avewright/proofwiki-math huggingface dataset:
Theorem: Primitive of Root of Function under Half its Derivative

Let $f$ be a real function which is integrable.
Then:
:$\ds \int \frac {\map {f'} x} {2 \sqrt {\map f x} } \rd x = \sqrt {\map f x} + C$
where $C$ is an arbitrary constant.

Proof:
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}}
Theorem: Primitive of Hyperbolic Secant Function/Arctangent of Half Hyperbolic Tangent Form

:$\ds \int \sech x \rd x =  2 \map \arctan {\tanh \dfrac x 2} + C$

Proof:
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}}
Theorem: First Order ODE/y' + 2 x y = 1

The first order ODE:
:$y' + 2 x y = 1$
has the general solution:
:$y = e^{-{x^2} } \ds \int_a^x e^{t^2} \rd t$
where $a$ is an arbitrary constant.

Proof:
This is a linear first order ODE in the form:
:$\dfrac {\d y} {\d x} + \map P x y = \map Q x$
where:
:$\map p x = 2 x$
:$\map Q x = 1$
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}}
Theorem: Equivalence of Definitions of Matroid Base Axioms/Formulation 1 Iff Formulation 4

Let $S$ be a finite set.
Let $\mathscr B$ be a non-empty set of subsets of $S$.
Then:
:$\mathscr B$ satisfies formulation $1$ of base axiom
{{Axiom:Base Axiom (Matroid)/Formulation 1}}
{{iff}}
:$\mathscr B$ satisfies formulation $4$ of base axiom
{{Axiom:Base Axiom (Matroid)/Formulation 4}}

Proof:
==== Necessary Condition ====
Follows immediately from:
:* Matroid Bases Iff Satisfies Formulation 1 of Matroid Base Axiom
:* Matroid Bases Satisfy Formulation 4 Base Axiom
{{qed|lemma}}
==== Sufficient Condition ====
Follows immediately from formulation $4$ and formulation $1$.
Theorem: Equivalence of Definitions of Matroid Base Axioms/Formulation 3 Iff Formulation 7

Let $S$ be a finite set.
Let $\mathscr B$ be a non-empty set of subsets of $S$.
Then:
:$\mathscr B$ satisfies formulation $3$ of base axiom:
{{Axiom:Base Axiom (Matroid)/Formulation 3}}
{{iff}}
:$\mathscr B$ satisfies formulation $7$ of base axiom:
{{Axiom:Base Axiom (Matroid)/Formulation 7}}

Proof:
==== Necessary Condition ====
Let $\mathscr B$ satisfy formulation $3$ of base axiom:
{{:Axiom:Base Axiom (Matroid)/Formulation 3}}
Let $B_1, B_2 \in \mathscr B$.
From $(\text B 3)$:
:$\exists \text{ a bijection } \pi : B_2 \to B_1 : \forall y \in B_2: \paren {B_2 \setminus \set y } \cup \set {\map \pi y} \in \mathscr B$
Let $\pi^{-1} : B_1 \to B_2$ denote the inverse mapping of $\pi$.
From Inverse of Bijection is Bijection, $\pi^{-1}$ is a bijection.
From Inverse Element of Bijection:
:$\forall x \in B_1: \map {\pi^{-1}} x = y \iff \map \pi y = x$
Hence:
:$\forall x \in B_1: \paren {B_2 \setminus \set{\map {\pi^{-1}} x} } \cup \set x  = \paren {B_2 \setminus \set y} \cup \set {\map \pi y} \in \mathscr B$
It follows that $\mathscr B$ satisfies formulation $7$ of base axiom:
{{:Axiom:Base Axiom (Matroid)/Formulation 7}}
==== Sufficient Condition ====
Let $\mathscr B$ satisfy formulation $7$ of base axiom:
{{:Axiom:Base Axiom (Matroid)/Formulation 7}}
Let $B_1, B_2 \in \mathscr B$.
From $(\text B 7)$:
:$\exists \text{ a bijection } \pi : B_2 \to B_1 : \forall y \in B_2 : \paren {B_1 \setminus \set {\map \pi y} } \cup \set y \in \mathscr B$
Let $\pi^{-1} : B_1 \to B_2$ denote the inyverse mapping of $\pi$.
From Inverse of Bijection is Bijection, $\pi^{-1}$ is a bijection.
From Inverse Element of Bijection:
:$\forall x \in B_1 : \map {\pi^{-1}} x = y \iff \map \pi y = x$
Hence:
:$\forall x \in B_1 : \paren {B_1 \setminus \set x } \cup \set{\map {\pi^{-1}} x} = \paren {B_1 \setminus \set {\map \pi y} } \cup \set y \in \mathscr B$
It follows that $\mathscr B$ satisfy formulation $3$ of base axiom:
{{:Axiom:Base Axiom (Matroid)/Formulation 3}}
Theorem: Equivalence of Definitions of Matroid Base Axioms/Formulation 1 Iff Formulation 7

