avewright/proofwiki-math huggingface dataset:
Theorem: Subset Relation is Ordering

Let $S$ be a set.
Let $\powerset S$ be the power set of $S$.
Let $\mathbb S \subseteq \powerset S$ be any subset of $\powerset S$, that is, an arbitrary set of subsets of $S$.
Then $\subseteq$ is an ordering on $\mathbb S$.

Proof:
To establish that $\subseteq$ is an ordering, we need to show that it is reflexive, antisymmetric and transitive.
So, checking in turn each of the criteria for an ordering:
Theorem: Principle of Mathematical Induction/Well-Ordered Set

Let $\struct {S, \preceq}$ be a well-ordered set.
Let $T \subseteq S$ be a subset of $S$ such that:
:$\forall s \in S: \paren {\forall t \in S: t \prec s \implies t \in T} \implies s \in T$
Then $T = S$.

Proof:
{{AimForCont}} that $T \ne S$.
From Set Difference is Subset, $S \setminus T \subset S$.
From Set Difference with Proper Subset, $S \setminus T \ne \O$.
By the definition of a well-ordered set, there exists a smallest element $s$ of $S \setminus T$.
As $s \in S$, it follows from the definition of $T$ that:
:$\forall t \in S: t \prec s \implies t \in T$
But then $s \in T$ by hypothesis, contradicting the definition of $s$.
Hence the result, by Proof by Contradiction.
{{qed}}
Theorem: Order Isomorphism from Woset onto Subset

Let $\struct {S, \preceq}$ be a woset.
Let $T \subseteq S$.
Let $f: S \to T$ be an order isomorphism.
Then $\forall x \in S: x \preceq \map f x$.

Proof:
Let $T = \set {x \in S: \map f x \prec x}$.
We are to show that $T = \O$.
{{AimForCont}} that $T \ne \O$.
Then as $\struct {S, \preceq}$ is a woset, by definition $T$ has a minimal element: call it $x_0$.
Since $x_0 \in T$, we have $\map f {x_0} \prec x_0$.
So, let $x_1 = \map f {x_0}$.
$f$ is an order isomorphism, so since $x_1 \prec x_0$, it follows that $\map f {x_1} \prec \map f {x_0} = x_1$.
So as $\map f {x_1} \prec x_1$ it follows that $x_1 \in T$.
But $x_0$ was chosen to be the minimal element of $T$.
From this contradiction, it follows that it can not be the case that $T \ne \O$.
So $T = \O$ and so $\forall x \in S: x \preceq \map f x$.
{{qed}}
Theorem: Order Isomorphism between Wosets is Unique

Let $\struct {S_1, \preceq_1}$ and $\struct {S_2, \preceq_2}$ be wosets.
Let $\struct {S_1, \preceq_1} \cong \struct {S_2, \preceq_2}$, that is, let $\struct {S_1, \preceq_1}$ and $\struct {S_2, \preceq_2}$ be order isomorphic.
Then there is exactly one mapping $f: S_1 \to S_2$ such that $f$ is an order isomorphism.

Proof:
Let $f: S_1 \to S_2$ and $g: S_1 \to S_2$ both be order isomorphisms.
By Inverse of Order Isomorphism is Order Isomorphism, the inverse $f^{-1}$ is also an order isomorphism.
Let $h = f^{-1} \circ g$ be the composition of $f^{-1}$ and $g$, which, by Composite of Order Isomorphisms is Order Isomorphism, is itself an order isomorphism.
So, by Order Isomorphism from Woset onto Subset:
:$\forall x \in S_1: x \preceq_1 \map h x$
Now we apply $f$ to $S_1$ and see that:
:$\forall x \in S_1: \map f x \preceq_2 \map f {\map h x} = \map g x$
In a similar way we can show that:
:$\forall x \in S_1: \map g x \preceq_2 \map f x$
Hence $f = g$ and the proof is complete.
{{qed}}
Theorem: Woset is Isomorphic to Set of its Initial Segments

Let $\struct {S, \preceq}$ be a well-ordered set.
Let:
:$A = \set {a^\prec: a \in S}$
where $a^\prec$ is the strict lower closure of $S$ determined by $a$.
Then:
:$\struct {S, \preceq} \cong \struct {A, \subseteq}$
where $\cong$ denotes order isomorphism.

