nyuuzyou/wikis huggingface dataset:
Second Socratic Paradox

== Paradox ==
 is supposed to have raised the question:
Is the following statement in true or false?
:This statement in italics is false.
Suppose the statement in italics is true.
Then it is false.
That is, what it says is different from the way things actually are.
So, the sentence is not false.
Therefore it is true.
But it cannot be both.
We can more formally demonstrate this as follows:
Let $\map T P$ be a modal operator interpreted as "The proposition $P$ is true."
Assume $P \iff \map T P$, or in other words, that declaring the proposition $P$ is equivalent to declaring that $P$ is true.
Let $L$ be a proposition representing the liar sentence, and define $L$ as $\neg \map T L$.
We thus have by definition:
:$\neg \map T L \iff L$
and by assumption:
:$L \iff \map T L$
By Biconditional is Transitive, we have $\neg \map T L \iff \map T L$, a contradiction.
== Resolution ==
There are multiple proposed resolutions to this paradox, and there is no consensus as to which one to choose.
One resolution is due to 's semantic theory of truth.
For any proposition $P$ in a language of truth, this theory requires that:
:$P$ is true $P$.
For instance:
:'Snow is white' is true snow is white.
This is analogous to the principle stated above:
:$P \iff \map T P$
 distinguishes object languages from metalanguages.
To talk about the truth of statements in language $A$, we must talk from the perspective of a higher language $B$.
The statement in language $A$ becomes an object of study, and not itself a vehicle with which to make truth assertions about statements in $A$.
Thus, the sentence "Snow is white is true snow is white" is composed of two languages:
:"Snow is white" is an assertion in English
and:
:"... is true snow is white" is an assertion in some metalanguage.
Thus, the paradox's resolution is that the sentence:
:"this sentence is not true"
is meaningless, as the language within which the sentence is expressed serves as its own metalanguage.
In other words, the sentence makes an assertion of truth about itself, which is disallowed.
An issue with 's resolution is that it imposes a severe limitation on our ability to coherently use the predicate "... is true".
For instance, consider the statement: "No language can talk about its own truth."
This statement expresses an assertion of truth about all languages, including itself.
This seemingly violates the very principle that uses to resolve the paradox.
== Also known as ==
== Also see ==
* First Socratic Paradox
* Epimenides Paradox
* Russell's Paradox
* Gödel's Incompleteness Theorems
== Historical Note ==
== Sources ==
* : 7. The Theory of Types and Symbolism: Classes
* : Chapter One: What is a Paradox?
* : $\text{I}$: 'NOT' and 'IF': $(3)$
* 
* 
* : Chapter $1$: Introduction: $\S 1.1$: The origins of mathematical logic
* 
* 
* 
Category:Socratic Paradoxes
Category:Semantic Paradoxes
Second Socratic Paradox/Also known as

== Second Socratic Paradox: Also known as == 
The second socratic paradox is often seen referred to as the liar paradox or liar's paradox.
== Sources ==
* 
* 
* : Chapter $1$: Introduction: $\S 1.1$: The origins of mathematical logic
* 
* 
* 
Category:Socratic Paradoxes
Second Socratic Paradox/Historical Note

== Historical Note on the Second Socratic Paradox ==
The Second Socratic Paradox supposedly originated from , who raised it in the $6$th century BCE in the form:
:All Cretans are liars.
This is known as the Epimenides Paradox.
In the basic stripped-down form as presented here, it is usually attributed to .
== Sources ==
* 
* 
Category:Historical Notes
Second Subsequence Rule

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Let $\sequence {x_n}$ be a sequence in $M$.
Suppose $\sequence {x_n}$ has a subsequence which is unbounded.
Then $\sequence {x_n}$ is divergent.
== Proof ==
Follows by the Rule of Transposition from Convergent Sequence is Bounded.
== Also see ==
* First Subsequence Rule
Category:Metric Spaces
Category:Named Theorems
Second Supplement to Law of Quadratic Reciprocity

