nyuuzyou/wikis huggingface dataset:
Equivalence of Definitions of Real Natural Logarithm/Proof 2

== Theorem ==
== Proof ==
Let $x \in \R$ be such that $x > 0$.
Let $y \in \R$ be the unique number such that:
:$e^y = x$
=== Definition 1 iff Definition 2 ===
=== Definition 3 iff Definition 2 ===
We shall show:
:$\ds \lim_{n \mathop \to \infty} n \paren {\sqrt [n] x - 1} = y$
If $y = 0$, then $x = e^0 = 1$.
Thus the claim is clear, as:
:$\forall n \in \N : n \paren {\sqrt [n] x - 1} = 0$
If $y \ne 0$, then:
Category:Natural Logarithms
Equivalence of Definitions of Reflexive Closure

== Theorem ==
Let $\RR$ be a relation on a set $S$.
=== Definition 1 ===
=== Definition 2 ===
=== Definition 3 ===
== Proof ==
Let $\RR$ be a relation on a set $S$.
=== Union with Diagonal is Smallest Reflexive Superset ===
Let $\Delta_S$ be the diagonal relation on $S$.
Let $\RR^= = \RR \cup \Delta_S$
By Smallest Element is Unique, at most one relation on $S$ can be the smallest reflexive superset of $\RR$.
From Subset of Union:
:$\RR \subseteq \RR^=$
:$\Delta_S \subseteq \RR^=$
By Relation Contains Diagonal Relation iff Reflexive, $\RR^=$ is reflexive.
Thus $\RR^=$ is a reflexive relation containing $\RR$.
Again by Relation Contains Diagonal Relation iff Reflexive, every reflexive relation containing $\RR$ must also contain $\Delta_S$.
From Union is Smallest Superset, it follows that $\RR^=$ is the smallest reflexive relation on $S$ which contains $\RR$.
=== Intersection of Reflexive Supersets is Union with Diagonal ===
Let $\QQ$ be the set of all reflexive relations containing $\RR$ as a subset.
Let $\RR^= = \bigcap \QQ$.
By the above proof that $\RR \cup \Delta_S$ is a reflexive relation containing $\RR$:
:$\RR \cup \Delta_S \in \QQ$
By Intersection is Subset:
:$\RR^= \subseteq \RR \cup \Delta_S$
By the above proof that $\RR \cup \Delta_S$ is the smallest reflexive relation containing $\RR$:
:$\forall \PP \in \QQ: \RR \cup \Delta_S \subseteq \PP$
By Intersection is Largest Subset:
:$\RR \cup \Delta_S \subseteq \RR^=$
Thus by definition of set equality:
:$\RR^= = \RR \cup \Delta_S$
Category:Reflexive Closures
Equivalence of Definitions of Reflexive Relation

== Theorem ==
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
=== Definition 1 implies Definition 2 ===
Let $\RR$ be a relation.
We use a Proof by Contraposition by showing that:
:$\Delta_S \nsubseteq \RR \implies \exists x \in S: \tuple {x, x} \notin \RR$
Thus, suppose:
:$\Delta_S \nsubseteq \RR$
Then by definition of Diagonal Relation:
:$\exists \tuple {x, x} \in S \times S: \tuple {x, x} \notin \RR$
Thus:
:$\exists x \in S: \tuple {x, x} \notin \RR$
From Rule of Transposition it follows that:
:$\forall x \in S: \tuple {x, x} \in \RR \implies \Delta_S \subseteq \RR$
=== Definition 2 implies Definition 1 ===
Let $\RR$ be a relation which fulfills the condition:
:$\Delta_S \subseteq \RR$
Then:
Thus $\RR$ is reflexive by Definition 1.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 10$: Equivalence Relations: Exercise $10.6 \ \text{(a)}$
* : Chapter $\text{I}$: Sets and Functions: Relations: Theorem $3$
* : $\S 6$. Indexed families; partitions; equivalence relations
Category:Reflexive Relations
Category:Proofs by Contraposition
Equivalence of Definitions of Reflexive Transitive Closure

