d t^2} + \dfrac c m \dfrac {\d \mathbf x} {\d t} + \dfrac k m \mathbf x = 0$ The result follows by setting $c = 0$. {{qed}}
We have that Between two Real Numbers exists Rational Number: :$\forall a, b \in \R : a < b : \exists r \in \Q : a < r < b$ Let $a := x$ with $x \in \R$. Let $\epsilon \in \R_{\mathop > 0} : r - a < \epsilon$. Let $b := x + \epsilon$. Then: {{begin-eqn}} {{eqn | l = x - \epsilon       | o = <       | r = x }} {{eqn | o = <       | r = r }} {{eqn | o = <       | r = x + \epsilon }} {{end-eqn}} Since this holds for all $x$, we have that: :$\forall x \in \R : \exists \epsilon \in \R_{\mathop > 0} : \exists r \in \Q : \size {x - r} < \epsilon$ By definition, $\Q$ is dense in $\R$.
By definition of the parameters of $M$: {{:Definition:Parameter of Autoregressive Model}} Thus: :there are $p$ parameters of the form $\phi_j$ :$1$ parameter $\mu$ :$1$ parameter $\sigma_a^2$. That is: $p + 1 + 1 = p + 2$ parameters. {{qed}}
We can eliminate $\tilde z_{t - 1}$ from the {{RHS}} of $(1)$ by substituting: :$\tilde z_{t - 1} = \phi_1 \tilde z_{t - 2} + \phi_2 \tilde z_{t - 3} + \dotsb + \phi_p \tilde z_{t - p - 1} + a_{t - 1}$ Similarly we can substitute for $\tilde z_{t - 2}$, and so on. Eventually we get an infinite series in $a_{t - j}$. Hence: :$\map \phi B \tilde z_t = a_t$ is equivalent to: :$\tilde z_t = \map \psi B a_t$ such that: :$\map \psi B = \map {\phi^{-1} } B$ Hence the result by definition of linear filter model. {{qed}}
Let $x \in \R$. Let $U \subseteq \R$ be an open set of $T$ such that $x \in U$. From Basis for Euclidean Topology on Real Number Line, there exists an open interval $V_0 = \openint {x - \epsilon} {x + \epsilon} \subseteq U$ for some $\epsilon > 0$ such that $x \in V_0$. From Between two Real Numbers exists Rational Number: :$\exists p \in \Q: p \in \openint {x - \epsilon} {x + \epsilon}$ Thus, we can define the open interval $V_1 = \openint x p \subseteq V_0$. Similarly: :$\exists q \in \Q: q \in \openint x p$ We can then define an open interval $V_2 = \openint q p \subseteq V_1$. We have $V_2 \subseteq V_1$, $V_1 \subseteq V_0$ and $V_0 \subseteq U$. By successively applying Subset Relation is Transitive, it follows that $V_2 \subseteq U$. Note that $x \notin V_2$, since $x < q < p < x + \epsilon$. From Between two Rational Numbers exists Irrational Number, there exists $y \in \R \setminus \Q: y \in \openint q p = V_2$. As $x \notin V_2$, it must be the case that $x \ne y$. Since $V_2 \subseteq U$, $U$ is an open set of $T$ containing $x$ which also contains an element of $\R \setminus \Q$ other than $x$. As $U$ is arbitrary, it follows that every open set of $T$ containing $x$ also contains an element of $\R \setminus \Q$ other than $x$. That is, $x$ is by definition a limit point of $\R \setminus \Q$. As $x$ is arbitrary, it follows that all elements of $\R$ are limit points of $\R \setminus \Q$. The result follows from the definition of everywhere dense. {{qed}} Category:Real Analysis Category:Real Number Line with Euclidean Topology Category:Irrational Number Space Category:Denseness 09xf05ij99hlu19r9bah8fmexbn8quc