Proof:
Define $f: S \to A$ as:
:$\forall a \in S: \map f a = a^\prec$
where $a^\prec$ is the initial segment determined by $a$.
Theorem: Minimal Element of an Ordinal

The minimal element of any nonempty ordinal is the empty set.
That is, if $S$ is a nonempty ordinal, $\bigcap S = \O$

Proof:
Let $S$ be an ordinal.
Let the minimal element of $S$ be $s_0$.
This exists by dint of an ordinal being a woset.
From Ordering on Ordinal is Subset Relation, $S$ is well-ordered by $\subseteq$.
So, by definition of an ordinal:
:$s_0 = \set {s \in S: s \subset s_0}$
But as $s_0$ is minimal, there ''are'' no elements of $S$ which are a subset of it.
So:
:$\set {s \in S: s \subset s_0} = \O$
Hence the result.
{{qed}}
Theorem: Ordering on Ordinal is Subset Relation

Let $\struct {S, \prec}$ be an ordinal.
Then $\forall x, y \in S:$
:$x \in y \iff x \prec y \iff S_x \subsetneqq S_y \iff x \subsetneqq y$
where $S_x$ and $S_y$ are the initial segments of $S$ determined by $x$ and $y$ respectively.
Thus there is no need to specify what the ordering on an ordinal is -- it is always the subset relation.

Proof:
The first equivalence is an immediate consequence of Equivalence of Definitions of Ordinal.
The second equivalence holds for any well-ordered set by Woset Isomorphic to Set of its Sections.
The third equivalence holds by definition of an ordinal.
It follows from Ordering is Equivalent to Subset Relation and Order Isomorphism between Wosets is Unique that this ordering is the only one.
{{qed}}
Theorem: Initial Segment of Ordinal is Ordinal

Let $S$ be an ordinal.
Let $a \in S$.
Then the initial segment $S_a = a$ of $S$ determined by $a$ is also an ordinal.
In other words, every element of a (non-empty) ordinal is also an ordinal.

Proof:
By Subset of Well-Ordered Set is Well-Ordered, $S_a$ is well-ordered.
Suppose that $b \in S_a$.
From Ordering on Ordinal is Subset Relation, and the definition of an initial segment, it follows that $b \subset a$.
Then:
{{begin-eqn}}
{{eqn | l = \paren {S_a}_b
      | r = \set {x \in S_a: x \subset b}
      | c = {{Defof|Initial Segment}}
}}
{{eqn | r = \set {x \in S: x \subset a \land x \subset b}
      | c = {{Defof|Initial Segment}}
}}
{{eqn | r = \set {x \in S: x \subset b}
      | c = as $b \subset a$
}}
{{eqn | r = S_b
      | c = {{Defof|Initial Segment}}
}}
{{eqn | r = b
      | c = as $S$ is an ordinal
}}
{{end-eqn}}
The result follows from the definition of an ordinal.
{{qed}}
Theorem: Ordinal Subset of Ordinal is Initial Segment

Let $S$ be an ordinal.
Let $T \subset S$ also be an ordinal.
Then $\exists a \in S: T = S_a$, where $S_a$ is the initial segment of $S$ determined by $a$.
That is, $T = S_a = a \in S$.

Proof:
Recall that the Ordering on Ordinal is Subset Relation.
Let $a$ be the minimal element of $S \setminus T$.
By definition of minimal element:
:$\forall x \in S: x \subset a$
Hence $x \notin S \setminus T$.
By definition of set difference:
:$x \in T$
Thus by definition of initial segment:
:$S_a \subseteq T$
Now let $b \in T$.
Then by definition of ordinal:
:$T_b = b = S_b$
Now, if $a \subset b$ then $a \in S_b$.
So $a \in T_b$ and hence $a \in T$.
But $a \in S \setminus T$, so by definition of set difference, this is not the case.
So $a \not \subset b$.
We have that an ordinal is well-ordered by $\subseteq$.
So by the definition of well-ordering, $\subseteq$ is a total ordering on $S$
So it follows that $b \subseteq a$.
But $b \ne a$ since $b \in T$.
Hence $b \subset a$, and so $b \in S_a$.
This proves that $T \subseteq S_a$.
Thus $T = S_a$.
{{qed}}
Theorem: Intersection of Two Ordinals is Ordinal

Let $S$ and $T$ be ordinals.
Then $S \cap T$ is an ordinal.

