hcju/proofwikiqa huggingface dataset:
{{begin-eqn}} {{eqn | l = 12^4 + 15^4 + 20^4       | r = 20 \, 736 + 50 \, 625 + 160 \, 000       | c =  }} {{eqn | r = 231 \, 361       | c =  }} {{eqn | r = 481^2       | c =  }} {{end-eqn}} The smallest solution must be fourth powers of coprime integers, otherwise dividing by their GCD would yield a smaller solution. By Fermat's Right Triangle Theorem, $x^4 + y^4 = z^2$ has no solutions. Thus none of the fourth powers is $0$. Consider the equation $x^4 + y^4 + z^4 = n^2$. By Square Modulo 4, since fourth powers are squares, at most one of $x, y, z$ is odd. If none of them are odd, the numbers are not coprime. Therefore there is exactly one odd number. {{qed|lemma}} By Fermat's Little Theorem, for $3 \nmid a$: :$a^2 \equiv 1 \pmod 3$ Therefore by Congruence of Powers: :$a^4 \equiv 1^2 \equiv 1 \pmod 3$ For $3 \divides a$: :$a^2 \equiv a^4 \equiv 0 \pmod 3$ Therefore we cannot have two of $x, y, z$ not divisible by $3$. By our coprimality condition we cannot have $3 \divides x, y, z$. Hence either $3 \nmid x, y, z$ or exactly one of $x, y, z$ is not divisible by $3$. {{qed|lemma}} Similarly, by Fermat's Little Theorem, for $5 \nmid a$: :$a^4 \equiv 1 \pmod 5$ For $5 \divides a$: :$a^4 \equiv 0 \pmod 5$ By Square Modulo 5: :$a^2 \equiv 0, 1, 4 \pmod 5$ Therefore there is exactly one number not divisible by $5$. {{qed|lemma}} Since $\sqrt {481} < 22$, we only need to check fourth powers of numbers less than $22$, satisfying the criteria above. Since there are exactly two multiples of $2$ and $5$ among $x, y, z$, at least one of them must be a multiple of $10$. Suppose $5 \nmid x$, $5 \divides y$ and $10 \divides z$. === Case $1$: $z = 10$ === The case $y = 20$ will be equivalent to Case $2$. Suppose $y = 15$. By the criteria above, $x$ is divisible by $2$ and $3$. Also $x < 22$, so we only need to check $x = 6, 12, 18$. {{qed|lemma}} Suppose $y = 10$. By the criteria above, $x$ is not divisible by $2$ or $3$. Also $x < 22$, so we only need to check $x = 1, 7, 11, 13, 17, 19$. {{qed|lemma}} Suppose $y = 5$. By the criteria above, $x$ is divisible by $2$ but not $3$. Also $x < 22$, so we only need to check $x = 2, 4, 8, 14, 16$. {{qed|lemma}} === Case $2$: $z = 20$ === Suppose $y = 20$. Then $x^4 + y^4 + z^4 > 20^4 + 20^4 = 320 \, 000 > 481^2$. Suppose $y = 15$. By the criteria above, $x$ is divisible by $2$ and $3$. Also $x < \sqrt [4] {481^2 - 20^4 - 15^4} = 12$, so we only need to check $x = 6$. {{qed|lemma}} Suppose $y = 10$. By the criteria above, $x$ is not divisible by $2$ or $3$. Also $x < \sqrt [4] {481^2 - 20^4} < 17$, so we only need to check $x = 1, 7, 11, 13$. {{qed|lemma}} Suppose $y = 5$. By the criteria above, $x$ is divisible by $2$ but not $3$. Also $x < \sqrt [4] {481^2 - 20^4} < 17$, so we only need to check $x = 2, 4, 8, 14, 16$. {{qed|lemma}} And we see that none of the above cases yield any smaller solutions. Hence the result. {{finish|Check the remaining $24$ candidates}}