Let $x \in \R$. Let $\epsilon \in \R_{\mathop > 0}$ Either $x \in \Q$ or $x \in \R \setminus \Q$. Suppose $x \in \R \setminus \Q$. Let $y := x$. Then: :$\size {x - y} < \epsilon$ Suppose $x \in \Q$. Let $\displaystyle n \in \N : n > \frac {\sqrt 2} \epsilon$ Let $\displaystyle y := x + \frac {\sqrt 2} n$ Then $y \in \R \setminus \Q$. Furthermore: {{begin-eqn}} {{eqn | l = \size {x - y}       | r = \size {\frac {\sqrt 2} n} }} {{eqn | o = <      | r = \epsilon }} {{end-eqn}} In both cases $x$ was arbitrary. Hence: :$\forall x \in \R : \exists \epsilon \in \R_{\mathop > 0} : \exists y \in \R \setminus \Q : \size {x - y} < \epsilon$ By definition, $\R \setminus \Q$ is dense in $\R$. {{qed}}
{{begin-eqn}} {{eqn | l = x - x       | r = x + \paren {-x}       | c = {{Defof|Real Subtraction}} }} {{eqn | r = 0       | c = Inverses for Real Addition }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = a < b       | o = \implies       | r = a \times c < b \times c       | c = Real Number Ordering is Compatible with Multiplication: Positive Factor as $c > 0$ }} {{eqn | l = c < d       | o = \implies       | r = b \times c < b \times d       | c = Real Number Ordering is Compatible with Multiplication: Positive Factor as $b > 0$ }} {{end-eqn}} The result follows by Real Number Ordering is Transitive. {{Qed}}
==== Lemma ==== {{:Leigh.Samphier/Sandbox/Equivalence of Definitions of Matroid Base Axiom/Lemma}}{{qed|lemma}} === Definition 1 iff Definition 2 === Definition 1 holds {{iff}} Definition 2 holds follows immediately from the lemma. {{qed|lemma}} === Definition 1 iff Definition 3 === {{:Leigh.Samphier/Sandbox/Equivalence of Definitions of Matroid Base Axiom/Definition 1 Iff Definition 3}}{{qed|lemma}} === Definition 1 iff Definition 4 === {{:Leigh.Samphier/Sandbox/Equivalence of Definitions of Matroid Base Axiom/Definition 1 Iff Definition 4}}{{qed|lemma}} === Definition 4 iff Definition 5 === {{:Leigh.Samphier/Sandbox/Equivalence of Definitions of Matroid Base Axiom/Definition 4 Iff Definition 5}}{{qed|lemma}} === Definition 5 iff Definition 6 === Definition 5 holds {{iff}} Definition 6 holds follows immediately from the lemma. {{qed|lemma}} === Definition 3 iff Definition 7 === {{:Leigh.Samphier/Sandbox/Equivalence of Definitions of Matroid Base Axiom/Definition 3 Iff Definition 7}}{{qed}} Category:Matroid Theory Category:Equivalence of Definitions of Matroid Base Axiom b2e3wn7jv8hbi7bkec9m0257n1rjulc