Proof:
Because $S$ and $T$ are ordinals, {{afortiori}} they are (strictly) well-ordered by the subset relation.
Let $a \in S \cap T$.
Then the initial segments $S_a$ and $T_a$ are such that:
:$S_a = a = T_a$
That is:
:$\set {x \in S: x \subset a} = a = \set {y \in T: y \subset a}$
So:
:$a = \set {z \in S \cap T: z \subset a} = \paren {S \cap T}_a$
Hence it is seen that $\paren {S \cap T}_a$ is an initial segment of both $S$ and $T$.
The result follows from Initial Segment of Ordinal is Ordinal.
{{qed}}
Theorem: Relation between Two Ordinals

Let $S$ and $T$ be ordinals.
Then either $S \subseteq T$ or $T \subseteq S$.

Proof:
{{improve|Worth removing the technical complexity to make it more accessible<br/>A better approach may be to write a second proof rather than change this one.}}
{{AimForCont}} the claim is false.
That is, by De Morgan's laws: Conjunction of Negations:
:$\paren {\neg \paren {S \subseteq T} } \land \paren {\neg \paren {T \subseteq S} }$
where $\neg$ denotes logical not and $\land$ denotes logical and.
Now from Intersection is Subset, we have $S \cap T \subseteq S$ and $S \cap T \subseteq T$.
Also, by Intersection with Subset is Subset, we have $S \cap T \ne S$ and $S \cap T \ne T$.
That is, $S \cap T \subsetneq S$ and $S \cap T \subsetneq T$.
Note that by Intersection of Two Ordinals is Ordinal, $S \cap T$ is an ordinal.
So by Transitive Set is Proper Subset of Ordinal iff Element of Ordinal:
:$S \cap T \in S$
:$S \cap T \in T$
But then $S \cap T \in S \cap T$, which contradicts Ordinal is not Element of Itself.
{{qed}}
Category:Relation between Two Ordinals
Category:Ordinals
ao691nsrv50mrkuapqaqgrot6ca2ef5
Theorem: Relation between Two Ordinals

Let $S$ and $T$ be ordinals.
Then either $S \subseteq T$ or $T \subseteq S$.

Proof:
By Ordinal Membership is Trichotomy, either $S \in T$ or $T \in S$.
By definition, every element of an ordinal is an initial segment.
Hence the result.
{{qed}}
Theorem: Relation between Two Ordinals

Let $S$ and $T$ be ordinals.
Then either $S \subseteq T$ or $T \subseteq S$.

Proof:
If either $S \subset T$ or $T \subset S$ then we invoke Ordinal Subset of Ordinal is Initial Segment, and the proof is complete.
{{AimForCont}} $S \not \subset T$ and $T \not \subset S$.
Now from Intersection is Subset, we have $S \cap T \subset T$ and $S \cap T \subset S$.
By Intersection of Two Ordinals is Ordinal, $S \cap T$ is an ordinal.
So by Ordinal Subset of Ordinal is Initial Segment, we have:
:$S \cap T = S_a$ for some $a \in S$
:$S \cap T = S_b$ for some $b \in T$
Then:
:$a = S_a = S \cap T = T_b = b$
But $a \in S, b \in T$.
Thus $a = b = S \cap T$.
But $S \cap T = S_a$, so:
:$x \in S \cap T \implies x \subset a$
In particular, this means $a \subset a$, which is a contradiction.
So either $S \subset T$ or $T \subset S$, and again we invoke Ordinal Subset of Ordinal is Initial Segment, and the proof is complete.
{{qed}}
Theorem: Relation between Two Ordinals

Let $S$ and $T$ be ordinals.
Then either $S \subseteq T$ or $T \subseteq S$.

Proof:
We have that $S \ne T$
Therefore, from Relation between Two Ordinals either $S \subset T$ or $T \subset S$.
By Ordering on Ordinal is Subset Relation or Transitive Set is Proper Subset of Ordinal iff Element of Ordinal, either $S \in T$ or $T \in S$.
By definition, every element of an ordinal is an initial segment; hence the result.
{{qed}}
Theorem: Isomorphic Ordinals are Equal

Let $A$ and $B$ be ordinals that are order isomorphic.
Then $A = B$.

