hcju/proofwikiqa huggingface dataset:
=== Necessary Condition === Let $\paren{B \setminus \set x} \cup \set y$ be a base of $M$. By definition of the fundamental circuit: :$\map C {y,B} \subseteq B \cup \set y$ and :$\map C {y,B}$ is dependent {{AimForCont}} $x \notin \map C {y,B}$. Then: :$\map C {y,B} \subseteq \paren{B \setminus \set x} \cup \set y$ By matroid axiom $( \text I 2)$: :$\map C {y,B}$ is independent This contradicts the fact that: :$\map C {y,B}$ is dependent It follows that: :$x \in \map C {y,B}$  {{qed|lemma}} === Sufficient Condition === We prove the contrapositive statement. Let $\paren{B \setminus \set x} \cup \set y$ not be a base of $M$. We have: {{begin-eqn}} {{eqn | l = \card{\paren{B \setminus \set x} \cup \set y}       | r = \card{B \setminus \set x} + \card{\set y}       | c = Corollary of Cardinality of Set Union }} {{eqn | r = \paren{\card B - \card{\set x} } + \card{\set y}       | c = Cardinality of Set Difference }} {{eqn | r = \paren{\card B - 1 } + 1       | c = Cardinality of Singleton }} {{eqn | r = \card B }} {{end-eqn}} From contrapositive statement of Independent Subset is Base if Cardinality Equals Cardinality of Base: :$\paren{B \setminus \set x} \cup \set y$ is a dependent subset of $M$ From Dependent Subset Contains a Circuit there exists a circuit $C'$: :$C' \subseteq \paren{B \setminus \set x} \cup \set y \subseteq B \cup \set y$ From Matroid Base Union External Element has Fundamental Circuit: :$C' = \map C {y,B}$ Hence: :$x \notin \map C {y, B}$ From Rule of Transposition we have: :$x \in \map C {y, B} \implies \paren{B \setminus \set x} \cup \set y$ is a base of $M$ {{qed}} Category:Matroid Theory nhqdzwrkojorfkazedzjs5tvqlktjlg
Let $P$ denote the simple statement ''{{AuthorRef|Socrates}} is a man.''. Let $Q$ denote the simple statement ''{{AuthorRef|Socrates}} is mortal.''. The argument can then be expressed as: {{begin-eqn}} {{eqn | n = 1       | l = P       | o = \implies       | r = Q       | c =  }} {{eqn | n = 2       | l = P       | o =        | c =  }} {{eqn | n = 3       | ll= \therefore       | l = Q       | o =        | c = Modus Ponendo Ponens }} {{end-eqn}} That is:  :''{{AuthorRef|Socrates}} is mortal.'' {{qed}}
We apply the Method of Truth Tables. As can be seen by inspection, the truth value under the main connective is true for all boolean interpretations. $\begin{array}{ccccc|c|ccccccc} (p & \implies & (q & \implies & r)) & \implies & ((p & \implies & q) & \implies & (p & \implies & r)) \\ \hline F & T & F & T & F & T & F & T & F & T & F & T & F \\ F & T & F & T & T & T & F & T & F & T & F & T & T \\ F & T & T & F & F & T & F & T & T & T & F & T & F \\ F & T & T & T & T & T & F & T & T & T & F & T & T \\ T & T & F & T & F & T & T & F & F & T & T & F & F \\ T & T & F & T & T & T & T & F & F & T & T & T & T \\ T & F & T & F & F & T & T & T & T & F & T & F & F \\ T & T & T & T & T & T & T & T & T & T & T & T & T \\ \end{array}$
It is to be demonstrated that $\struct {S, +}$ does not satisfy the group axioms. Let $a$ and $b$ be odd integers. Then  $a = 2 n + 1$ and $b = 2 m + 1$ for some $m, n \in \Z$. Then: {{begin-eqn}} {{eqn | l = a + b       | r = 2 n + 1 + 2 m + 1       | c =  }} {{eqn | r = 2 \paren {n + m + 1}       | c =  }} {{end-eqn}} and it is seen that $a + b$ is even. That is: :$a + b \notin S$ Thus $\struct {S, +}$ does not fulfil {{GroupAxiom|0}}. Hence the result. {{qed}}
