nyuuzyou/wikis huggingface dataset:
Exchange of Rows as Sequence of Other Elementary Row Operations

== Theorem ==
Let $\mathbf A$ be an $m \times n$ matrix.
Let $i, j \in \closedint 1 m: i \ne j$
Let $r_k$ denote the $k$th row of $\mathbf A$ for $1 \le k \le m$:
:$r_k = \begin {pmatrix} a_{k 1} & a_{k 2} & \cdots & a_{k n} \end {pmatrix}$
Let $e$ be the elementary row operation acting on $\mathbf A$ as:
Then $e$ can be expressed as a finite sequence of exactly $4$ instances of the other two elementary row operations.
== Proof ==
In the below:
:$r_i$ denotes the initial state of row $i$
:$r_j$ denotes the initial state of row $j$
:$r_i'$ denotes the state of row $i$ after having had the latest elementary row operation applied
:$r_j'$ denotes the state of row $j$ after having had the latest elementary row operation applied.
$(1)$: Apply $\text {ERO} 2$ to row $j$ for $\lambda = 1$:
:$r_j \to r_j + r_i$
After this operation:
$(2)$: Apply $\text {ERO} 2$ to row $i$ for $\lambda = -1$:
:$r_i \to r_i + \paren {-r_j}$
After this operation:
$(3)$: Apply $\text {ERO} 2$ to row $j$ for $\lambda = 1$:
:$r_j \to r_j + r_i$
After this operation:
$(4)$: Apply $\text {ERO} 1$ to row $i$ for $\lambda = -1$:
:$r_i \to -r_i$
After this operation:
Thus, after all the $4$ elementary row operations have been applied, we have:
Hence the result.
== Also see ==
* Exchange of Columns as Sequence of Other Elementary Column Operations
Category:Elementary Row Operations
Excluded Point Space is Compact/Proof 2

== Theorem ==
== Proof ==
By definition of excluded point space, the only open set of $T$ which contains $p$ is $S$.
So any open cover $\CC$ of $T$ must have $S$ in it.
So $\set S$ will be a subcover of $\CC$, whatever $\CC$ may be.
And $\set S$ (having only one set in it) is trivially a finite cover of $T$.
Hence the result, by definition of compact topological space.
== Sources ==
* : Part $\text {II}$: Counterexamples: $13 \text { - } 15$. Excluded Point Topology: $3$
Category:Excluded Point Space is Compact
Excluded Point Space is Connected

== Theorem ==
Let $T = \left({S, \tau_{\bar p}}\right)$ be an excluded point space.
Then $T^*_{\bar p}$ is a connected space.
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
Category:Connected Spaces
Category:Excluded Point Topology
Category:Excluded Point Space is Connected
Excluded Point Space is Connected/Proof 2

== Theorem ==
== Proof ==
The only open set of $T$ which contains $p$ is $S$.
Therefore it is impossible to set up a separation of $T$, as $S$ will always need to be an element of such a separation.
== Sources ==
* : Part $\text {II}$: Counterexamples: $13 \text { - } 15$. Excluded Point Topology: $3$
Category:Excluded Point Space is Connected
Excluded Point Space is First-Countable

== Theorem ==
Let $T = \struct {S, \tau_{\bar p} }$ be an excluded point space.
Then $T$ is first-countable.
== Proof ==
Let $x \in S$ such that $x \ne p$.
By definition, the set $\set x$ is open in $T$.
Let $U \in \tau_{\bar p}: x \in U$.
Then $\set x \subseteq U$ and so $\set {\set x}$ is a local basis at $x$ which is trivially countable.
Now if $x = p$ there is only one $U \in \tau_{\bar p}: p \in U$, and that is $S$.
So $\set S$ is a local basis at $x$ which is trivially countable.
Hence the result, by definition of first-countable.
== Sources ==
* : Part $\text {II}$: Counterexamples: $13 \text { - } 15$. Excluded Point Topology: $6$
Category:First-Countable Spaces
Category:Excluded Point Topology
Excluded Point Space is Locally Path-Connected

