hcju/proofwikiqa huggingface dataset:
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}}
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}}
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}}
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}}
=== 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}} :$\displaystyle \lim_{\size h \mathop \to 0} \frac {\phi_1 \sqbrk {y; h} - \phi_2 \sqbrk {y; h} } {\size h} = 0$ 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}}
By definition of ordered subset: :$S' \subseteq S$ === Reflexivity === Let $x \in S'$. By definition of subset: :$x \in S$ By definition of reflexivity: :$x \preceq x$ Thus by definition of ordered subset: :$x \preceq' x$ {{qed|lemma}} === Transitivity === Let $x, y, z \in S'$ such that: :$x \preceq' y$ and $y \preceq' z$ By definition of ordered subset: :$x \preceq  y$ and $y \preceq z$ By definition of transitivity: :$x \preceq z$ Thus by definition of ordered subset: :$x \preceq' z$ {{qed|lemma}} === Antisymmetry === Let $x, y \in S'$ such that: :$x \preceq' y$ and $y \preceq' x$ By definition of ordered subset: :$x \preceq  y$ and $y \preceq x$ Thus by definition of antisymmetry: :$x = y$ {{qed}}
=== Non-Empty === By Empty Set is Subset of All Sets: :$\varnothing \subseteq X$ abd $\varnothing$ is finite. By definition of $\mathit{Fin}$: :$\varnothing \in \mathit{Fin}\left({X}\right)$ Thus by definition: :$\mathit{Fin}\left({X}\right)$ is non-empty. {{qed|lemma}} === Directed === Yhis follows from Finite Subsets form Directed Set. {{qed|lemma}} === Lower === Let $x \in \mathit{Fin}\left({X}\right)$, $y \in \mathcal P\left({X}\right)$ such that :$y \subseteq x$ By definition of $\mathit{Fin}$: :$x$ is a finite set. By Subset of Finite Set is Finite: :$y$ is a finite set. By definition of power set: :$y \subseteq X$ Thus by definition of $\mathit{Fin}$: :$y \in \mathit{Fin}\left({X}\right)$ {{qed|lemma}} Hence $\mathit{Fin}\left({X}\right)$ is ideal in $\left({\mathcal P\left({X}\right), \subseteq}\right)$ {{qed}}
Let $\map D n$ denote the product of '''all''' the divisors of $n$. From Product of Divisors: :$\map D n = n^{\map \tau n / 2}$ The proper divisors of $n$ are defined as being the divisors of $n$ excluding $n$ itself. Thus: :$\map P n = \dfrac {\map D n} n = \dfrac {n^{\map \tau n / 2} } n = n^{\map \tau n / 2 - 1}$ {{qed}}
Let $p$ be a prime number. From Sigma Function of Prime Number: :$\sigma \left({p}\right) = p + 1$ and so: :$\dfrac {\sigma \left({p}\right)} p = \dfrac {p + 1} p = 1 + \dfrac 1 p$ As $p > 1$ it follows that $\dfrac 1 p < 1$. Hence: :$\dfrac {\sigma \left({p}\right)} p < 2$ The result follows by definition of deficient. {{qed}}
=== 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}} === Necessary Condition === Let $g \sqbrk S \subseteq f \sqbrk S$. Let $x \in S$. By definition of image of mapping: :$\map g x \in g \sqbrk S$ By definition of subset: :$\map g x \in f \sqbrk S$ By definition of image of mapping: :$\exists a \in S: \map f a = \map g x$ By definition of closure operator/idempotent: :$\map f {\map g x} = \map g x$ By definition of closure operator/inflationary: :$x \preceq \map g x$ Thus by definition of closure operator/increasing: :$\map f x \preceq \map g x$ {{qed}}
We have by definition of $\sigma$ function and perfect number that: :$\dfrac {\map \sigma n} n = 2$ But from Abundancy Index of Product is greater than Abundancy Index of Proper Factors: :$\dfrac {\map \sigma {m n} } {m n} > 2$ Hence the result by definition of abundant. {{qed}}
We have by definition of $\sigma$ function and perfect number that: :$\dfrac {\map \sigma n} n > 2$ But from Abundancy Index of Product is greater than Abundancy Index of Proper Factors: :$\dfrac {\map \sigma {m n} } {m n} > 2$ Hence the result by definition of abundant. {{qed}}
