avewright/proofwiki-math huggingface dataset:
Theorem: Ordered Set of All Mappings is Ordered Set

Let $L = \struct {S, \preceq}$ be an ordered set.
Let $X$ be a set.
Then $L^X$ is also an ordered set.

Proof:
By definition of ordered set of all mappings:
:$L^X = \struct {S^X, \precsim}$
where
:$\forall f, g \in S^X: f \precsim g \iff f \preceq g$
:$\preceq$ denotes the ordering on mappings,
:$S^X$ denotes the set of all mappings from $X$ into $S$.
Theorem: Closed Form for Lucas Numbers

The Lucas numbers have a closed-form solution:
:$L_n = \phi^n + \paren {1 - \phi}^n = \paren {\dfrac {1 + \sqrt 5} 2}^n + \paren {\dfrac {1 - \sqrt 5} 2}^n$
where $\phi$ is the golden mean.
Putting $\hat \phi = 1 - \phi = -\dfrac 1 \phi$ this can be written:
:$L_n = \phi^n + \hat \phi^n$

Proof:
Proof by induction:
For all $n \in \N$, let $\map P n$ be the proposition:
:$L_n = \phi^n + \paren {1 - \phi}^n = \paren {\dfrac {1 + \sqrt 5} 2})^n + \paren {\dfrac {1 - \sqrt 5} 2}^n = \phi^n + \hat \phi^n$
Theorem: Fibonacci Number 2n equals Fibonacci Number n by Lucas Number n

Let $F_n$ denote the $n$th Fibonacci number.
Let $L_n$ denote the $n$th Lucas number.
Then:
:$F_{2 n} = F_n L_n$

Proof:
Let:
:$\phi = \dfrac {1 + \sqrt 5} 2$
:$\hat \phi = \dfrac {1 - \sqrt 5} 2$
Then:
{{begin-eqn}}
{{eqn | l = F_{2 n}
      | r = \dfrac {\phi^{2 n} - \hat \phi^{2 n} } {\sqrt 5}
      | c = Euler-Binet Formula
}}
{{eqn | r = \dfrac {\paren {\phi^n + \hat \phi^n} \paren {\phi^n - \hat \phi^n} } {\sqrt 5}
      | c = Difference of Two Squares
}}
{{eqn | r = L_n \dfrac {\phi^n - \hat \phi^n} {\sqrt 5}
      | c = Closed Form for Lucas Numbers
}}
{{eqn | r = L_n F_n
      | c = Euler-Binet Formula
}}
{{end-eqn}}
{{qed}}
Theorem: Fibonacci Number 3n in terms of Fibonacci Number n and Lucas Number 2n

Let $F_n$ denote the $n$th Fibonacci number.
Let $L_n$ denote the $n$th Lucas number.
Then:
:$F_{3 n} = F_n \paren {L_{2 n} + \paren {-1}^n}$

Proof:
Let:
:$\phi = \dfrac {1 + \sqrt 5} 2$
:$\hat \phi = \dfrac {1 - \sqrt 5} 2$
Then:
{{begin-eqn}}
{{eqn | l = F_{3 n}
      | r = \dfrac {\phi^{3 n} - \hat \phi^{3 n} } {\sqrt 5}
      | c = Euler-Binet Formula
}}
{{eqn | r = \dfrac {\paren {\phi^n - \hat \phi^n} \paren {\phi^{2 n} + \phi^n \hat \phi^n + \hat \phi^{2 n} } } {\sqrt 5}
      | c = Difference of Two Cubes
}}
{{eqn | r = F_n \paren {\phi^{2 n} + \phi^n \hat \phi^n + \hat \phi^{2 n} }
      | c = Euler-Binet Formula
}}
{{eqn | r = F_n \paren {L_{2 n} + \phi^n \hat \phi^n}
      | c = Closed Form for Lucas Numbers
}}
{{end-eqn}}
Then we note:
{{begin-eqn}}
{{eqn | l = \phi \hat \phi
      | r = \dfrac {1 + \sqrt 5} 2 \dfrac {1 - \sqrt 5} 2
      | c = 
}}
{{eqn | r = \dfrac {1 - 5} 4
      | c = Difference of Two Squares
}}
{{eqn | r = -1
      | c = 
}}
{{end-eqn}}
The result follows.
{{qed}}
Theorem: Relation between Square of Fibonacci Number and Square of Lucas Number

