hcju/proofwikiqa huggingface dataset:
Take $K_{II}$ to be the set of all limit ordinals. Then set $y = \bigcup \set {w \in K_{II}: w \le x}$ The set $\set {w \in K_{II}: w \le x}$ is non-empty because $\omega \subseteq x$. By Union of Ordinals is Least Upper Bound, $y \in K_{II}$ and $y \le x$. By Ordinal Subtraction when Possible is Unique, there is a unique $z$ such that $x = \paren {y + z}$ Assume $\omega \le z$. Then, again by Ordinal Subtraction when Possible is Unique: :$z = \paren {\omega + w}$ and so: {{begin-eqn}} {{eqn | l = x       | r = \paren {y + \paren {\omega + w} }       | c = Equality is Transitive }} {{eqn | r = \paren {\paren {y + \omega} + w}       | c = Ordinal Addition is Associative }} {{end-eqn}} But $y + \omega$ is a limit ordinal by Limit Ordinals Preserved Under Ordinal Addition: :$\O < \omega \implies y < y + \omega$ This contradicts the fact that $y$ is the largest limit ordinal smaller than $x$. Therefore, $z \in \omega$. Thus we have proven that such a selection of $y$ and $z$ exists. Suppose $z$ and $w$ both satisfy: :$\paren {y + w} = \paren {y + z}$ By Ordinal Addition is Left Cancellable, we have $w = z$. Thus $z$ is unique. It remains to prove uniqueness for $y$. Suppose that $x = \paren {y + u}$ and $x = \paren {w + z}$. {{WLOG}}, assume further that $y \le w$. Then: {{begin-eqn}} {{eqn | l = y       | o = \le       | r = w       | c = }} {{eqn | lll= \leadsto       | lo= \exists n:       | l = w       | r = \paren {y + n}       | c = Ordinal Subtraction when Possible is Unique }} {{eqn | lll= \leadsto       | l = \paren {y + u}       | r = \paren {\paren {y + n} + z}       | c = Substitutivity of Equality }} {{eqn | lll= \leadsto       | l = \paren {y + u}       | r = \paren {y + \paren {n + z} }       | c = Ordinal Addition is Associative }} {{eqn | lll= \leadsto       | l = u       | r = \paren {n + z}       | c = Ordinal Addition is Left Cancellable }} {{eqn | lll= \leadsto       | l = \paren {n + z}       | o = \in       | r = \omega       | c = as $u \in \omega$ }} {{eqn | lll= \leadsto       | l = n       | o = \in       | r = \omega       | c = Ordinal is Less than Sum }} {{eqn | lll= \leadsto       | l = n       | o = \notin       | r = K_{II}       | c = as $\omega$ is the smallest limit ordinal }} {{eqn | lll= \leadsto       | l = n       | r = \O       | c = {{Defof|Limit Ordinal}} }} {{eqn | ll= \lor       | lo= \exists m:       | l = n       | r = m^+       | c =  }} {{end-eqn}} {{AimForCont}} that $\exists m: n = m^+$. {{begin-eqn}} {{eqn | l = n       | r = m^+       | c =  }} {{eqn | ll= \leadsto       | l = w       | r = \paren {y + m^+}       | c = Definition of $n$ }} {{eqn | ll= \leadsto       | l = w       | r = \paren {y + m}^+       | c = {{Defof|Ordinal Addition}} }} {{eqn | ll= \leadsto       | l = w       | o = \notin       | r = K_{II}       | c = {{Defof|Limit Ordinal}} }} {{end-eqn}} This is clearly a contradiction. Hence: :$n = \O$ and so: :$w = y$ The result follows. {{qed}}