== Theorem ==
Let $T = \struct {S, \tau_{\bar p} }$ be an excluded point space.
Then $T$ is locally path-connected.
== Proof ==
Consider the set $\BB$ defined as:
:$\BB = \set {\set x: x \in S \setminus \set p} \cup \set S$
Then $\BB$ is a basis for $T$.
Let $H \in \BB$.
Then $\exists x \in S: H = \set x$.
From Point is Path-Connected to Itself we have that $H$ is path-connected.
Now consider the open set $S \in \BB$.
Let $a, b \in S$.
Let $\mathbb I$ be the closed unit interval in $\R$.
Let $f: \mathbb I \to S$ be the mapping defined as:
:$\forall x \in \mathbb I: \map f x = \begin{cases}
a & : x \in \hointr 0 1 \\
b & : x = 1
\end{cases}$
Then $f^{-1} \sqbrk S = \closedint 0 1$ which is open in $\mathbb I$ because $\closedint 0 1 = \mathbb I$.
As $a$ and $b$ are any two points of $S$ it follows that $S$ is path-connected.
So we have shown that all elements of a basis of $T$ are path-connected.
Hence by definition $T$ is locally path-connected.
== Sources ==
* : Part $\text {II}$: Counterexamples: $13 \text { - } 15$. Excluded Point Topology: $3$
Category:Locally Path-Connected Spaces
Category:Excluded Point Topology
Excluded Point Space is Scattered

== Theorem ==
Let $T = \struct {S, \tau_{\bar p} }$ be an excluded point space.
Then $T$ is a scattered space.
== Proof ==
We have that Subset of Excluded Point Space is not Dense-in-itself.
So, by definition, $T$ is scattered.
== Sources ==
* : Part $\text {II}$: Counterexamples: $13 \text { - } 15$. Excluded Point Topology: $5$
Category:Scattered Spaces
Category:Excluded Point Topology
Excluded Point Space is Sequentially Compact

== Theorem ==
Let $T = \struct {S, \tau_{\bar p} }$ be an excluded point space.
Then $T$ is a sequentially compact space.
== Proof ==
:Excluded Point Space is Compact
:Compact Space is Countably Compact
:Excluded Point Space is First-Countable
:First-Countable Space is Sequentially Compact iff Countably Compact
== Sources ==
* : Part $\text {II}$: Counterexamples: $13 \text { - } 15$. Excluded Point Topology: $6$
Category:Sequentially Compact Spaces
Category:Excluded Point Topology
Excluded Point Space is T0

== Theorem ==
Let $T = \struct {S, \tau_{\bar p} }$ be an excluded point space.
Then $T$ is a $T_0$ (Kolmogorov) space.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Part $\text {II}$: Counterexamples: $13 \text { - } 15$. Excluded Point Topology: $2$
Category:T0 Spaces
Category:Excluded Point Topology
Category:Excluded Point Space is T0
Excluded Point Space is T0/Proof 1

== Theorem ==
== Proof ==
Let $T$ be a trivial space.
That is, let $S = \set p$.
Then the result holds vacuously, as there are no two distinct points in $T$.
Now suppose $T$ is not trivial.
Then $\exists x \in S: x \ne p$.
Now we have that $\set x \subseteq T$ is open in $T$ such that $p \notin \set x$ but $x \in \set x$.
Finally, suppose that $x, y \in S: x \ne y, x \ne p, y \ne p$.
Then we have that (for example) $\set x \subseteq T$ is open in $T$ such that $x \in \set x$ but $y \notin \set x$
Hence the result.
Category:Excluded Point Space is T0
Excluded Point Space is T0/Proof 2

== Theorem ==
== Proof ==
We have:
: Excluded Point Topology is Open Extension Topology of Discrete Topology
: Discrete Space satisfies all Separation Properties (including being a $T_0$ space)
Then by Condition for Open Extension Space to be $T_0$ Space, as a discrete space is $T_0$ then so is its open extension.
Category:Excluded Point Space is T0
Excluded Point Space is T5