Proof:
Let $S \cong T$.
{{AimForCont}} that $S \ne T$.
Then from {{Corollary|Relation between Two Ordinals}}, either:
:$S$ is an initial segment of $T$
or:
:$T$ is an initial segment of $S$.
But as $S \cong T$, from Well-Ordered Class is not Isomorphic to Initial Segment, neither $S$ nor $T$ can be an initial segment of the other.
From this contradiction it follows that $S = T$.
{{qed}}
Theorem: Isomorphic Ordinals are Equal

Let $A$ and $B$ be ordinals that are order isomorphic.
Then $A = B$.

Proof:
From Well-Ordered Class is not Isomorphic to Initial Segment, neither $A$ nor $B$ can be an initial segment of the other.
By definition, every element of an ordinal is an initial segment of it.
Hence, neither $A$ nor $B$ can be an element of the other.
By Ordinal Membership is Trichotomy, it follows that $A = B$.
{{qed}}
Theorem: Condition for Woset to be Isomorphic to Ordinal

Let $\struct {S, \preceq}$ be a woset.
Let $\struct {S, \preceq}$ be such that $\forall a \in S$, the initial segment $S_a$ of $S$ determined by $a$ is order isomorphic to some ordinal.
Then $\struct {S, \preceq}$ itself is order isomorphic to an ordinal.

Proof:
For each $a \in S$, let $g_a: S_a \to \map Z a$ be an order isomorphism from $S_a$ to an ordinal $\map Z a$.
By Isomorphic Ordinals are Equal and Order Isomorphism between Wosets is Unique, both $\map Z a$ and $g_a$ are unique.
So this defines a mapping $Z$ on $S$.
Let the image of $Z$ be $W$:
:$W = \set {\map Z a: a \in S}$
Now we define $f: S \to W$ as:
:$\map f a = \map Z a$
Now we show that $x, y \in S, x \prec y \implies \map Z x \subset \map Z y$.
So, let $x, y \in S$ such that $x \prec y$.
Then:
:$(1) \quad g_x: S_x \cong \map Z x$
Also, because:
{{begin-eqn}}
{{eqn | l = S_x
      | r = \set {z \in S: z \prec x}
      | c = Initial Segment of Ordinal is Ordinal
}}
{{eqn | r = \set {z \in S: z \prec y \land z \prec x}
      | c = 
}}
{{eqn | r = \set {z \in S_y: z \prec x}
      | c = 
}}
{{eqn | r = \paren {S_y}_x
      | c = 
}}
{{end-eqn}}
we have:
:$(2): \quad \paren {g_y \restriction S_x}: S_x \to \paren {\map Z y}_{\map {g_y} x}$ is an order isomorphism
So by Isomorphic Ordinals are Equal, we have:
:$(3): \quad \map Z x = \paren {\map Z y}_{\map {g_y} x}$
So, in particular:
:$\map Z x \subset \map Z y$
By this result, $f: S \to W$ is a bijection.
Also by this result, $f: S \to \struct {W, \subseteq}$ is an order isomorphism.
This means that $W$ is well-ordered by $\subseteq$.
Now we show that $W$ is an ordinal.
Let $y \in S$.
Since $\map Z y$ is an ordinal, we have:
:$x \prec y \implies \paren {\map Z y}_{\map {g_y} x} = \map {g_y} x$
So by $(3)$ above, we have:
:$(4) \quad x \prec y \implies \map Z x = \map {g_y} x$
Hence:
{{begin-eqn}}
{{eqn | l = W_{\map Z y}
      | r = \set {\map Z x: \map Z x \subset \map Z y}
      | c = 
}}
{{eqn | r = \set {\map Z x: x \prec y}
      | c = 
}}
{{eqn | r = \set {\map {g_y} x: x \prec y}
      | c = 
}}
{{eqn | r = \map {g_y} {S_y}
      | c = 
}}
{{eqn | r = \map Z y
      | c = 
}}
{{end-eqn}}
As $y$ is an arbitrary element of $S$, it follows that $\map Z y$ is an arbitrary element of $W$.
So $W$ is an ordinal, as we wanted to prove.
{{qed}}
Theorem: Counting Theorem

Every well-ordered set is order isomorphic to a unique ordinal.