Let $F_n$ denote the $n$th Fibonacci number.
Let $L_n$ denote the $n$th Lucas number.
Then:
:$5 {F_n}^2 + 4 \paren {-1}^n = {L_n}^2$

Proof:
Let:
:$\phi = \dfrac {1 + \sqrt 5} 2$
:$\hat \phi = \dfrac {1 - \sqrt 5} 2$
Note that we have:
{{begin-eqn}}
{{eqn | l = \phi \hat \phi
      | r = \dfrac {1 + \sqrt 5} 2 \dfrac {1 - \sqrt 5} 2
      | c = 
}}
{{eqn | r = \dfrac {1 - 5} 4
      | c = Difference of Two Squares
}}
{{eqn | r = -1
      | c = 
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = 5 {F_n}^2
      | r = 5 \paren {\dfrac {\phi^n - \hat \phi^n} {\sqrt 5} }^2
      | c = Euler-Binet Formula
}}
{{eqn | r = \phi^{2 n} - 2 \phi^n \hat \phi^n + \hat \phi^{2 n}
      | c = simplifying
}}
{{eqn | r = \phi^{2 n} + 2 \phi^n \hat \phi^n + \hat \phi^{2 n} - 4 \paren {\phi \hat \phi}^n 
      | c = 
}}
{{eqn | r = \paren {\phi^n + \hat \phi^n}^2 - 4 \paren {-1}^n
      | c = simplifying, and from above: $\phi \hat \phi = -1$
}}
{{eqn | r = {L_n}^2 - 4 \paren {-1}^n
      | c = Closed Form for Lucas Numbers
}}
{{end-eqn}}
Hence the result.
{{qed}}
Theorem: Lucas Number 2n in terms of Square of Lucas Number n

Let $L_n$ denote the $n$th Lucas number.
Then:
:$L_{2 n} = {L_n}^2 + 2 \left({-1}\right)^n$

Proof:
Let:
:$\phi = \dfrac {1 + \sqrt 5} 2$
:$\hat \phi = \dfrac {1 - \sqrt 5} 2$
Note that we have:
{{begin-eqn}}
{{eqn | l = \phi \hat \phi
      | r = \dfrac {1 + \sqrt 5} 2 \dfrac {1 - \sqrt 5} 2
      | c = 
}}
{{eqn | r = \dfrac {1 - 5} 4
      | c = Difference of Two Squares
}}
{{eqn | r = -1
      | c = 
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = L_{2 n}
      | r = \phi^{2 n} + \hat \phi^{2 n}
      | c = Closed Form for Lucas Numbers
}}
{{eqn | r = \phi^{2 n} + 2 \phi^n \hat \phi^n + \hat \phi^{2 n} - 2 \phi^n \hat \phi^n
      | c = 
}}
{{eqn | r = \paren {\phi^n + \hat \phi^n}^2 - 2 \paren {-1}^n
      | c = simplifying, and from above: $\phi \hat \phi = -1$
}}
{{eqn | r = {L_n}^2 - 2 \paren {-1}^n
      | c = Closed Form for Lucas Numbers
}}
{{end-eqn}}
Hence the result.
{{qed}}
Theorem: 12 Pentominoes

There exist $12$ distinct free pentominoes:
:600px

Proof:
{{ProofWanted|Work to be done yet to establish method of creation}}
Theorem: 18 Fixed Pentominoes

There exist $18$ distinct fixed pentominoes:
:600px

Proof:
{{ProofWanted|Work to be done yet to establish method of creation}}
Theorem: Differential of Differentiable Functional is Unique

The differential of a differentiable functional is unique.

