nyuuzyou/wikis huggingface dataset:
Equivalence of Definitions of Martingale in Discrete Time

== Theorem ==
Let $\struct {\Omega, \Sigma, \sequence {\FF_n}_{n \mathop \ge 0}, \Pr}$ be a filtered probability space.
Let $\sequence {X_n}_{n \mathop \ge 0}$ be an adapted stochastic process.
=== Definition 1 ===
 
=== Definition 2 ===
 
== Proof ==
=== Definition 1 implies Definition 2 ===
Suppose that:
:$(1): \quad$ $X_n$ is integrable for each $n \in \Z_{\ge 0}$
:$(2): \quad \forall n \in \Z_{\ge 0}: \expect {X_{n + 1} \mid \FF_n} = X_n$.
We prove that:
:$\forall n \in \Z_{\ge 0}, \, \forall m \ge n: \expect {X_m \mid \FF_n} = X_n$.
Fix $n \in \Z_{\ge 0}$.
We induct on $m$. 
For all $m \in \Z_{\ge 0}$ with $m \ge n$, let $\map P m$ be the proposition:
:$\expect {X_m \mid \FF_n} = X_n$ almost surely.
==== Basis for Induction ====
The case $m = n$ is immediate from Conditional Expectation of Measurable Random Variable.
So $\map P n$ is seen to hold.
This is our basis for the induction.
==== Induction Hypothesis ====
Now we need to show that, if $\map P m$ is true, where $m \ge n$, then it logically follows that $\map P {m + 1}$ is true.
So this is our induction hypothesis:
:$\expect {X_m \mid \FF_n} = X_n$ almost surely.
Then we need to show:
:$\expect {X_{m + 1} \mid \FF_n} = X_n$ almost surely.
==== Induction Step ====
This is our induction step.
From $(2)$, we have: 
:$X_m = \expect {X_{m + 1} \mid \FF_m}$ almost surely.
So we have:
:$\expect {X_m \mid \FF_n} = \expect {\expect {X_{m + 1} \mid \FF_m} \mid \FF_n}$ almost surely.
Since $n \le m$, we have $\FF_n \subseteq \FF_m$ since $\sequence {\FF_n}_{n \mathop \ge 0}$ is a filtration.
Then, by the Tower Property of Conditional Expectation, we have:
:$\expect {\expect {X_{m + 1} \mid \FF_m} \mid \FF_n} = \expect {X_{m + 1} \mid \FF_n}$ almost surely.
So:
:$\expect {X_{m + 1} \mid \FF_n} = \expect {X_m \mid \FF_n} = X_n$ almost surely.
So $\map P m \implies \map P {m + 1}$ and the result follows by the Principle of Mathematical Induction.
=== Definition 2 implies Definition 1 ===
Suppose that: 
:$(1): \quad$ $X_n$ is integrable for each $n \in \Z_{\ge 0}$
:$(2): \quad \forall n \in \Z_{\ge 0}, \, \forall m \ge n: \expect {X_m \mid \FF_n} = X_n$.
We need to prove that:
:$\forall n \in \Z_{\ge 0}: \expect {X_{n + 1} \mid \FF_n} = X_n$
This is immediate from setting $m = n + 1$ in $(2)$. 
Category:Martingales
Equivalence of Definitions of Matrix Equivalence

== Theorem ==
Let $R$ be a ring with unity.
Let $\mathbf A, \mathbf B$ be $m \times n$ matrices over $R$.
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
This is specifically demonstrated in Change of Basis Matrix under Linear Transformation.
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 29$. Matrices
Category:Matrix Equivalence
Equivalence of Definitions of Matrix Similarity

== Theorem ==
Let $R$ be a ring with unity.
Let $n \in \N_{>0}$ be a natural number.
Let $\mathbf A, \mathbf B$ be square matrices of order $n$ over $R$.
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
This is specifically demonstrated in the corollary to Change of Basis Matrix under Linear Transformation.
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 29$. Matrices
Category:Matrix Similarity
Equivalence of Definitions of Matroid