Let $n$ have $d$ digits when expressed in base $b$. Then $n$ is expressed as: :$n = \sqbrk {n_{d - 1} n_{d - 2} \dotsm d_1 d_0}$ where: :$n = \displaystyle \sum_{k \mathop = 0}^{d - 1} n_k b^k$ Thus: :$b^{d - 1} \le n < b^d$ Thus we have: {{begin-eqn}} {{eqn | l = b^{d - 1}       | o = \le       | m = n       | mo= <       | r = b^d       | c =  }} {{eqn | ll= \leadsto       | l = d - 1       | o = \le       | m = \log_b n       | mo= <       | r = d       | c =  }} {{end-eqn}} By Integer equals Floor iff Number between Integer and One More: :$d - 1 \le \log_b n < d \iff \floor {\log_b n} = d - 1$ and the result follows. {{Qed}} Category:Number Theory l1hru3y24mg0nhjqmv5bmly8y48rz1o
{{begin-eqn}} {{eqn | l = 2^{10 x}       | r = \paren {2^{10} }^x       | c = Power of Power }} {{eqn | r = 1024^x       | c = as $2^{10} = 1024$ }} {{eqn | r = \paren {1000 + 24}^x       | c = rewriting $1024$ as the sum of a power of $10$ and some integer }} {{eqn | r = \sum_{k \mathop = 0}^n 1000^{x - k} \, 24^k       | c = Binomial Theorem }} {{eqn | n = 1       | r = 1000^x + \sum_{k \mathop = 1}^n 1000^{x - k} \, 24^k       | c = extracting first term from summation }} {{eqn | r = 1000^x + 24 \paren {\sum_{k \mathop = 1}^n 1000^{x - k} \, 24^{k - 1} }       | c = extracting $24$ as a divisor }} {{eqn | ll= \leadsto       | l = 2^{10 x} - 1000^x       | r = 24 i       | c = setting $i = \displaystyle \sum_{k \mathop = 1}^n 1000^{x - k} \, 24^{k - 1}$ }} {{eqn | ll= \leadsto       | l = 2^{10 x} - 10^{3 x}       | r = 24 i       | c = rewriting $1000^x$ as a power of $10$ }} {{eqn | r = 3 k       | c = setting $k = 8 i$ }} {{eqn | ll= \leadsto       | l = 2^{10x} - 10^{3 x}       | o = \equiv       | r = 0 \pmod 3       | c = {{Defof|Congruence Modulo Integer}} }} {{end-eqn}} {{qed}}
Let $\mathbf x = \sequence {x_n}_{n \mathop \in \N} \in \ell^2$. By definition of $\ell^2$: :$\displaystyle \sum_{i \mathop = 0}^\infty \size {x_i}^2 < \infty$ Let $\displaystyle s_n := \sum_{i \mathop = 0}^n \size {x_i}^2$ be a sequence of partial sums of $\displaystyle s = \sum_{i \mathop = 0}^\infty \size {x_i}^2$. We have that $s$ is a convergent sequence: :$\forall \epsilon \in \R_{>0}: \exists N \in \N: \forall n \in \N: n > N \implies \size {s_n - s} < \epsilon$ Note that: {{begin-eqn}} {{eqn | l = \size {s_n - s}       | r = \size {\sum_{i \mathop = 0}^n \size {x_i}^2 - \sum_{i \mathop = 0}^\infty \size {x_i}^2} }} {{eqn | r = \size {\sum_{i \mathop = n \mathop + 1}^\infty \size {x_i}^2} }} {{end-eqn}} Let $\displaystyle N \in \N : \sum_{n \mathop = N + 1}^\infty \size {x_n}^2 < \epsilon^2$ Let $\mathbf y := \tuple {x_1 \ldots, x_N, 0, \ldots}$. By definition, $\mathbf y \in c_{00}$. We have that: {{begin-eqn}} {{eqn | l = \norm {\mathbf x - \mathbf y}_2^2       | r = \sum_{i \mathop = 0}^\infty \size {x_i - y_i}^2       | c = {{defof|Euclidean Norm}} }} {{eqn | r = \sum_{i \mathop = N + 1}^\infty \size {x_i}^2 }} {{eqn | o = <       | r = \epsilon^2 }} {{eqn | ll = \leadsto       | l = \norm {\mathbf x - \mathbf y}_2       | o = <       | r = \epsilon }} {{end-eqn}} By definition, $c_{00}$ is dense in $\ell^2$. {{qed}}