Proof:
=== Lemma ===
{{:Differential of Differentiable Functional is Unique/Lemma}}{{qed|lemma}}
Let $J \sqbrk y$ be a differentiable functional.
Suppose the differential of $J \sqbrk y$ is not uniquely defined. 
Then at least $2$ different forms of this exist:
:$\Delta J \sqbrk {y; h} = \phi_1 \sqbrk {y; h} + \epsilon_1 \size h$
:$\Delta J \sqbrk {y; h} = \phi_2 \sqbrk {y; h} + \epsilon_2 \size h$
where:
:$\phi_1 \sqbrk {y; h}$ and $\phi_2 \sqbrk {y; h}$ are linear functionals
:$\epsilon_1, \epsilon_2 \to 0$ as $\size h \to 0$.
Because $\size h \to 0$, the whole expression $\to 0$ faster than $\size h$ due to constraints on $\epsilon$.
{{handwaving|the above statement}}
Hence $\epsilon_1 \size h$, $\epsilon_2 \size h$ and $\paren {\epsilon_1 - \epsilon_2} \size h$ are all infinitesimals of order higher than $1$ relative to $\size h$.
Consequently, subtraction of one from the other leads to:
:$(1): \quad \phi_1 \sqbrk {y; h} - \phi_2 \sqbrk {y; h} = \paren {\epsilon_2 - \epsilon_1} \size h$
Therefore, $\phi_1 \sqbrk {y; h} - \phi_2 \sqbrk {y; h}$ is an infinitesimal of order higher than $1$ relative to $\size h$.
Since each of the members are linear functionals, the whole term keeps the same property. 
Rearranging the terms of $(1)$:
:$\dfrac {\phi_1 \sqbrk {y; h} - \phi_2 \sqbrk {y; h} } {\size h} = \epsilon_2 - \epsilon_1$
Taking the limit and recalling the constraint on both $\epsilon_1$ and $\epsilon_2$:
{{explain|"recalling the constraint on both $\epsilon_1$ and $\epsilon_2$" needs to be more explicit}}
:$\ds \lim_{\size h \mathop \to 0} \frac {\phi_1 \sqbrk {y; h} - \phi_2 \sqbrk {y; h} } {\size h} = 0$
The given limit with the linearity of the term in the numerator allows us to use the lemma. 
{{explain|Demonstrate linearity}}
Hence:
:$\phi_1 \sqbrk {y; h} - \phi_2 \sqbrk {y; h} = 0$
That is, there is only one form the differential of a differentiable functional can take.
{{qed}}
Theorem: Ordered Subset of Ordered Set is Ordered Set

Let $L = \struct {S, \preceq}$ be an ordered set.
Let $\struct {S', \preceq'}$ be an ordered subset of $L$.
Then $\struct {S', \preceq'}$ is an ordered set.

Proof:
By definition of ordered subset:
:$S' \subseteq S$
Theorem: Finite Subsets form Ideal

Let $X$ be a set.
Let $\map {\operatorname {Fin} } X$ be the set of all finite subsets of $X$.
Then $\map {\operatorname {Fin} } X$ is ideal in $\struct {\powerset X, \subseteq}$
where $\powerset X$ denotes the power set of $X$.

Proof:
=== Non-Empty ===
By Empty Set is Subset of All Sets:
:$\O \subseteq X$ and $\O$ is finite.
By definition of $\operatorname {Fin}$:
:$\O \in \map {\operatorname {Fin} } X$
Thus by definition:
:$\map {\operatorname {Fin} } X$ is non-empty.
{{qed|lemma}}
Theorem: Product of Proper Divisors

Let $n$ be an integer such that $n \ge 1$.
Let $\map P n$ denote the product of the proper divisors of $n$.
Then:
:$\map P n = n^{\map {\sigma_0} n / 2 - 1}$
where $\map {\sigma_0} n$ denotes the divisor count Function of $n$.