== Theorem ==
Let $\RR$ be a relation on a set $S$.
=== Smallest Reflexive Transitive Superset ===
=== Reflexive Closure of Transitive Closure ===
=== Transitive Closure of Reflexive Closure ===
== Proof ==
The result follows from:
:Transitive Closure of Reflexive Relation is Reflexive
:Reflexive Closure of Transitive Relation is Transitive
:Composition of Compatible Closure Operators
Category:Reflexive Closures
Category:Transitive Closures
Equivalence of Definitions of Removable Discontinuity of Real Function

== Theorem ==
Let $A \subseteq \R$ be a subset of the real numbers.
Let $f: A \to \R$ be a real function.
Let $f$ be discontinuous at $a \in A$.
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
=== Lemma ===
Let $A \subseteq \R$ be a subset of the real numbers.
Let $f: A \to \R$ be a real function.
Let $f$ be discontinuous at $a \in A$.
For any $b \in \R$, define the function $f_b$ by:
:$\map {f_b} x = \begin {cases} \map f x &: x \ne a \\ b &: x = a \end {cases}$
Then $\map {f_b} x = \map f x$ for every $x \ne a$.
=== Definition 1 implies Definition 2 ===
Suppose the limit $\ds \lim_{x \mathop \to a} \map f x$ exists.
Let $\ds \lim_{x \mathop \to a} \map f x = b$.
Then:
By definition, $f_b$ is continuous at $a$. 
=== Definition 2 implies Definition 1 ===
Suppose there exists $b \in \R$ such that $f_b$ is continuous at $a$.
By definition of continuity:
:$\ds \lim_{x \mathop \to a} \map {f_b} x = \map {f_b} a = b $
By the lemma:
:$\ds \lim_{x \mathop \to a} \map f x = \lim_{x \mathop \to a} \map {f_b} x = b$
It follows that the limit $\ds \lim_{x \mathop \to a} \map f x$ exists.
Category:Removable Discontinuities (Real Analysis)
Category:Equivalence of Definitions of Removable Discontinuity of Real Function
Equivalence of Definitions of Removable Discontinuity of Real Function/Lemma

== Lemma for Equivalence of Definitions of Removable Discontinuity of Real Function ==
Let $A \subseteq \R$ be a subset of the real numbers.
Let $f, g: A \to \R$ be real functions.
Let $a \in A$.
Suppose $\map f x = \map g x$ for every $x \ne a$.
Suppose the limit $\ds \lim_{x \mathop \to a} \map f x$ exists.
Then the limit $\ds \lim_{x \mathop \to a} \map g x$ exists and is equal to $\ds \lim_{x \mathop \to a} \map f x$.
== Proof ==
This can be regarded as a corollary of the Squeeze Theorem for Functions.
Let $\ds \lim_{x \mathop \to a} \map f x = L$.
We have:
:$\forall x \ne a: \map f x \le \map g x \le \map f x$
 
Hence by the Squeeze Theorem for Functions:
:$\ds \lim_{x \mathop \to a} \map g x = L$
Category:Equivalence of Definitions of Removable Discontinuity of Real Function
Equivalence of Definitions of Right Quasi-Reflexive Relation

== Theorem ==
Let $\RR \subseteq S \times S$ be a relation in $S$.
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
=== $(1)$ implies $(2)$ ===
Let $\RR$ be a right quasi-reflexive relation by definition $1$.
Then by definition:
:$\forall x, y \in S: \tuple {x, y} \in \RR \implies \tuple {y, y} \in \RR$
Let $y \in \Img \RR$ be arbitrary.
Then by definition of image set:
:$\exists x \in S: \tuple {x, y} \in \RR$
Hence :
:$\tuple {y, y} \in \RR$
As $y$ is arbitrary:
:$\forall y \in \Img \RR: \tuple {y, y} \in \RR$
Thus $\RR$ is a right quasi-reflexive relation by definition $2$.
=== $(2)$ implies $(1)$ ===
Let $\RR$ be a right quasi-reflexive relation by definition $2$.
Then by definition:
:$\forall y \in \Img \RR: \tuple {y, y} \in \RR$
Let $x, y \in S$ be arbitrary such that $\tuple {x, y} \in \RR$.
Then by definition of image set:
:$y \in \Img \RR$
Hence :
:$\tuple {y, y} \in \RR$
As $\tuple {x, y} \in \RR$ is arbitrary:
:$\forall x, y \in S: \tuple {x, y} \in \RR \implies \tuple {y, y} \in \RR$
Thus $\RR$ is a right quasi-reflexive relation by definition $1$.
Category:Right Quasi-Reflexive Relations
Equivalence of Definitions of Ring of Sets

