hcju/proofwikiqa huggingface dataset:
Let $X$ be an $F_\sigma$ set of $T$. Then $X = \displaystyle \bigcup \VV$ where $\VV$ is a countable union of closed sets in $T$. Then from De Morgan's Laws: Difference with Union we have: :$\displaystyle S \setminus X = S \setminus \bigcup \VV = \bigcap_{V \mathop \in \VV} \paren {S \setminus V}$ By definition of closed set, each of the $S \setminus V$ are open sets. So $\displaystyle \bigcap_{V \mathop \in \VV} \paren {S \setminus V}$ is a countable intersection of open sets in $T$. Hence $S \setminus X$ is, by definition, a $G_\delta$ set of $T$. {{qed}}
$V$ is the union of a singleton. So $V$ is trivially the union of a countable number of closed sets of $T$. The result follows by definition of $F_\sigma$ set. {{qed}}
$U$ is the intersection of a singleton. So $U$ is trivially the intersection of a countable number of open sets of $T$. The result follows by definition of $G_\delta$ set. {{qed}}
=== Necessary Condition === Let $V$ be open in $T$. Let $z \in V$. By definition, a neighborhood of $z$ is any subset of $S$ containing an open set which itself contains $z$. But $V$ is itself an open set which itself contains $z$. Hence by Set is Subset of Itself, $V$ is a subset of $S$ which contains an open set which itself contains $z$. So for all points of $z \in V$, $V$ is a neighborhood of $z$. {{qed|lemma}} === Sufficient Condition === Suppose that for all points of $z \in V$, $V$ is a neighborhood of $z$. That is, for all $z \in V$ there exists an open set $T_z \subseteq V$ of $T$ such that $z \in T_z$. Now by Union is Smallest Superset: Family of Sets: :$\displaystyle \bigcup_{z \mathop \in V} T_z \subseteq V$ as $\forall z \in V: T_z \subseteq V$. If $z \in V$, then $z \in T_z$ by definition of $T_z$. So: :$\displaystyle z \in \bigcup_{z \mathop \in V} T_z$ Thus we also have: :$\displaystyle V \subseteq \bigcup_{z \mathop \in V} T_z$ Hence by definition of set equality: :$\displaystyle V = \bigcup_{z \mathop \in V} T_z$ Thus $V$ can be expressed as a union of open sets. Hence $V$ is open in $T$, by axiom $(\text O 1)$ of a topological space. {{qed}}
Let $x \in C$. By definition of condensation point, every open set of $T$ containing $x$ also contains an uncountable number of points of $A$. As an uncountable number is also an infinite number, we could also say that every open set of $T$ containing $x$ also contains an infinite number of points of $A$. That is, $x$ is also by definition an $\omega$-accumulation point. So $x \in W$ and by definition of subset: :$C \subseteq W$ Note that if $x \in W$ then it could be that there exists an open set $U$ of $T$ containing $x$ with a countably infinite number of points of $A$. In that case $x \notin C$. That is, not every $\omega$-accumulation point is necessarily a condensation point. {{qed|lemma}} Let $x \in W$. By definition of $\omega$-accumulation point, every open set $U$ of $T$ containing $x$ also contains an infinite number of points of $A$. So every open set $U$ of $T$ such that $x \in U$ contains some point of $A$ other than $x$ (an infinite number, indeed). That is, $x$ is also by definition a limit point. So $x \in L$ and by definition of subset: :$W \subseteq L$ <!-- Note that if $x \in L$ then it could be that there exists an open set $U$ of $T$ containing $x$ which contains only a finite number of points of $A$. In that case $x \notin W$. --> Let $T = \struct {S, \tau_p}$ be a particular point space. From Limit Points in Particular Point Space, every point $x \ne p$ is a limit point of $T$. From Point in Particular Point Space is not Omega-Accumulation Point, $x$ is not an definition of $\omega$-accumulation point of $T$.  So it is seen that not every limit point is necessarily an $\omega$-accumulation point. {{qed|lemma}} Let $x \in L$. By definition of limit point, every open set $U$ of $T$ containing $x$ also contains some point of $A$ other than $x$. So every open set $U$ of $T$ such that $x \in U$ contains some point of $A$. That is, $x$ is also by definition an adherent point. So $x \in L$ and by definition of subset: :$L \subseteq D$ Note that if $x \in D$ then it could be that there exists an open set $U$ of $T$ containing $x$ in which the only point of $A$ is $x$ itself. In that case $x \notin L$. That is, not every adherent point is necessarily a limit point. {{qed|lemma}} Hence the result. {{qed}}
Let $\alpha$ be a limit point of $\sequence {x_n}$. Then by definition of limit, $\sequence {x_n}$ converges to $\alpha$. By definition of convergence that means: :for any open set $U \subseteq T$ such that $\alpha \in U$: $\exists N \in \R: n > N \implies x_n \in U$. As there is an infinite number of values of $n > N$, there are an infinite number of terms of $\sequence {x_n}$. Hence the result, from definition of accumulation point. {{qed}}