Proof:
Let $\map D n$ denote the product of '''all''' the divisors of $n$.
From Product of Divisors:
:$\map D n = n^{\map {\sigma_0} n / 2}$
The proper divisors of $n$ are defined as being the divisors of $n$ excluding $n$ itself.
Thus:
:$\map P n = \dfrac {\map D n} n = \dfrac {n^{\map {\sigma_0} n / 2} } n = n^{\map {\sigma_0} n / 2 - 1}$
{{qed}}
Theorem: Product of Divisors

Let $n$ be an integer such that $n \ge 1$.
Let $\map D n$ denote the product of the divisors of $n$.
Then:
:$\map D n = n^{\map {\sigma_0} n / 2}$
where $\map {\sigma_0} n$ denotes the divisor count function of $n$.

Proof:
We have by definition that:
:$\map D n = \ds \prod_{d \mathop \divides n} d$
Also by definition, $\map {\sigma_0} n$ is the number of divisors of $n$.
Suppose $n$ is not a square number.
Let $p \divides n$, where $\divides$ denotes divisibility.
Then:
:$\exists q \divides n : p q = n$
Thus the divisors of $n$ come in pairs whose product is $n$.
From Divisor Count Function is Odd Iff Argument is Square, $\map {\sigma_0} n$ is even.
Thus $\dfrac {\map {\sigma_0} n} 2$ is an integer.
Thus there are exactly $\dfrac {\map {\sigma_0} n} 2$ pairs of divisors of $n$ whose product is $n$.
Thus the product of the divisors of $n$ is:
:$\ds \prod_{d \mathop \divides n} d = n^{\map {\sigma_0} n / 2}$
Now suppose $n$ is square such that $n = r^2$.
Then from Divisor Count Function is Odd Iff Argument is Square, $\map {\sigma_0} n$ is odd.
Hence the number of divisors of $n$ not including $r$ is $\map {\sigma_0} n - 1$.
As before, these exist in pairs whose product is $n$.
Thus:
{{begin-eqn}}
{{eqn | l = \prod_{d \mathop \divides n} d
      | r = r \times n^{\paren {\map {\sigma_0} n - 1} / 2}
      | c = 
}}
{{eqn | r = r \times n^{\map {\sigma_0} n / 2 - 1 / 2}
      | c = 
}}
{{eqn | r = n^{\map {\sigma_0} n / 2 - 1 / 2 + 1 / 2}
      | c = as $r = n^{1/2}$
}}
{{eqn | r = n^{\map {\sigma_0} n / 2}
      | c = 
}}
{{end-eqn}}
Hence the result.
{{qed}}
Theorem: Product of Divisors

Let $n$ be an integer such that $n \ge 1$.
Let $\map D n$ denote the product of the divisors of $n$.
Then:
:$\map D n = n^{\map {\sigma_0} n / 2}$
where $\map {\sigma_0} n$ denotes the divisor count function of $n$.

Proof:
{{begin-eqn}}
{{eqn | l = \map D n^2
      | r = \paren {\prod_{d \mathop \divides n} d}^2
      | c = by definition
}}
{{eqn | r = \paren {\prod_{d \mathop \divides n} d} \paren {\prod_{d \mathop \divides n} \dfrac n d}
      | c = Sum Over Divisors Equals Sum Over Quotients
}}
{{eqn | r = \prod_{d \mathop \divides n} n
      | c = combining the two products
}}
{{eqn | r = n^{\map {\sigma_0} n}
      | c = {{Defof|Divisor Count Function}}
}}
{{end-eqn}}
The result follows by taking the (positive) square root of both sides.
{{qed}}
Theorem: Square of Reversal of Small-Digit Number

Let $n$ be an integer whose decimal representation consists of sufficiently small digits.
Then the reversal of the square of $n$ is the square of the reversal of $n$.

Proof:
{{ProofWanted|Needs to be formulated properly.}}
Theorem: 12 times Divisor Sum of 12 equals 14 times Divisor Sum of 14

$x = 12$ and $y = 14$ are solutions to the indeterminate equation:
:$x \, \map {\sigma_1} x = y \, \map {\sigma_1} y$
where $\sigma_1$ denotes the divisor sum function.