Proof:
=== Existence ===
Let $\struct {S, \preceq}$ be a woset.
From Condition for Woset to be Isomorphic to Ordinal, it is enough to show that for every $a \in S$, the initial segment $S_a$ of $S$ determined by $a$ is order isomorphic to some ordinal.
Let:
:$E = \set {a \in S: S_a \text{ is not isomorphic to an ordinal} }$
We will show that $E = \O$.
{{AimForCont}} that $E \ne \O$.
Let $a$ be the minimal element of $E$.
This is bound to exist by definition of woset.
So, if $x \prec a$, it follows that $S_x$ is isomorphic to an ordinal.
But for $x \prec a$, we have $S_x = \paren {S_a}_x$ from definition of an ordinal.
So every segment of $S_a$ is isomorphic to an ordinal.
Hence from Condition for Woset to be Isomorphic to Ordinal, $S_a$ itself is isomorphic to an ordinal.
This contradicts the supposition that $a \in E$.
Hence $E = \O$ and existence has been proved.
{{qed|lemma}}
=== Uniqueness ===
Uniqueness follows from Isomorphic Ordinals are Equal.
Hence the result.
{{qed}}
Theorem: Counting Theorem

Every well-ordered set is order isomorphic to a unique ordinal.

Proof:
Let $A$ be a properly well-ordered class.
Let $\On$ denote the class of all ordinals.
By the Comparability Theorem, either:
:$A$ is order isomorphic to a lower section of $\On$
or:
:$\On$ is order isomorphic to a lower section of $A$.
Let $A$ be a set.
From Well-Ordering on Set is Proper Well-Ordering, $A$ is a properly well-ordered class.
As $A$ is a set, every lower section of $A$ is a set.
{{AimForCont}} $\On$ is order isomorphic to a lower section $L$ of $A$.
Then $\On$ would then be in one-to-one correspondence with that lower section.
Thus there would be a mapping $\phi: L \to \On$.
By the Axiom of Replacement, $\phi \sqbrk L = \On$ is then a set.
But from Class of All Ordinals is Proper Class, $\On$ is a proper class.
From this contradiction it follows that $\On$ cannot be order isomorphic to a lower section $L$ of $A$.
Hence $A$ is order isomorphic to a lower section of $\On$.
From Lower Section of Class of All Ordinals is Ordinal, that means $A$ is order isomorphic to an ordinal.
From Distinct Ordinals are not Order Isomorphic it follows that $A$ is order isomorphic to exactly one such ordinal.
{{qed}}
Theorem: Product of Incidence Matrix of BIBD with its Transpose

Let $A$ be the block incidence matrix for a BIBD with parameters $v, k, \lambda$.
Then:
:$A^\intercal \cdot A = \sqbrk {a_{ij} } = \paren {r - \lambda} I_v + \lambda J_v$
where:
:$A$ is $v \times b$
:$A^\intercal$ is the transpose of $A$
:$J_v$ is the all $v \times v$ $1$'s matrix
:$I_v$ is the $v \times v$ identity matrix.
That is:
:<nowiki>$A^\intercal \cdot A = \begin{bmatrix}
r & \lambda & \cdots & \lambda \\
\lambda & r & \cdots & \lambda \\
\vdots & \vdots & \ddots & \vdots \\
\lambda & \lambda & \cdots & r \\
\end{bmatrix}$</nowiki>

Proof:
Let row $i$ of $A$ be multiplied by column $i$ of $A^\intercal$.
This is the same as multiplying row $i$ of $A$ by row $i$ of $A$.
Each row of $A$ has $r$ entries (since any point must be in $r$ blocks).
Then:
:$\sqbrk {a_{ii} } = r = \sum $ of the all the $1$'s in row $i$
This completes the main diagonal.
Let row $i$ of $A$ be multiplied by column $j$ of $A^\intercal$.
This is the same as multiplying row $i$ of $A$ by row $j$ of $A$.
This will give the number of times point $i$ is the same block as point $j$.
Therefore:
:$i \ne j \implies \sqbrk {a_{ij} } = \lambda$
So:
:$A^\intercal \cdot A = \sqbrk {a_{ij} } = \paren {r - \lambda} I_v + \lambda J_v$
{{qed}}
{{explain|The order of this matrix is not apparent - it needs to be made more explicit.}}
{{explain|Better idea - write a separate page to express this result for a general logical matrix. I think this is a valid construction. Establish the result as the fact that it is a combinatorial matrix.}}
Category:Balanced Incomplete Block Designs
j299bb32eu24zcsy67j37n7ve0xwj5i
Theorem: Fisher's Inequality