By Transfinite Induction on $x$. The proof will use $<$, $\in$, and $\subset$ interchangeably.  This is justified by Transitive Set is Proper Subset of Ordinal iff Element of Ordinal. === Base Case === By our hypothesis, $\omega \le x$, so $x \not < \omega$, so we may begin our induction at $\omega$. {{begin-eqn}} {{eqn | l = \left({n + \omega}\right)       | r = \bigcup_{y \mathop \in \omega} \left({n + y}\right)       | c = Definition:Ordinal Addition }} {{eqn | l = \forall y \in \omega: \left({n + y}\right) \le \omega       | o = \implies       | r = \bigcup_{y \mathop \in \omega} \left({n + y}\right) \le \omega       | c = Natural Number Addition is Closed and Indexed Union Subset }} {{eqn | l = \forall y \in \omega: y \le \left({n + y}\right)       | o = \implies       | r = \bigcup_{y \mathop \in \omega} y \le \bigcup_{y \mathop \in \omega} \left({n + y}\right)       | c = Ordinal is Less than Sum and Indexed Union Subset }} {{eqn | o = \implies       | r = \omega \le \bigcup_{y \mathop \in \omega} \left({n + y}\right)       | c = Union of Limit Ordinal }} {{end-eqn}} From these conclusions, we may deduce that: : $\displaystyle \omega = \bigcup_{y \mathop \in \omega} \left({n + y}\right)$ === Inductive Case === {{begin-eqn}} {{eqn | l = \left({n + x}\right) = x       | o = \implies       | r = \left({n + x}\right)^+ = x^+       | c = Equality of Successors }} {{eqn | o = \implies       | r = \left({n + x^+}\right) = x^+       | c = Definition:Ordinal Addition }} {{end-eqn}} === Limit Case === {{begin-eqn}} {{eqn | l = \forall y \in x: \left({n + y}\right) = y       | o = \implies       | r = \bigcup_{y \mathop \in x} \left({n + y}\right) = \bigcup_{y \mathop \in x} y       | c = Indexed Union Equality }} {{eqn | o = \implies       | r = \left({n + x}\right) = \bigcup_{y \mathop \in x} y       | c = Definition:Ordinal Addition }} {{eqn | o = \implies       | r = \left({n + x}\right) = x       | c = Union of Limit Ordinal }} {{end-eqn}} {{qed}}
=== Necessary Condition === {{:Class is Transitive iff Union is Subset/Necessary Condition}}{{qed|lemma}} === Sufficient Condition === {{:Class is Transitive iff Union is Subset/Sufficient Condition}}{{qed}}
{{begin-eqn}} {{eqn | l = \map \bigcup {x^+}       | r = \map \bigcup {x \cup \set x}       | c = {{Defof|Successor Set}} }} {{eqn | r = \paren {\bigcup x \cup \bigcup \set x}       | c = Union Distributes over Union/Sets of Sets }} {{eqn | r = \paren {\bigcup x \cup x}       | c = Union of Singleton }} {{eqn | r = x       | c = Class is Transitive iff Union is Subset }} {{end-eqn}} {{qed}} Category:Ordinals s9lv42cpbdx9ne3g864307yy5yphl2p
We have: * Union is Associative * Union is Commutative * Union is Idempotent The result follows from Associative Commutative Idempotent Operation is Distributive over Itself. {{qed}} Category:Set Union Category:Distributive Operations Category:Union Distributes over Union 7es6ntapftyilyqcnc8jf4c5smumevl
{{begin-eqn}} {{eqn | l = A = \O       | o = \implies        | r = A \subseteq \O       | c = Definition of Set Equality }} {{end-eqn}} Conversely: {{begin-eqn}} {{eqn | l = A \subseteq \O       | o = \implies       | r = A \subseteq \O \land \O \subseteq A       | c = Empty Set is Subset of All Sets }} {{eqn | o = \implies       | r = A = \O       | c = Definition of Set Equality }} {{end-eqn}} {{qed}} Category:Set Theory kqrh1diwhxhoognphe1sh0e1brfoe71