== Theorem ==
$\quad \paren {\dfrac 2 p} = \paren {-1}^{\paren {p^2 - 1} / 8} = \begin {cases}
+1 & : p \equiv \pm 1 \pmod 8 \\
-1 & : p \equiv \pm 3 \pmod 8
\end {cases}$
where $\paren {\dfrac 2 p}$ is defined as the Legendre symbol.
== Proof ==
Consider the numbers in the set $S = \set {2 \times 1, 2 \times 2, 2 \times 3, \dots, 2 \times \dfrac {p - 1} 2} = \set {2, 4, 6, \dots, p - 1}$.
From Gauss's Lemma:
:$\paren {\dfrac 2 p} = \paren {-1}^n$
where $n$ is the number of elements in $S$ whose least positive residue modulo $p$ is greater than $\dfrac p 2$.
As they are, the elements of $S$ are already least positive residues of $p$ (as they are all less than $p$).
What we need to do is count how many are greater than $\dfrac p 2$.
We see that:
:$2 k > \dfrac p 2 \iff k > \dfrac p 4$
So the first $\floor {\dfrac p 4}$ elements of $S$ are not greater than $\dfrac p 2$, where $\floor {\dfrac p 4} $ is the floor function of $\dfrac p 4$.
The rest of the elements of $S$ are greater than $\dfrac p 2$.
So we have:
:$n = \dfrac {p - 1} 2 - \floor {\dfrac p 4}$
Consider the four possible residue classes modulo $8$ of the odd prime $p$.
$p = 8 k + 1$:
$p = 8 k + 3$:
$p = 8 k + 5$:
$p = 8 k + 7$:
We see that $n$ is even when $p = 8 k + 1$ or $p = 8 k + 7$ and odd in the other two cases.
So from Gauss's Lemma, we have:
As $7 \equiv -1 \pmod 8$ and $5 \equiv -3 \pmod 8$ the result follows.
== Also see ==
* Law of Quadratic Reciprocity
* First Supplement to Law of Quadratic Reciprocity
== Sources ==
* : $\S 1.2.4$: Integer Functions and Elementary Number Theory: Exercise $47 \ \text{b)}$
* 
* 
Category:Legendre Symbol
Category:Named Theorems
Category:Law of Quadratic Reciprocity
Second Sylow Theorem

== Theorem ==
Let $P$ be a Sylow $p$-subgroup of the finite group $G$.
Let $Q$ be any $p$-subgroup of $G$.
Then $Q$ is a subset of a conjugate of $P$. 
== Proof ==
Let $P$ be a Sylow $p$-subgroup of $G$.
Let $\mathbb S$ be the set of all distinct $G$-conjugates of $P$:
:$\mathbb S = \set {g P g^{-1}: g \in G}$
Let $h * S$ be the conjugacy action:
:$\forall h \in P, S \in \mathbb S: h * S = h S h^{-1}$
From Conjugacy Action on Subgroups is Group Action, this is a group action for $S \le G$.
To show it is closed for $S \in \mathbb S$:
So, consider the orbits and stabilizers of $\mathbb S$ under this group action.
Since $\forall S \in \mathbb S: \Stab S \le P$, we have that:
:$\size {\Stab S} \divides \order P$
Therefore, by the Orbit-Stabilizer Theorem, these orbit lengths are all congruent to either $0$ or $1$ modulo $p$, since $P$ is a Sylow $p$-subgroup of $G$.
Note that this will imply, as we shall mark later on:
:$\size {\mathbb S} \equiv 1 \pmod p$
Now, $h * P = h P h^{-1} = P$, so:
:$\Orb P = \set P$
We now show that $P$ is the only element of $\mathbb S$ such that $\size {\Orb S} = 1$.
If $g P g^{-1}$ has one element in its orbit, then:
:$\forall x \in P: x \paren {g P g^{-1} } x^{-1} = g P g^{-1}$
Thus $\forall x \in P$ we have that:
:$g^{-1} x g \in \map {N_G} P$
From Order of Conjugate Element equals Order of Element, we have that:
:$\order {g^{-1} x g} = \order x$
Thus $P_1 = g^{-1} P g$ is a $p$-subgroup of $\map {N_G} P$.
As $P$ and $P_1$ have the same number of elements, $P_1$ is a Sylow $p$-subgroup of $\map {N_G} P$.
Hence $P_1 = P$ by Normalizer of Sylow p-Subgroup, so $g P g^{-1} = P$.
Thus $P$ is the only element of $\mathbb S$ whose orbit has length $1$.
From Stabilizer of Coset Action on Set of Subgroups, $P = \map {N_G} P$.
Thus, for any $g \notin P$, $\size {\Orb {g P g^{-1} } }$ under conjugation by elements of $P$ has orbit greater than $1$.
Hence:
: $\size {\mathbb S} \equiv 1 \pmod p$
as promised.
Next we consider orbits of $\mathbb S$ under conjugation by elements of $Q$.
Since every orbit has length a power of $p$, the above conclusion shows there is at least one orbit of length $1$.
So there is an element $g$ such that:
:$\forall x \in Q: x \paren {g P g^{-1} } x^{-1} = g P g^{-1}$
As previously:
:$g^{-1} Q g \subseteq \map {N_G} P$
So by Normalizer of Sylow p-Subgroup:
:$g^{-1} Q g \subseteq P$
Thus $Q \subseteq g P g^{-1}$ as required.
== Also known as ==
Some sources call this the third Sylow theorem.
== Also see ==
* Sylow Theorems
== Historical Note ==
== Sources ==
* : Chapter $11$: The Sylow Theorems: Theorem $11.10$
Category:Sylow Theorems
Segment of Auxiliary Relation Mapping is Element of Increasing Mappings Satisfying Inclusion in Lower Closure