Let $n = \sqbrk {abc}_{10}$ denote a $3$-digit integer whose digits are $a, b, c$. If $a = b = c$ then Kaprekar's process trivially results in $0$ after the first iteration. {{WLOG}}, let $a \ge b \ge c$ but such that $a \ne c$. By the Basis Representation Theorem: :$n = 10^2 a + 10 b + c$ Let $n' = 10^2 a' + 10 b' + c'$ be the result of Kaprekar's process after the $1$st iteration on $n$. We have: {{begin-eqn}} {{eqn | l = n'       | r = \paren {100 a + 10 b + c} - \paren {100 c + 10 b + a}       | c =  }} {{eqn | r = 100 \paren {a - c} + \paren {c - a}       | c =  }} {{eqn | r = 100 \paren {a - c} - 10 + \paren {10 + c - a}       | c = as $c < a$ }} {{eqn | r = 100 \paren {a - c - 1} + \paren {100 - 10} + \paren {10 + c - a}       | c = as $c < a$ }} {{eqn | r = 10^2 \paren {a - c - 1} + 10 \times 9 + \paren {10 + c - a}       | c = as $c < a$ }} {{end-eqn}} Thus we have that: {{begin-eqn}} {{eqn | l = a'       | r = a - c - 1 }} {{eqn | l = b'       | r = 9 }} {{eqn | l = c'       | r = 10 + c - a }} {{end-eqn}} Hence we have that: {{begin-eqn}} {{eqn | l = b'       | o = \ge       | r = a' }} {{eqn | l = b'       | o = \ge       | r = c' }} {{eqn | l = a' + c'       | r = 9 }} {{end-eqn}} There are in fact only the following possibilities for $n'$: {{begin-eqn}} {{eqn | l = n'       | r = 099       | c = note that Kaprekar's process retains leading zeroes }} {{eqn | l = n'       | r = 198 }} {{eqn | l = n'       | r = 297 }} {{eqn | l = n'       | r = 396 }} {{eqn | l = n'       | r = 495 }} {{eqn | l = n'       | r = 594 }} {{eqn | l = n'       | r = 693 }} {{eqn | l = n'       | r = 792 }} {{eqn | l = n'       | r = 891 }} {{end-eqn}} {{WLOG}} we inspect  Kaprekar's process on $198$: {{begin-eqn}} {{eqn | l = 099       | o = \to       | r = 990 - 099       | c =  }} {{eqn | l = 891       | o = \to       | r = 981 - 189       | c =  }} {{eqn | o = \to       | r = 792       | c =  }} {{eqn | o = \to       | r = 972 - 279       | c =  }} {{eqn | o = \to       | r = 693       | c =  }} {{eqn | o = \to       | r = 963 - 369       | c =  }} {{eqn | o = \to       | r = 594       | c =  }} {{eqn | o = \to       | r = 954 - 459       | c =  }} {{eqn | o = \to       | r = 495       | c =  }} {{end-eqn}} and the result is seen to follow. {{qed}}
Let $N$ be the subset of nilpotent elements. Because $0_R$ is nilpotent, $0_R \in N$ and so $N$ is non-empty. Let $x \in N$ and $a \in R$. We have: {{begin-eqn}} {{eqn | lo= \exists n \in \Z_{>0}:       | l = x^n       | r = 0_R       | c = {{Defof|Nilpotent Ring Element}} }} {{eqn | ll= \leadsto       | l = a^n \circ x^n       | r = 0_R       | c = {{Defof|Ring Zero}} }} {{eqn | ll= \leadsto       | l = \paren {a \circ x}^n       | r = 0_R       | c = Power of Product of Commutative Elements in Semigroup }} {{eqn | ll= \leadsto       | l = a x       | o = \in       | r = N       | c = Definition of $N$ }} {{end-eqn}} Let $x, y \in N$. Let $x^n = 0$ and $y^m = 0$. By the Binomial Theorem, $\paren {x - y}^p = 0$ for $p \ge n + m - 1$. Thus $x - y \in N$. Thus from Test for Ideal, $N$ is an ideal. {{qed}}
By Nilpotent Element is Contained in Prime Ideals, $\Nil A$ is contained in the intersection of all prime ideals. It remains to prove the other inclusion. Let $f \in A$ be not nilpotent. Let $S$ be the set of ideals of $A$ that are disjoint from $\set {f^n: n \in \N}$. By Zorn's Lemma, $S$ has a maximal element $P$. In particular, $f \notin P$. We want to show that $P$ is prime. Let $a, b \in A$ with $a, b \notin P$. Then the sums of ideals $\ideal a + P$ and $\ideal b + P$ contain $P$ strictly. By the maximality of $P$, there exist $n, m \in \N$ with $f^n \in \ideal a + P$ and $f^m \in \ideal b + P$. Then: :$f^{m + n} \in \ideal {a b} + P$ Thus: :$\ideal {a b} + P \notin S$ In particular: :$a b \notin P$ Thus $P$ is prime, and so: :$\displaystyle f \notin \bigcap_{\mathfrak p \mathop \in \Spec A} \mathfrak p$ {{qed}} {{AoC|Zorn's Lemma}}