== Theorem ==
Let $T = \struct {S, \tau_{\bar p} }$ be an excluded point space.
Then $T$ is a $T_5$ space.
== Proof ==
Let $A, B \subseteq S$ such that:
:$A^- \cap B = A \cap B^- = \O$
where $A^-$ denotes the closure of $A$.
Every closed set of $T$ contains $\set p$.
From Limit Points in Excluded Point Space and the definition of closure:
:$A^- = A \cup \set p, B^- = B \cup \set p$
So if $A^- \cap B = A \cap B^- = \O$ then $p \in A^-$ and $p \in B^-$.
Thus, $p \notin A$ and $p \notin B$ and so by definition $A$ and $B$ are open sets in $T$.
It follows that $A \cap B = \O$.
Hence the result, as we have shown that:
:$\forall A, B \subseteq X, A^- \cap B = A \cap B^- = \O: \exists U, V \in \tau_{\bar p}: A \subseteq U, B \subseteq V, U \cap V = \O$
where $U = A$ and $V = B$.
== Sources ==
* : Part $\text {II}$: Counterexamples: $13 \text { - } 15$. Excluded Point Topology: $2$
Category:T5 Spaces
Category:Excluded Point Topology
Excluded Point Space is Ultraconnected/Proof 2

== Theorem ==
== Proof ==
Apart from $S$, every open set of $T$ does not contain $p$, by definition of excluded point space.
So, apart from $\O$, every closed set of $T$ does contain $p$, by definition of closed set.
So every pair of closed sets of $T$ has an intersection which contains at least $p$.
So there are no non-empty disjoint closed sets of $T$.
Hence the result, by definition of ultraconnected.
== Sources ==
* : Part $\text {II}$: Counterexamples: $13 \text { - } 15$. Excluded Point Topology: $3$
Category:Excluded Point Space is Ultraconnected
Excluded Point Space is not Arc-Connected

== Theorem ==
Let $S$ be a set with at least two distinct elements.
Let $T = \struct {S, \tau_{\bar p} }$ be an excluded point space.
Then $T$ is not arc-connected.
== Proof ==
Let $q \in S$ be such that $q \ne p$.
Let $f: \closedint 0 1 \to T$ be an injection such that:
:$\map f 0 = p$
:$\map f 1 = q$
Because $f$ is an injection, it must be that:
:$f^{-1} \sqbrk {\set q} = \set 1$
where $f^{-1}$ denotes the preimage under $f$.
Now since $p \notin \set q$, $\set q$ is open in $\tau_{\bar p}$, by definition of the excluded point topology.
By Closed Real Interval is not Open Set, $\set 1 = \closedint 1 1$ is not open.
Thus we have exhibited an open set whose preimage under $f$ is not open.
Hence $f$ is not continuous.
Since $f$ was an arbitrary injection, no arc between $p$ and $q$ can exist.
Hence $T$ is not arc-connected.
== Sources ==
* : Part $\text {II}$: Counterexamples: $13 \text { - } 15$. Excluded Point Topology: $3$
Category:Arc-Connected Spaces
Category:Excluded Point Topology
Excluded Point Space is not Irreducible

== Theorem ==
Let $T = \struct {S, \tau_{\bar p} }$ be an excluded point space with at least three points.
Then $T^*_{\bar p}$ is not irreducible.
== Proof ==
By definition, open sets of $S$ are precisely the open sets of $S \setminus \set p$ under the discrete topology.
Let $x, y \in S \setminus \set p: x \ne y$.
Then $\set x$ and $\set y$ are both open sets of $T$ such that $\set x \cap \set y = \O$.
Hence the result, by definition of irreducible.
== Also see ==
* Sierpiński Space is Irreducible
== Sources ==
* : Part $\text {II}$: Counterexamples: $13 \text { - } 15$. Excluded Point Topology: $3$
Category:Irreducible Spaces
Category:Excluded Point Topology
Excluded Point Space is not Locally Arc-Connected