Let $A \subset B$ denote that $A$ is a proper subset of $B$. Let $A \in B$ denote that $A$ is an element of $B$. From Transitive Set is Proper Subset of Ordinal iff Element of Ordinal, $\subset$ and $\in$ can be used interchangeably. Thus: {{begin-eqn}} {{eqn | l = x \subseteq y       | o = \implies       | r = x \subseteq y^+       | c = Ordinal is Less than Successor }} {{eqn | l = x = y^+       | o = \implies       | r = x \subseteq y^+       | c = {{Defof|Set Equality|index = 2}} }} {{end-eqn}} Conversely: {{begin-eqn}} {{eqn | l = x \subseteq y^+       | o = \implies       | r = \left({x \subset y^+ \lor x = y^+}\right) }} {{eqn | l = x \subset y^+       | o = \implies       | r = x \in y^+       | c = Transitive Set is Proper Subset of Ordinal iff Element of Ordinal }} {{eqn | o = \implies       | r = \left({x = y \lor x \in y}\right)       | c = {{Defof|Successor Set}} }} {{eqn | o = \implies       | r = x \subseteq y       | c = Transitive Set is Proper Subset of Ordinal iff Element of Ordinal }} {{end-eqn}} {{qed}} Category:Ordinals csh1k3z2zeonzncbp8m6fkbfxz0vtbu
Proof follows from Indexed Union Subset and definition of set equality. {{handwaving}} Category:Set Union dduegthppwj0bh28jrjl7czbgq63tfe
By Ordinal Addition by Zero: :$x = \left({x + \varnothing}\right) = \left({\varnothing + x}\right)$ By Membership is Left Compatible with Ordinal Addition: :$\varnothing < y \implies x < \left({x + y}\right)$ But if $y = \varnothing$, then it is clear the inequality holds as well. {{explain|How? If $y {{=}} \varnothing$ then $\varnothing < y$ certainly does not hold. Clarification needed as to what exactly is meant here.}} One of these cases must hold by Empty Set is Subset of All Sets. So in either case: :$x \le \left({x + y}\right)$ Similarly, by Subset is Right Compatible with Ordinal Addition: :$\varnothing \le y \implies x \le \left({y + x}\right)$ The fact that $\varnothing \le y$ is clear from Empty Set is Subset of All Sets. Therefore: :$x \le \left({y + x}\right)$ {{qed}} Category:Ordinal Arithmetic 5zkmy5h6cj61hi57ep4yao04wop3qan
{{begin-eqn}} {{eqn | l = x \in K_{II}       | o = \implies       | r = x \ne \varnothing       | c = Definition of Limit Ordinal }} {{eqn | o = \implies       | r = \left({y + x}\right) \ne \varnothing       | c = Ordinal is Less than Sum }} {{eqn | o = \implies       | r = \left({y + x}\right) \in K_{II} \lor \left({y + x}\right) = z^+       | c = Definition of Limit Ordinal }} {{end-eqn}} The result is now obtained by Proof by Contradiction: Assume that $\left({y + x}\right) = z^+$. {{begin-eqn}} {{eqn | l = \left({y + x}\right) = z^+       | o = \implies       | r = \bigcup_{w \mathop \in x} \left({y + w}\right) = z^+       | c = Definition of Ordinal Addition }} {{eqn | o = \implies       | r = z \in \bigcup_{w \mathop \in x} \left({y + w}\right)       | c = Ordinal is Less than Successor }} {{eqn | o = \implies       | r = \exists w \in x: z \in \left({y + w}\right)       | c = Definition of Set Union }} {{end-eqn}} But $w \in x \implies w^+ \in x$ by Successor in Limit Ordinal. {{begin-eqn}} {{eqn | l = z \in \left({y + w}\right)       | o = \implies       | r = z^+ \in \left({y + w}\right)^+       | c = Successor is Less than Successor }} {{eqn | o = \implies       | r = z^+ \in \left({y + w}\right)       | c = Definition of Ordinal Addition }} {{eqn | o = \implies       | r = z^+ \in z^+        | c = Assumption }} {{end-eqn}} But $z^+ \in z^+$ is clearly a membership loop, and therefore, our assumption must be wrong. By disjunctive syllogism: : $\left({y + x}\right) \in K_{II}$ {{qed}}