Let $S$ be a finite set.
Let $\mathscr B$ be a non-empty set of subsets of $S$.
Then:
:$\mathscr B$ satisfies formulation $1$ of base axiom:
{{:Axiom:Base Axiom (Matroid)/Formulation 1}}
{{iff}}
:$\mathscr B$ satisfies formulation $7$ of base axiom:
{{:Axiom:Base Axiom (Matroid)/Formulation 7}}

Proof:
==== Necessary Condition ====
Follows immediately from:
:* Matroid Bases Iff Satisfies Formulation 1 of Matroid Base Axiom
:* Matroid Bases Satisfy Formulation 7 of Matroid Base Axiom
{{qed|lemma}}
==== Sufficient Condition ====
From Formulation 3 Iff Formulation 7:
:$\mathscr B$ satisfies formulation $7$ of base axiom:
{{:Axiom:Base Axiom (Matroid)/Formulation 7}}
{{iff}}
:$\mathscr B$ satisfies formulation $3$ of base axiom:
{{:Axiom:Base Axiom (Matroid)/Formulation 3}}
By choosing $y = \map \pi x$ in formulation $3$, formulation $1$ follows immediately.
Theorem: Number of Permutations with Repetition

Set $S$ be a set of $n$ elements.
Let $\sequence T_m$ be a sequence of $m$ terms of $S$.
Then there are $n^m$ different instances of $\sequence T_m$.

Proof:
Let $N_m$ denote the set $\set {1, 2, \ldots, m}$.
Let $f: N_m \to S$ be the mapping defined as:
:$\forall k \in N_m: \map f t = t_m$
By definition, $f$ corresponds to one of the specific instances of $\sequence T_m$.
Hence the number of different instances of $\sequence T_m$ is found from Cardinality of Set of All Mappings:
:$\card S^{\card {N_m} }$
The result follows.
{{qed}}
Theorem: Definite Integral to Infinity of Reciprocal of x Squared plus a Squared/Corollary

:$\ds \int_0^\infty \dfrac {\d x} {1 + x^2} = \frac \pi 2$

Proof:
From Definite Integral to Infinity of Reciprocal of x Squared plus a Squared:
:$\ds \int_0^\infty \dfrac {\d x} {x^2 + a^2} = \frac \pi {2 a}$
which holds for for $a \ne 0$. 
The result follows by setting $a = 1$.
{{qed}}
Theorem: P-Norm is Norm/P-Sequence Space

The $p$-norm on the $p$-sequence space is a vector space norm.

Proof:
{{refactor|This should be presented in the context of a general Banach space.|level = advanced}}
Theorem: P-Norm is Norm/Complex Numbers

The $p$-norm on the complex numbers is a norm.

Proof:
Let $K \in \C^d$, where $d \in \N_{>0}$.
Theorem: P-Norm is Norm/Real Numbers

The $p$-norm on the real numbers is a norm.