== Theorem ==
Let $M = \struct {S, \mathscr I}$ be an independence system.
=== Definition 1 ===
=== Definition 2 ===
=== Definition 3 ===
=== Definition 4 ===
== Proof ==
=== Definition 1 implies Definition 2 ===
=== Definition 2 implies Definition 3 ===
=== Definition 3 implies Definition 1 ===
=== Definition 1 implies Definition 4 ===
=== Definition 4 implies Definition 1 ===
== Sources ==
* Chapter $1.$ $\S 5.$ Properties of independent sets and bases, Theorem $2$
* Chapter $13$ Matroids $\S 13.1$ Independence Systems and Matroids, Theorem $13.5$
Category:Matroid Theory
Category:Equivalence of Definitions of Matroid
Equivalence of Definitions of Matroid/Definition 1 implies Definition 2

== Theorem ==
Let $M = \struct {S, \mathscr I}$ be an independence system.
Let $M$ also satisfy:
Then $M$ satisfies:
== Proof ==
Since:
:$\forall U, V \in \mathscr I : \size U = \size V + 1 \implies \size V < \size U$
If follows that if $M$ satisfies condition $(\text I 3)$ then $M$ satisfies condition $(\text I 4)$.
Category:Equivalence of Definitions of Matroid
Equivalence of Definitions of Matroid/Definition 1 implies Definition 4

== Theorem ==
Let $M = \struct {S, \mathscr I}$ be an independence system.
Let $M$ also satisfy:
Then $M$ satisfies:
== Proof ==
Let $M$ satisfy condition $(\text I 3)$.
Let $A \subseteq S$.
Let $Y_1, Y_2$ be maximal independent subsets of $A$.
, let:
:$\size {Y_2} \le \size {Y_1}$
:
:$\size {Y_2} < \size {Y_1}$
By condition $(\text I 3)$:
:$\exists y \in Y_1 \setminus Y_2 : Y_2 \cup \set y \in \mathscr I$
From Union of Subsets is Subset:
:$Y_2 \cup \set y \subseteq A$
This contradicts the maximality of $Y_2$.
Then:
:$\size {Y_2} = \size {Y_1}$
It follows that $M$ satisifies $(\text I 6)$.
Category:Equivalence of Definitions of Matroid
Equivalence of Definitions of Matroid/Definition 2 implies Definition 3

== Theorem ==
Let $M = \struct {S, \mathscr I}$ be an independence system.
Let $M$ also satisfy:
Then $M$ satisfies:
== Proof ==
From Independent Set can be Augmented by Larger Independent Set it follows that if $M$ satisfies condition $(\text I 4)$ then $M$ satisfies condition $(\text I 5)$.
Category:Equivalence of Definitions of Matroid
Equivalence of Definitions of Matroid/Definition 3 implies Definition 1

== Theorem ==
Let $M = \struct {S, \mathscr I}$ be an independence system.
Let $M$ also satisfy:
Then $M$ satisfies:
== Proof ==
Let $M$ satisfy condition $(\text I 5)$.
Let $U, V \in \mathscr I$ such that $\size V < \size U$.
By condition $(\text I 5)$:
:$\exists Z : \exists Z \subseteq U \setminus V : \paren {V \cup Z \in \mathscr I} \land \paren {\size {V \cup Z} = \size U}$
Then:
:$V \cup Z \ne V$
From Union with Empty Set:
:$Z \ne \O$
Then:
:$\exists x : x \in Z$
From Singleton of Element is Subset:
:$\set x \subseteq Z$
From Set Union Preserves Subsets:
:$V \cup \set x \subseteq V \cup Z$
From independence system axiom $(\text I 2)$:
:$V \cup \set x \in \mathscr I$
By definition of a subset:
:$x \in U \setminus V$
It follows that $M$ satisfies condition $(\text I 3)$.
Category:Equivalence of Definitions of Matroid
Equivalence of Definitions of Matroid/Definition 4 implies Definition 1