== Theorem ==
Let $R = \struct {S, \preceq}$ be an ordered set.
Let $\map {\mathit {Ids} } R$ be the set of all ideals in $R$.
Let $L = \struct {\map {\mathit {Ids} } R, \precsim}$ be an ordered set where $\precsim \mathop = \subseteq {\restriction_{\map {\mathit {Ids} } R \times \map {\mathit {Ids} } R} }$.
Let $r$ be an auxiliary relation on $S$.
Let $M = \struct {F, \preccurlyeq}$ be the ordered set of increasing mappings $g$ satisfying $\forall x \in S: \map g x \subseteq x^\preceq$.
Let $f: S \to \map {\mathit {Ids} } R$ be a mapping such that:
:$\forall x \in S: \map f x = x^r$
where $x^r$ denotes the $r$-segment of $x$.
Then:
:$(1): \quad f \in F$
Let $h: S \to \map {\mathit {Ids} } R: x \mapsto x^\preceq$
Then:
:$(2): \quad h \in F \land f \preccurlyeq h$
Let $k: S \to \map {\mathit {Ids} } R: x \mapsto \set \bot$
where $\bot$ denotes the smallest element in $L$.
Then:
:$(3): \quad k \in F$ and $k \preccurlyeq f$
== Proof ==
=== Condition $(1)$ ===
By Segment of Auxiliary Relation Mapping is Increasing:
:$f$ is an increasing mapping.
By Segment of Auxiliary Relation is Subset of Lower Closure:
:$\forall x \in S: x^r \subseteq x^\preceq$
By definition of $f$:
:$\forall x \in S: \map f x \subseteq x^\preceq$
Thus
:$f \in F$
=== Condition $(2)$ ===
By Lower Closure is Increasing:
:$\forall x, y \in S: x \preceq y \implies x^\preceq \subseteq y^\preceq$
By definitions of $h$ and $\precsim$:
:$\forall x, y \in S: x \preceq y \implies \map h x \precsim \map h y$
By definition:
:$h$ is an increasing mapping.
By definition of $h$:
:$\forall x \in S: \map h x \subseteq x^\preceq$
Thus
:$h \in F$
By Segment of Auxiliary Relation is Subset of Lower Closure:
:$\forall x \in S: x^r \subseteq x^\preceq$
By definitions of $f$ and $h$:
:$\forall x \in S: \map f x \subseteq \map h x$
By definition of $\precsim$:
:$\forall x \in S: \map f x \precsim \map h x$
Thus by definition of ordering of $M$:
:$f \preccurlyeq h$
=== Condition $(3)$ ===
$k$ is well-defined because by Singleton of Bottom is Ideal:
:$\set \bot$ is an ideal in $R$.
By Mapping is Constant iff Increasing and Decreasing:
:$k$ is an increasing mapping.
By definition of smallest element:
:$\forall x \in S: \bot \preceq x$
By definition of lower closure of element:
:$\forall x \in S: \bot \in x^\preceq$
By definition of subset:
:$\forall x \in S: \set \bot \subseteq x^\preceq$
By definition of $k$:
:$\forall x \in S: \map k x \subseteq x^\preceq$
Thus
:$k \in F$
By definition of auxiliary relation:
:$\forall x \in S: \tuple {\bot, x} \in r$
By definition of $r$-segment:
:$\forall x \in S: \bot \in x^r$
By definition of subset:
:$\forall x \in S: \set \bot \subseteq x^r$
By definitions of $k$ and $f$:
:$\forall x \in S: \map k x \subseteq \map f x$
By definition of $\precsim$:
:$\forall x \in S: \map k x \precsim \map f x$
Thus by definition of ordering of $M$:
:$k \preccurlyeq f$
== Sources ==
* 
* 
* 
* 
* 
Category:Auxiliary Relations
Segment of Auxiliary Relation Mapping is Increasing