From the formal definition of integers, $\eqclass {a, b} {}$ is an equivalence class of ordered pairs of natural numbers. Let $x = \eqclass {a, b} {}$ and $y = \eqclass {c, d} {}$ for some $x, y \in \Z$. We have: {{begin-eqn}} {{eqn | l = x       | o = >       | r = y       | c =  }} {{eqn | ll= \leadsto       | l = \eqclass {a, b} {}       | o = >       | r = \eqclass {c, d} {}       | c = {{Defof|Integer|subdef = Formal Definition}} }} {{eqn | ll= \leadsto       | l = a + d       | o = >       | r = b + c       | c = {{Defof|Strict Ordering on Integers}} }} {{eqn | ll= \leadsto       | l = b + c       | o = <       | r = a + d       | c =  }} {{eqn | ll= \leadsto       | l = \eqclass {b, a} {}       | o = <       | r = \eqclass {d, c} {}       | c =  }} {{eqn | ll= \leadsto       | l = -\eqclass {a, b} {}       | o = <       | r = -\eqclass {c, d} {}       | c = Negative of Integer }} {{eqn | ll= \leadsto       | l = -x       | o = <       | r = -y       | c = {{Defof|Integer|subdef = Formal Definition}} }} {{end-eqn}} {{qed}}
Let $x, y \in \Z$ such that $x \le y$. Let $x$ and $y$ be defined as from the formal definition of integers: :$x = \eqclass {x_1, x_2} {}$ and $y = \eqclass {y_1, y_2} {}$ where $x_1, x_2, y_1, y_2 \in \N$. === $(1)$ implies $(2)$ === Let $\le$ be an ordering on integers by definition $1$. Then by definition: :$y - x$ is non-negative  That is: :$\map \PP {y - x}$ where $\PP$ is the positivity property. Thus: :$\eqclass {y_1, y_2} {} - \eqclass {x_1, x_2} {}$ is non-negative By definition of integer subtraction: :$\eqclass {y_1, y_2} {} + \eqclass {x_2, x_1} {}$ is non-negative and by the formal definition of integers: :$\eqclass {y_1 + x_2, y_2 + x_1} {}$ is non-negative We have that $y_1 + x_2$ and $y_2 + x_1$ are natural numbers. Thus by definition of natural number ordering: :$y_1 + x_2 \ge y_2 + x_1$ Thus $\le$ is an ordering on integers by definition $2$. {{qed|lemma}} === $(2)$ implies $(1)$ === Let $\le$ be an ordering on integers by definition $2$. Then by definition: :$x_1 + y_2 \le x_2 + y_1$ That is: :$x_2 + y_1 \ge x_1 + y_2$ Hence by the formal definition of integers: :$\eqclass {y_1 + x_2, y_2 + x_1} {}$ is non-negative By definition of integer addition: :$\eqclass {y_1, y_2} {} + \eqclass {x_2, x_1} {}$ is non-negative By definition of integer subtraction: :$\eqclass {y_1, y_2} {} - \eqclass {x_1, x_2} {}$ is non-negative That is: :$\map \PP {y - x}$ where $\PP$ is the positivity property. Thus $\le$ is an ordering on integers by definition $1$. {{qed}} Category:Orderings on Integers 02kdwdfrxraur2rcdib88vt3lwow89j
Let $x, y \in \Z$ such that $x < y$. Let $x$ and $y$ be defined as from the formal definition of integers: :$x = \eqclass {x_1, x_2} {}$ and $y = \eqclass {y_1, y_2} {}$ where $x_1, x_2, y_1, y_2 \in \N$. === $(1)$ implies $(2)$ === Let $<$ be a strict ordering on integers by definition $1$. Then by definition: :$y - x$ is non-negative  That is: :$\map \PP {y - x}$ where $\PP$ is the strict positivity property. Thus: :$\eqclass {y_1, y_2} {} - \eqclass {x_1, x_2} {}$ is strictly positive By definition of integer subtraction: :$\eqclass {y_1, y_2} {} + \eqclass {x_2, x_1} {}$ is strictly positive and by the formal definition of integers: :$\eqclass {y_1 + x_2, y_2 + x_1} {}$ is strictly positive We have that $y_1 + x_2$ and $y_2 + x_1$ are natural numbers. Thus by definition of natural number ordering: :$y_1 + x_2 > y_2 + x_1$ Thus $<$ is a strict ordering on integers by definition $2$. {{qed|lemma}} === $(2)$ implies $(1)$ === Let $<$ be a strict ordering on integers by definition $2$. Then by definition: :$x_1 + y_2 < x_2 + y_1$ That is: :$x_2 + y_1 > x_1 + y_2$ Hence by the formal definition of integers: :$\eqclass {y_1 + x_2, y_2 + x_1} {}$ is strictly positive By definition of integer addition: :$\eqclass {y_1, y_2} {} + \eqclass {x_2, x_1} {}$ is strictly positive By definition of integer subtraction: :$\eqclass {y_1, y_2} {} - \eqclass {x_1, x_2} {}$ is strictly positive That is: :$\map \PP {y - x}$ where $\PP$ is the strict positivity property. Thus $\le$ is a strict ordering on integers by definition $1$. {{qed}} Category:Orderings on Integers bmloxwlrq81j9xcmpt5gs7efxak3a5m