== Theorem ==
=== Definition 1 ===
=== Definition 2 ===
=== Definition 3 ===
== Proof ==
=== Definition 1 implies Definition 2 ===
Let $\RR$ be a system of sets such that for all $A, B \in \RR$:
:$(\text {RS} 1_1): \quad \RR \ne \O$
:$(\text {RS} 2_1): \quad A \cap B \in \RR$
:$(\text {RS} 3_1): \quad A \symdif B \in \RR$
As $\RR$ is non-empty, there exists some $A \in \RR$.
From Symmetric Difference with Self is Empty Set:
:$A \symdif A = \O$
By hypothesis $A \symdif A \in \RR$ and so $\O \in \RR$.
Thus criterion $(\text {RS} 1_2)$ is fulfilled.
From Closure of Intersection and Symmetric Difference imply Closure of Set Difference it follows that criterion $(\text {RS} 2_2)$ is fulfilled.
From Closure of Intersection and Symmetric Difference imply Closure of Union it follows that criterion $(\text {RS} 3_2)$ is fulfilled.
=== Definition 2 implies Definition 1 ===
Let $\RR$ be a system of sets such that for all $A, B \in \RR$:
:$(\text {RS} 1_2): \quad \O \in \RR$
:$(\text {RS} 2_2): \quad A \setminus B \in \RR$
:$(\text {RS} 3_2): \quad A \cup B \in \RR$
We have that $\O \in \RR$ and so $\RR$ is non-empty.
Thus criterion $(\text {RS} 1_1)$ is fulfilled.
By hypothesis, $\RR$ is closed under $\setminus$ and $\cup$.
Thus:
:$\forall A, B \in \RR: \paren {A \setminus B} \cup \paren {B \setminus A} \in \RR$
But by the definition of symmetric difference:
:$A \symdif B := \paren {A \setminus B} \cup \paren {B \setminus A}$
Thus:
:$\forall A, B \in \RR: A \symdif B \in \RR$
and so $\RR$ is closed under symmetric difference.
Thus criterion $(\text {RS} 3_1)$ is fulfilled.
From Union minus Symmetric Difference equals Intersection:
:$\forall A, B \in \RR: \paren {A \cup B} \setminus \paren {A \symdif B} = A \cap B$
Thus $\RR$ is closed under set intersection.
Thus criterion $(\text {RS} 2_1)$ is fulfilled.
=== Definition 2 iff Definition 3 ===
Let $\RR$ be a system of sets such that for all $A, B \in \RR$:
:$(\text {RS} 1_2): \quad \O \in \RR$
:$(\text {RS} 2_2): \quad A \setminus B \in \RR$
:$(\text {RS} 3_2): \quad A \cup B \in \RR$
Criteria $(\text {RS} 1_3)$ and $(\text {RS} 2_3)$ are fulfilled immediately.
Consider $A, B \in \RR: A \cap B = \O$.
Then as $A, B \in \RR$ it follows by $(\text {RS} 3_2)$ that $A \cup B \in \RR$ and so $(\text {RS} 3_3)$ is fulfilled.
Now let $\RR$ be a system of sets such that for all $A, B \in \RR$:
:$(\text {RS} 1_3): \quad \O \in \RR$
:$(\text {RS} 2_3): \quad A \setminus B \in \RR$
:$(\text {RS} 3_3): \quad A \cap B = \O \implies A \cup B \in \RR$
Again, criteria $(\text {RS} 1_2)$ and $(\text {RS} 2_2)$ are fulfilled immediately.
Let $A, B \in \RR$.
Then from Set Difference Union Second Set is Union:
:$A \cup B = \paren {A \setminus B} \cup B$
From Set Difference Intersection with Second Set is Empty Set:
:$\paren {A \setminus B} \cap B = \O$
Thus from $(\text {RS} 3_3)$:
::$A \cup B = \paren {A \setminus B} \cup B \in \RR$
== Sources ==
* : $\S 1.1$: Rings and Algebras
Category:Rings of Sets
Equivalence of Definitions of Saturated Set Under Equivalence Relation