Proof:
{{begin-eqn}}
{{eqn | l = 12 \, \map {\sigma_1} {12}
      | r = 12 \times 28
      | c = {{DSFLink|12}}
}}
{{eqn | r = 12 \times 2 \times 14
      | c = 
}}
{{eqn | r = 14 \times 24
      | c = 
}}
{{eqn | r = 14 \, \map {\sigma_1} {14}
      | c = {{DSFLink|14}}
}}
{{end-eqn}}
{{qed}}
Theorem: 12 Knights to Attack or Occupy All Squares on Chessboard

On a standard chessboard, a minimum of $12$ knights are needed to ensure all squares are either occupied or under attack.

Proof:
First, we show that fewer than $12$ knights are not enough to occupy or attack each square.
Consider the $12$ squares $\text a 1$, $\text a 2$, $\text b 2$, $\text a 8$, $\text b 8$, $\text b 7$, $\text h 8$, $\text h 7$, $\text g 7$, $\text h 1$, $\text g 1$ and $\text g 2$.
{{ChessDiagram|
|
|xx|xx|  |  |  |  |  |xx 
|  |xx|  |  |  |  |xx|xx
|  |  |  |  |  |  |  |   
|  |  |  |  |  |  |  |  
|  |  |  |  |  |  |  |  
|  |  |  |  |  |  |  |  
|xx|xx|  |  |  |  |xx|   
|xx|  |  |  |  |  |xx|xx
|
}}
No knight can occupy or attack more than one of these squares.
Hence, fewer than $12$ knights are not sufficient.
{{qed|lemma}}
Second, we show that $12$ knights can be placed to ensure all squares are either occupied or under attack.
For example, with the knights being on $\text b 3$, $\text c 3$, $\text c 4$, $\text c 7$, $\text c 6$, $\text d 6$, $\text g 6$, $\text f 6$, $\text f 5$, $\text f 2$, $\text f 3$ and $\text e 3$, all squares are either occupied or under attack:
{{ChessDiagram|
|
|  |  |  |  |  |  |  |   
|  |  |nd|  |  |  |  |  
|  |  |nd|nd|  |nd|nd|   
|  |  |  |  |  |nd|  |  
|  |  |nd|  |  |  |  |  
|  |nd|nd|  |nd|nd|  |  
|  |  |  |  |  |nd|  |   
|  |  |  |  |  |  |  |  
|
}}
{{qed}}
Theorem: Prime Number is Deficient

Let $p$ be a prime number.
Then $p$ is deficient.

Proof:
A specific instance of Power of Prime is Deficient.
{{qed}}
Theorem: Prime Number is Deficient

Let $p$ be a prime number.
Then $p$ is deficient.

Proof:
Let $p$ be a prime number.
From Divisor Sum of Prime Number:
:$\map {\sigma_1} p = p + 1$
and so:
:$\dfrac {\map {\sigma_1} p} p = \dfrac {p + 1} p = 1 + \dfrac 1 p$
As $p > 1$ it follows that $\dfrac 1 p < 1$.
Hence:
:$\dfrac {\map {\sigma_1} p} p < 2$
The result follows by definition of deficient.
{{qed}}
Theorem: Ordering on Closure Operators iff Images are Including

Let $L = \struct {S, \vee, \wedge, \preceq}$ be a complete lattice.
Let $f, g: S \to S$ be closure operators on $L$.
Then:
:$f \preceq g$ {{iff}} $g \sqbrk S \subseteq f \sqbrk S$
where
:$\preceq$ denotes the ordering on mappings
:$f \sqbrk S$ denotes the image of $f$.

Proof:
=== Sufficient Condition ===
Let $f \preceq g$.
Let $x \in g \sqbrk S$
By definition of image of mapping:
:$\exists y \in S: \map g y = x$
By definition of closure operator/idempotent:
:$\map g {\map g y} = \map g y$
By definition of ordering on mappings:
:$\map f {\map g y} \preceq \map g {\map g y}$
By definition of closure operator/inflationary:
:$\map g y \preceq \map f {\map g y}$
By definition of antisymmetry:
:$\map f x = x$
Thus by definition of image of mapping:
:$x \in f \sqbrk S$
{{qed|lemma}}
Theorem: Multiple of Perfect Number is Abundant

Let $n$ be a perfect number.
Let $m$ be a positive integer such that $m > 1$.
Then $m n$ is abundant.