{{WLOG}}, consider $r$. We have by definition of $\sigma$ function and perfect number that: :$\dfrac {\map \sigma {r s} } {r s} = 2$ But from Abundancy Index of Product is greater than Abundancy Index of Proper Factors: :$\dfrac {\map \sigma {r s} } {r s} > \dfrac {\map \sigma r} r$ That is: :$\dfrac {\map \sigma r} r < 2$ Mutatis mutandis: :$\dfrac {\map \sigma r} s < 2$ Hence the result by definition of deficient. {{qed}}
We have by definition of $\sigma$ function and perfect number that: :$\dfrac {\map \sigma {k d} } {k d} < 2$ But from Abundancy Index of Product is greater than Abundancy Index of Proper Factors: :$\dfrac {\map \sigma {k d} } {k d} > \dfrac {\map \sigma k} k$ That is: :$\dfrac {\map \sigma k} k < 2$ Hence the result by definition of deficient. {{qed}}
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 r       | r = \sum_{d \mathop \divides r} d       | c =  }} {{eqn | ll= \leadsto       | l = s \map \sigma r       | r = \sum_{d s \mathop \divides n} d s       | c =  }} {{eqn | ll= \leadsto       | l = \map \sigma 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 n} n       | o = >       | r = \dfrac {s \map \sigma r} n       | c =  }} {{eqn | r = \dfrac {\map \sigma r} {n / s}       | c =  }} {{eqn | r = \dfrac {\map \sigma r} r       | c =  }} {{end-eqn}} Similarly for $s$. {{qed}} Category:Abundancy 3h0co61bs51yg70qn6eko5l4o898lmh
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:   :$\displaystyle \lim_{\size h \mathop \to 0} \epsilon = 0$ 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}}
{{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}}
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 $L$: $s_1 \precsim s_2 \iff T_1 \subseteq T_2$ === Reflexivity === Let $s = \left({T, \preceq'}\right)$ be a closure system. By Set is Subset of Itself: :$T \subseteq T$ By definition of $\precsim$: :$s \precsim s$ Thus by definition :$\precsim$ is reflexive. {{qed|lemma}} === Transitivity === Let $s_1 = \left({T_1, \preceq_1}\right)$, $s_2 = \left({T_2, \preceq_2}\right)$, $s_3 = \left({T_3, \preceq_3}\right)$ be closure systems such that :$s_1 \precsim s_2$ and $s_2 \precsim s_3$ By definition of $\precsim$: :$T_1 \subseteq T_2$ and $T_2 \subseteq T_3$ By Subset Relation is Transitive: :$T_1 \subseteq T_3$ By definition of $\precsim$: :$s_1 \precsim s_3$ Thus by definition: :$\precsim$ is transitive. {{qed|lemma}} === Antisymmetry === Let $s_1 = \left({T_1, \preceq_1}\right)$, $s_2 = \left({T_2, \preceq_2}\right)$ be closure systems such that :$s_1 \precsim s_2$ and $s_2 \precsim s_1$ By definition of $\precsim$: :$T_1 \subseteq T_2$ and $T_2 \subseteq T_1$ By definition of set equality: :$T_1 = T_2$ By definition of ordered subset: :$\mathord\preceq_1 = \mathord\preceq \cap \left({T_1 \times T_1}\right) = \mathord\preceq_2$ Then :$s_1 = s_2$ Thus by definition: :$\precsim$ is antisymmetric {{qed|lemma}} So, by definition: :$\precsim$ is ordering. Hence $\operatorname{ClSystems}\left({L}\right)$ is an ordered set, {{qed}}
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 $h'$: :$\map F {x, y + h, y' + h'} = \bigvalueat {\map F {x, y + h, y' + h'} } {h \mathop = 0, \, h' \mathop = 0} + \valueat {\dfrac {\partial {\map F {x, y + h, y' + h'} } } {\partial y} } {h \mathop = 0, \, h' \mathop = 0} h + \bigvalueat {\dfrac {\partial {\map F {x, y + h, y'+ h'} } } {\partial y'} } {h \mathop = 0, \, h' \mathop = 0} h' + \map \OO {h^2, h h', h'^2}$ 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}}
Let $X$ be subset of $f\left[{S}\right]$ such that :$X$ admits an infimum in $L$. By Closure Operator does not Change Infimum of Subset of Image: :$f\left({\inf_L X}\right) = \inf_L X$ By definition of image of mapping: :$\inf_L X \in f\left[{S}\right]$ Thus by Infimum in Ordered Subset: :$X$ admits an infimum in $R$ and $\inf_R X = \inf_L X$ {{qed}}