== Theorem ==
Let $\sim$ be an equivalence relation on a set $S$.
Let $T \subset S$ be a subset.
=== Definition 1 ===
=== Definition 2 ===
=== Definition 3 ===
== Proof ==
=== 1 implies 2 ===
Let $T = \overline T$.
By definition of saturation:
:$T = \ds \bigcup_{t \mathop \in T} \eqclass t {}$
so we can take $U = T$.
=== 1 implies 3 ===
Let $T = \overline T$.
By definition of saturation:
:$T = q^{-1} \sqbrk {q \sqbrk T}$
so we can take $V = q \sqbrk T$.
=== 2 implies 1 ===
Let $T = \ds \bigcup_{u \mathop \in U} \eqclass u {}$ with $U \subset S$.
Let $s \in S$ and $t \in T$ such that $s \sim t$.
By definition of union:
:$\exists u \in U : t \in \eqclass u {}$
By definition of equivalence class:
:$t \sim u$
Because $\sim$ is transitive:
:$s \sim u$
By definition of equivalence class:
:$s \in \eqclass u {}$
Thus:
:$s \in T$
Because $s$ was arbitrary:
:$\overline T \subset T$
By Set is Contained in Saturation Under Equivalence Relation:
:$T \subset \overline T$
Thus:
:$T = \overline T$
=== 3 implies 1 ===
Let $V$ be a subset of the quotient mapping of $S$ by $\sim$:
:$V \subset S / \sim$
Let $T$ be the preimage of $V$ under $q$:
:$T = q^{-1} \sqbrk V$
By Quotient Mapping is Surjection and Image of Preimage of Subset under Surjection equals Subset:
:$q \sqbrk {q^{-1} \sqbrk V} = V$
Thus:
Thus $T$ equals its saturation.
Category:Equivalence Relations
Equivalence of Definitions of Saturation Under Equivalence Relation

== Theorem ==
Let $\sim$ be an equivalence relation on a set $S$.
Let $T \subset S$ be a subset.
=== Definition 1 ===
=== Definition 2 ===
=== Definition 3 ===
== Proof ==
=== Definitions 1 and 2 are equivalent ===
=== Definitions 1 and 3 are equivalent ===
Category:Equivalence Relations
Equivalence of Definitions of Scalar Triple Product

== Theorem ==
Let $\mathbf a$, $\mathbf b$ and $\mathbf c$ be vectors in a Cartesian $3$-space:
where $\tuple {\mathbf i, \mathbf j, \mathbf k}$ is the standard ordered basis.
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
Then:
Hence the result.
== Sources ==
* : Chapter $\text {II}$: The Products of Vectors: $7$. Products of Three Vectors
* : $\S 22$: Miscellaneous Formulas involving Dot and Cross Products: $22.16$
Category:Scalar Triple Product
Equivalence of Definitions of Scattered Space

== Definition ==
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
=== Definition 1 implies Definition 2 ===
Let $T$ be defined as in definition 1.
That is, $T$ contains no non-empty subset which is dense-in-itself.
Let $H \subseteq T$.
Then whether $H$ is closed or not, it is not dense-in-itself.
In particular, if $H$ is closed, then it is not dense-in-itself.
Hence $T$ satisfies definition 2 .
=== Definition 2 implies Definition 1 ===
Now let $T$ be defined as in definition 2.
Let $H \subseteq T$.
Then $H^-$ is closed, where $H^-$ denotes the closure of $H$.
From Topological Closure is Closed, $H^-$ is closed.
Because $T$ satisfies definition $(2)$, $H^-$ has an isolated point by definition.
From Isolated Point of Closure of Subset is Isolated Point of Subset, we conclude that $H$ also has an isolated point.
Hence $T$ satisfies definition 2.
Category:Scattered Spaces
Equivalence of Definitions of Schauder Basis

== Theorem ==
Let $\Bbb F \in \set {\R, \C}$. 
Let $\struct {X, \norm \cdot}$ be a normed vector space over $\Bbb F$.
Let $\set {e_n : n \in \N}$ be a countable subset of $X$. 
 