Follows immediately from Finite Multiplicative Subgroup of Field is Cyclic. {{qed}} Category:Galois Fields jjwa16ge80xdni2x02rvbql2smj1753
Thus {{begin-eqn}} {{eqn | l = i_S\left[{Z}\right] | r = \left\{ {i_S\left({z}\right): z \in Z}\right\} | c = definition of image of set }} {{eqn | r = \left\{ {z: z \in Z}\right\} | c = definition of inclusion mapping }} {{eqn | r = Z | c = definition of set equality }} {{end-eqn}} {{qed}}
We will prove that :(lemma): $\forall j \in D: \inf_L\left({\preceq'\left({j}\right)}\right) = j$ Let $j \in D$. By definitions of image of element and upper closure of element: :$\preceq'\left({j}\right) = j^{\succeq'}$ By Upper Closure in Ordered Subset is Intersection of Subset and Upper Closure: :$j^{\succeq'} = D \cap j^\succeq$ By Intersection is Subset: :$j^{\succeq'} \subseteq j^\succeq$ By Infimum of Subset and Infimum of Upper Closure of Element: :$j \preceq \inf_L\left({\preceq'\left({j}\right)}\right)$ By definition of reflexivity: :$j \preceq' j$ By definition of image of element: :$j \in \preceq'\left({j}\right)$ By definitions of infimum and lower bound: :$\inf_L\left({\preceq'\left({j}\right)}\right) \preceq j$ Thus by definition of antisymmetry: :$\inf_L\left({\preceq'\left({j}\right)}\right) = j$ {{qed|lemma}} Thus {{begin-eqn}} {{eqn | l = \liminf i_D | r = {\sup}_L \left\{ { {\inf}_L\left({i_D\left[{\preceq'\left({j}\right)}\right]}\right): j \in D}\right\} | c = definition of limit inferior of Moore-Smith sequence }} {{eqn | r = {\sup}_L \left\{ { {\inf}_L\left({\preceq'\left({j}\right)}\right): j \in D}\right\} | c = Image under Inclusion Mapping }} {{eqn | r = {\sup}_L \left\{ { j: j \in D}\right\} | c = (lemma) }} {{eqn | r = {\sup}_L D | c = definition of set equality }} {{end-eqn}} {{qed}}
For $g\in G$, define the mapping $\phi_g : X \to X$ as: :$\map {\phi_g} x = \map \phi {g, x}$ Then $\rho$ is the permutation representation associated to $\phi$ {{iff}}: :$\forall g \in G : \map \rho g  = \phi_g$ By Equality of Mappings, this is equivalent to: :$\forall g \in G : \forall x \in X : \map {\map \rho g} x = \map {\phi_g} x$ $\phi$ is the group action associated to $\rho$ {{iff}}: :$\forall g \in G : \forall x \in X : \map \phi {g, x} = \map {\map \rho g} x$. Because $\map {\phi_g} x = \map \phi {g, x}$, they are equivalent. {{qed}} Category:Group Actions hdpiy278iyjot15nkzz1r1ebt9q1c89
Follows from Closed Fields in Galois Connection for Field Extension (and does not use Fundamental Theorem of Galois Theory). {{proof wanted}} Category:Galois Theory 5f8c1ibg4vz9kywca8kf6h0d0lu6e4m
Let $\sigma \in \operatorname{Aut}(E/F)$ stabilize $S$. Then $S$ is contained in the fixed field of $\sigma$. By definition of generating set, $\sigma$ fixes $E$. Thus $\operatorname{Aut}(E/F)$ acts faithfully. {{qed}} Category:Field Extensions g4ymi5pk0s0rofvoxzbkfzzn6cods98
By definition of ordered subset: :$T \subseteq S$ We will prove that :$t^\succsim \subseteq T \cap t^\succeq$ Let $x \in t^\succsim$ By definition of upper closure of element: :$x \in T$ and $t \precsim x$ By definition of ordered subset: :$t \preceq x$ By definition of upper closure of element: :$x \in t^\succeq$ Thus by definition of intersection: :$x \in T \cap t^\succeq$ {{qed|lemma}} We will prove that :$T \cap t^\succeq \subseteq t^\succsim$ Let $x \in T \cap t^\succeq$ By definition of intersection: :$x \in T$ and $x \in t^\succeq$ By definition of upper closure of element: :$t \preceq x$ By definition of ordered subset: :$t \precsim x$ Thus by definition of upper closure of element: :$x \in t^\succsim$ {{qed|lemma}} By definition of set equality: :$t^\succsim = T \cap t^\succeq$ {{qed}} Category:Upper Closures kb3ltlc37a3kdtoi35lw7nsmwueduh2