== Theorem ==
Let $T = \struct {S, \tau_{\bar p} }$ be an excluded point space.
Then $T$ is not locally arc-connected.
== Proof ==
Let $\BB \subseteq \tau_{\bar p}$ be a basis for $\tau_{\bar p}$.
Since $\BB$ covers $S$, there must be an open set $B \in \BB$ such that $p \in B$.
By definition of the excluded point topology, the only open set containing $p$ is $S$ itself.
Hence necessarily $S \in \BB$.
But by Excluded Point Space is not Arc-Connected, $S$ is not arc-connected.
Hence $\BB$ does not consist only of arc-connected sets.
Because $\BB$ was arbitrary, there cannot exist a basis for $\tau_{\bar p}$ comprising only arc-connected sets.
Hence, by definition, $T$ is not locally arc-connected.
== Sources ==
* : Part $\text {II}$: Counterexamples: $13 \text { - } 15$. Excluded Point Topology: $3$
Category:Locally Arc-Connected Spaces
Category:Excluded Point Topology
Excluded Point Space is not Perfectly T4

== Theorem ==
Let $T = \struct {S, \tau_{\bar p} }$ be an excluded point space.
Then $T$ is not a perfectly $T_4$ space.
== Proof ==
:Excluded Point Topology is Open Extension Topology of Discrete Topology
:Open Extension Topology is not Perfectly T4
== Sources ==
* : Part $\text {II}$: Counterexamples: $13 \text { - } 15$. Excluded Point Topology: $4$
Category:Perfectly T4 Spaces
Category:Excluded Point Topology
Excluded Point Topology is Open Extension Topology of Discrete Topology

== Theorem ==
Let $S$ be a set and let $p \in S$.
Let $\tau_{\bar p}$ be the excluded point topology on $S$.
Let $T = \struct {S \setminus \set p, \tau_D}$ be the discrete topological space on $S \setminus \set p$.
Then $T^* = \struct {S, \tau_{\bar p} }$ is an open extension space of $T$.
== Proof ==
Directly apparent from the definitions of excluded point topology, discrete topological space and open extension space.
== Sources ==
* : Part $\text {II}$: Counterexamples: $16$. Open Extension Topology: $8$
Category:Open Extension Topology
Category:Excluded Point Topology
Category:Discrete Topology
Excluded Point Topology is T4

== Theorem ==
Let $T = \struct {S, \tau_{\bar p} }$ be an excluded point space.
Then $T$ is a $T_4$ space.
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
Category:T4 Spaces
Category:Excluded Point Topology
Category:Excluded Point Topology is T4
Excluded Point Topology is T4/Proof 1

== Theorem ==
== Proof ==
We have that an Excluded Point Space is Ultraconnected.
That means none of its closed sets are disjount.
Hence, vacuously, any two of its disjoint closed subsets of $S$ are separated by neighborhoods.
The result follows by definition of $T_4$ space.
== Sources ==
* : Part $\text {II}$: Counterexamples: $13 \text { - } 15$. Excluded Point Topology: $2$
Category:Excluded Point Topology is T4
Excluded Point Topology is Topology

== Theorem ==
Let $T = \struct {S, \tau_{\bar p} }$ be an excluded point space.
Then $\tau_{\bar p}$ is a topology on $S$, and $T$ is a topological space.
== Proof ==
We have by definition that $S \in \tau_{\bar p}$, and as $p \notin \O$ we have that $\O \in \tau_{\bar p}$.
Now let $U_1, U_2 \in \tau_{\bar p}$.
By definition $p \notin U_1$ and $p \notin U_2$.
By definition of set intersection:
:$p \notin U_1 \cap U_2$
So $U_1 \cap U_2 \in \tau_{\bar p}$.
Now let $\UU \subseteq \tau_{\bar p}$.
We have that $\forall U \in \UU: p \notin U$.
Hence from Subset of Union $p \notin \bigcup \UU$.
So all the properties are fulfilled for $\tau_{\bar p}$ to be a topology on $S$.
== Sources ==
* : Part $\text {II}$: Counterexamples: $13 \text { - } 15$. Excluded Point Topology
Category:Excluded Point Topology
Excluded Point Topology is not T3