Proof:
We have that $p$-norm is a norm on complex numbers.
Since real numbers are wholly real complex numbers, the same result holds.
{{qed}}
Category:P-Norms
49fea5yjsxkrx7plgbhhne7b5myaxez
Theorem: P-Sequence Space with P-Norm forms Normed Vector Space

A $p$-Sequence Space with a $p$-norm forms a normed vector space.

Proof:
We have that:
:a $p$-sequence space is a vector space
:the $p$-norm on the $p$-sequence space is a norm
By definition, $\struct {\ell^p, \norm {\, \cdot \,}_p}$ is a normed vector space.
{{qed}}
Theorem: Space of Bounded Sequences with Pointwise Addition and Pointwise Scalar Multiplication on Ring of Sequences forms Vector Space

Let $\map {\ell^\infty} \C$ be the space of bounded sequences on $\C$.
Let $\struct {\C, +_\C, \times_\C}$ be the field of complex numbers.
Let $\paren +$ be the pointwise addition on the ring of sequences.
Let $\paren {\, \cdot \,}$ be the pointwise multiplication on the ring of sequences.
Then $\struct {\map {\ell^\infty} \C, +, \, \cdot \,}_\C$ is a vector space.

Proof:
Let $\sequence {a_n}_{n \mathop \in \N}, \sequence {b_n}_{n \mathop \in \N}, \sequence {c_n}_{n \mathop \in \N} \in \map {\ell^\infty} \C$.
Let $\lambda, \mu \in \C$.
Let $\sequence 0 := \tuple {0, 0, 0, \dots}$ be a complex-valued function.
Let us use real number addition and multiplication.
Define pointwise addition as:
:$\sequence {a_n + b_n}_{n \mathop \in \N} := \sequence {a_n}_{n \mathop \in \N} +_\C \sequence {b_n}_{n \mathop \in \N}$.
Define pointwise scalar multiplication as:
:$\sequence {\lambda \cdot a_n}_{n \mathop \in \N} := \lambda \times_\C \sequence {a_n}_{n \mathop \in \N}$
Let the additive inverse be $\sequence {-a_n} := - \sequence {a_n}$.
Theorem: Zero Vector has no Direction

A zero vector has no direction.

Proof:
Let $\mathbf 0$ denote a zero vector.
{{AimForCont}} $\mathbf 0$ has a direction.
Then $\mathbf 0$ can be represented as an arrow in a real vector space $\R^n$ with a Cartesian frame.
Let $\mathbf 0$ be so embedded.
Thus it consists of a line segment between two points with an initial point $A$ and a terminal point $B$.
The initial point and a terminal point are distinct from each other.
Let these points be identified as:
{{begin-eqn}}
{{eqn | l = A
      | r = \tuple {a_1, a_2, \ldots, a_n}
}}
{{eqn | l = B
      | r = \tuple {b_1, b_2, \ldots, b_n}
}}
{{end-eqn}}
Hence we have that the length of $\mathbf 0$ is defined as:
:$\norm {\mathbf 0} = \ds \sqrt {\sum_{i \mathop = 1}^n \paren {a_i - b_i}^2} > 0$
which means that at least one of $a_i - b_i$ is non-zero.
But this contradicts the definition of $\mathbf 0$ being the zero vector.
It follows by Proof by Contradiction that our assumption that $\mathbf 0$ has a direction must be false.
Hence the result.
{{qed}}
Theorem: Power Set is Closed under Countable Unions

Let $S$ be a set.
Let $\powerset S$ be the power set of $S$.
Then:
:$\forall A_n \in \powerset S: n = 1, 2, \ldots: \ds \bigcup_{n \mathop = 1}^\infty A_n \in \powerset S$