An intuition is given as follows: Suppose for some $m, n \in \N$: :$7^m = c \cdot 10^n$, where $c$ is very close to $1$. Taking common logarithm: :$m \log 7 = \log c + n$ Which leads to: :$\log 7 = \dfrac n m + \dfrac {\log c} m$ where $\dfrac {\log c} m$ is very close to $0$. To make a good approximation is to minimize $\size {\log c}$. One may calculate the continued fraction of $\log 7$. We use continued fractions because Convergents are Best Approximations. One obtains $\sqbrk {0; 1, 5, 2, 5, 6, 1, 4813, 1, 1, 2, 2, \dots}$. {{OEIS|A082571}} Let $\dfrac {p_n} {q_n}$ be the convergent $\sqbrk {0; a_1, \dots, a_n}$. Then: {{begin-eqn}} {{eqn | l = \size {\log c}       | r = q_k \size {\log 7 - \frac {p_k} {q_k} } }} {{eqn | o = <       | r = q_k \paren {\frac 1 {q_k q_{k + 1} } }       | c = Accuracy of Convergents of Continued Fraction }} {{eqn | r = \frac 1 {q_{k + 1} } }} {{eqn | r = \frac 1 {a_{k + 1} q_k + q_{k - 1} }       | c = Recursive Definition of Denominators of Convergents }} {{end-eqn}} Thus, at least intuitively, the larger the value of the next $a_i$, the smaller $\size {\log c}$ is. {{finish|I am not familiar with this, this is just some justification}} $4813$ is the largest number in the expansion among at least the first $100$ numbers. Note that $\sqbrk {0; 1, 5, 2, 5, 6, 1} = \dfrac {431} {510}$ This gives the estimate $7^{510} \approx 10^{431}$. {{qed}}
We will prove that :(lemma): $\forall j \in \N: f\left[{\le\left({j}\right)}\right] = \left\{ {a, b}\right\}$ Let $j \in \N$. Let $x \in S$. Assume that :$x \in f\left[{\le\left({j}\right)}\right]$ By definition of image of set: :$\exists i \in \le\left({j}\right): x = f\left({i}\right)$ By definition of $f$: :$x = a$ or $x = b$ Thus by definition of unordered tuple: :$x \in \left\{ {a, b}\right\}$ {{qed|lemma}} Assume that :$x \in \left\{ {a, b}\right\}$ We have: :$j \le j$ and $j \le j+1$ By definition of image of element: :$j, j+1 \in \le\left({j}\right)$ By definition of $f$: :$\left\{ {f\left({j}\right), f\left({j+1}\right)}\right\} = \left\{ {a, b}\right\}$ By definition of image of set: :$f\left({j}\right), f\left({j+1}\right) \in f\left[{\le\left({j}\right)}\right]$ Thus by definition of unordered tuple: $x \in f\left[{\le\left({j}\right)}\right]$ {{qed|lemma}} Thus by definition of set equality: :$f\left[{\le\left({j}\right)}\right] = \left\{ {a, b}\right\}$ {{qed|lemma}} Thus {{begin-eqn}} {{eqn | l = \liminf \left({c_i}\right)_{i \mathop\in \N} | r = \sup \left\{ {\inf \left({f\left[{\mathord\le\left({j}\right)}\right]}\right): j \in \N}\right\} | c = definition of limit inferior }} {{eqn | r = \sup \left\{ {\inf \left\{ {a, b}\right\}: j \in \N}\right\} | c = (lemma) }} {{eqn | r = \sup \left\{ {a \wedge b: j \in \N}\right\} | c = definition of meet }} {{eqn | r = a \wedge b | c = Supremum of Singleton }} {{end-eqn}} {{qed}}