Proof:
We have by definition of divisor sum function and perfect number that:
:$\dfrac {\map {\sigma_1} n} n = 2$
But from Abundancy Index of Product is greater than Abundancy Index of Proper Factors:
:$\dfrac {\map {\sigma_1} {m n} } {m n} > 2$
Hence the result by definition of abundant.
{{qed}}
Theorem: Multiple of Abundant Number is Abundant

Let $n$ be an abundant number.
Let $m$ be a positive integer such that $m > 1$.
Then $m n$ is abundant.

Proof:
We have by definition of divisor sum function and abundant number that:
:$\dfrac {\map {\sigma_1} n} n > 2$
But from Abundancy Index of Product is greater than Abundancy Index of Proper Factors:
:$\dfrac {\map {\sigma_1} {m n} } {m n} > 2$
Hence the result by definition of abundant.
{{qed}}
Theorem: Divisor of Perfect Number is Deficient

Let $n$ be a perfect number.
Let $n = r s$ where $r$ and $s$ are positive integers such that $r > 1$ and $s > 1$.
Then $r$ and $s$ are both deficient.

Proof:
{{WLOG}}, consider $r$.
We have by definition of divisor sum function and perfect number that:
:$\dfrac {\map {\sigma_1} {r s} } {r s} = 2$
But from Abundancy Index of Product is greater than Abundancy Index of Proper Factors:
:$\dfrac {\map {\sigma_1} {r s} } {r s} > \dfrac {\map {\sigma_1} r} r$
That is:
:$\dfrac {\map {\sigma_1} r} r < 2$
And {{mutatis}}:
:$\dfrac {\map {\sigma_1} r} s < 2$
Hence the result by definition of deficient.
{{qed}}
Theorem: Divisor of Deficient Number is Deficient

Let $n$ be a perfect number.
Let $n = k d$ where $r$ is a positive integer.
Then $k$ is deficient.

Proof:
We have by definition of divisor sum function and perfect number that:
:$\dfrac {\map {\sigma_1} {k d} } {k d} < 2$
But from Abundancy Index of Product is greater than Abundancy Index of Proper Factors:
:$\dfrac {\map {\sigma_1} {k d} } {k d} > \dfrac {\map {\sigma_1} k} k$
That is:
:$\dfrac {\map {\sigma_1} k} k < 2$
Hence the result by definition of deficient.
{{qed}}
Theorem: Abundancy Index of Product is greater than Abundancy Index of Proper Factors

Let $n \in \Z_{>0}$ be a composite number such that $n = r s$, where $r, s \in \Z_{>1}$.
Then:
:$\dfrac {\map {\sigma_1} n} n > \dfrac {\map {\sigma_1} r} r$
and consequently also:
:$\dfrac {\map {\sigma_1} n} n > \dfrac {\map {\sigma_1} s} s$
where $\sigma_1$ denotes the divisor sum function.
That is, the abundancy index of a composite number is strictly greater than the abundancy index of its proper divisors.

Proof:
Consider the divisors of $r$.
Let $d \divides r$, where $\divides$ indicates divisibility.
We have that:
:$d \divides n$
and also that:
:$d s \divides n$
Thus:
{{begin-eqn}}
{{eqn | l = \map {\sigma_1} r
      | r = \sum_{d \mathop \divides r} d
      | c = 
}}
{{eqn | ll= \leadsto
      | l = s \map {\sigma_1} r
      | r = \sum_{d s \mathop \divides n} d s
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \map {\sigma_1} n
      | o = >
      | r = \sum_{d s \mathop \divides n} d s
      | c = as numbers of the form $d s$ do not exhaust divisors of $n$: note $1 \divides n$
}}
{{eqn | ll= \leadsto
      | l = \dfrac {\map {\sigma_1} n} n
      | o = >
      | r = \dfrac {s \map {\sigma_1} r} n
      | c = 
}}
{{eqn | r = \dfrac {\map {\sigma_1} r} {n / s}
      | c = 
}}
{{eqn | r = \dfrac {\map {\sigma_1} r} r
      | c = 
}}
{{end-eqn}}
Similarly for $s$.
{{qed}}
Category:Abundancy
5xxnxpt27r4hnkzeh4826bwubefra3n
Theorem: Condition for Differentiable Functional to have Extremum