From Characterization of Multivariate Gaussian Distribution, the Gaussian distribution is completely characterized by its expectation and its variance. The result follows. {{qed}}
By definition of a Gaussian process, the probability distribution of $T$ be a multivariate Gaussian distribution. By definition, a Gaussian distribution is characterized completely by its expectation and its variance. That is, its $1$st and $2$nd moments. The result follows. {{qed}}
By the formal definition of integers, we have that $a, b, c, d, e, f$ are all natural numbers. To eliminate confusion between integer ordering and the ordering on the natural numbers, let $a \preccurlyeq b$ denote that the natural number $a$ is less than or equal to the natural number $b$. We have: {{begin-eqn}} {{eqn | l = \eqclass {a, b} {}       | o = \le       | r = \eqclass {c, d} {}       | c =  }} {{eqn | n = 1       | ll= \leadsto       | l = a + d       | o = \preccurlyeq       | r = b + c       | c = {{Defof|Ordering on Integers}} }} {{eqn | l = \eqclass {c, d} {}       | o = <       | r = \eqclass {e, f} {}       | c =  }} {{eqn | n = 2       | ll= \leadsto       | l = c + f       | o = \preccurlyeq       | r = d + e       | c = {{Defof|Ordering on Integers}} }} {{eqn | ll= \leadsto       | l = a + d + f       | o = \preccurlyeq       | r = b + c + f       | c = adding $f$ to both sides of $(1)$ }} {{eqn | ll= \leadsto       | l = a + d + f       | o = \preccurlyeq       | r = b + d + e       | c = from $(2)$: $b + \paren {c + f} \preccurlyeq b + \paren {d + e}$ }} {{eqn | ll= \leadsto       | l = a + f       | o = \preccurlyeq       | r = b + e       | c = subtracting $d$ from both sides }} {{eqn | ll= \leadsto       | l = \eqclass {a, b} {}       | o = \le       | r = \eqclass {e, f} {}       | c = {{Defof|Ordering on Integers}} }} {{end-eqn}} {{qed}} Category:Orderings on Integers 6ks88hv97gg4h02vn1o4gocu699l1ci
By the formal definition of integers, we have that $a, b, c, d, e, f$ are all natural numbers. To eliminate confusion between integer ordering and the ordering on the natural numbers, let $a \prec b$ denote that the natural number $a$ is less than the natural number $b$. We have: {{begin-eqn}} {{eqn | l = \eqclass {a, b} {}       | o = <       | r = \eqclass {c, d} {}       | c =  }} {{eqn | ll= \leadsto       | l = a + d       | o = \prec       | r = b + c       | c = {{Defof|Strict Ordering on Integers}} }} {{eqn | ll= \leadsto       | l = b + c       | o = \not \prec       | r = a + d       | c = {{Defof|Ordering on Natural Numbers}} }} {{eqn | ll= \leadsto       | l = \eqclass {c, d} {}       | o = \not <       | r = \eqclass {a, b} {}       | c = {{Defof|Strict Ordering on Integers}} }} {{end-eqn}} {{qed}}