Proof:
Let $\sequence {A_i}$ be a countably infinite sequence of sets in $\powerset S$.
Consider an element of the union of all the sets in this $\sequence {A_i}$:
:$\ds x \in \bigcup_{i \mathop \in \N} A_i$
By definition of union:
:$\exists i \in \N: x \in A_i$
But $A_i \in \powerset S$ and so by definition $A_i \subseteq S$.
By definition of subset, it follows that $x \in S$.
Hence, again by definition of subset:
:$\ds \bigcup_{i \mathop \in \N} A_i \subseteq S$
Finally, by definition of power set:
:$\ds \bigcup_{i \mathop \in \N} A_i \in \powerset S$
{{qed}}
Theorem: Set of Elementary Events belonging to k Events is Event

Let $\EE$ be an experiment with a probability space $\struct {\Omega, \Sigma, \Pr}$.
Let $A_1, A_2, \ldots, A_m$ be events in the event space $\Sigma$ of $\EE$.
Let $S$ denote the set of all elementary events of $\EE$ which are elements of exactly $k$ of the events $A_1, A_2, \ldots, A_m$.
Then $S$ is an event of $\Sigma$.

Proof:
Let $r_1, r_2, \ldots r_k$ be a set of $k$ elements of $\set {1, 2, \ldots, m}$.
Then:
:$\paren {A_{r_1} \cap A_{r_2} \cap \cdots \cap A_{r_k} } \setminus \paren { A_{r_{k + 1} } \cup A_{r_{k + 2} } \cup \cdots \cup A_{r_m} }$
contains exactly those elements of $\Omega$ which are contained in exactly those events $A_{r_1}, A_{r_2}, \ldots, \cap A_{r_k}$.
By Elementary Properties of Event Space, that last set is an event of $\EE$.
By selecting all combinations of $\set {1, 2, \ldots, m}$, we can form the subsets of $\Sigma$ containing all those elementary events of exactly $k$ events.
Each one is an element of $\Sigma$.
We then form the union $S_k$ of each of those element of $\Sigma$.
By Elementary Properties of Event Space, $S_k$ is an event of $\EE$.
By construction, $S_k$ contains all and only the elementary events of $\EE$ which are elements of exactly $k$ of the events $A_1, A_2, \ldots, A_m$.
{{qed}}
Theorem: Event Space of Experiment with Final Sample Space has Even Cardinality

Let $\EE$ be an experiment with a probability space $\struct {\Omega, \Sigma, \Pr}$.
Let $\Omega$ be a finite set.
Then the event space $\Sigma$ consists of an even number of subsets of $\Omega$.

Proof:
Let $A \in \Sigma$ be one of the events of $\EE$.
We have by definition that $\Omega$ is itself an events of $\EE$.
Hence by Set Difference of Events is Event, $\Omega \setminus A$ is also an event of $\EE$.
As $A$ is arbitrary, the same applies to all events of $\EE$.
Thus all events of $\EE$ come in pairs: $A$ and $\Omega \setminus A$.
Hence the result.
{{qed}}
Theorem: Probability of Union of Disjoint Events is Sum of Individual Probabilities

Let $\EE$ be an experiment.
Let $\struct {\Omega, \Sigma, \Pr}$ be a probability measure on $\EE$.
Then:
:$\forall A, B \in \Sigma: A \cap B = \O \implies \map \Pr {A \cup B} = \map \Pr A + \map \Pr B$
where $A \cap B$ denotes the '''union''' of $A$ and $B$.

Proof:
From the Kolmogorov Axioms:
:$\ds \map \Pr {\bigcup_{i \mathop \ge 1} A_i} = \sum_{i \mathop \ge 1} \map \Pr {A_i}$
where $\set {A_1, A_2, \ldots}$ is a countable set of pairwise disjoint events of $\EE$.
This applies directly to $\map \Pr {A \cup B}$ where $A \cap B = \O$.
{{qed}}
Theorem: Probability of Union of Disjoint Events is Sum of Individual Probabilities

Let $\EE$ be an experiment.
Let $\struct {\Omega, \Sigma, \Pr}$ be a probability measure on $\EE$.
Then:
:$\forall A, B \in \Sigma: A \cap B = \O \implies \map \Pr {A \cup B} = \map \Pr A + \map \Pr B$
where $A \cap B$ denotes the '''union''' of $A$ and $B$.