Let $a, b \in S$ such that :$a \preceq_1 b$ Define $M = \left({\N, \le}\right)$ being an ordered set. We will prove that :$M$ is a directed set. Let $x, y \in \N$. Thus by definition of max operation: :$\max\left({x, y}\right) \in \N$ Thus by definition of max operation: :$x \le \max\left({x, y}\right)$ and $y \le \max\left({x, y}\right)$ {{qed|lemma}} Define $g = \left({c_i}\right)_{i \in \N} = \left({a, b, a, b, \dots}\right):\N \to S$ being a Moore-Smith sequence in $S$. By Limit Inferior of Repetition Moore-Smith Sequence: :$\liminf \left({c_i}\right)_{i \in \N} = a \wedge_1 b$ By Preceding iff Meet equals Less Operand: :$f\left({\liminf \left({c_i}\right)_{i \in \N} }\right) = f\left({a}\right)$ By definition of composition of mappings: :$f \circ g = \left({f\left({c_i}\right)}\right)_{i \in \N} = \left({f\left({a}\right), f\left({b}\right), f\left({a}\right), f\left({b}\right), \dots}\right)$ By Limit Inferior of Repetition Moore-Smith Sequence: :$\liminf \left({f \circ g}\right) = f\left({a}\right) \wedge_2 f\left({b}\right)$ By assumption: :$f\left({a}\right) \preceq_2 f\left({a}\right) \wedge_2 f\left({b}\right)$ By Meet Precedes Operands: :$f\left({a}\right) \wedge_2 f\left({b}\right) \preceq_2 f\left({a}\right)$ By definition of antisymmetry: :$f\left({a}\right) = f\left({a}\right) \wedge_2 f\left({b}\right)$ Thus by Preceding iff Meet equals Less Operand: :$f\left({a}\right) \preceq_2 f\left({b}\right)$ {{qed}}
By Mapping at Limit Inferior Precedes Limit Inferior of Composition Mapping and Sequence implies Mapping is Increasing: :$f$ is an increasing mapping. By Image of Directed Subset under Increasing Mapping is Directed: :$f\left[{D}\right]$ is directed. By definition of up-complete: :$D$ and $f\left[{D}\right]$ admit suprema. By Subset and Image Admit Suprema and Mapping is Increasing implies Supremum of Image Precedes Mapping at Supremum: :$\sup \left({f\left[{D}\right]}\right) \preceq_2 f\left({\sup D}\right)$ By Limit Inferior of Inclusion Moore-Smith Sequence is Supremum of Directed Subset: :$\sup D = \liminf i_D$ By assumption: :$f\left({\sup D}\right) \preceq_2 \liminf \left({f \circ i_D}\right)$ By Composition of Mapping and Inclusion is Restriction of Mapping: :$f \circ i_D = f\restriction{D}$ By Limit Inferior of Restriction Moore-Smith Sequence is Supremum of Image of Directed Subset: :$\sup \left({f\left[{D}\right]}\right) = \liminf \left({f \circ i_D}\right)$ Then :$f\left({\sup D}\right) \preceq_2 \sup \left({f\left[{D}\right]}\right)$ Thus by definition of antisymmetry: :$\sup \left({f\left[{D}\right]}\right) = f\left({\sup D}\right)$ {{qed}}
We have that: {{begin-eqn}} {{eqn | l = 0!! + 1       | r = 1 + 1       | c = {{Defof|Double Factorial}} }} {{eqn | r = 2       | c = which is prime }} {{end-eqn}} {{begin-eqn}} {{eqn | l = 1!! + 1       | r = 1 + 1       | c = {{Defof|Double Factorial}} }} {{eqn | r = 2       | c = which is prime }} {{end-eqn}} {{begin-eqn}} {{eqn | l = 2!! + 1       | r = 2 \times 0!! + 1       | c = {{Defof|Double Factorial}} }} {{eqn | r = 2 \times 1 + 1       | c = {{Defof|Double Factorial}} }} {{eqn | r = 3       | c = which is prime }} {{end-eqn}} {{ProofWanted|It remains to be shown that the other integers are the only other ones with this property. Have fun.}}