By the formal definition of integers, we have that $a, b, c, d, e, f$ are all natural numbers. To eliminate confusion between integer ordering and the ordering on the natural numbers, let $a \prec b$ denote that the natural number $a$ is less than the natural number $b$. We have: {{begin-eqn}} {{eqn | l = \eqclass {a, b} {}       | o = <       | r = \eqclass {c, d} {}       | c =  }} {{eqn | n = 1       | ll= \leadsto       | l = a + d       | o = \prec       | r = b + c       | c = {{Defof|Strict Ordering on Integers}} }} {{eqn | l = \eqclass {c, d} {}       | o = <       | r = \eqclass {e, f} {}       | c =  }} {{eqn | n = 2       | ll= \leadsto       | l = c + f       | o = \prec       | r = d + e       | c = {{Defof|Strict Ordering on Integers}} }} {{eqn | ll= \leadsto       | l = a + d + f       | o = \prec       | r = b + c + f       | c = adding $f$ to both sides of $(1)$ }} {{eqn | ll= \leadsto       | l = a + d + f       | o = \prec       | r = b + d + e       | c = from $(2)$: $b + \paren {c + f} \prec b + \paren {d + e}$ }} {{eqn | ll= \leadsto       | l = a + f       | o = \prec       | r = b + e       | c = subtracting $d$ from both sides }} {{eqn | ll= \leadsto       | l = \eqclass {a, b} {}       | o = <       | r = \eqclass {e, f} {}       | c = {{Defof|Strict Ordering on Integers}} }} {{end-eqn}} {{qed}}
Follows from Inverses for Integer Addition. {{finish|The whole area of construction of integers from the inverse completion needs to be reviewed}} Category:Integers c8490t2014ko917aj6xcdymx2ob1gqk
From Integers form Ordered Integral Domain, $\Z$ is an ordered integral domain. The result follows from Product of Absolute Values on Ordered Integral Domain. {{qed}}
This is a specific example of Cardinality of Set of All Mappings where $S = T$. {{qed}}
{{begin-eqn}} {{eqn | l = \mathbf F       | r = \map {\dfrac \d {\d t} } {m \mathbf v}       | c = Newton's Second Law of Motion }} {{eqn | r = \map {\dfrac \d {\d t} } {m \dfrac {\d \mathbf r} {\d t} }       | c = {{Defof|Velocity}} }} {{eqn | r = m \map {\dfrac \d {\d t} } {\dfrac {\d \mathbf r} {\d t} }       | c = Derivative of Constant Multiple }} {{eqn | r = m \dfrac {\d^2 \mathbf r} {\d t^2}       | c = {{Defof|Second Derivative}} }} {{end-eqn}} {{qed}}
Let $D$ be the set of all finitely supported sequences with rational terms: :$\displaystyle D = \set {\sequence {q_i}_{i \mathop \in \N} : n \in \N : i \le n : q_i \in \Q}$ We have that: :Rational Numbers are Countably Infinite :A finite set is countable :$D$ is a union of finite sets indexed by $n$, which is countable By Countable Union of Countable Sets is Countable, $D$ is countable. Let $\mathbf x := \tuple {x_n}_{n \mathop \in \N} \in \ell^1$ By definition of $\ell^1$: :$\displaystyle \sum_{n \mathop = 1}^\infty \size {x_n} < \infty$ Let $\displaystyle s_n := \sum_{i \mathop = 0}^n \size {x_i}$ be a sequence of partial sums of $\displaystyle s = \sum_{i \mathop = 0}^\infty \size {x_i}$. Then $s$ is a convergent sequence: :$\forall \epsilon' \in \R_{>0}: \exists N \in \N: \forall n \in \N: n > N \implies \size {s_n - s} < \epsilon'$ Note that: {{begin-eqn}} {{eqn | l = \size {s_n - s}       | r = \size {\sum_{i \mathop = 0}^n \size {x_i} - \sum_{i \mathop = 0}^\infty \size {x_i} } }} {{eqn | r = \size {\sum_{i \mathop = n \mathop + 1}^\infty \size {x_i} } }} {{eqn | r = \sum_{i \mathop = n \mathop + 1}^\infty \size {x_i}       | c = {{defof|Absolute Value}} }} {{end-eqn}} Let $\displaystyle \epsilon' = \frac \epsilon 2$. Let $\displaystyle N \in \N : \sum_{n \mathop = N + 1}^\infty \size {x_n} < \frac \epsilon 2$. We have that $\Q$ is dense in $\R$. :$\forall x \in \R : \exists \epsilon'' \in \R_{\mathop > 0} : \exists q \in \Q : \size {x - q} < \epsilon''$ Then: :$\displaystyle \forall i \in \N : 1 \le i \le N : q_i \in \Q : \sum_{n \mathop = 1}^N \size {x_n - q_n} < \frac \epsilon 2$ where :$\displaystyle \frac \epsilon 2 = \sum_{i \mathop = 1}^N \epsilon_i''$ Let $\mathbf x' := \tuple {q_1, \ldots, q_N, 0, \ldots} \in D$. Then: :$\displaystyle \norm {\mathbf x - \mathbf x '} = \sum_{n \mathop = 1}^N \size {x_n - q_n} + \sum_{n \mathop = N + 1}^\infty \size {x_n} < \epsilon$ {{qed}}