Since $x$ is a limit ordinal: : $x \ne y^+$ Moreover by Successor of Element of Ordinal is Subset: : $y \in x \implies y^+ \subseteq x$ Therefore by Transitive Set is Proper Subset of Ordinal iff Element of Ordinal: : $y^+ \subset x$ and $y^+ \in x$ {{qed}} Category:Ordinals alnkhhfhchk499c90s39l96im9nav7m
{{begin-eqn}} {{eqn | l = x \in y       | o = \implies       | r = x^+ \in y^+       | c = Subset is Compatible with Ordinal Successor }} {{eqn | l = x \in y       | o = \impliedby       | r = x^+ \in y^+       | c = Sufficient Condition }} {{eqn | ll = \implies       | l = x \in y       | o = \iff       | r = x^+ \in y^+       | c =  }} {{end-eqn}} {{qed}} Category:Ordinals 3ybapwazqv3gj3077nq7eh4lzn0vjqb
By Transfinite Induction on $z$. === Basis for the Induction === {{begin-eqn}} {{eqn|l = \left({x + y}\right) + \varnothing      |r = x + y      |c = Definition of ordinal addition }} {{eqn|r = x + \left({y + \varnothing}\right)      |c = Definition of ordinal addition }} {{end-eqn}} This proves the basis for the induction. {{qed|lemma}} === Induction Step === Suppose that: :$x + \left({y + z}\right) = \left({x + y}\right) + z$ Then: {{begin-eqn}} {{eqn|l = \left({ \left({x + y}\right) + z}\right)^+      |r = \left({x + \left({y + z}\right) }\right)^+      |c = Equality of Successors }} {{eqn|ll= \implies      |l = \left({x + y}\right) + z^+      |r = x + \left({y + z}\right)^+      |c = Definition of ordinal addition }} {{eqn|r = x + \left({y + z^+}\right)      |c = Definition of ordinal addition }} {{end-eqn}} This proves the induction step. {{qed|lemma}} === Limit Case === Let $z \in K_{II}$ be a limit ordinal, and suppose that: :$\forall w \in z: x + \left({y + w}\right) = \left({x + y}\right) + w$ Then it follows that: {{begin-eqn}} {{eqn|l = \bigcup_{w \mathop \in z} \left({x + \left({y + w}\right) }\right)      |r = \bigcup_{w \mathop \in z} \left({ \left({x + y}\right) + w}\right)      |c = Indexed Union Equality }} {{eqn|r = \left({x + y}\right) + z      |c = Definition of ordinal addition }} {{end-eqn}} By the hypothesis, we have that $z \in K_{II}$. From Limit Ordinals Preserved Under Ordinal Addition, also $y + z \in K_{II}$. Therefore, by definition of ordinal addition: :$x + \left({y + z}\right) = \displaystyle \bigcup_{n \mathop \in y + z} \left({x + n}\right)$ So it is sufficient to prove that: :$\displaystyle \bigcup_{w \mathop \in z} \left({x + \left({y + w}\right) }\right) = \bigcup_{n \mathop \in y + z} \left({x + n}\right)$ Take any $w \in z$.  Then by Membership is Left Compatible with Ordinal Addition, $y + w < y + z$. Setting $n = y + w$, we now have that: :$x + \left({y + w}\right) \le x + n$ By Supremum Inequality for Ordinals, it now follows that: :$\displaystyle \bigcup_{w \mathop \in z} \left({x + \left({y + w}\right) }\right) \le \bigcup_{n \mathop \in y + z} \left({x + n}\right)$ To prove the converse inequality, take any $n \in y + z$. By definition of ordinal addition, this means: :$n \in \displaystyle \bigcup_{w \mathop \in z} \left({y + w}\right)$ and so, for some $w \in z$, we have $n \in y + w$. By Membership is Left Compatible with Ordinal Addition, this yields: :$x + n \in x + \left({y + w}\right)$ and whence by Supremum Inequality for Ordinals: :$\displaystyle \bigcup_{n \mathop \in y + z} \left({x + n}\right) \le \bigcup_{w \mathop \in z} \left({x + \left({y + w}\right) }\right)$ By definition of set equality: :$\displaystyle \bigcup_{n \mathop \in y + z} \left({x + n}\right) = \bigcup_{w \mathop \in z} \left({x + \left({y + w}\right) }\right)$ This proves the limit case. {{qed|lemma}} Hence the result, by Transfinite Induction. {{qed}}