We will prove that :(lemma): $\forall j \in D: \inf_L\left({\left({f \restriction D}\right)\left[{\preceq'\left({j}\right)}\right]}\right) = f\left({j}\right)$ Let $j \in D$. By definitions of image of element and upper closure of element: :$\preceq'\left({j}\right) = j^{\succeq'}$ By Upper Closure in Ordered Subset is Intersection of Subset and Upper Closure: :$j^{\succeq'} = D \cap j^{\succeq_1}$ By Intersection is Subset: :$j^{\succeq'} \subseteq j^{\succeq_1}$ By Image of Subset under Relation is Subset of Image/Corollary 2: :$\left({f \restriction D}\right)\left[{j^{\succeq'} }\right] \subseteq \left({f \restriction D}\right)\left[{j^{\succeq_1} }\right]$ By Infimum of Subset and Infimum of Image of Upper Closure of Element under Increasing Mapping: :$f\left({j}\right) \preceq_2 \inf_L\left({\left({f \restriction D}\right)\left[{\preceq'\left({j}\right)}\right]}\right)$ By definition of reflexivity: :$j \preceq' j$ By definition of image of element: :$j \in \preceq'\left({j}\right)$ By definition of image of set: :$f\left({j}\right) \in \left({f \restriction D}\right)\left[{\preceq'\left({j}\right)}\right]$ By definitions of infimum and lower bound: :$\inf_L\left({\left({f \restriction D}\right)\left[{\preceq'\left({j}\right)}\right]}\right) \preceq_2 f\left({j}\right)$ Thus by definition of antisymmetry: :$\inf_L\left({\left({f \restriction D}\right)\left[{\preceq'\left({j}\right)}\right]}\right) = f\left({j}\right)$ {{qed|lemma}} Thus {{begin-eqn}} {{eqn | l = \liminf \left({f \restriction D}\right) | r = {\sup} \left\{ { {\inf}\left({\left({f \restriction D}\right)\left[{\preceq'\left({j}\right)}\right]}\right): j \in D}\right\} | c = definition of limit inferior of Moore-Smith sequence }} {{eqn | r = {\sup} \left\{ { f\left({j}\right): j \in D}\right\} | c = (lemma) }} {{eqn | r = {\sup} \left({f\left[{D}\right]}\right) | c = definition of image of set }} {{end-eqn}} {{qed}}
By Infimum of Upper Closure of Element: :$\inf x^\succeq = x$ By definition of infimum: :$x$ is lower bound for $x^\succeq$ Thus by Increasing Mapping Preserves Lower Bounds: :$f\left({x}\right)$ is lower bound for $f\left[{x^\succeq}\right]$ By definition of reflexivity: :$x \preceq x$ By definition of upper closure of element: :$x \in x^\succeq$ By definition of image of set: :$f\left({x}\right) \in f\left[{x^\succeq}\right]$ Thus by definition: :$\forall y \in T: y$ is lower bound for $f\left[{x^\succeq}\right] \implies y \precsim f\left({x}\right)$ Thus by definition of infimum: :$\inf \left({f\left[{x^\succeq}\right]}\right) = f\left({x}\right)$ {{qed}}
From the sequence of Poulet numbers, these are Poulet numbers: :$1105, 2465, \ldots$ Then we have: {{begin-eqn}} {{eqn | l = 1105       | r = \dfrac {13 \paren {13^2 + 1} } 2       | c = so $1105$ is the magic constant of the order $13$ magic square }} {{eqn | l = 2465       | r = \dfrac {17 \paren {17^2 + 1} } 2       | c = so $2465$ is the magic constant of the order $17$ magic square }} {{end-eqn}} {{expand|Further investigation needed. I have not seen this correspondence explored on the internet. I note though that both $1105$ and $2465$ are also Carmichael.}} Category:Poulet Numbers Category:Magic Squares ahytsnqo4qufaah5amkvd28zlhygl7u
By definition of inclusion mapping: :$i_A: A \to S$ By definitions of composition of mappings and restriction of mapping: :$f \circ i_A: A \to T$ and $f \restriction A: A \to T$ Let $a \in A$. Thus {{begin-eqn}} {{eqn | l = \left({f \circ i_A}\right)\left({a}\right) | r = f \left({i_A \left({a}\right)}\right) | c = definition of composition of mappings }} {{eqn | r = f \left({a}\right) | c = definition of inclusion mapping }} {{eqn | r = \left({f \restriction A}\right)\left({a}\right) | c = definition of restriction of mapping }} {{end-eqn}} {{qed}}