== Theorem ==
Let $M = \struct {S, \mathscr I}$ be an independence system.
Let $M$ also satisfy:
Then $M$ satisfies:
== Proof ==
Let $M$ satisfy condition $(\text I 6)$.
Let $U, V \in \mathscr I$ such that $\size V < \size U$.
Let $W$ be a maximal independent subset of $U \cup V$ containing $U$.
Then:
:$\size U \le \size W$
By condition $(\text I 6)$:
:$V$ is not a maximal independent subset of $U \cup V$
Then:
:$\exists x \in \paren {U \cup V} \setminus V$ such that $V \cup \set x \in \mathscr I$
From Set Difference with Union is Set Difference:
:$\exists x \in U \setminus V$ such that $V \cup \set x \in \mathscr I$
It follows that $M$ satisifies $(\text I 3)$
Category:Equivalence of Definitions of Matroid
Equivalence of Definitions of Matroid Base Axioms

== Theorem ==
Let $S$ be a finite set.
Let $\mathscr B$ be a non-empty set of subsets of $S$.
== Proof ==
=== Formulation $1$ iff Formulation $2$ ===
Formulation $1$ holds formulation $2$ holds follows immediately from the Corollary of Set Difference Then Union Equals Union Then Set Difference.
=== Formulation $1$ iff Formulation $4$ ===
=== Formulation $1$ iff Formulation $5$ ===
=== Formulation $1$ iff Formulation $7$ ===
=== Formulation $3$ iff Formulation $7$ ===
=== Formulation $5$ iff Formulation $6$ ===
Formulation $5$ holds formulation $6$ holds follows immediately from the Corollary of Set Difference Then Union Equals Union Then Set Difference.
Category:Matroid Bases
Category:Equivalence of Definitions of Matroid Base Axioms
Equivalence of Definitions of Matroid Base Axioms/Formulation 1 Iff Formulation 4

== Theorem ==
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
:$\mathscr B$ satisfies formulation $4$ of base axiom
== Proof ==
==== Necessary Condition ====
Follows immediately from:
:* Matroid Bases Iff Satisfies Formulation 1 of Matroid Base Axiom
:* Matroid Bases Satisfy Formulation 4 Base Axiom
==== Sufficient Condition ====
Follows immediately from formulation $4$ and formulation $1$.
Category:Equivalence of Definitions of Matroid Base Axioms
Equivalence of Definitions of Matroid Base Axioms/Formulation 1 Iff Formulation 5

== Theorem ==
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:
:$\mathscr B$ satisfies formulation $5$ of base axiom:
== Proof ==
=== Necessary Condition ===
From:
:* Matroid Bases Iff Satisfies Formulation 1 of Matroid Base Axiom
:* Matroid Bases Satisfy Formulation 4 Base Axiom
formulation $4$ follows from formulation $1$.
We have formulation $5$ follows from formulation $4$.
=== Sufficient Condition ===
Let $\mathscr B$ satisfy formulation $5$ base axiom.
Let $\mathscr B^* = \set{S \setminus B : B \in \mathscr B}$.
From Subsets Satisfy Formulation 5 Matroid Base Axiom Iff Complements Satisfy Formulation 1:
:$\mathscr B^*$ satisfies formulation 1 base axiom.
From Matroid Bases Iff Satisfies Formulation 1 of Matroid Base Axiom:
:there exists a matroid $M^* = \struct{S, \mathscr I^*}$ whose bases is the set:
::$\mathscr B^* = \set{S \setminus B : B \in \mathscr B}$.
From Matroid Bases Satisfy Formulation 4 Base Axiom:
:$\mathscr B^*$ satisfies formulation $4$ base axiom
By formulation $5$ base axiom:
:$\mathscr B^*$ satisfies formulation $5$ base axiom
From Subsets Satisfy Formulation 1 Matroid Base Axiom Iff Complements Satisfy Formulation 5:
:$\mathscr B$ satisfies formulation 1 base axiom.
Category:Equivalence of Definitions of Matroid Base Axioms
Equivalence of Definitions of Matroid Base Axioms/Formulation 1 Iff Formulation 7