{{begin-eqn}} {{eqn | l = \frac {\d y} {\d x} \frac {\d x} {\d u}       | r = \frac {\d y} {\d u}       | c = Derivative of Composite Function }} {{eqn | ll= \leadsto       | l = \frac {\d y} {\d x}       | r = \frac {\paren {\dfrac {\d y} {\d u} } } {\paren {\dfrac {\d x}{\d u} } }       | c = divide both sides by $\dfrac {\d x} {\d u}$ }} {{end-eqn}} {{qed}}
Let $X = \set x$. We have from Product of Subset with Intersection that: : $X \circ \paren {Y \cap Z} \subseteq \paren {X \circ Y} \cap \paren {X \circ Z}$ Let $g \in \paren {X \circ Y} \cap \paren {X \circ Z}$. Then by definition of subset product: : $\exists y \in Y, z \in Z: g = x \circ y = x \circ z$ By the Cancellation Laws it follows that $y = z$. But: : $g = x \circ y \implies g \in X \circ Y$ : $g = x \circ z \implies g \in X \circ Z$ and so as $x \circ y = x \circ z$ it follows that $g \in X \circ \paren {Y \cap Z}$. Thus: : $X \circ \paren {Y \cap Z} \supseteq \paren {X \circ Y} \cap \paren {X \circ Z}$ It follows by definition of set equality that: :$X \circ \paren {Y \cap Z} = \paren {X \circ Y} \cap \paren {X \circ Z}$ By similar reasoning: :$\paren {Y \cap Z} \circ X = \paren {Y \circ X} \cap \paren {Z \circ X}$ Hence the result. {{qed}}
The zero function is computed by the following URM program: {| |- ! align="right" | Line !! ! align="left" | Command |- | align="right" | $1$ || | align="left" | $\map Z 1$ |} This sets the value $0$ into $R_1$ and then stops. The output $0$ is in $R_1$ when the program terminates. {{qed}} Category:URM Programs Category:Primitive Recursive Functions j8wnk5c37v24jr0wfwo7ehgessgcfj2
The successor function is computed by the following URM program: {| |- ! align="right" | Line !! ! align="left" | Command |- | align="right" | $1$ || | align="left" | $\map S 1$ |} The input $n$ is in $R_1$ when the program starts. The program adds $1$ to $r_1$ and then stops. The output $n + 1$ is in $R_1$ when the program terminates. {{qed}} Category:URM Programs Category:Primitive Recursive Functions 4uvfxbgayev57oipuc6m5u5z7apfukr
The projection functions are computed by the following URM program: {| |- ! align="right" | Line !! ! align="left" | Command |- | align="right" | $1$ || | align="left" | $\map C {j, 1}$ |} The input $\tuple {n_1, n_2, \ldots, n_j, \ldots, n_k}$ is in $R_1, R_2, \ldots, R_j, \ldots, R_k$ when the program starts. The program copies $r_j$ to $r_1$ and then stops. The output $n_j$ is in $R_1$ when the program terminates. {{qed}} Category:URM Programs Category:Primitive Recursive Functions d2vaqix20wd0djv8t6s5w14bb7z4trw
{{AimForCont}} $a \in S$ such that $a$ is idempotent under $\circ$. That is: :$a \circ a = a$ Then: :$\paren {a \circ a} \circ a = a \circ a$  and  :$a \circ \paren {a \circ a} = a \circ a$ This contradicts our assumption that $\circ$ is antiassociative on $S$. {{qed}} Category:Antiassociative Structures tncjel62vl0ccibyi69iuwofmsvsf74