For a BIBD $\struct {v, k, \lambda}$, the number of blocks $b$ must be greater than or equal to the number of points $v$:
:$ b \ge v$

Proof:
Let $A$ be the incidence matrix.
By Product of Incidence Matrix of BIBD with its Transpose, we have that:
:$A^\intercal \cdot A = \begin{bmatrix} 
r & \lambda & \cdots & \lambda \\
\lambda & r & \cdots & \lambda \\
\vdots & \vdots & \ddots & \vdots \\
\lambda & \lambda & \cdots & r \\
\end{bmatrix}$
From Necessary Condition for Existence of BIBD:
:$r > \lambda$
So we can write $r = \lambda + \mu$ for some $\mu > 0$ and so:
:$A^\intercal \cdot A = \begin{bmatrix} 
\lambda + \mu & \lambda & \cdots & \lambda \\
\lambda & \lambda + \mu & \cdots & \lambda \\
\vdots & \vdots & \ddots & \vdots \\
\lambda & \lambda & \cdots & \lambda + \mu \\
\end{bmatrix}$
This is a combinatorial matrix of order $v$.
So:
{{begin-eqn}}
{{eqn | l = \map \det {A^\intercal \cdot A}
      | r = \mu^{v - 1} \paren {\mu + v \lambda}
      | c = Determinant of Combinatorial Matrix
}}
{{eqn | r = \paren {r + \paren {v - 1} \lambda} \paren {r - \lambda}^{v - 1}
      | c = 
}}
{{eqn | r = r k \paren {r - \lambda}^{v - 1}
      | c = Necessary Condition for Existence of BIBD
}}
{{end-eqn}}
{{handwaving|obvious}}
Now, since $k < v$ (this is obvious) and using the properties of a BIBD, we get that $r > \lambda$. 
So $\map \det {A^\intercal A} \ne 0$.
Since $A^\intercal A$ is a $v \times v$ matrix, then the rank, $\rho$, of $A^\intercal A = v$.
Using the facts that $\map \rho {A^\intercal A} \le \map \rho A$, and $\map \rho A \le b$ (since $A$ only has $b$ columns), we have that:
:$v \le \map \rho A \le b$
{{qed}}
{{explain|Some "basic properties" here need to be linked, in order that the general thrust of this can be understood.}}
{{explain|We can probably lose most of the fiddly stuff at the end by noting that as $\lambda$ and $\mu$ are both $>0$ then so is the determinant, and go straight to the last-line result about rank.}}
{{proofread}}
{{Namedfor|Ronald Aylmer Fisher|cat = Fisher}}
Category:Balanced Incomplete Block Designs
mtnxvi5352el1ij9daoa4u22buqjnqy
Theorem: Self-Distributive Law for Conditional

The following is known as the Self-Distributive Law:
=== Formulation 1 ===
{{:Self-Distributive Law for Conditional/Formulation 1}}
=== Formulation 2 ===
{{:Self-Distributive Law for Conditional/Formulation 2}}

Proof:
{{BeginTableau|p \implies \paren {q \implies r} \vdash \paren {p \implies q} \implies \paren {p \implies r} }}
{{Premise|1|p \implies \paren {q \implies r} }}
{{Assumption|2|p \implies q}}
{{Assumption|3|p}}
{{ModusPonens|4|1, 3|q \implies r|1|3}}
{{ModusPonens|5|2, 3|q|2|3}}
{{ModusPonens|6|1, 2, 3|r|4|5}}
{{Implication|7|1, 2|p \implies r|3|6}}
{{Implication|8|1|\paren {p \implies q} \implies \paren {p \implies r}|2|7}}
{{EndTableau|qed}}
Theorem: Self-Distributive Law for Conditional