== Theorem ==
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:
:$\mathscr B$ satisfies formulation $7$ of base axiom:
== 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
==== Sufficient Condition ====
From Formulation 3 Iff Formulation 7:
:$\mathscr B$ satisfies formulation $7$ of base axiom:
:$\mathscr B$ satisfies formulation $3$ of base axiom:
By choosing $y = \map \pi x$ in formulation $3$, formulation $1$ follows immediately. 
Category:Equivalence of Definitions of Matroid Base Axioms
Equivalence of Definitions of Matroid Base Axioms/Formulation 3 Iff Formulation 7

== Theorem ==
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:
:$\mathscr B$ satisfies formulation $7$ of base axiom:
== Proof ==
==== Necessary Condition ====
Let $\mathscr B$ satisfy formulation $3$ of base axiom:
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:
==== Sufficient Condition ====
Let $\mathscr B$ satisfy formulation $7$ of base axiom:
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:
Category:Equivalence of Definitions of Matroid Base Axioms
Equivalence of Definitions of Matroid Circuit Axioms

== Theorem ==
Let $S$ be a finite set.
Let $\mathscr C$ be a non-empty set of subsets of $S$.
=== Formulation 1 ===
$\mathscr C$ satisfies the circuit axioms:
=== Formulation 2 ===
$\mathscr C$ satisfies the circuit axioms:
=== Formulation 3 ===
$\mathscr C$ satisfies the circuit axioms:
== Proof ==
=== Formulation 1 implies Formulation 2 ===
=== Formulation 2 implies Formulation 1 ===
=== Formulation 1 implies Formulation 3 ===
=== Formulation 3 implies Formulation 1 ===
== Also see ==
* Circuits of Matroid iff Matroid Circuit Axioms
Category:Matroid Circuits
Category:Equivalence of Definitions of Matroid Circuit Axioms
Equivalence of Definitions of Matroid Circuit Axioms/Formulation 1 Implies Formulation 2

== Theorem ==
Let $S$ be a finite set.
Let $\mathscr C$ be a non-empty set of subsets of $S$.
Let $\mathscr C$ satisfy the circuit axioms (formulation 1):
Then:
:$\mathscr C$ satisfies the circuit axioms (formulation 2):
== Proof ==
Let $\mathscr C$ satisfy the circuit axioms $(\text C 1)$, $(\text C 2)$ and $(\text C 3)$.
It has only to be shown that circuit axiom $(\text C 4)$ is satisfied by $\mathscr C$.
Let:
To show that $\mathscr C$ satisfies circuit axiom $(\text C 4)$, it needs to be shown that $F = \O$.
 :
:$F \neq \O$
Let $\tuple{C_1, C_2, z, w} \in F$ :
:$\size{C_1 \cup C_2} = \min \set{\size{C \cup D} : \tuple{C, D, x, y} \in F}$
By circuit axiom $(\text C 3)$:
:$\exists C_3 \in \mathscr C : C_3 \subseteq \paren{C_1 \cup C_2} \setminus \set z$
By assumptiom:
:$w \notin C_3$
Consider $C_3 \cap \paren{C_2 \setminus C_1}$.
By circuit axiom $(\text C 2)$:
:$C_3 \nsubseteq C_1$
From Set Difference and Intersection form Partition:
:$C_3 \cap \paren{C_2 \setminus C_1} \neq \O$
Let $x \in C_3 \cap \paren{C_2 \setminus C_1}$.
We have:
:$x \in C_3 \cap C_2$
and
:$z \in C_2 \setminus C_3$
and
:$w \notin C_2 \cup C_3$
From Set is Subset of Union and Union of Subsets is Subset:
:$C_2 \cup C_3 \subseteq \C_1 \cup C_2$
Since $w \notin C_2 \cup C_3$:
:$C_2 \cup C_3$ is a proper subset of $C_1 \cup C_2$
By circuit axiom $(\text C 3)$ and the minimality of $C_1 \cup C_2$:
:$\exists C_4 \in \mathscr C : z \in C_4 \subseteq \paren{C_2 \cup C_3} \setminus \set{x}$
Now consider $C_1$ and $C_4$, we have:
:$z \in C_1 \cap C_4$
Since $w \notin C_2 \cup C_3$ then:
:$w \in C_1 \setminus C_4$
We have:
:$C_4 \subset C_2 \cup C_3 \subset C_1 \cup C_2$
From Set is Subset of Union and Union of Subsets is Subset:
:$C_1 \cup C_4 \subseteq C_1 \cup C_2$
Recall $x \in C_3 \cap \paren{C_2 \setminus C_1}$, then:
:$x \in C_2$
and
:$x \notin C_1$
Since $C_4 \subseteq \paren{C_2 \cup C_3} \setminus \set{x}$, then:
:$x \notin C_4$
It follows that:
:$x \notin C_1 \cup C_4$
and
:$x \in C_1 \cup C_2$
Hence:
:$C_1 \cup C_4$ is a proper subset of $C_1 \cup C_2$
By circuit axiom $(\text C 3)$ and the minimality of $C_1 \cup C_2$:
:$\exists C_5 \in \mathscr C : w \in C_5 \subseteq \paren{C_1 \cup C_4} \setminus \set{z}$
Since $C_1 \cup C_4 \subset C_1 \cup C_2$, then we have found $C_5$ such that:
:$w \in C_5 \subseteq \paren{C_1 \cup C_2} \setminus \set{z}$
This contradicts the fact that $\tuple{C_1, C_2, z, w} \in F$.
It follows that circuit axiom $(\text C 4)$ is satisfied.
== Sources ==
* Chapter $1.$ $\S 9.$ Circuits, Theorem $2$
Category:Equivalence of Definitions of Matroid Circuit Axioms
Equivalence of Definitions of Matroid Circuit Axioms/Formulation 1 Implies Formulation 3