== Theorem ==
Let $R = \left({S, \preceq}\right)$ be an ordered set.
Let ${\it Ids}\left({R}\right)$ be the set of all ideals in $R$.
Let $L = \left({ {\it Ids}\left({R}\right), \precsim}\right)$ be an ordered set
where $\precsim \mathop = \subseteq\restriction_{ {\it Ids}\left({R}\right) \times {\it Ids}\left({R}\right)}$
Let $r$ be an auxiliary relation on $S$.
Let $f: S \to {\it Ids}\left({R}\right)$ be a mapping such that
:$\forall x \in S: f\left({x}\right) = x^r$
where $x^r$ denotes the $r$-segment of $x$.
Then
:$f$ is increasing mapping.
== Proof ==
$f$ is well-defined because by Relation Segment of Auxiliary Relation is Ideal:
:$\forall x \in S: x^r$ is ideal in $L$
Let $x, y \in S$ such that
:$x \preceq y$
By Preceding implies Inclusion of Segments of Auxiliary Relation:
:$x^r \subseteq y^r$
Thus by definitions of $\precsim$ and $f$:
:$f\left({x}\right) \precsim f\left({y}\right)$
Thus by definition:
:$f$ is increasing mapping.
== Sources ==
* 
* 
Category:Auxiliary Relations
Segment of Auxiliary Relation is Subset of Lower Closure

== Theorem ==
Let $\struct {S, \vee, \preceq}$ be a bounded below join semilattice.
Let $R$ be auxiliary relation on $S$.
Let $x \in S$.
Then
:$x^R \subseteq x^\preceq$
where
:$x^R$ denotes the $R$-segment of $x$,
:$x^\preceq$ denotes the lower closure of $x$.
== Proof ==
Let $a \in x^R$.
By definition of $R$-segment of $x$:
:$\tuple {a, x} \in R$
By definition of auxiliary relation:
:$a \preceq x$
Thus by definition of lower closure of element:
:$a \in x^\preceq$
== Sources ==
* 
* 
Category:Auxiliary Relations
Category:Lower Closures
Segment on Given Base Unique

== Theorem ==
== Proof ==
:
Suppose it were possible to construct two similar and unequal segments $ACB$ and $ADB$ on the same base $AB$.
Let $ACD$ be drawn through, and join $CB$ and $CB$.
We have that segment $ACD$ is similar to $ADB$.
We also have by that similar segments admit equal angles.
So $\angle ACB = \angle ADB$, which from External Angle of Triangle is Greater than Internal Opposite is impossible.
== Sources ==
* : Book $\text{III}$. Propositions
Category:Circles
Segments of Rational Straight Line cut in Extreme and Mean Ratio are Apotome