Proof:
From the Addition Law of Probability, the union of $A$ and $B$ can be evaluated as:
:$\map \Pr {A \cup B} = \map \Pr A + \map \Pr B - \map \Pr {A \cap B}$
From the definition of disjoint events:
:$\map \Pr {A \cap B} = 0$
Hence the resullt.
{{qed}}
Theorem: Discrete Uniform Distribution gives rise to Probability Measure

Let $\EE$ be an experiment.
Let the probability space $\struct {\Omega, \Sigma, \Pr}$ be defined as:
:$\Omega = \set {\omega_1, \omega_2, \ldots, \omega_n}$
:$\Sigma = \powerset \Omega$
:$\forall A \in \Sigma: \map \Pr A = \dfrac 1 n \card A$
where:
:$\powerset \Omega$ denotes the power set of $\Omega$
:$\card A$ denotes the cardinality of $A$.
Then $\Pr$ is a probability measure on $\struct {\Omega, \Sigma}$.

Proof:
From Power Set of Sample Space is Event Space we have that $\Sigma$ is an event space.
{{qed|lemma}}
We check the axioms defining a probability measure:
{{begin-axiom}}
{{axiom | n = \text I
        | q = \forall A \in \Sigma
        | ml= \map \Pr A
        | mo= \ge
        | mr= 0
}}
{{axiom | n = \text {II}
        | ml= \map \Pr \Omega
        | mo= =
        | mr= 1
}}
{{axiom | n = \text {III}
        | q = \forall A \in \Sigma
        | ml= \map \Pr A
        | mo= =
        | mr= \sum_{\bigcup \set e \mathop = A} \map \Pr {\set e}
        | rc= where $e$ denotes the elementary events of $\EE$
}}
{{end-axiom}}
Axiom $\text I$ is seen to be satisfied by the observation that the cardinality of a set is never negative.
Hence $\map \Pr A \ge 0$.
{{qed|lemma}}
Then we have:
{{begin-eqn}}
{{eqn | l = \map \Pr \Omega
      | r = \dfrac 1 n \card \Omega
      | c = 
}}
{{eqn | r = \dfrac 1 n \times n
      | c = Definition of $\Omega$: it has been defined as having $n$ elements
}}
{{eqn | r = 1
      | c = 
}}
{{end-eqn}}
Axiom $\text {II}$ is thus seen to be satisfied.
{{qed|lemma}}
Let $A = \set {\omega_{r_1}, \omega_{r_2}, \ldots, \omega_{r_k} }$ where $k = \card A$.
Then by Union of Set of Singletons:
:$A = \set {\omega_{r_1} } \cup \set {\omega_{r_2} } \cup \cdots \cup \set {\omega_{r_k} }$
Hence:
{{begin-eqn}}
{{eqn | l = \map \Pr A
      | r = \dfrac 1 n \card A
      | c = 
}}
{{eqn | r = \dfrac 1 n \card {\set {\omega_{r_1}, \omega_{r_2}, \ldots, \omega_{r_k} } }
      | c = 
}}
{{eqn | r = \dfrac 1 n \card {\set {\omega_{r_1} } \cup \set {\omega_{r_2} } \cup \cdots \cup \set {\omega_{r_k} } }
      | c = 
}}
{{eqn | r = \dfrac 1 n \paren {\underbrace {1 + 1 + \cdots + 1}_{\text {$k$ times} } }
      | c = 
}}
{{eqn | r = \underbrace {\dfrac 1 n + \dfrac 1 n + \cdots + \dfrac 1 n}_{\text {$k$ times} }
      | c = 
}}
{{eqn | r = \map \Pr {\set {\omega_{r_1} } } + \map \Pr {\set {\omega_{r_2} } } + \cdots + \map \Pr {\set {\omega_{r_k} } }
      | c = 
}}
{{eqn | r = \sum_{\bigcup \set e \mathop = A} \map \Pr {\set e}
      | c = 
}}
{{end-eqn}}
Hence Axiom $\text {III}$ is thus seen to be satisfied.
{{qed|lemma}}
All axioms are seen to be satisfied.
Hence the result.
{{qed}}
Theorem: Probability Measure on Finite Sample Space