== Theorem ==
Let $T = \struct {S, \tau_{\bar p} }$ be a excluded point space.
Then $T$ is not a $T_3$ space.
== Proof ==
:Excluded Point Topology is Open Extension Topology of Discrete Topology
:Open Extension Topology is not T3
== Sources ==
* : Part $\text {II}$: Counterexamples: $13 \text { - } 15$. Excluded Point Topology: $2$
Category:T3 Spaces
Category:Excluded Point Topology
Excluded Set Topology is Topology

== Theorem ==
Let $T = \struct {S, \tau_{\bar H} }$ be an excluded set space.
Then $\tau_{\bar H}$ is a topology on $S$, and $T$ is a topological space.
== Proof ==
We have by definition that $S \in \tau_{\bar H}$.
Also, as $H \cap \O = \O$, we have that $\O \in \tau_{\bar H}$.
Now let $U_1, U_2 \in \tau_{\bar H}$.
By definition:
:$H \cap U_1 = \O$
and:
:$H \cap U_2 = \O$
and so by definition of set intersection:
:$H \cap \paren {U_1 \cap U_2} = \O$
So:
:$U_1 \cap U_2 \in \tau_{\bar H}$
Now let $\UU \subseteq \tau_{\bar H}$.
We have that:
:$\forall U \in \UU: H \cap U = \O$
Hence from Subset of Union:
:$H \cap \bigcup \UU = \O$
So all the conditions are fulfilled for $\tau_{\bar H}$ to be a topology on $S$.
== Sources ==
* : Part $\text {II}$: Counterexamples: $13 \text { - } 15$. Excluded Point Topology: $7$
Category:Excluded Set Topology
Excluded Set Topology is not T0

== Theorem ==
Let $T = \struct {S, \tau_{\bar H} }$ be an excluded set space where $H$ has at least two distinct points.
Then $T$ is not a $T_0$ (Kolmogorov) space.
== Proof ==
Let $x, y \in H$ such that $x \ne y$.
Then $x, y \in S$, but by definition $x$ and $y$ are in no other open sets of $T$.
Hence there is no $U \in \tau_{\bar H}$ such that $x \in U, y \notin U$ or $y \in U, x \notin U$.
Hence the result, by definition of $T_0$ (Kolmogorov) space.
== Sources ==
* : Part $\text {II}$: Counterexamples: $13 \text { - } 15$. Excluded Point Topology: $7$
Category:T0 Spaces
Category:Excluded Set Topology
Exclusive Or/Examples/Monday or Not Monday

== Example of Exclusive Or ==
The following is an example of an exclusive or statement:
:Either today is Monday or today is not Monday.
== Proof ==
By the Law of Excluded Middle, today is Monday or today is not Monday is a truth.
By the Principle of Non-Contradiction, today is Monday and today is not Monday is a falsehood.
Hence the result by definition of exclusive or.
== Sources ==
* : Chapter $1$: Elementary, my dear Watson: $\S 1.1$: You have a logical mind if... Example $1.1.1$
Category:Examples of Exclusive Or
Exclusive Or/Examples/One of Five Statements is True

== Example of Exclusive Or ==
Which of the following $5$ statements is true?
:$(1): \quad$ Exactly one of these statements is false.
:$(2): \quad$ Exactly two of these statements are false.
:$(3): \quad$ Exactly three of these statements are false.
:$(4): \quad$ Exactly four of these statements are false.
:$(5): \quad$ Exactly five of these statements are false.
== Solution ==
 more than $1$ of these statements is true.
That would mean both that:
:$\text{(a)}: \quad$ Exactly $n$ of these statements is false.
and:
:$\text{(b)}: \quad$ Exactly $m$ of these statements is false.
where $m \ne n$.
But that would mean that $\size {n - m}$ of the statements is both true and false at the same time.
That would contradict the Principle of Non-Contradiction.
Hence no more than $1$ of these statements can be true.
That would mean either $(4)$ or $(5)$ must be true.
 $(5)$ is true.
But then that would mean that $(5)$ itself must be false.
Again, that would contradict the Principle of Non-Contradiction.
Hence it follows that $(4)$ must be true, and all the others false.
== Sources ==
* : Chapter $1$: Elementary, my dear Watson: $\S 1.1$: You have a logical mind if...: Ponderable $1.1.3$
Category:Examples of Exclusive Or
Exclusive Or/Examples/Tea or Coffee