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
=== Definition 1 implies Definition 2 ===
Suppose that: 
:for each $x \in X$, there exists a unique sequence $\sequence {\map {\alpha_j} x}_{j \mathop \in \N}$ in $\Bbb F$ such that: 
::$\ds x = \sum_{j \mathop = 1}^\infty \map {\alpha_j} x e_j$
Clearly $\set {e_n : n \in \N}$ then satisfies $(1)$ of Definition 2.
Suppose the sequence $\sequence {\alpha_j}_{j \mathop \in \N}$ is such that: 
:$\ds 0 = \sum_{j \mathop = 1}^\infty \alpha_j e_j$
Note that: 
:$\ds 0 = \sum_{j \mathop = 1}^\infty 0 e_j$
By hypothesis, this expansion is unique, so we have: 
:$\alpha_j = 0$ 
for each $j \in \N$, as required. 
So $\set {e_n : n \in \N}$ also satisfies $(2)$ of Definition 2.
=== Definition 2 implies Definition 1 ===
Suppose that:
:$(1): \quad$ for each $x \in X$, there exists a sequence $\sequence {\alpha_j}_{j \mathop \in \N}$ in $\Bbb F$ such that: 
::$\ds x = \sum_{j \mathop = 1}^\infty \alpha_j e_j$
:$(2): \quad$ whenever $\sequence {\alpha_j}_{j \mathop \in \N}$ is a sequence in $\Bbb F$ such that:
::$\ds \sum_{j \mathop = 1}^\infty \alpha_j e_j = 0$
:we have $\alpha_j = 0$ for each $j \in \N$
Let $x \in X$. 
From $(1)$, there exists a sequence $\sequence {\alpha_j}_{j \mathop \in \N}$ in $\Bbb F$ such that: 
:$\ds x = \sum_{j \mathop = 1}^\infty \alpha_j e_j$
We show that this expansion is unique.
Let $\sequence {\beta_j}_{j \mathop \in \N}$ be another sequence in $\Bbb F$ such that: 
:$\ds x = \sum_{j \mathop = 1}^\infty \beta_j e_j$
Subtracting, we obtain: 
:$\ds 0 = \sum_{j \mathop = 1}^\infty \paren {\beta_j - \alpha_j} e_j$ 
Using $(2)$, we then have: 
:$\beta_j - \alpha_j = 0$ for each $j \in \N$
That is: 
:$\beta_j = \alpha_j$ for each $j \in \N$.
So the expansion is unique, and $\set {e_n : n \mathop \in \N}$ satisfies the requirement of Definition 1.
Category:Schauder Bases
Equivalence of Definitions of Secant of Angle

== Theorem ==
Let $\theta$ be an angle.
=== Definition from Triangle ===
=== Definition from Circle ===
== Proof ==
=== Definition from Triangle implies Definition from Circle ===
Let $\sec \theta$ be defined as $\dfrac {\text{Hypotenuse}} {\text{Adjacent}}$ in a right triangle.
Consider the triangle $\triangle OAB$.
By construction, $\angle OAB$ is a right angle.
Thus:
That is:
: $\sec \theta = OB$
=== Definition from Circle implies Definition from Triangle ===
Let $\sec \theta$ be defined as the length of $OB$ in the triangle $\triangle OAB$.
Compare $\triangle OAB$ with $\triangle ABC$ in the diagram above.
We have that:
:$\angle CAB = \angle BOA = \theta$
:$\angle ABC = \angle OAB$ which is a right angle
Therefore by Triangles with Two Equal Angles are Similar it follows that $\triangle OAB$ and $\triangle ABC$ are similar.
By definition of similarity:
That is:
:$\dfrac {\text{Hypotenuse} } {\text{Adjacent} } = \sec \theta$
Category:Secant Function
Equivalence of Definitions of Second Chebyshev Function