By Integration by Substitution (with appropriate renaming of variables): :$\ds \int \map g u \rd u = \int \map g {\map f x} \map {f'} x \rd x$ Let $\map u x = \sqrt {\map f x}$ {{begin-eqn}} {{eqn | l = \map u x       | r = \sqrt {\map f x}       | c =  }} {{eqn | ll= \leadsto       | l = \dfrac {\d u} {\d x}       | r = \dfrac {\map {f'} x} {2 \sqrt {\map f x} }       | c = Chain Rule for Derivatives, Derivative of Power }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \int \frac {\map {f'} x} {2 \sqrt {\map f x} } \rd x       | r = \int \rd u       | c = Integration by Substitution: putting $\map u x = \sqrt {\map f x}$ }} {{eqn | r = u + C       | c = Primitive of Constant }} {{eqn | r = \sqrt {\map f x} + C       | c = Definition of $u$ }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \int \csc x \rd x       | r = -\ln \size {\csc x + \cot x} + C       | c = Primitive of $\csc x$: Cosecant plus Cotangent Form }} {{eqn | r = \ln \size {\frac 1 {\csc x + \cot x} } + C       | c = Logarithm of Reciprocal }} {{eqn | r = \ln \size {\frac 1 {\frac 1 {\sin x} + \frac {\cos x} {\sin x} } } + C       | c = {{Defof|Cosecant}} and {{Defof|Cotangent}} }} {{eqn | r = \ln \size {\frac {\sin x} {1 + \cos x} } + C       | c =  }} {{eqn | r = \ln \size {\tan \frac x 2} + C       | c = Half Angle Formula for Tangent: Corollary 1 }} {{end-eqn}} {{qed}}
Let $u = \tanh \dfrac x 2$. Then: {{begin-eqn}} {{eqn | l = \int \sech x \rd x       | r = \int \frac {1 - u^2} {1 + u^2} \frac {2 \rd u} {1 - u^2}       | c = Hyperbolic Tangent Half-Angle Substitution }} {{eqn | r = \int \frac {2 \rd u} {1 + u^2}       | c = simplifying }} {{eqn | r = 2 \arctan u + C       | c = Primitive of $\dfrac 1 {x^2 + a^2}$: Arctangent Form }} {{eqn | r = 2 \map \arctan {\tanh \dfrac x 2} + C       | c = substituting for $u$ }} {{end-eqn}} {{qed}}
This is a linear first order ODE with constant coefficents in the form: :$\dfrac {\d y} {\d x} + a y = \map Q x$ where: :$a = 2 x$ :$\map Q x = 1$ From Solution to Linear First Order Ordinary Differential Equation: :$\ds y = e^{-\int P \rd x} \paren {\int Q e^{\int P \rd x} \rd x + C}$ Thus  {{begin-eqn}} {{eqn | l = y       | r = e^{-\int 2 x \rd x} \int e^{\int 2 x \rd x} \rd x + C       | c =  }} {{eqn | r = e^{- {x^2} } \int e^{x^2} \rd x + C       | c = Primitive of Power }} {{end-eqn}} {{finish|Get this into the form as given -- the technique will be provided later in the book, one expects}} Further work on this is not trivial, as $\ds \int e^{x^2} \rd x$ has no solution in elementary functions. {{qed}}