== Example of Exclusive Or ==
Consider the sentence that may be found on the table d'hote menu of a restaurant:
:Breakfast includes either tea or coffee.
It is usually taken for granted that this is an example of an exclusive or statement:
:The diner may include either tea with breakfast, or coffee with breakfast, but not both.
The usual understanding is that if the diner wishes both, then extra is to be paid.
== Sources ==
* : $2$ Arguments Containing Compound Statements: $2.1$: Simple and Compound Statements
Category:Examples of Exclusive Or
Exclusive Or as Conjunction of Disjunctions

== Theorem ==
: $p \oplus q \dashv \vdash \left({p \lor q}\right) \land \left({\neg p \lor \neg q}\right)$
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $\S 2.5$: Further Logical Constants
Category:Exclusive Or
Category:Disjunction
Category:Conjunction
Category:Exclusive Or as Conjunction of Disjunctions
Exclusive Or as Conjunction of Disjunctions/Proof 2

== Theorem ==
== Proof ==
We apply the Method of Truth Tables.
As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations.
$\begin{array}{|ccc||ccccccccc|} \hline
p & \oplus & q & (p & \lor & q) & \land & (\neg & p & \lor & \neg & q) \\
\hline
F & F & F & F & F & F & F & T & F & T & T & F \\
F & T & T & F & T & T & T & T & F & T & F & T \\
T & T & F & T & T & F & T & F & T & T & T & F \\
T & F & T & T & T & T & F & F & T & F & F & T \\
\hline
\end{array}$
Category:Exclusive Or as Conjunction of Disjunctions
Category:Truth Table Proofs
Exclusive Or as Disjunction of Conjunctions

== Theorem ==
:$p \oplus q \dashv \vdash \paren {\neg p \land q} \lor \paren {p \land \neg q}$
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $\S \text{I}: 12$: Material Equivalence and Alternation
Category:Exclusive Or
Category:Disjunction
Category:Conjunction
Category:Exclusive Or as Disjunction of Conjunctions
Exclusive Or as Disjunction of Conjunctions/Proof 2

== Theorem ==
== Proof ==
We apply the Method of Truth Tables.
As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations.
$\begin{array}{|ccc||ccccccccc|} \hline
p & \oplus & q & (\neg & p & \land & q) & \lor & (p & \land & \neg & q) \\
\hline
F & F & F & T & F & F & F & F & F & F & T & F \\
F & T & T & T & F & T & T & T & F & F & F & T \\
T & T & F & F & T & F & F & T & T & T & T & F \\
T & F & T & F & T & F & T & F & T & F & F & T \\
\hline
\end{array}$
Category:Exclusive Or as Disjunction of Conjunctions
Category:Truth Table Proofs
Exclusive Or is Associative

== Theorem ==
Exclusive or is associative:
:$p \oplus \paren {q \oplus r} \dashv \vdash \paren {p \oplus q} \oplus r$
== Proof ==
We apply the Method of Truth Tables.
As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations.
$\begin{array}{|ccccc||ccccc|} \hline
p & \oplus & (q & \oplus & r) & (p & \oplus & q) & \oplus & r \\
\hline
\F & \F & \F & \F & \F & \F & \F & \F & \F & \F \\
\F & \T & \F & \T & \T & \F & \F & \F & \T & \T \\
\F & \T & \T & \T & \F & \F & \T & \T & \T & \F \\
\F & \F & \T & \F & \T & \F & \T & \T & \F & \T \\
\T & \T & \F & \F & \F & \T & \T & \F & \T & \F \\
\T & \F & \F & \T & \T & \T & \T & \F & \F & \T \\
\T & \F & \T & \T & \F & \T & \F & \T & \F & \F \\
\T & \T & \T & \F & \T & \T & \F & \T & \T & \T \\
\hline
\end{array}$
== Sources ==
* : $\S 2.3.3$
Category:Exclusive Or
Category:Truth Table Proofs
Exclusive Or is Commutative/Proof 2