== Theorem ==
== Proof ==
:
Let $AB$ be a rational straight line.
Let $AB$ be cut in extreme and mean ratio at the point $C$.
Let $AC$ be the greater segment.
It is to be demonstrated that each of $AC$ and $CB$ is the irrational straight line known as apotome.
Let $BA$ be produced.
Let $AD = \dfrac {BA} 2$.
We have that $AB$ is cut in extreme and mean ratio at $C$.
From :
:$CD^2 = 5 \cdot DA^2$
Therefore $CD^2 : DA^2$ is the ratio that a number has to a number.
Therefore by :
:$CD^2$ is commensurable with $DA^2$.
But by :
:$DA^2$ is rational.
Therefore by :
:$CD^2$ is also rational.
Therefore $CD$ is rational.
But $CD^2 : DA^2$ is not the ratio that a square number has to a square number.
Therefore by :
:$CD$ and $DA$ are incommensurable in length.
Thus $CD$ and $DA$ are commensurable in square only.
Therefore by definition, $AC$ is an apotome.
We have that $AB$ is cut in extreme and mean ratio at $C$.
Therefore from:
:
and:
:
it follows that:
:$AB \cdot BC = AC^2$
The square on the apotome $AC$, applied to the rational straight line $AB$, produces $BC$ as breadth.
Therefore from :
:$CB$ is a first apotome.
== Sources ==
* : Book $\text{XIII}$. Propositions
Category:Golden Mean
Seifert-van Kampen Theorem

== Theorem ==
The functor $\pi_1 : \mathbf{Top_\bullet} \to \mathbf{Grp}$ preserves pushouts of inclusions.
== Proof ==
Let $\struct {X, \tau}$ be a topological space.
Let $U_1, U_2 \in \tau$ such that:
: $U_1 \cup U_2 = X$
: $U_1 \cap U_2 \ne \O$ is connected
Let $\ast \in U_1 \cap U_2$.
Let:
: $i_k : U_1 \cap U_2 \hookrightarrow U_k$
: $j_k : U_k \hookrightarrow U_1 \cup U_2$
be inclusions.
For simplicity, let:
:$\map {\pi_1} X = \map {\pi_1} {X, \ast}$
It is to be shown that $\map {\pi_1} X$ is the amalgamated free product:
:$\map {\pi_1} {U_1} *_{\map {\pi_1} {U_1 \cap U_2} } \map {\pi_1} {U_2}$
Category:Category Theory
Self-Adjoint Densely-Defined Linear Operator has Empty Residual Spectrum

== Theorem ==
Let $\struct {\HH, \innerprod \cdot \cdot}$ be a Hilbert space.
Let $\struct {\map D T, T}$ be a self-adjoint densely defined linear operator.
Then the residual spectrum $\map {\sigma_r} T$ is empty.
== Proof ==
Let $\struct {\map D {T^\ast}, T^\ast}$ be the adjoint of $\struct {\map D T, T}$. 
Since $\struct {\map D T, T}$ is self-adjoint, we have: 
:$\struct {\map D {T^\ast}, T^\ast} = \struct {\map D T, T}$
Suppose that $\map {\sigma_r} T$ is non-empty.
Then there exists $\lambda \in \map {\sigma_r} T$.
That is, there exists $\lambda \in \C$ such that $T - \lambda I$ is injective, but such that: 
:$\map {\paren {T - \lambda I} } {\map D T}$ is not everywhere dense in $\HH$.
Let:
:$R = \overline {\map {\paren {T - \lambda I} } {\map D T} }$
Then from Linear Subspace Dense iff Zero Orthocomplement, we have: 
:$R^\bot \ne \set 0$
where $R^\bot$ is the orthocomplement of $R$.
Then there exists some $y \in R^\bot$ such that $y \ne 0$. 
Then: 
:$\innerprod {\paren {T - \lambda I} x} y = 0$ for all $x \in \map D T$
from the definition of the orthocomplement.
From Inner Product is Sesquilinear, we have: 
:$\innerprod {T x} y = \lambda \innerprod x y$ for each $x \in \map D T$.
As shown in Riesz Representation Theorem (Hilbert Spaces):
:$x \mapsto \innerprod x y$ is a bounded linear functional.
So:
:$x \mapsto \innerprod {T x} y$ is a bounded linear functional.
So from the definition of the adjoint, we have: 
:$y \in \map D {T^\ast}$
We therefore have: 
Since $\map D T$ is everywhere dense in $H$, there exists a sequence in $\map D T$ with: 
:$x_n \to T y - \overline \lambda y$ 
Then: 
:$\innerprod {x_n} {T y - \overline \lambda y}$ for each $n \in \N$.
So, taking $n \to \infty$ and using Inner Product is Continuous:
:$\innerprod {T y - \overline \lambda y} {T y - \overline \lambda y} = 0$
So from the definition of the inner product, we have: 
:$T y = \overline \lambda y$ 
So $\lambda$ is in the point spectrum $\map {\sigma_p} T$ of $T$. 
From Point Spectrum of Symmetric Densely-Defined Linear Operator is Real, we have: 
:$\overline \lambda \in \R$
so from Complex Number equals Conjugate iff Wholly Real:
:$\overline \lambda = \lambda$ 
So $\lambda \in \map {\sigma_p} T$.
But then: 
:$\paren {T - \lambda I} y = \map {\paren {T - \lambda I} } 0$
with $y \ne 0$.
So $T - \lambda I$ is not injective.
So we have a contradiction, and so $\map {\sigma_r} T = \O$.
== Sources ==
* $25.3$: The Spectrum of Closed Unbounded Self-Adjoint Operators
Category:Self-Adjoint Densely-Defined Linear Operators
Category:Residual Spectra (Densely-Defined Linear Operators)
Self-Adjoint Densely-Defined Linear Operator is Maximally Symmetric