We will show there are two elements in $S$ that do not commute. Let $a \in S$. From Antiassociative Operation has no Idempotent Elements: :$a \circ a \ne a$ So for some $b \in S$:  :$a \circ a = b$ Then: :$\left({a \circ a}\right) \circ a = b \circ a$ and: :$a \circ \left({a \circ a}\right) = a \circ b$ From our assumption, $\circ$ is antiassociative. So: :$\left({a \circ a}\right) \circ a \ne a \circ \left({a \circ a}\right)$ Hence for some $a, b \in S$: :$a \circ b \ne b \circ a$ {{qed}} Category:Antiassociative Structures l9m4ygj9k0dih8kp5opewenkd490cv5
Let $S = \left\{{a_0, \ldots, a_{n - 1}}\right\}$. Let $\circ$ be a binary operation defined on $S$ such that: :$\forall x \in S: x \circ a_i = a_{\left({i + 1}\right) \pmod n}$ Then $\forall j, k, m \in \left[{0 \,.\,.\, n-1}\right]$: {{begin-eqn}} {{eqn | l = \left({a_j \circ a_k}\right) \circ a_m       | r = a_{\left({m + 1}\right) \pmod n} }} {{end-eqn}} and: {{begin-eqn}} {{eqn | l = a_j \circ \left({a_k \circ a_m}\right)       | r = a_j \circ a_{\left({m + 1}\right) \pmod n} }} {{eqn | r = a_{\left({m + 2}\right) \pmod n} }} {{end-eqn}} As $n \ge 2$ it follows that: :$m + 1 \ne m + 2 \pmod n$ and so $\forall j, k, m \in \left[{0 \,.\,.\, n - 1}\right]$: :$\left({a_j \circ a_k}\right) \circ a_m \ne a_j \circ \left({a_k \circ a_m}\right)$ Hence the result. {{qed}} Category:Antiassociative Structures 8t3e3r774sn1qorbevb0l7t7cwpxcpg
Follows from Reciprocal Function is Strictly Decreasing and from Restriction of Monotone Function is Monotone. {{qed}}
Suppose that $(1)$ and $(2)$ hold. From $(1)$, we obtain for every $\lambda \in K$ and $u \in U$ that $\lambda u \in U$. An application of $(2)$ yields the condition of the One-Step Vector Subspace Test. Hence $U$ is a vector subspace of $V$. {{qed}}
Let $x \in \left ({0 \,.\,.\, +\infty} \right)$. By the definition of negative powers: :$\dfrac 1 x = x^{-1}$ From Power Rule for Derivatives: {{begin-eqn}} {{eqn | l = \frac {\mathrm d} {\mathrm d x} x^{-1}       | r = -x^{-2} }} {{end-eqn}} From Square of Real Number is Non-Negative: : $-x^{-2} < 0$ for all $x$ within the domain. Thus from Derivative of Monotone Function, $\operatorname{recip}$ is strictly decreasing. The proof for $x \in \left ({-\infty \,.\,.\, 0} \right)$ is similar. {{qed}}
=== Strictly Increasing on $(0 \,.\,.\, \to)$ === Let $a,b \in \operatorname{Dom}\left({\operatorname{recip}}\right)$ such that $a$ and $b$ are both positive. Let $a < b$. Then $0 < a < b$. By Properties of Totally Ordered Field: :$0 < b^{-1} < a^{-1}$ That is, $\dfrac 1 b < \dfrac 1 a$. {{qed|lemma}} === Strictly Increasing on $(\gets \,.\,.\, 0)$ === Let $a,b \in \operatorname{Dom}\left({\operatorname{recip}}\right)$ such that $a$ and $b$ are both negative. Let $a < b$. Then $a < b < 0$. By Inversion Mapping Reverses Ordering in Ordered Group: :$0 < -b < -a$ By Properties of Totally Ordered Field: :$0 < (-a)^{-1} < (-b)^{-1}$ By Negative of Product Inverse: :$(-b)^{-1} = -b^{-1}$ :$(-a)^{-1} = -a^{-1}$ Thus: :$0 < -a^{-1} < -b^{-1}$ By Inversion Mapping Reverses Ordering in Ordered Group: :$-(-b^{-1}) < -(-a^{-1}) < 0$ By Inverse of Group Inverse: :$b^{-1} < a^{-1} < 0$ Thus in particular: :$\dfrac 1 b < \dfrac 1 a$ {{qed}}