Using Integration by Parts allows us to factor out $\map h x$: {{begin-eqn}} {{eqn | l = \int_a^b \paren {\map \alpha x \map h x + \map \beta x \map {h'} x} \rd x       | r = \int_a^b \map \alpha x \map h x \rd x + \int_a^b \map \beta x \rd \map h x       | c = where $\d \map h x = \map {h'} x \rd x$ }} {{eqn | r = \int_a^b \map \alpha x \map h x \rd x + \bigintlimits {\map \beta x \map h x} a b - \int_a^b \map h x \rd \map \beta x       | c = Integration by Parts }} {{eqn | r = \int_a^b \paren {\map \alpha x - \map {\beta'} x} \map h x \rd x       | c = as $\map h a = 0$, $\map h b = 0$ }} {{end-eqn}} Hence the problem has been reduced to: :$\displaystyle \int_a^b \paren {\map \alpha x - \map {\beta'} x} \map h x \rd x = 0$ Since If Definite Integral of a(x)h(x) vanishes for any C^0 h(x) then C^0 a(x) vanishes, the conclusion is that in the considered interval $\closedint a b$ it holds that: :$\map \alpha x = \map {\beta'} x$ {{qed}}
{{AimForCont}} the real function $\map \alpha x$ is nonzero at some point in $\closedint a b$ for some arbitrary $\map h x$.  Due to belonging to $C^0$ it is also nonzero in some interval $\closedint {x_1} {x_2}$ contained in $\closedint a b$. Let us choose $\map h x$ to be of a specific form, while still satisfying the requirements in the statement of the theorem: :$\map h x = \begin {cases} \map \sgn {\map \alpha x} \paren {x - x_1} \paren {x_2 - x} & : x \in \closedint {x_1} {x_2} \\ 0 & : x \notin \closedint {x_1} {x_2} \end {cases}$ Then: :$\displaystyle \int_a^b \map \alpha x \map h x \rd x = \int_{x_1}^{x_2} \size {\map \alpha x} \paren {x - x_1} \paren {x_2 - x} \rd x$ where we used the fact that: :$\map \alpha x = \map \sgn {\map \alpha x} \size {\map \alpha x}$ as well as: :$\map {\sgn^2} x = 1$ if $x \ne 0$ and $x \in \R$. The integrand is positive for $x \in \closedint {x_1} {x_2}$, whence the integral is positive. However, that contradicts the condition on the integral in the statement of the theorem. Thus, with the provided assumption the condition for the integral does not hold for all $\map h x$ with aforementioned conditions. Hence the result by Proof by Contradiction. {{qed}}
We will prove that :$c\left({\inf X}\right)$ is lower bound for $X$. Let $x \in X$. By definition of subset: :$x \in c\left[{S}\right]$ By definition of image of mapping: :$\exists y \in S: x = c\left({y}\right)$ By definition of closure operator/idempotent: :$x = c\left({x}\right)$ By definition of infimum: :$\inf X$ is lower bound for $X$. By definition of lower bound: :$\inf X \preceq x$ Thus by definition of closure operator/increasing: :$c\left({\inf X}\right) \preceq x$ {{qed|lemma}} By definition of infimum: :$c\left({\inf X}\right) \preceq \inf X$ By definition of closure operator/inflationary: :$c\left({\inf X}\right) \succeq \inf X$ Thus by definition of antisymmetry: :$c\left({\inf X}\right) = \inf X$ {{qed}}
By definition of ordered subset: :$T \subseteq S$ and :$\forall x, y \in T: x \preceq' y \iff x \preceq y$ === Sufficient Condition === Let $\inf_L X \in T$. By definition of infimum: :$\inf_L X$ is lower bound for $X$ in $L$. By definition of $\preceq'$: :$\inf_L X$ is lower bound for $X$ in $R$. We will prove that: :$\forall x \in T: x$ is lower bound for $X$ in $R \implies x \preceq' \inf_L X$ Let $x \in T$ such that: :$x$ is lower bound for $X$ in $R$. By definition of $\preceq'$: :$x$ is lower bound for $X$ in $L$. By definition of infimum: :$x \preceq \inf_L X$ Thus by definition of $\preceq'$: :$x \preceq' \inf_L X$ {{qed|lemma}} Hence $X$ admits an infimum in $R$ and $\inf_R X = \inf_L X$. {{qed|lemma}} === Necessary Condition === Assume that: :$X$ admits an infimum in $R$ and $\inf_R X = \inf_L X$ By definition of infimum: :$\inf_R X \in T$ Thus by assumption: :$\inf_L X \in T$ {{qed}}