== Theorem ==
Let $\struct {\HH, \innerprod \cdot \cdot}$ be a Hilbert space over $\C$.
Let $\tuple {\map D T, T}$ be a self-adjoint densely-defined linear operator.
Then $\tuple {\map D T, T}$ is maximally symmetric.
== Proof ==
Let $\tuple {\map D S, S}$ be a symmetric densely-defined linear operator extending $\tuple {\map D T, T}$.
Let $\tuple {\map D {S^\ast}, S^\ast}$ and $\tuple {\map D {T^\ast}, T^\ast}$ be the adjoints of $\tuple {\map D S, S}$ and $\tuple {\map D T, T}$ respectively.
From Adjoint of Densely-Defined Linear Operator reverses Extension of Mapping:
:$\tuple {\map D {T^\ast}, T^\ast}$ extends $\tuple {\map D {S^\ast}, S^\ast}$.
Since $\tuple {\map D T, T}$ is self-adjoint, we have.
:$\tuple {\map D T, T}$ extends $\tuple {\map D {S^\ast}, S^\ast}$.
From Adjoint of Symmetric Densely-Defined Linear Operator Extends Operator, we have:
:$\tuple {\map D {S^\ast}, S^\ast}$ extends $\tuple {\map D S, S}$.
Hence $\tuple {\map D T, T}$ extends $\tuple {\map D S, S}$.
By assumption $\tuple {\map D S, S}$ extends $\tuple {\map D T, T}$.
Hence $\tuple {\map D T, T} = \tuple {\map D S, S}$.
So $\tuple {\map D T, T}$ is maximally symmetric.
== Sources ==
* : $13.15$: Theorem
Category:Maximally Symmetric Densely-Defined Linear Operators
Self-Conjugate Triangle needs Two Sides to be Specified

== Definition ==
Let $\CC$ be a circle.
Let $\triangle PQR$ be a triangle such that:
:$PR$ is the polar of $Q$
:$QR$ is the polar of $P$
with respect to $\CC$.
Then
:$PQ$ is the polar of $R$
and so $\triangle PQR$ is conjugate to itself.
Hence a self-conjugate triangle requires just two sides to be specified as polars of their opposite vertices.
The third side is the polar of its opposite vertex as a consequence.
== Proof ==
We have that:
:the polar of $P$ is $QR$
:the polar of $Q$ is $PR$ 
and so both polars pass through $R$.
Therefore:
:the polar of $R$ is $PQ$.
== Also see ==
* Definition:Conjugate Triangles
* Definition:Self-Conjugate Triangle
== Sources ==
* : Chapter $\text {III}$. The Circle: $9$. Conjugate triangles
Category:Conjugate Triangles
Self-Distributive Law for Conditional/Formulation 1