From Singleton of Element is Subset: :$\set x \subseteq B_1 \setminus B_2$ and :$\set y \subseteq B_2 \setminus B_1$ From Set Difference is Disjoint with Reverse: :$\paren{B_1 \setminus B_2} \cap \paren{B_2 \setminus B_1} = \O$ From Subsets of Disjoint Sets are Disjoint: :$\set x \cap \set y = \O$ We have: {{begin-eqn}} {{eqn | l = \paren{B_1 \cup \set y} \setminus \set x       | r = \paren {B_1 \setminus \set x} \cup \paren {\set y \setminus \set x}       | c = Set Difference is Right Distributive over Union }} {{eqn | r = \paren {B_1 \setminus \set x} \cup \set y       | c = Set Difference with Disjoint Set }} {{end-eqn}} {{qed}} Category:Equivalence of Definitions of Matroid Base Axiom qzqxo0oq4xm2gd56eptqyn4fx03vi6k
By definition of the parameters of $M$: {{:Definition:Parameter of Moving Average Model}} Thus: :there are $q$ parameters of the form $\theta_j$ :$1$ parameter $\mu$ :$1$ parameter $\sigma_a^2$. That is: $q + 1 + 1 = q + 2$ parameters. {{qed}}
By definition of the parameters of $M$: {{:Definition:Parameter of ARMA Model}} Thus: :there are $p$ parameters of the form $\phi_i$ :there are $q$ parameters of the form $\theta_j$ :$1$ parameter $\mu$ :$1$ parameter $\sigma_a^2$. That is: $p + q + 1 + 1 = p + q + 2$ parameters. {{qed}}
Let the zero of $F$ be $0_F$ and the unity of $F$ be $1_F$. By definition of characteristic, $\Char F = 0$ {{iff}}: :$\not \exists n \in \Z, n > 0: \forall r \in F: n \cdot r = 0_F$ That is, there exists no $n \in \Z, n > 0$ such that $n \cdot r = 0_F$ for all $r \in F$. But note that $\forall r \in F: 0 \cdot r = 0_F$ by definition of integral multiple. {{AimForCont}} there exists a non-zero element of $\map {\mathrm {Ann} } F$. From Non-Trivial Annihilator Contains Positive Integer, $\map {\mathrm {Ann} } F$ must contain a (strictly) positive integer. But this would contradict the statement that $\Char F = 0$. So it follows that: :$\map {\mathrm {Ann} } F = \set 0 \iff \Char F = 0$ {{qed|lemma}}
Let $A := \map {\mathrm {Ann} } F$. We are told that: :$\exists n \in A: n \ne 0$ Consider the set $A^+ \set {n \in A: n > 0}$. From Non-Trivial Annihilator Contains Positive Integer we have that $A^+ \ne \O$. As $A^+ \subseteq \N$ it follows from the well-ordering principle that $A^+$ has a least value $p$, say. {{AimForCont}} $p$ is not a prime number. Then $p$ can be expressed as $p = a b$ where $1 < a, b < p$. {{begin-eqn}} {{eqn | l = 0_R       | r = p \cdot 1_F       | c = {{Defof|Annihilator of Ring}} }} {{eqn | r = \paren {a b} \cdot 1_F       | c =  }} {{eqn | r = \paren {a \cdot 1_F} \times \paren {b \cdot 1_F}       | c = Product of Integral Multiples }} {{eqn | r = \paren {a \cdot 1_F} = 0_F \lor \paren {b \cdot 1_F} = 0_F       | c = Field has no Proper Zero Divisors }} {{end-eqn}} But then either $a \in A$ or $b \in A$, and so $p$ is not the minimal positive element of $A$ after all. So from this contradiction it follows that $p$ is necessarily prime. Next let $n \in \Z$. Then: {{begin-eqn}} {{eqn | l = \paren {n p} \cdot 1_F       | r = n \cdot \paren {p \cdot 1_F}       | c = Integral Multiple of Integral Multiple }} {{eqn | r = n \cdot 0_F       | c = as $p$ is in the annihilator of $F$ }} {{eqn | r = 0_F       | c = {{Defof|Integral Multiple}} }} {{end-eqn}} So all multiples of $p$ are in $A$. Finally, suppose $k \in A$. By the Division Theorem $k = q p + r$ where $0 \le r < p$. Then: {{begin-eqn}} {{eqn | l = 0_F       | r = k \cdot 0_F       | c = {{Defof|Integral Multiple}} }} {{eqn | r = \paren {q p + r} \cdot 1_F       | c =  }} {{eqn | r = \paren {q p} \cdot 1_F + r \cdot 1_F       | c = Integral Multiple Distributes over Ring Addition }} {{eqn | r = 0_F + r \cdot 1_F       | c = from above: $q p$ is a multiple of $p$ }} {{eqn | r = r \cdot 1_F       | c = {{Defof|Field Zero}} }} {{end-eqn}} So $r \in A$ contradicting the stipulation that $p$ is the smallest positive element of $A$. Hence all and only multiples of $p$ are in $A$. {{qed}}
Let the values of $S$ at timestamps $t, t - 1, t - 2, \dotsc$ be $z_t, z_{t - 1}, z_{t - 2}, \dotsc$ Let $\tilde z_t, \tilde z_{t - 1}, \tilde z_{t - 2}, \dotsc$ be deviations from a constant mean level $\mu$: :$\tilde z_t = z_t - \mu$ Let $a_t, a_{t - 1}, a_{t - 2}, \dotsc$ be a sequence of independent shocks at timestamps $t, t - 1, t - 2, \dotsc$ By definition of ARMA model, $M$ is implemented as: :$(1): \quad \tilde z_t = \phi_1 \tilde z_{t - 1} + \phi_2 \tilde z_{t - 2} + \dotsb + \phi_p \tilde z_{t - p} + a_t - \theta_1 a_{t - 1} - \theta_2 a_{t - 2} - \dotsb - \theta_q a_{t - q}$ The general ARIMA model is implemented as: :$w_t = \phi_1 w_{t - 1} + \phi_2 w_{t - 2} + \dotsb + \phi_p w_{t - p} + a_t - \theta_1 a_{t - 1} - \theta_2 a_{t - 2} - \dotsb - \theta_q a_{t - q}$ where: :$w_t = \nabla^d z_t$ Letting $d = 0$ we have: :$w_t = z_t$ Setting $w_t = z_t - \mu = \tilde z$, we recover $(1)$. {{qed}}
Let the values of $S$ at timestamps $t, t - 1, t - 2, \dotsc$ be $z_t, z_{t - 1}, z_{t - 2}, \dotsc$ Let $\tilde z_t, \tilde z_{t - 1}, \tilde z_{t - 2}, \dotsc$ be deviations from a constant mean level $\mu$: :$\tilde z_t = z_t - \mu$ Let $a_t, a_{t - 1}, a_{t - 2}, \dotsc$ be a sequence of independent shocks at timestamps $t, t - 1, t - 2, \dotsc$ By definition of autoregressive model, $M$ is implemented as: :$(1): \quad \tilde z_t = \phi_1 \tilde z_{t - 1} + \phi_2 \tilde z_{t - 2} + \dotsb + \phi_p \tilde z_{t - p} + a_t$ The general ARIMA model is implemented as: :$w_t = \phi_1 w_{t - 1} + \phi_2 w_{t - 2} + \dotsb + \phi_p w_{t - p} + a_t - \theta_1 a_{t - 1} - \theta_2 a_{t - 2} - \dotsb - \theta_q a_{t - q}$ where: :$w_t = \nabla^d z_t$ Let $\theta_j = 0$ for all $j$. Letting $d = 0$ we have: :$w_t = z_t$ Setting $w_t = z_t - \mu = \tilde z$, we recover $(1)$. {{qed}}
Let the values of $S$ at timestamps $t, t - 1, t - 2, \dotsc$ be $z_t, z_{t - 1}, z_{t - 2}, \dotsc$ Let $\tilde z_t, \tilde z_{t - 1}, \tilde z_{t - 2}, \dotsc$ be deviations from a constant mean level $\mu$: :$\tilde z_t = z_t - \mu$ Let $a_t, a_{t - 1}, a_{t - 2}, \dotsc$ be a sequence of independent shocks at timestamps $t, t - 1, t - 2, \dotsc$ By definition of moving average model, $M$ is implemented as: :$(1): \quad \tilde z_t = a_t - \theta_1 a_{t - 1} - \theta_2 a_{t - 2} - \dotsb - \theta_q a_{t - q}$ The general ARIMA model is implemented as: :$w_t = \phi_1 w_{t - 1} + \phi_2 w_{t - 2} + \dotsb + \phi_p w_{t - p} + a_t - \theta_1 a_{t - 1} - \theta_2 a_{t - 2} - \dotsb - \theta_q a_{t - q}$ where: :$w_t = \nabla^d z_t$ Let $\phi_i = 0$ for all $i$. Letting $d = 0$ we have: :$w_t = z_t$ Setting $w_t = z_t - \mu = \tilde z$, we recover $(1)$. {{qed}}