== Theorem ==
== Proof ==
We apply the Method of Truth Tables.
As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations.
$\begin{array}{|ccc||ccc|} \hline
p & \oplus & q & q & \oplus & p \\
\hline
F & F & F & F & F & F \\
F & T & T & T & T & F \\
T & T & F & F & T & T \\
T & F & T & T & F & T \\
\hline
\end{array}$
Category:Exclusive Or is Commutative
Exclusive Or is Negation of Biconditional

== Theorem ==
Exclusive or is equivalent to the negation of the biconditional:
:$p \oplus q \dashv \vdash \neg \paren {p \iff q}$
== Proof ==
== Sources ==
* : Chapter $1$: Elementary, my dear Watson: $\S 1.1$: You have a logical mind if...: Ponderable $1.1.1$
Category:Exclusive Or
Category:Biconditional
Exclusive Or is Self-Inverse

== Theorem ==
:$\paren {p \oplus q} \oplus q \dashv \vdash p$
where $\oplus$ denotes the exclusive or operator.
== Proof ==
We apply the Method of Truth Tables.
As can be seen by inspection, the truth values under the main connective on the match those for $p$ on the for all boolean interpretations:
$\begin{array}{|ccccc||c|} \hline
(p & \oplus & q) & \oplus & q & p \\
\hline
\F & \F & \F & \F & \F & \F \\
\F & \T & \T & \F & \T & \F \\
\T & \T & \F & \T & \F & \T \\
\T & \F & \T & \T & \T & \T \\
\hline
\end{array}$
== Sources ==
* : Chapter $2$: Propositional Calculus: $\S 2.1$: Boolean operators
Category:Exclusive Or
Exclusive Or with Contradiction/Proof 2

== Theorem ==
== Proof ==
We apply the Method of Truth Tables to the proposition.
As can be seen by inspection, in each case, the truth values in the appropriate columns match for all boolean interpretations.
$\begin{array}{|c|ccc||c|ccc|} \hline
p & p & \oplus & \bot & p \\
\hline
F & F & F & F & F \\
T & T & T & F & T \\
\hline
\end{array}$
Category:Exclusive Or with Contradiction
Exclusive Or with Itself

== Theorem ==
Exclusive or destroys copies of itself:
:$p \oplus p \dashv \vdash \bot$
== Proof ==
We apply the Method of Truth Tables.
As can be seen by inspection, the truth values under the main connective match for each boolean interpretations.
$\begin{array}{|ccc|} \hline
p & \oplus & p \\
\hline
\F & \F & \F \\
\T & \F & \T \\
\hline
\end{array}$
== Sources ==
* : $\S 2.4$: Logical Equivalence and Substitution
* : $\S 2.3.3$
Category:Exclusive Or
Category:Truth Table Proofs
Exclusive Or with Tautology

== Theorem ==
An exclusive or with a tautology:
:$p \oplus \top \dashv \vdash \neg p$
== Proof by Natural Deduction ==
== Proof by Truth Table ==
== Sources ==
* : $\S 2.3.3$
Category:Tautology
Category:Exclusive Or
Category:Exclusive Or with Tautology
Exclusive Or with Tautology/Proof 2

== Theorem ==
== Proof ==
We apply the Method of Truth Tables to the proposition.
As can be seen by inspection, in each case, the truth values in the appropriate columns match for all boolean interpretations.
$\begin{array}{|c|ccc||cc|} \hline
p & p & \oplus & \top & \neg & p \\
\hline
F & F & T & T & T & F \\
T & T & F & T & F & T \\
\hline
\end{array}$
Category:Exclusive Or with Tautology