==== Necessary Condition ==== Let $\mathscr B$ satisfy the base axiom: {{:Leigh.Samphier/Sandbox/Definition:Base Axiom (Matroid)/Definition 1}} From Leigh.Samphier/Sandbox/Matroid Satisfies Base Axiom: :there exists a matroid $M = \struct{S, \mathscr I}$ such that $\mathscr B$ is the set of bases of $M$. Let $B_1, B_2 \in \mathscr B$. Let $x \in B_1 \setminus B_2$. From Matroid Base Union External Element has Fundamental Circuit: :there exists a fundamental circuit $\map C {x, B_2}$ of $M$ such that $x \in \map C {x, B_2} \subseteq B_2 \cup \set x$ By definition of set intersection: :$x \in B_1 \cap \map C {x, B_2}$ From Element of Matroid Base and Circuit has a Substitute: :$\exists y \in \map C {x, B_2} \setminus B_1 : \paren{B_1 \setminus \set x} \cup \set y \in \mathscr B$ We have: {{begin-eqn}} {{eqn | l = y       | o = \in       | r = \map C {x, B_2} \setminus B_1 }} {{eqn | o = \subseteq       | r = \map C {x, B_2} \setminus \set x       | c = Set Difference with Subset is Superset of Set Difference }} {{eqn | o = \subseteq       | r = \paren{B_2 \cup \set x} \setminus \set x       | c = Set Difference over Subset }} {{eqn | o = \subseteq       | r = B_2 \setminus \set x       | c = Set Difference with Union is Set Difference }} {{eqn | o = \subseteq       | r = B_2       | c = Set Difference is Subset }} {{end-eqn}} From Leigh.Samphier/Sandbox/Matroid Base Substitution From Fundamental Circuit: :$\paren{B_2 \setminus \set y} \cup \set x \in \mathscr B$ It follows that $\mathscr B$ satisfies the base axiom: {{:Leigh.Samphier/Sandbox/Definition:Base Axiom (Matroid)/Definition 4}}{{qed|lemma}} ==== Sufficient Condition ==== Follows immediately from Definition 4 and Definition 1.
==== Necessary Condition ==== Follows immediately from Definition 4 and Definition 5. {{qed|lemma}} ==== Sufficient Condition ==== Let $\mathscr B$ satisfy the base axiom: {{:Leigh.Samphier/Sandbox/Definition:Base Axiom (Matroid)/Definition 5}}
==== Necessary Condition ==== Let $\mathscr B$ satisfy the base axiom: {{:Leigh.Samphier/Sandbox/Definition:Base Axiom (Matroid)/Definition 3}} Let $B_1, B_2 \in \mathscr B$. From $(\text B 3)$: :$\exists \text{ a bijection } \pi : B_2 \setminus B_1 \to B_1 \setminus B_2 : \forall y \in B_2 \setminus B_1 : \paren {B_2 \setminus \set y } \cup \set {\map \pi y} \in \mathscr B$ Let $\pi^{-1} : B_1 \setminus B_2 \to B_2 \setminus B_1$ denote the inverse mapping of $\pi$. From Inverse of Bijection is Bijection, $\pi^{-1}$ is a bijection. From Inverse Element of Bijection: :$\forall x \in B_1 \setminus B_2 : \map {\pi^{-1}} x = y \iff \map \pi y = x$ Hence: :$\forall x \in B_1 \setminus B_2 : \paren {B_1 \setminus \set{\map {\pi^{-1}} x} } \cup \set x  = \paren {B_1 \setminus \set y} \cup \set {\map \pi y} \in \mathscr B$ It follows that $\mathscr B$ satisfies the base axiom: {{:Leigh.Samphier/Sandbox/Definition:Base Axiom (Matroid)/Definition 7}}{{qed|lemma}} ==== Sufficient Condition ==== Let $\mathscr B$ satisfy the base axiom: {{:Leigh.Samphier/Sandbox/Definition:Base Axiom (Matroid)/Definition 7}} Let $B_1, B_2 \in \mathscr B$. From $(\text B 7)$: :$\exists \text{ a bijection } \pi : B_2 \setminus B_1 \to B_1 \setminus B_2 : \forall y \in B_2 \setminus B_1 : \paren {B_1 \setminus \set {\map \pi y} } \cup \set y \in \mathscr B$ Let $\pi^{-1} : B_1 \setminus B_2 \to B_2 \setminus B_1$ denote the inverse mapping of $\pi$. From Inverse of Bijection is Bijection, $\pi^{-1}$ is a bijection. From Inverse Element of Bijection: :$\forall x \in B_1 \setminus B_2 : \map {\pi^{-1}} x = y \iff \map \pi y = x$ Hence: :$\forall x \in B_1 \setminus B_2 : \paren {B_1 \setminus \set x } \cup \set{\map {\pi^{-1}} x} = \paren {B_1 \setminus \set {\map \pi y} } \cup \set y \in \mathscr B$ It follows that $\mathscr B$ satisfies the base axiom: {{:Leigh.Samphier/Sandbox/Definition:Base Axiom (Matroid)/Definition 3}}