== Theorem ==
Let $S$ be a finite set.
Let $\mathscr C$ be a non-empty set of subsets of $S$.
Let $\mathscr C$ satisfy the circuit axioms (formulation 1):
Then:
:$\mathscr C$ satisfies the circuit axioms (formulation 3):
== Proof ==
Let $\mathscr C$ satisfy the circuit axioms $(\text C 1)$, $(\text C 2)$ and $(\text C 3)$.
We need to show that $\mathscr C$ satisfies circuit axiom:
Let $X \subset S : \forall C \in \mathscr C : C \nsubseteq X$.
Let $x \in S$.
:
:$\exists C_1, C_2 \in \mathscr C : C_1 \neq C_2 : C_1, C_2 \subseteq X \cup \set x$.
Since $C_1, C_2 \nsubseteq X$ then:
:$x \in C_1 \cap C_2$
From circuit axiom $(\text C 3)$:
:$\exists C_3 \in \mathscr C : C_3 \subseteq \paren{C_1 \cup C_2} \setminus \set x$
We have:
:$\paren{C_1 \cup C_2} \setminus \set x \subseteq X$
Hence:
:$C_3 \subseteq X$
This contradicts the assumption that:
:$X \subset S : \forall C \in \mathscr C : C \nsubseteq X$
It follows that:
:$\exists \text{ at most one } C \in \mathscr C : C \subseteq X \cup \set x$
It follows that $\mathscr C$ satisfies circuit axiom $(\text C 5)$.
Category:Equivalence of Definitions of Matroid Circuit Axioms
Equivalence of Definitions of Matroid Circuit Axioms/Formulation 2 Implies Formulation 1

== Theorem ==
Let $S$ be a finite set.
Let $\mathscr C$ be a non-empty set of subsets of $S$.
Let $\mathscr C$ satisfy the circuit axioms (formulation 2):
Then:
:$\mathscr C$ satisfies the circuit axioms (formulation 1):
== Proof ==
Let $\mathscr C$ satisfy the circuit axioms $(\text C 1)$, $(\text C 2)$ and $(\text C 4)$.
We need to show that $\mathscr C$ satisfies circuit axiom:
Let $C_1, C_2 \in \mathscr C : C_1 \ne C_2$.
Let $z \in C_1 \cap C_2$.
From circuit axiom $(\text C 2)$:
:$C_2 \nsubseteq C_1$
By definition of subset and set difference:
:$\exists w \in C_2 \setminus C_1$
From circuit axiom $(\text C 4)$:
:$\exists C_3 \in \mathscr C : w \in C_3 \subseteq \paren{C_1 \cup C_2} \setminus \set z$
It follows that $\mathscr C$ satisfies circuit axiom $(\text C 3)$.
Category:Equivalence of Definitions of Matroid Circuit Axioms
Equivalence of Definitions of Matroid Circuit Axioms/Formulation 3 Implies Formulation 1