Let $S$ be a set of mappings.
Let $y, h \in S: \R \to \R$ be real functions. 
Let $J \sqbrk y: S \to \R$ be a differentiable functional.
Then a necessary condition for the differentiable functional $J \sqbrk {y; h}$ to have an extremum for $y = \hat y$ is:
:$\bigvalueat {\delta J \sqbrk {y; h} } {y \mathop = \hat y} = 0$

Proof:
Suppose $J \sqbrk {y; h}$ attains a minimum for $y = \hat y$.
Then:
:$\Delta J \sqbrk {\hat y; h} \ge 0$
By definition of the differentiable functional:
:$\Delta J \sqbrk {y; h} = \delta J \sqbrk {y; h} + \epsilon \size h$
where:
:$\ds \lim_{\size h \mathop \to 0} \epsilon = 0$
Hence, there exists $\size h$ small enough, that signs of $\Delta J \sqbrk {y; h}$ and $\delta J \sqbrk {y; h}$ match.
Therefore, for $\size h$ small enough it holds that $\delta J \sqbrk {\hat y; h}\ge 0$.
{{AimForCont}} $\delta J \sqbrk {y; h_0} \ne 0$ for some $h_0 \in S$.
Then: 
:$\forall \alpha > 0: \delta J \sqbrk {y; -\alpha h_0} = -\delta J \sqbrk {y; \alpha h_0}$
Therefore, for $\size h$ however small, $\Delta J \sqbrk {y; h}$ can be made to have any sign.
{{Explain|Clarify the part with $\alpha$}}
However, by assumption that $J \sqbrk {y; h}$ has a minimum, for sufficiently small $\size h$ it holds that:
:$\Delta J \sqbrk {\hat y; h} = J \sqbrk {\hat y + h} - J \sqbrk {\hat y} \ge 0$
This automatically fixes the sign of $\delta J \sqbrk {y; h}$.
This is in contradiction with the previous statement.
Hence for all $h$ it holds that:
:$\bigvalueat {\delta J \sqbrk {y; h} } {y \mathop = \hat y} = 0$ 
{{qed}}
Theorem: Superabundant Numbers are Infinite in Number

There are infinitely many superabundant numbers.

Proof:
{{AimForCont}} the set $S$ of superabundant numbers is finite.
Let $m$ be the greatest element of $S$.
By definition of superabundant, $m$ has the largest abundancy index of all the elements of $S$.
Consider the integer $2 m$.
From Abundancy Index of Product is greater than Abundancy Index of Proper Factors, $2 m$ has a higher abundancy index than $m$.
There are two possibilities:
:$(1): \quad 2 m$ is the smallest integer greater that $n$ which has a higher abundancy index than $m$.
By definition, that would make $m$ superabundant.
:$(2) \quad$ There exists a finite set $T := \set {n \in \Z: m < n < 2 m: \map A n > \map A m}$, where $\map A n$ denotes the abundancy index of $n$.
The smallest element $t$ of $T$ therefore has an abundancy index greater than all smaller positive integers.
Thus by definition $t$ is superabundant.
In either case, there exists a superabundant number not in $S$.
Thus $S$ cannot contain all superabundant numbers.
But this contradicts our initial assumption that the set $S$, containing all superabundant numbers is finite.
It follows by Proof by Contradiction that $S$ is infinite.
Hence the result.
{{qed}}
Theorem: Ordered Set of Closure Systems is Ordered Set

Let $L = \left({S, \preceq}\right)$ be an ordered set.
Then $\operatorname{ClSystems}\left({L}\right)$ is an ordered set,
where $\operatorname{ClSystems}\left({L}\right)$ denotes the ordered set of closure systems.