The following is known as the Self-Distributive Law:
=== Formulation 1 ===
{{:Self-Distributive Law for Conditional/Formulation 1}}
=== Formulation 2 ===
{{:Self-Distributive Law for Conditional/Formulation 2}}

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.
$\begin{array}{|ccccc||ccccccc|} \hline
p & \implies & (q & \implies & r) & (p & \implies & q) & \implies & (p & \implies & r) \\
\hline
\F & \T & \F & \T & \F & \F & \T & \F & \T & \F & \T & \F \\
\F & \T & \F & \T & \T & \F & \T & \F & \T & \F & \T & \T \\
\F & \T & \T & \F & \F & \F & \T & \T & \T & \F & \T & \F \\
\F & \T & \T & \T & \T & \F & \T & \T & \T & \F & \T & \T \\
\T & \T & \F & \T & \F & \T & \F & \F & \T & \T & \F & \F \\
\T & \T & \F & \T & \T & \T & \F & \F & \T & \T & \T & \T \\
\T & \F & \T & \F & \F & \T & \T & \T & \F & \T & \F & \F \\
\T & \T & \T & \T & \T & \T & \T & \T & \T & \T & \T & \T \\
\hline
\end{array}$
{{qed}}
Theorem: Self-Distributive Law for Conditional

The following is known as the Self-Distributive Law:
=== Formulation 1 ===
{{:Self-Distributive Law for Conditional/Formulation 1}}
=== Formulation 2 ===
{{:Self-Distributive Law for Conditional/Formulation 2}}

Proof:
{{BeginTableau|\paren {p \implies q} \implies \paren {p \implies r} \vdash p \implies \paren {q \implies r} }}
{{Premise|1|\paren {p \implies q} \implies \paren {p \implies r} }}
{{Assumption|2|p}}
{{Assumption|3|q}}
{{SequentIntro|4|3|p \implies q|3|True Statement is implied by Every Statement}}
{{ModusPonens|5|1, 3|p \implies r|1|4}}
{{ModusPonens|6|1, 2, 3|r|5|2}}
{{Implication|7|1, 2|q \implies r|3|6}}
{{Implication|8|1|p \implies \paren {q \implies r}|2|7}}
{{EndTableau}}
{{qed}}
Theorem: Self-Distributive Law for Conditional

The following is known as the Self-Distributive Law:
=== Formulation 1 ===
{{:Self-Distributive Law for Conditional/Formulation 1}}
=== Formulation 2 ===
{{:Self-Distributive Law for Conditional/Formulation 2}}

Proof:
{{BeginTableau|\paren {p \implies \paren {q \implies r} } \implies \paren {\paren {p \implies q} \implies \paren {p \implies r} } }}
{{Assumption  |1|p \implies \paren {q \implies r} }}
{{SequentIntro|2|1|\paren {p \implies q} \implies \paren {p \implies r}|1|Self-Distributive Law for Conditional: Formulation 1}}
{{Implication |3||\paren {p \implies \paren {q \implies r} } \implies \paren {\paren {p \implies q} \implies \paren {p \implies r} }|1|2}}
{{EndTableau}}
{{Qed}}
Theorem: Self-Distributive Law for Conditional

The following is known as the Self-Distributive Law:
=== Formulation 1 ===
{{:Self-Distributive Law for Conditional/Formulation 1}}
=== Formulation 2 ===
{{:Self-Distributive Law for Conditional/Formulation 2}}

Proof:
{{BeginTableau|\paren {p \implies \paren {q \implies r} } \implies \paren {\paren {p \implies q} \implies \paren {p \implies r} } }}
{{Assumption  |1|p \implies \paren {q \implies r} }}
{{Assumption  |2|p \implies q}}
{{Assumption  |3|p}}
{{ModusPonens |4|1,3|q \implies r|1|3}}
{{ModusPonens |5|2,3|q|2|3}}
{{ModusPonens |6|1,2,3|r|4|5}}
{{Implication |7|1,2|p \implies r|3|6}}
{{Implication |8|1|\paren {p \implies q} \implies \paren {p \implies r}|2|7}}
{{Implication |9||\paren {p \implies \paren {q \implies r} } \implies \paren {\paren {p \implies q} \implies \paren {p \implies r} }|1|8}}
{{EndTableau}}
{{Qed}}