We proceed by induction over $n \ge 1$. === Basis for the Induction === If $n = 1$ the result is trivially true. This establishes the basis for the induction. === Induction Hypothesis === This is our induction hypothesis: :$\displaystyle \prod_{i \mathop = 1}^{n - 1} p_i = \displaystyle \sum_{k \mathop \in Z} c_d X^k$ where: :$\displaystyle c_d := \sum_{k_1 + \cdots + k_{n - 1} = d} \paren {\prod_{i \mathop = 1}^{n - 1} a_{i, k_i} }$ Now we need to show that the result is true for the product $\displaystyle \prod_{i \mathop = 1}^n p_i$. === Induction Step === This is our induction step: Let $b_k$ be the coefficient of $X^k$ in $\displaystyle \prod_{i \mathop = 1}^n p_i$. Then: {{begin-eqn}} {{eqn | l = b_k       | r = \sum_{d + k_n \mathop = k} c_d a_{n, k_n}       | c = {{Defof|Multiplication of Polynomial Forms}} }} {{eqn | r = \sum_{d + k_n \mathop = k}\ \sum_{k_1 + \cdots + k_{n - 1} \mathop = d} \paren {\prod_{i \mathop = 1}^{n-1} a_{i, k_i} } a_{n, k_n}       | c = Induction Hypothesis }} {{eqn | r = \sum_{d + k_n \mathop = k}\ \sum_{k_1 + \cdots + k_{n - 1} \mathop = d} \paren {\prod_{i \mathop = 1}^n a_{i, k_i} }       | c =  }} {{eqn | r = \sum_{\substack {k_1 + \dotsb + k_{n - 1} \mathop = d \\ d + k_n \mathop = k} } \paren {\prod_{i \mathop = 1}^n a_{i, k_i} }       | c =  }} {{eqn | r = \sum_{k_1 + \cdots + k_n = k} \paren {\prod_{i \mathop = 1}^n a_{i, k_i} }       | c =  }} {{end-eqn}} The result follows by induction. {{qed}} Category:Polynomial Theory Category:Proofs by Induction nzqrwdbhmmu7s566vxk03zz9gmx4ept
Suppose $H \circ K = K \circ H$. Then: {{begin-eqn}} {{eqn | l = \paren {H \circ K} \circ \paren {H \circ K}^{-1}       | r = H \circ K \circ K^{-1} \circ H^{-1}       | c = Inverse of Product of Subsets of Group }} {{eqn | r = H \circ K \circ K \circ H       | c = Inverse of Subgroup }} {{eqn | r = H \circ K \circ H       | c = Product of Subgroup with Itself }} {{eqn | r = H \circ H \circ K       | c = by hypothesis }} {{eqn | r = H \circ K       | c = Product of Subgroup with Itself }} {{end-eqn}} That is: :$\paren {H \circ K} \circ \paren {H \circ K}^{-1} = H \circ K$ Thus by definition of set equality: :$\paren {H \circ K} \circ \paren {H \circ K}^{-1} \subseteq H \circ K$ So from One-Step Subgroup Test using Subset Product: : $H \circ K \le G$ {{qed}}
Suppose $H \circ K = K \circ H$. First note that $H \circ K \ne \O$, as $e_G = e_G \circ e_G \in H \circ K$, from Identity of Subgroup. Suppose $a_1, a_2 \in H, b_1, b_2 \in K$. Then: :$\paren {a_1 \circ b_1} \circ \paren {a_2 \circ b_2} = a_1 \circ \paren {b_1 \circ a_2} \circ b_2$. Since $H \circ K = K \circ H$, we see that, for some $a \in H, b \in K$: :$b_1 \circ a_2 = a \circ b$ Thus: :$\paren {a_1 \circ b_1} \circ \paren {a_2 \circ b_2} = \paren {a_1 \circ a} \circ \paren {b \circ b_2}$ As $H, K \le G$, we have $a_1 \circ a \in H$ and $b \circ b_2 \in K$, hence: :$\paren {a_1 \circ b_1} \circ \paren {a_2 \circ b_2} \in H \circ K$ thus demonstrating closure of $H \circ K$ under $\circ$. Finally, if $a \circ b \in H \circ K$, then by Inverse of Group Product: :$\paren {a \circ b}^{-1} = b^{-1} \circ a^{-1}$ Since $b^{-1} \in K$ and $a^{-1} \in H$, we have: :$\paren {a \circ b}^{-1} \in K \circ H$ and hence $H \circ K$ is shown to be closed under inverses. Thus, from the Two-Step Subgroup Test, $H \circ K$ is a subgroup of $G$. {{qed}}
Suppose $H \circ K$ is a subgroup of $G$. Then: {{begin-eqn}} {{eqn | l = H \circ K       | r = \paren {H \circ K}^{-1}       | c = Inverse of Subgroup }} {{eqn | r = K^{-1} \circ H^{-1}       | c = Inverse of Product of Subsets of Group }} {{eqn | r = K \circ H       | c = Inverse of Subgroup }} {{end-eqn}} {{qed}}