== Theorem ==
=== Definition 1 ===
=== Definition 2 ===
=== Definition 3 ===
== Proof ==
=== Definition 1 equivalent to Definition 2 ===
The equivalence:
:$\ds \sum_{k \mathop \ge 1} \sum_{p^k \mathop \le x} \ln p \equiv \sum_{1 \mathop \le n \mathop \le x} \map \Lambda n$
follows directly from the definition of the von Mangoldt function.
=== Definition 1 equivalent to Definition 3 ===
If $x < 2$, then: 
:$\ds \sum_{k \mathop \ge 1} \sum_{p^k \mathop \le x} \ln p = \sum_{p \mathop \le x} \floor {\log_p x} \ln p = 0$
since the sums are empty.
So take $x \ge 2$. 
Consider the sum: 
:$\ds \sum_{k \mathop \ge 1} \sum_{p^k \mathop \le x} \ln p$
This sum runs over the pairs of natural numbers $\tuple {k, p}$ such that $p$ is a prime number with $p^k \le x$. 
Equivalently, for each prime $p$, we sum over the natural numbers $k$ with $p^k \le x$. 
Since $p^k \le x$ implies that $p \le x$, such $k$ only exists if $p \le x$. 
Note that for a fixed prime number $p$, we have $p^k \le x$ :
:$k \ln p \le \ln x$
by Logarithm of Power and Logarithm is Strictly Increasing. 
Since $p \ge 2$, this is equivalent to: 
:$\ds k \le \frac {\ln x} {\ln p} = \log_p x$ 
So, we can equivalently sum $\ln p$ over the pairs of natural numbers $\tuple {p, k}$ such that $p \le x$ is a prime number and $k \le \log_p x$. 
That is: 
for $x \ge 2$. 
Hence:
:$\ds \forall x \in \R: \sum_{k \mathop \ge 1} \sum_{p^k \mathop \le x} \ln p = \sum_{p \mathop \le x} \floor {\log_p x} \ln p$
Category:Number Theory
Category:Von Mangoldt Function
Equivalence of Definitions of Second Pentagonal Number

== Theorem ==
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
Let $\map {P'} n$ denote the $n$th second pentagonal number.
We have:
But:
:$\map {P'} n = \dfrac {n \paren {3 n + 1} } 2$
is the definition of a second pentagonal Number by definition 2.
Hence the result.
Category:Second Pentagonal Numbers
Equivalence of Definitions of Semantic Equivalence for Boolean Interpretations

== Theorem ==
Let $\mathbf A, \mathbf B$ be WFFs of propositional logic.
=== Definition 1 ===
=== Definition 2 ===
=== Definition 3 ===
== Proof ==
=== Definition 1 implies Definition 2 ===
Let $\mathbf A, \mathbf B$ be equivalent according to definition 1.
Let $v$ be a boolean interpretation.
By definition. either $v \left({\mathbf A}\right) = T$ or $v \left({\mathbf A}\right) = F$.
In the first case, it follows by hypothesis that $v \left({\mathbf B}\right) = T$.
In particular, then:
:$v \left({\mathbf A}\right) = v \left({\mathbf B}\right)$
In the second case, it must be that $v \left({\mathbf B}\right) \ne T$.
That is, $v \left({\mathbf B}\right) = F$, so that:
:$v \left({\mathbf A}\right) = v \left({\mathbf B}\right)$
Hence $\mathbf A$ and $\mathbf B$ are also equivalent in the sense of definition 2.
=== Definition 2 implies Definition 3 ===
Let $\mathbf A, \mathbf B$ be equivalent according to definition 2.
By definition of the boolean interpretation of $\iff$:
:$v \left({\mathbf A \iff \mathbf B}\right)= T$ $v \left({\mathbf A}\right) = v \left({\mathbf B}\right)$
Therefore, by hypothesis and definition of tautology:
:$\mathbf A \iff \mathbf B$
is a tautology.
=== Definition 3 implies Definition 1 ===
Let $\mathbf A, \mathbf B$ be equivalent according to definition 3.
That is, let $\mathbf A \iff \mathbf B$ be a tautology.
From the boolean interpretation of $\iff$, we have:
:$v \left({\mathbf A}\right) = v \left({\mathbf B}\right)$
for every boolean interpretation $v$.
Therefore it immediately follows that:
:$v \left({\mathbf A}\right) = T$ $v \left({\mathbf B}\right) = T$
i.e. $\mathbf A$ and $\mathbf B$ are equivalent in the sense of definition 1.
== Sources ==
* : $\S 2.3.1$: Theorem $2.29$ (proves equivalence of Definition 2 and Definition 3)
Category:Boolean Interpretations
Equivalence of Definitions of Semantic Equivalence for Predicate Logic