== Theorem ==
Let $S$ be a finite set.
Let $\mathscr C$ be a non-empty set of subsets of $S$.
Let $\mathscr C$ satisfy the circuit axioms (formulation 3):
Then:
:$\mathscr C$ satisfies the circuit axioms (formulation 1):
== Proof ==
Let $\mathscr C$ satisfy the circuit axioms $(\text C 1)$, $(\text C 2)$ and $(\text C 3)$.
We need to show that $\mathscr C$ satisfies circuit axiom:
In fact we prove the contrapositive statement:
Let:
:$C_1, C_2 \in \mathscr C$
:$z \in C_1 \cap C_2$
:$\forall C \in \mathscr C : C \nsubseteq \paren{C_1 \cup C_2} \setminus \set z$
From circuit axiom $(\text C 5)$:
:$\exists \text{ at most one } C \in \mathscr C : C \subseteq \paren{\paren{C_1 \cup C_2} \setminus \set z} \cup \set z = C_1 \cup C_2$
From Set is Subset of Union:
:$C_1, C_2 \subseteq C_1 \cup C_2$
Hence:
:$C_1 = C_2$
It follows that $\mathscr C$ satisfies circuit axiom $(\text C 3)$.
Category:Equivalence of Definitions of Matroid Circuit Axioms
Equivalence of Definitions of Matroid Circuit Axioms/Lemma 2

== Theorem ==
Let $S$ be a finite set.
Let $\mathscr C$ be a non-empty set of subsets of $S$ that satisfies the circuit axioms:
For any ordered tuple $\tuple{x_1, \ldots, x_q}$ of elements of $S$, let $\map \theta {x_1, \ldots, x_q}$ be the ordered tuple defined by:
:$\forall i \in \set{1, \ldots, q} : \map \theta {x_1, \ldots, x_q}_i = \begin{cases} 0 & : \exists C \in \mathscr C : x_i \in C \subseteq \set{x_1, \ldots, x_i}\\ 1 & : \text {otherwise} \end{cases}$
Let $t$ be the mapping from the set of ordered tuple of $S$ defined by:
:$\map t {x_1, \ldots, x_q} = \ds \sum_{i = 1}^q \map \theta {x_1, \ldots, x_q}_i$
Let $\rho : \powerset S \to \Z$ be the mapping defined by:
:$\forall A \subseteq S$:
::$\map \rho A = \begin{cases} 0 & : \text{if } A = \O \\ \map t {x_1, \ldots, x_q } & : \text{if } A = \set{x_1, \ldots, x_q} \end{cases}$
Let $X \subseteq S$ and $y \in S \setminus X$.
Then:
:$\map \rho {X \cup \set y} = \map \rho X$ $\exists C \in \mathscr C : y \in C \subseteq X \cup \set y$
== Proof ==
Let $X = \set{x_1, \ldots, x_q}$
We have:
Hence:
Category:Equivalence of Definitions of Matroid Circuit Axioms
Equivalence of Definitions of Matroid Rank Axioms

== Theorem ==
Let $S$ be a finite set.
Let $\rho : \powerset S \to \Z$ be a mapping from the power set of $S$ to the integers.
=== Formulation 1 ===
$\rho$ is said to satisfy the rank axioms 
=== Formulation 2 ===
$\rho$ is said to satisfy the rank axioms 
== Proof ==
=== Formulation 1 implies Formulation 2 ===
=== Formulation 2 implies Formulation 1 ===
Let $\rho$ satisfy formulation 2 of the rank axioms:
== Sources ==
* Chapter $1.$ $\S 6.$ Properties of the rank function
* Chapter $13$ Matroids $\S 13.2$ Other Matroid Axion, Theorem $13.10$
Category:Matroid Rank Functions
Category:Equivalence of Definitions of Matroid Rank Axioms
Equivalence of Definitions of Matroid Rank Axioms/Formulation 1 Implies Formulation 2