Let $D$ be a directed subset of $S$. Assume that :$D$ admits a supremum. Thus by definition of complete lattice: :$f \sqbrk D$ admits a supremum. Thus by Mapping at Limit Inferior Precedes Limit Inferior of Composition Mapping and Sequence implies Supremum of Image is Mapping at Supremum of Directed Subset: :$\map \sup {f \sqbrk D} = \map f {\sup D}$ {{qed}}
Let $P$ be a closed subset of $T$. By Closed Set iff Lower and Closed under Directed Suprema in Scott Topological Ordered Set: :$P$ is lower and closed under directed suprema. We will prove that :for all directed subset $D$ of $S$: $D \subseteq f^{-1}\left[{P}\right] \implies \sup D \in f^{-1}\left[{P}\right]$ Let $D$ be a directed subset of $S$ such that :$D \subseteq f^{-1}\left[{P}\right]$ By definition of mapping preserves directed suprema: :$f$ preserves the supremum of $D$. By definition of up-complete: :$D$ admits a supremum. By definition of mapping preserves the supremum: :$ \sup \left({f\left[{D}\right]}\right) = f \left({\sup D}\right)$ By Directed Suprema Preserving Mapping is Increasing: :$f$ is increasing. By Image of Directed Subset under Increasing Mapping is Directed: :$f\left[{D}\right]$ is directed. By Preimage is Subset of Set implies Image of Set is Subset: :$f\left[{D}\right] \subseteq P$ By definition of closed under directed suprema: :$\sup \left({f\left[{D}\right]}\right) \in P$ Thus by definition of preimage of set: :$\sup D \in f^{-1}\left[{P}\right]$ {{qed|lemma}} By definition: :$f^{-1}\left[{P}\right]$ is closed under directed suprema. By Preimage of Lower Set under Increasing Mapping is Lower: :$f^{-1}\left[{P}\right]$ is lower. Thus by Closed Set iff Lower and Closed under Directed Suprema in Scott Topological Ordered Set: :$f^{-1}\left[{P}\right]$ is closed. {{qed}}
{{begin-eqn}} {{eqn | l = \sec \paren {a + b i}       | r = \frac 1 {\cos \paren {a + b i} }       | c = {{Defof|Complex Secant Function}} }} {{eqn | r = \dfrac 1 {\cos a \cosh b - i \sin a \sinh b}       | c = Cosine of Complex Number }} {{eqn | r = \dfrac {\cos a \cosh b + i \sin a \sinh b} {\paren {\cos a \cosh b + i \sin a \sinh b} \paren {\cos a \cosh b - i \sin a \sinh b} }       | c = multiplying denominator and numerator by $\cos a \cosh b + i \sin a \sinh b$ }} {{eqn | r = \dfrac {\cos a \cosh b + i \sin a \sinh b} {\cos^2 a \cosh^2 b - i^2 \sin^2 a \sinh^2 b}       | c = Difference of Two Squares }} {{eqn | r = \dfrac {\cos a \cosh b + i \sin a \sinh b} {\cos^2 a \cosh^2 b + \sin^2 a \sinh^2 b}       | c = {{Defof|Imaginary Unit}} }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \tan \paren {a + b i}       | r = \frac {\sin \paren {a + b i} } {\cos \paren {a + b i} }       | c = {{Defof|Complex Tangent Function}} }} {{eqn | r = \dfrac {\sin a \cosh b + i \cos a \sinh b} {\cos a \cosh b - i \sin a \sinh b}       | c = Sine of Complex Number and Cosine of Complex Number }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \tan \paren {a + b i}       | r = \dfrac {\sin a \cosh b + i \cos a \sinh b} {\cos a \cosh b - i \sin a \sinh b}       | c = Tangent of Complex Number: Formulation 1 }} {{eqn | r = \dfrac {\tan a \cosh b + i \sinh b} {\cosh b - i \tan a \sinh b}       | c = multiplying denominator and numerator by $\dfrac 1 {\cos a}$ }} {{eqn | r = \frac {\tan a + i \tanh b} {1 - i \tan a \tanh b}       | c = multiplying denominator and numerator by $\dfrac 1 {\cosh b}$ }} {{end-eqn}} {{qed}}