== Theorem ==
Let $\mathbf A, \mathbf B$ be WFFs of predicate logic.
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
Let $\AA$ be a structure for predicate logic.
Let $\sigma$ be an assignment for $\mathbf A \iff \mathbf B$ in $\AA$.
Then the value of $\mathbf A \iff \mathbf B$ under $\sigma$ is given by:
:$\map {f^\leftrightarrow} {\map {\operatorname{val}_\AA} {\mathbf A} \sqbrk \sigma, \map {\operatorname{val}_\AA} {\mathbf B} \sqbrk \sigma}$
and from the definition of $f^\leftrightarrow$ we see that $\AA, \sigma \models_{\mathrm{PL_A}} \mathbf A \iff \mathbf B$ :
:$\map {\operatorname{val}_\AA} {\mathbf A} \sqbrk \sigma = \map {\operatorname{val}_\AA} {\mathbf B} \sqbrk \sigma$
Because the possible values are just $\T$ and $\F$, this is equivalent to:
:$\map {\operatorname{val}_\AA} {\mathbf A} \sqbrk \sigma = \T$ $\map {\operatorname{val}_\AA} {\mathbf B} \sqbrk \sigma = \T$
which by definition of $\mathrm{PL_A}$-model amounts to:
:$\AA, \sigma \models_{\mathrm{PL_A}} \mathbf A$ $\AA, \sigma \models_{\mathrm{PL_A}} \mathbf B$
Because $\AA$ and $\sigma$ were arbitrary, the above equivalence holds for all such $\AA$ and $\sigma$.
The result follows by definition of tautology.
Category:Definitions/Model Theory for Predicate Logic
Equivalence of Definitions of Semiring of Sets/Definition 1 implies Definition 2

== Theorem ==
Let $\SS$ be a system of sets satisfying the semiring of sets axioms:
Then $\SS$ satisfies the semiring of sets axioms:
== Proof ==
Let $\SS$ be a system of sets satisfying the axioms:
It remains to be shown that $\SS$ satisfies the axiom 
Let $A, B \in \SS$. 
Let $A_1 = A \cap B$.
By axiom $(2)$:
:$A_1 \in \SS$
From Intersection is Subset:
:$A_1 \subseteq A$
By axiom $(3)$:
:$\exists$ a finite sequence of pairwise disjoint sets $A_2, A_3, \ldots, A_n \in \SS : \ds A = \bigcup_{k \mathop = 1}^n A_k$
Then:
As $A$ and $B$ were arbitrary, then $\SS$ satisfies axiom $(3')$
The result follows
Category:Equivalence of Definitions of Semiring of Sets
Equivalence of Definitions of Semiring of Sets/Definition 2 implies Definition 1

== Theorem ==
Let $\SS$ be a system of sets satisfying the semiring of sets axioms:
Then $\SS$ satisfies the semiring of sets axioms:
== Proof ==
Let $\SS$ be a system of sets satisfying the axioms:
It remains to be shown that $\SS$ satisfies the axiom 
Let $A, B \in \SS : B \subseteq A$.
By axiom $(3)$:
:$\exists$ a finite sequence of pairwise disjoint sets $A_1, A_2, \ldots, A_n \in \SS : \ds A \setminus B = \bigcup_{k \mathop = 1}^n A_k$.
Then $B$ is disjoint with each of the sets $A_k$.
Then:
As $A$ and $B$ were arbitrary, then $\SS$ satisfies axiom $(3)$
The result follows
Category:Equivalence of Definitions of Semiring of Sets