== Theorem ==
Let $S$ be a finite set.
Let $\rho : \powerset S \to \Z$ be a mapping from the power set of $S$ to the integers.
Let $\rho$ satisfy formulation 1 of the rank axioms:
Then $\rho$ satisfies formulation 2 of the rank axioms:
== Proof ==
Follows immediately from:
:* Formulation 1 Rank Axioms Implies Rank Function of Matroid
:* Rank Function of Matroid Satisfies Formulation 2 Rank Axioms
Category:Equivalence of Definitions of Matroid Rank Axioms
Equivalence of Definitions of Matroid Rank Axioms/Formulation 2 Implies Formulation 1

== Theorem ==
Let $S$ be a finite set.
Let $\rho : \powerset S \to \Z$ be a mapping from the power set of $S$ to the integers.
Let $\rho$ satisfy formulation 2 of the rank axioms:
Then $\rho$ satisfies formulation 1 of the rank axioms:
== Proof ==
==== $\rho$ satisfies $(\text R 1)$ ====
We have:
Hence:
:$\map \rho \O = 0$
==== $\rho$ satisfies $(\text R 2)$ ====
Let $X \subseteq S$.
Let $y \in S$.
We have:
This proves rank axiom $(\text R 2)$
==== $\rho$ satisfies $(\text R 3)$ ====
Let $X \subseteq S$.
Let $x, y \in S$.
Let $\map \rho {X \cup \set x} = \map \rho {X \cup \set y} = \map \rho X$.
We have:
Hence:
:$\map \rho {X \cup \set x \cup \set y} = \map \rho X$
This proves rank axiom $(\text R 3)$
== Sources ==
* Chapter $1.$ $\S 6.$ Properties of the rank function, Proof of Theorem $2.3$
* Chapter $13$ Matroids $\S 13.2$ Other Matroid Axioms, Theorem $13.10$
Category:Equivalence of Definitions of Matroid Rank Axioms
Equivalence of Definitions of Maximal Element

== Theorem ==
Let $\struct {S, \RR}$ be a relational structure.
Let $T \subseteq S$ be a subset of $S$.
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
=== Definition 1 implies Definition 2 ===
Let $x$ be an maximal element by definition 1.
That is:
:$(1): \quad \forall y \in T: x \mathrel \RR y \implies x = y$
 that:
:$\exists y \in T: x \mathrel {\RR^\ne} y$
Then by definition:
:$x \mathrel \RR y \land x \ne y$
which contradicts $(1)$.
Thus by Proof by Contradiction:
:$\nexists y \in T: x \mathrel {\RR^\ne} y$
That is $x$ is a maximal element by definition 2.
=== Definition 2 implies Definition 1 ===
Let $x$ be a maximal element by definition 2.
That is:
:$(2): \quad \nexists y \in T: x \mathrel {\RR^\ne} y$
 that:
:$\exists y \in T: x \mathrel \RR y: x \ne y$
That is:
:$\exists y \in T: x \mathrel {\RR^\ne} y$
which contradicts $(2)$.
Thus:
:$\forall y \in T: x \mathrel \RR y \implies x = y$
Thus $x$ is a maximal element by definition 1.
== Sources ==
* : $\S 14$: Order
Category:Maximal Elements
Equivalence of Definitions of Meet Semilattice

== Theoremm ==
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
Definition 1 of meet semilattice is the dual statement of definition 1 of join semilattice by Dual Pairs (Order Theory).
Definition 2 of meet semilattice is the dual statement of definition 2 of join semilattice by Dual Pairs (Order Theory).
Hence this theorem is the dual statement of Equivalence of Definitions of Join Semilattice by Dual Pairs (Order Theory).
The result follows from the Duality Principle.
Category:Meet Semilattices