== Theorem ==
The following is known as the Self-Distributive Law:
:$p \implies \paren {q \implies r} \dashv \vdash \paren {p \implies q} \implies \paren {p \implies r}$
This can of course be expressed as two separate theorems:
=== Forward Implication ===
=== Reverse Implication ===
== Proof 1 ==
== Proof by Truth Table ==
== Also see ==
* Conditional is not Right Self-Distributive where it is shown that while:
:$\paren {p \implies q} \implies r \vdash \paren {p \implies r} \implies \paren {q \implies r}$
it is not the case that:
:$\paren {p \implies r} \implies \paren {q \implies r} \not \vdash \paren {p \implies q} \implies r$
Category:Self-Distributive Law for Conditional
Self-Distributive Law for Conditional/Formulation 1/Proof by Truth Table

== Theorem ==
== Proof ==
We apply the Method of Truth Tables to the proposition:
$p \implies \paren {q \implies r} \dashv \vdash \paren {p \implies q} \implies \paren {p \implies r}$
As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations.
$\begin{array}{|ccccc||ccccccc|} \hline
p & \implies & (q & \implies & r) & (p & \implies & q) & \implies & (p & \implies & r) \\
\hline
\F & \T & \F & \T & \F & \F & \T & \F & \T & \F & \T & \F \\
\F & \T & \F & \T & \T & \F & \T & \F & \T & \F & \T & \T \\
\F & \T & \T & \F & \F & \F & \T & \T & \T & \F & \T & \F \\
\F & \T & \T & \T & \T & \F & \T & \T & \T & \F & \T & \T \\
\T & \T & \F & \T & \F & \T & \F & \F & \T & \T & \F & \F \\
\T & \T & \F & \T & \T & \T & \F & \F & \T & \T & \T & \T \\
\T & \F & \T & \F & \F & \T & \T & \T & \F & \T & \F & \F \\
\T & \T & \T & \T & \T & \T & \T & \T & \T & \T & \T & \T \\
\hline
\end{array}$
Category:Truth Table Proofs
Category:Self-Distributive Law for Conditional
Self-Distributive Law for Conditional/Formulation 2

== Theorem ==
The following is known as the Self-Distributive Law:
:$\vdash \paren {p \implies \paren {q \implies r} } \iff \paren {\paren {p \implies q} \implies \paren {p \implies r} }$
== Proof ==
=== Proof of Forward Implication ===
=== Proof of Reverse Implication ===
== Also see ==
* Conditional is not Right Self-Distributive where it is shown that while:
:$\paren {p \implies q} \implies r \vdash \paren {p \implies r} \implies \paren {q \implies r}$
it is not the case that:
:$\paren {p \implies r} \implies \paren {q \implies r} \vdash \paren {p \implies q} \implies r$
== Sources ==
* : $\text{II}$: 'AND', 'OR', 'IF AND ONLY IF': $\S 5$: Theorem $\text{T106}$
* : $\S 1.14$: Exercise $12 \ (8)$
Category:Self-Distributive Law for Conditional
Self-Distributive Law for Conditional/Forward Implication/Formulation 1

== Theorem ==
:$p \implies \paren {q \implies r} \vdash \paren {p \implies q} \implies \paren {p \implies r}$
== Proof ==
== Sources ==
* : Chapter $1$: The Propositional Calculus $1$: $2$ Conditionals and Negation: Exercise $1 \ \text{(j)}$
Category:Self-Distributive Law for Conditional
Self-Distributive Law for Conditional/Forward Implication/Formulation 2

== Theorem ==
:$\paren {p \implies \paren {q \implies r} } \implies \paren {\paren {p \implies q} \implies \paren {p \implies r} }$
== Proof 1 ==
== Proof by Truth Table ==
== Sources ==
* : Chapter $2$: The Propositional Calculus $2$: $2$: Theorems and Derived Rules: Theorem $43$
* : Chapter $1$: Sets and mappings: $\S 1.1$: The need for logic: Exercise $(2) \ \text{(ii)}$
Category:Self-Distributive Law for Conditional