Let $\Omega = \set {\omega_1, \omega_2, \ldots, \omega_n}$ be a finite set.
Let $\Sigma$ be a $\sigma$-algebra on $\Omega$.
Let $p_1, p_2, \ldots, p_n$ be non-negative real numbers such that:
:$p_1 + p_2 + \cdots + p_n = 1$
Let $Q: \Sigma \to \R$ be the mapping defined as:
:$\forall A \in \Sigma: \map Q A = \ds \sum_{i: \omega_i \in A} p_i$
Then $\struct {\Omega, \Sigma, Q}$ constitutes a probability space.
That is, $Q$ is a probability measure on $\struct {\Omega, \Sigma}$.

Proof:
Recall the Kolmogorov axioms:
{{:Axiom:Kolmogorov Axioms}}
First we determine that $\Pr$ as defined is actually a probability measure.
By definition, we have that $\map \Pr A$ is the sum of some subset of $\set {p_1, p_2, \ldots, p_n}$.
Thus $0 \le \map \Pr A \le 1$ and Axiom $(1)$ is fulfilled trivially by definition.
Let $A \in \Sigma$ be such that:
:$A = \set {\omega_{r_1}, \omega_{r_2}, \ldots, \omega_{r_k} }$
We have that:
:$A = \set {\omega_{r_1} } \cup \set {\omega_{r_2} } \cup \cdots \cup \set {\omega_{r_k} }$
From Simple Events are Mutually Exclusive, $\set {\set {\omega_{r_1} }, \set {\omega_{r_2} }, \ldots, \set {\omega_{r_k} } }$ constitutes a set of pairwise disjoint events.
Hence:
:$\map \Pr {\set {\omega_{r_1} } \cup \set {\omega_{r_2} } \cup \cdots \cup \set {\omega_{r_k} } } = \ds \sum_{i \mathop = 1}^k \map \Pr {\omega_{r_1} }$
and it is seen that axiom $(3)$ is fulfilled.
Then we have that:
{{begin-eqn}}
{{eqn | l = \map \Pr \Omega
      | r = \map \Pr {\bigcup_{i \mathop = 1}^n \omega_i}
      | c = 
}}
{{eqn | r = \ds \sum_{i \mathop = 1}^n \map \Pr {\omega_i}
      | c = 
}}
{{eqn | r = \ds \sum_{i \mathop = 1}^n p_i
      | c = 
}}
{{eqn | r = 1
      | c = 
}}
{{end-eqn}}
Hence axiom $(2)$ is satisfied.
{{Qed}}
Theorem: Probability Measure on Single-Subset Event Space

Let $\EE$ be an experiment whose sample space is $\Omega$.
Let $\O \subsetneqq A \subsetneqq \Omega$.
Let $\Sigma := \set {\O, A, \Omega \setminus A, \Omega}$ be the event space of $\EE$.
Let $\Pr: \Sigma \to \R$ be a probability measure on $\struct {\Omega, \Sigma}$.
Then $\Pr$ has the form:
{{begin-eqn}}
{{eqn | n = Pr 1
      | l = \map \Pr \O
      | r = 0
}}
{{eqn | n = Pr 2
      | l = \map \Pr A
      | r = p
}}
{{eqn | n = Pr 3
      | l = \map \Pr {\Omega \setminus A}
      | r = 1 - p
}}
{{eqn | n = Pr 4
      | l = \map \Pr \Omega
      | r = 1
}}
{{end-eqn}}
for some $p \in \R$ satisfying $0 \le p \le 1$.