By definition, a field is a ring whose ring product less zero is an abelian group. The result follows from Ring Zero is Unique. {{qed}}
Let $0_1$ and $0_2$ both be elements of $F$ such that: :$\forall a \in F: a + 0_1 = a$ :$\forall a \in F: a + 0_2 = a$ Then: :$0_1 + 0_2 = 0_2$ because $0_1$ is a zero element :$0_1 + 0_2 = 0_1$ because $0_2$ is a zero element Hence: :$0_1 = 0_2$ and the two zero elements are the same. {{qed}}
By definition, a field is a ring whose ring product less zero is an abelian group. The result follows from Ring Negative is Unique. {{qed}}
Let $b, c \in F$ such that both $a + b = 0$ and $a + c = 0$. Thus: {{begin-eqn}} {{eqn | l = b + \paren {a + c}       | r = b + 0       | c = as $c$ is a negative of $a$ }} {{eqn | r = b       | c = {{Defof|Field Zero}} }} {{end-eqn}} But also: {{begin-eqn}} {{eqn | l = \paren {b + a} + c       | r = \paren {a + b} + c       | c = {{Field-axiom|A1}} }} {{eqn | r = 0 + c       | c = as $b$ is a negative of $a$ }} {{eqn | r = c       | c = {{Defof|Field Zero}} }} {{end-eqn}} So $b = c$ and the result follows. {{qed}}
{{AimForCont}}: :$\exists B_1, B_2 \in \mathscr B : \card{B_1} \ne \card{B_2}$ {{WLOG}} and from Max Equals an Operand let: :$\card{B_1 \cap B_2} = \max \set{\card{C_1 \cap C_2} : C_1, C_2 \in \mathscr B : \card{C_1} \ne \card{C_2}}$ {{WLOG}} let: :$\card{B_1} > \card{B_2}$ From Set Difference of Larger Set with Smaller is Not Empty: :$\exists x \in B_1 \setminus B_2$ From base axiom $(\text B 1)$: :$\exists y \in B_2 \setminus B_1 : \paren {B_1 \setminus \set x} \cup \set y \in \mathscr B$ Let $B_3 = \paren {B_1 \setminus \set x} \cup \set y$. We have: {{begin-eqn}} {{eqn | l = \card{B_3 \cap B_2}       | r = \card{\paren{\paren {B_1 \setminus \set x} \cup \set y} \cap B_2} }} {{eqn | r = \card{\paren{\paren {B_1 \setminus \set x} \cap B_2} \cup \paren {\set y \cap B_2} }       | c = Intersection Distributes over Union }} {{eqn | r = \card{\paren{\paren {B_1 \setminus \set x} \cap B_2} \cup \set y }       | c = Intersection with Subset is Subset }} {{eqn | r = \card{\paren{\paren {B_1 \cap B_2 } \setminus \set x } \cup \set y }       | c = Intersection with Set Difference is Set Difference with Intersection }} {{eqn | r = \card{\paren {B_1 \cap B_2 } \cup \set y }       | c = Set Difference with Disjoint Set }} {{eqn | o = >       | r = \card{B_1 \cap B_2 }       | c = Cardinality of Proper Subset of Finite Set }} {{end-eqn}} This contradicts the choice of $B_1$ and $B_2$ as: :$\card{B_1 \cap B_2} = \max \set{\card{C_1 \cap C_2} : C_1, C_2 \in \mathscr B : \card{C_1} \ne \card{C_2}}$ It follows that: :$\forall B_1, B_2 \in \mathscr B : \card{B_1} = \card{B_2}$ {{qed}} Category:Matroid Theory t0zxd9t5pt6bppfs5fvirdnv4yao2uk