Proof:
By definition of ordered set of closure systems:
:$\operatorname{ClSystems}\left({L}\right) = \left({X, \precsim}\right)$
where
:$X$ is the set of all closure systems of $L$,
:dor all closure systems $s_1 = \left({T_1, \preceq_1}\right), s_2 = \left({T_2, \preceq_2}\right)$ of $L$: $s_1 \precsim s_2 \iff T_1 \subseteq T_2$
Theorem: Necessary Condition for Integral Functional to have Extremum for given function

Let $S$ be a set of real mappings such that:
:$S = \set {\map y x: \paren {y: S_1 \subseteq \R \to S_2 \subseteq \R}, \paren {\map y x \in C^1 \closedint a b}, \paren {\map y a = A, \map y b = B} }$
Let $J \sqbrk y: S \to S_3 \subseteq \R$ be a functional of the form:
:$\ds \int_a^b \map F {x, y, y'} \rd x$ 
Then a necessary condition for $J \sqbrk y$ to have an extremum (strong or weak) for a given function $\map y x$ is that $\map y x$ satisfy Euler's equation:
:$F_y - \dfrac \d {\d x} F_{y'} = 0$

Proof:
From Condition for Differentiable Functional to have Extremum we have
:$\delta J \sqbrk {y; h} \bigg \rvert_{y = \hat y} = 0$
The variation exists if $J$ is a differentiable functional.
The endpoints of $\map y x$ are fixed.
Hence:
:$\map h a = 0$
:$\map h b = 0$.
From the definition of increment of a functional:
{{begin-eqn}}
{{eqn | l = \Delta J \sqbrk {y; h}
      | r = J \sqbrk {y + h} - J \sqbrk y
      | c = definition
}}
{{eqn | r = \int_a^b \map F {x, y + h, y' + h'} \rd x - \int_a^b \map F {x, y, y'} \rd x
      | c = form of considered functional
}}
{{eqn | r = \int_a^b \paren {\map F {x, y + h, y' + h'} - \map F {x, y, y'} } \rd x
      | c = bringing under the same integral
}}
{{end-eqn}}
Using multivariate Taylor's Theorem, expand $\map F {x, y + h, y' + h'}$ {{WRT}} $h$ and $h'$:
:$\map F {x, y + h, y' + h'} = \bigvalueat {\map F {x, y + h, y' + h'} } {h \mathop = 0, \, h' \mathop = 0} + \valueat {\dfrac {\partial {\map F {x, y + h, y' + h'} } } {\partial y} } {h \mathop = 0, \, h' \mathop = 0} h + \valueat {\dfrac {\partial {\map F {x, y + h, y'+ h'} } } {\partial y'} } {h \mathop = 0, \, h' \mathop = 0} h' + \map \OO {h^2, h h', h'^2}$
Substitute this back into the integral: 
:$\ds \Delta J \sqbrk {y; h} = \int_a^b \paren {\map F {x, y, y'}_y h + \map F {x, y, y'}_{y'} h' + \map \OO {h^2, h h', h'^2} } \rd x$
Terms in $\map \OO {h^2, h'^2}$ represent terms of order higher than 1 with respect to $h$ and $h'$.
Suppose we expand $\ds \int_a^b \map \OO {h^2, h h', h'^2} \rd x$. 
Every term in this expansion will be of the form:
:$\ds \int_a^b \map A {m, n} \frac {\partial^{m + n} \map F {x, y, y'} } {\partial y^m \partial y'^n} h^m h'^n \rd x$
where $m, n \in \N: m + n \ge 2$
{{Explain|How to convert powers of $h'$ into $h$ ? Integration by parts is the only obvious candidate, but how precisely? Also check, if this is necessary}}
By definition, the integral not counting in $\map \OO {h^2, h h', h'^2}$ is a variation of functional:
:$\ds \delta J \sqbrk {y; h} = \int_a^b \paren {F_y h + F_{y'} h'} \rd x$
Use lemma. 
Then for any $\map h x$ variation vanishes if:
:$F_y - \dfrac \d {\d x} F_{y'} = 0$
{{qed}}