Proof:
From Event Space from Single Subset of Sample Space, we have that $\Sigma$ is an event space.
Recall the Kolmogorov axioms:
{{:Axiom:Kolmogorov Axioms}}
First we determine that $\Pr$ as defined is actually a probability measure.
Axiom $(1)$ and axiom $(2)$ are fulfilled trivially by definition.
Then we note that, apart from $\O$, $\set {A, \Omega \setminus A}$ are the only pairwise disjoint events whose union is $\Omega$.
Hence by definition of $\Pr$ we see that axiom $(3)$ degenerates to:
{{begin-eqn}}
{{eqn | l = \map \Pr {A \cup \paren {\Omega \setminus A} }
      | r = \map \Pr \Omega
      | c = 
}}
{{eqn | r = 1
      | c = $\text {Pr 4}$
}}
{{eqn | r = p + \paren {1 - p}
      | c = Definition of $p$
}}
{{eqn | r = \map \Pr A + \map \Pr {\Omega \setminus A}
      | c = $\text {Pr 2}$ and $\text {Pr 3}$
}}
{{end-eqn}}
and thus axiom $(3)$ is fulfilled.
Thus $\Pr$ is indeed a probability measure.
{{qed|lemma}}
Now it is established that a probability measure on $\Sigma$ is of the form of $\Pr$ as defined.
First we note that from Probability of Empty Event is Zero:
:$\map \Pr \O  = 0$
Hence $\text {Pr 1}$ is satisfied.
From Axiom $(2)$ we have that $\map \Pr \Omega = 1$
Hence $\text {Pr 4}$ is satisfied.
By definition of probability measure, $\map \Pr A$ is between $0$ and $1$ inclusive.
That is:
:$\map \Pr A = p$
for some $p \in \R$ such that $0 \le p \le 1$.
This is exactly $\text {Pr 2}$, which is hence seen to be satisfied.
Then we note that from Probability of Event not Occurring:
:$\map \Pr {\Omega \setminus A} = 1 - \map \Pr A$
That is:
:$\map \Pr {\Omega \setminus A} = 1 - p$
Hence $\text {Pr 3}$ is satisfied.
Thus it has been proved that a probability measure on such an an event space $\Sigma$ has to be of the form $\Pr$ as defined.
{{Qed}}
Theorem: Probability of Set Difference of Events

Let $\EE$ be an experiment with probability space $\struct {\Omega, \Sigma, \Pr}$.
Let $A, B \in \Sigma$ be events of $\EE$.
Let $\map \Pr A$ denote the probability of event $A$ occurring.
Then:
:$\map \Pr {A \setminus B} = \map \Pr A - \map \Pr {A \cap B}$

Proof:
From Set Difference and Intersection form Partition:
:$A$ is the union of the two disjoint sets $A \setminus B$ and $A \cap B$
So, by the definition of probability measure:
:$\map \Pr A = \map \Pr {A \setminus B} + \map \Pr {A \cap B}$
Hence the result.
{{qed}}
Theorem: Equivalence of Definitions of Matroid Rank Axioms

Let $S$ be a finite set.
Let $\rho : \powerset S \to \Z$ be a mapping from the power set of $S$ to the integers.
{{TFAE|axiom = Rank Axioms (Matroid)|view = rank axioms}}
=== Formulation 1 ===
$\rho$ is said to satisfy the '''rank axioms''' {{iff}}
{{:Axiom:Rank Axioms (Matroid)/Definition 1}}
=== Formulation 2 ===
$\rho$ is said to satisfy the '''rank axioms''' {{iff}}
{{:Axiom:Rank Axioms (Matroid)/Definition 2}}

Proof:
=== Formulation 1 implies Formulation 2 ===
{{:Equivalence of Definitions of Matroid Rank Axioms/Formulation 1 Implies Formulation 2}}{{qed|lemma}}