Define $f = \operatorname{operator}\left({S}\right)$. === Inflationary === By Intersection is Subset: :$x^\succeq \cap T \subseteq x^\succeq$ By Infimum of Subset: :$\inf x^\succeq \preceq \inf\left({x^\succeq \cap T}\right)$ By Infimum of Upper Closure of Element: :$x \preceq \inf\left({x^\succeq \cap T}\right)$ Thus by definition of operator generated by $S$: :$x \preceq f\left({x}\right)$ {{qed|lemma}} === Increasing === Let $x, y \in X$ such that :$x \preceq y$ By Upper Closure is Decreasing: :$y^\succeq \subseteq x^\succeq$ By Set Intersection Preserves Subsets/Corollary: :$y^\succeq \cap T \subseteq x^\succeq \cap T$ By Infimum of Subset: :$\inf\left({x^\succeq \cap T}\right) \preceq \inf\left({y^\succeq \cap T}\right)$ Thus by definition of operator generated by $S$: :$f\left({x}\right) \preceq f\left({y}\right)$ {{qed|lemma}} === Idempotent === Let $x \in X$. By Intersection is Subset: :$x^\succeq \cap T \subseteq T$ By definition of complete lattice: :$x^\succeq \cap T$ admits an infimum in $L$. By definition of infima inheriting: :$x^\succeq \cap T$ admits an infimum in $S$ and $\inf_S\left({x^\succeq \cap T}\right) = \inf_L \left({x^\succeq \cap T}\right)$ Then by definition of operator generated by $S$: :$f\left({x}\right) \in T$ and :$f\left({f\left({x}\right)}\right) = \inf_L \left({f\left({x}\right)^\succeq \cap T}\right)$ By definitions of upper closure of element and reflexivity: :$f\left({x}\right) \in f\left({x}\right)^\succeq$ By definition of intersection: :$f\left({x}\right) \in f\left({x}\right)^\succeq \cap T$ By definitions of infimum and lower bound: :$\inf_L \left({f\left({x}\right)^\succeq \cap T}\right) \preceq f\left({x}\right)$ By Intersection is Subset: :$f\left({x}\right)^\succeq \cap T \subseteq f\left({x}\right)^\succeq$ By Infimum of Subset: :$\inf_L \left({f\left({x}\right)^\succeq}\right) \preceq \inf_L \left({f\left({x}\right)^\succeq \cap T}\right)$ By Infimum of Upper Closure of Element: :$f\left({x}\right) \preceq \inf_L \left({f\left({x}\right)^\succeq \cap T}\right)$ Thus by definition of antisymmetry: :$f\left({f\left({x}\right)}\right) = f\left({x}\right)$ {{qed|lemma}} Hence $f$ is closure operator. {{qed}}
Define $f = \operatorname{operator}\left({S}\right)$. === $\subseteq$ === Let $x \in f\left[{X}\right]$. By definition of image of mapping: :$\exists y \in X: x = f\left({y}\right)$ Define $Y = y^\succeq \cap T$ By definition of complete lattice: :$Y$ admits an infimum in $L$. By Intersection is Subset: :$Y \subseteq T$ By definition of closure system: :$S$ inherits infima. By definition of infima inheriting: :$Y$ admits an infimum in $S$ and $\inf_S Y = \inf_L Y$ Thus by definition of operator generated by $S$: :$x \in T$ {{qed|lemma}} === $\supseteq$ === Let $x \in T$. By definition of ordered subset: :$T \subseteq X$ By definition of subset: :$x \in X$ Define $Y = x^{\succeq_L} \cap T$ By definitions of upper closure of element and reflexivity: :$x \in x^{\succeq_L}$ By definition of intersection: :$x \in Y$ By definitions of infimum and lower bound: :$\inf_L Y \preceq x$ By Intersection is Subset: :$Y \subseteq x^{\succeq_L}$ By Infimum of Subset: :$\inf_L x^{\succeq_L} \preceq \inf_L Y$ By Infimum of Upper Closure of Element: :$x \preceq \inf_L Y$ By definition of antisymmetry: :$x = \inf_L Y$ By definition of operator generated by $S$: :$x = f\left({x}\right)$ Thus by definition of image of mapping: :$x \in f\left[{X}\right]$ {{qed|lemma}} Hence by definition of set equality: :$f\left[{X}\right] = T$ {{qed}}