{{AimForCont}}: :$\exists B_1, B_2 \in \mathscr B : \card{B_1} \ne \card{B_2}$ {{WLOG}} and from Max Equals an Operand let: :$\card{B_1 \cap B_2} = \max \set{\card{C_1 \cap C_2} : C_1, C_2 \in \mathscr B : \card{C_1} \ne \card{C_2}}$ {{WLOG}} let: :$\card{B_1} > \card{B_2}$ From Set Difference of Larger Set with Smaller is Not Empty: :$\exists x \in B_1 \setminus B_2$ From base axiom $(\text B 5)$: :$\exists y \in B_2 \setminus B_1 : \paren {B_2 \setminus \set y} \cup \set x \in \mathscr B$ Let $B_3 = \paren {B_2 \setminus \set y} \cup \set x$. We have: {{begin-eqn}} {{eqn | l = \card{B_3 \cap B_1}       | r = \card{\paren{\paren {B_2 \setminus \set y} \cup \set x} \cap B_1} }} {{eqn | r = \card{\paren{\paren {B_2 \setminus \set y} \cap B_1} \cup \paren {\set x \cap B_1} }       | c = Intersection Distributes over Union }} {{eqn | r = \card{\paren{\paren {B_2 \setminus \set y} \cap B_1} \cup \set x }       | c = Intersection with Subset is Subset }} {{eqn | r = \card{\paren{\paren {B_2 \cap B_1 } \setminus \set y } \cup \set x }       | c = Intersection with Set Difference is Set Difference with Intersection }} {{eqn | r = \card{\paren {B_2 \cap B_1 } \cup \set x }       | c = Set Difference with Disjoint Set }} {{eqn | o = >       | r = \card{B_1 \cap B_2 }       | c = Cardinality of Proper Subset of Finite Set }} {{end-eqn}} This contradicts the choice of $B_1$ and $B_2$ as: :$\card{B_1 \cap B_2} = \max \set{\card{C_1 \cap C_2} : C_1, C_2 \in \mathscr B : \card{C_1} \ne \card{C_2}}$ It follows that: :$\forall B_1, B_2 \in \mathscr B : \card{B_1} = \card{B_2}$ {{qed}} Category:Equivalence of Definitions of Matroid Base Axiom pyv08opjnvc6w5sxlt3xrcjg4lii8vj
{{begin-eqn}} {{eqn | l = a       | r = a \times 1_F       | c = {{Field-axiom|M3}}: $1_F$ is the unity of $F$ }} {{eqn | r = a \times \paren {0_F + 1_F}       | c = {{Field-axiom|A3}} }} {{eqn | r = a \times 0_F + a \times 1_F       | c = {{Field-axiom|D}} }} {{eqn | r = a \times 0_F + a       | c = {{Field-axiom|M3}} }} {{eqn | ll= \leadsto       | l = \paren {a \times 0_F + a} + \paren {-a}       | r = a + \paren {-a}       | c = adding $-a$ to both sides }} {{eqn | ll= \leadsto       | l = a \times 0_F + \paren {a + \paren {-a} }       | r = a + \paren {-a}       | c = {{Field-axiom|A1}} }} {{eqn | ll= \leadsto       | l = a \times 0_F + 0_F       | r = 0_F       | c = {{Field-axiom|A4}} }} {{eqn | ll= \leadsto       | l = a \times 0_F       | r = 0_F       | c = {{Field-axiom|A3}} }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = a \times b + a \times \paren {-b}       | r = a \times \paren {b + \paren {-b} }       | c = {{Field-axiom|A1}} }} {{eqn | r = a \times 0_F       | c = {{Field-axiom|A4}} }} {{eqn | r = 0_F       | c = Field Product with Zero }} {{eqn | ll= \leadsto       | l = -\paren {a \times b}       | r = a \times \paren {-b}       | c = {{Defof|Ring Negative}} }} {{end-eqn}} Similarly: {{begin-eqn}} {{eqn | l = \paren {-a} \times b + a \times b       | r = \paren {\paren {-a} + a} \times b       | c = {{Field-axiom|A1}} }} {{eqn | r = 0_F \times a       | c = {{Field-axiom|A4}} }} {{eqn | r = 0_F       | c = Field Product with Zero }} {{eqn | ll= \leadsto       | l = -\paren {a \times b}       | r = \paren {-a} \times b       | c = {{Defof|Ring Negative}} }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = a + \paren {-1_F} \times a       | r = 1_F \times a + \paren {-1_F} \times a       | c = {{Field-axiom|M3}} }} {{eqn | r = \paren {1_F + \paren {-1_F} } \times a       | c = {{Field-axiom|D}} }} {{eqn | r = 0_F \times a       | c = {{Field-axiom|A4}} }} {{eqn | r = 0_F       | c = Field Product with Zero }} {{eqn | ll= \leadsto       | l = \paren {-a}       | r = \paren {-1_F} \times a       | c = {{Defof|Ring Negative}} }} {{end-eqn}} {{qed}}
Let $a \times b = a \times c$. Then: {{begin-eqn}} {{eqn | l = a       | o = \ne       | r = 0_F       | c =  }} {{eqn | ll= \leadsto       | lo = \exists a^{-1} \in F:       | l = a^{-1} \times a       | r = 1_F       | c =  }} {{eqn | ll= \leadsto       | l = a^{-1} \times \paren {a \times b}       | r = a^{-1} \times \paren {a \times c}       | c =  }} {{eqn | ll= \leadsto       | l = \paren {a^{-1} \times a} \times b       | r = \paren {a^{-1} \times a} \times c       | c = {{Field-axiom|M1}} }} {{eqn | ll= \leadsto       | l = 1_F \times b       | r = 1_F \times c       | c = {{Field-axiom|M4}} }} {{eqn | ll= \leadsto       | l = b       | r = c       | c = {{Field-axiom|M3}} }} {{end-eqn}} That is: :$a \ne 0 \implies b = c$ {{qed|lemma}} Suppose $b \ne c$. Then by Rule of Transposition: :$\neg \paren {a \ne 0}$ that is: :$a = 0_F$ and we note that in this case: :$a \times b = 0_F = a \times c$ {{qed}}
From Cardinality of Subset of Finite Set: :$S \not \subseteq T$ From the contrapositive statement of Set Difference with Superset is Empty Set: :$S \setminus T \ne \O$. {{qed}} Category:Set Difference Category:Cardinality ot8xec777hfs32dsdmty8u4wqsoqvgy
From Autocorrelation of Strictly Stationary Stochastic Process: :$\rho_k = \dfrac {\gamma_k} {\gamma_0}$ Then from Autocovariance at Zero Lag for Strictly Stationary Stochastic Process is Variance: :$\gamma_0 = \sigma_z^2$ Hence: :$\gamma_k = \sigma_z^2 \rho_k$ and the result follows. {{qed}}
We have that the Autocorrelation Matrix is Positive Definite. The result follows from Determinant of Positive Definite Matrix is Strictly Positive. {{qed}}
From the contrapositive statement of Cardinality of Proper Subset of Finite Set: :$S \not \subset T$ Because $S \ne T$: :$S \not \subseteq T$ From the contrapositive statement of Set Difference with Superset is Empty Set: :$S \setminus T \ne \O$. {{qed}} Category:Set Difference Category:Cardinality 4weoeikehps5mmuvv65hf05sttmp3up