From Moment Generating Function of Poisson Distribution, the moment generating functions $X$ and $Y$, $M_X$ and $M_Y$, are given by:  :$\displaystyle \map {M_X} t = e^{\lambda_1 \paren {e^t - 1} }$ and :$\displaystyle \map {M_Y} t = e^{\lambda_2 \paren {e^t - 1} }$ As $X$ and $Y$ are independent, we may apply Moment Generating Function of Linear Combination of Independent Random Variables, to give:  {{begin-eqn}} {{eqn	| l = \map {M_{X + Y} } t  	| r = \map {M_X} t \map {M_Y} t 	| c = Moment Generating Function of Linear Combination of Independent Random Variables }} {{eqn	| r = e^{\lambda_1 \paren {e^t - 1} } e^{\lambda_2 \paren {e^t - 1} } }} {{eqn	| r = e^{\paren {\lambda_1 + \lambda_2} \paren {e^t - 1} } 	| c = Exponential of Sum }} {{end-eqn}} This is the moment generating function of a random variable with distribution $\Poisson {\lambda_1 + \lambda_2}$.  So, $X + Y \sim \Poisson {\lambda_1 + \lambda_2}$. {{qed}}
From the general definition of a filter, we have: A '''filter on $\left({S, \preccurlyeq}\right)$''' is a subset $\mathcal F \subseteq S$ which satisfies the following conditions: : $\mathcal F \ne \varnothing$ : $x, y \in \mathcal F \implies \exists z \in \mathcal F: z \preccurlyeq x, z \preccurlyeq y$ : $\forall x \in \mathcal F: \forall y \in S: x \preccurlyeq y \implies y \in \mathcal F$ A filter $\mathcal F$ is proper if it does not equal $S$ itself. From the definition of a filter on a set, we have: A '''filter on $T$''' is a set $\mathcal F \subset \mathcal P \left({T}\right)$ which satisfies the following conditions: : $T \in \mathcal F$ : $\varnothing \notin \mathcal F$ : $U, V \in \mathcal F \implies U \cap V \in \mathcal F$ : $\forall U \in \mathcal F: U \subseteq V \subseteq T \implies V \in \mathcal F$ We can identify: : $\mathcal P \left({T}\right)$ with $S$ : $\subseteq$ with $\preccurlyeq$. === Filter Not Empty === We have that $T \in \mathcal F$ and so $\mathcal F \ne \varnothing$. === Preceding Elements in Filter === We have that: :$U, V \in \mathcal F \implies U \cap V \in \mathcal F$ From Intersection is Subset, we have that $U \cap V \subseteq U$ and $U \cap V \subseteq V$. So identifying $U$ with $x$, $V$ with $y$ and $U \cap V$ with $z$ it is clear that: : $x, y \in \mathcal F \implies \exists z \in \mathcal F: z \preccurlyeq x, z \preccurlyeq y$ === Succeeding Elements in Filter === We have that: : $\forall U \in \mathcal F: U \subseteq V \subseteq T \implies V \in \mathcal F$ This can be rewritten: : $\forall U \in \mathcal F, V \in \mathcal P \left({T}\right): U \subseteq V \implies V \in \mathcal F$ Identifying $U$ with $x$ and $V$ with $y$, this translates as: : $\forall x \in \mathcal F, y \in S: x \preccurlyeq y \implies y \in \mathcal F$ === Proper Filter === For $\mathcal F$ to be a proper filter on $\left({\mathcal P \left({T}\right), \subseteq}\right)$, it must not equal $\mathcal P \left({T}\right)$. This is seen to be satisfied by the axiom $\varnothing \notin \mathcal F$. All axioms are fulfilled, hence the result. {{qed}}
=== Law of Excluded Middle implies Peirce's Law === Let the truth of the Law of Excluded Middle be assumed. Then: :$\paren {p \lor \neg p} \vdash \paren {\paren {p \implies q} \implies p} \implies p$ is demonstrated, as follows. {{:Law of Excluded Middle implies Peirce's Law}} The result follows by an application of Modus Ponendo Ponens: :$\paren {\paren {p \implies q} \implies p} \implies p, \paren {p \implies q} \implies p \vdash p$ {{qed|lemma}} === Peirce's Law implies Law of Excluded Middle === {{:Peirce's Law implies Law of Excluded Middle}} Category:Peirce's Law Category:Law of Excluded Middle lccwfn88z3nvrhfp9lbdjoszhi64i4h
A propositional variable is already trivially in negation normal form (NNF). So we consider the general propositional formula $S$. In the following, $P$ and $Q$ are to stand for general propositional formulas. First, from Functionally Complete Logical Connectives: Conjunction, Negation and Disjunction we have that: :$\left\{{\neg, \land, \lor}\right\}$ forms a functionally complete set of logical connectives. So we can convert $S$ so that: * By the definition of the biconditional, instances of $P \iff Q$ can be converted into $\left({P \implies Q}\right) \land \left({Q \implies P}\right)$; * By the Rule of Material Implication, instances of $P \implies Q$ can be converted into $\neg P \lor Q$. Other connectives can likewise be treated appropriately. Thus $S$ will contain nothing but connectives from $\left\{{\neg, \land, \lor}\right\}$. Next we can replace: * Instances of $\neg \left({P \land Q}\right)$ with $\neg P \lor \neg Q$, by De Morgan's Laws: Disjunction of Negations; * Instances of $\neg \left({P \lor Q}\right)$ with $\neg P \land \neg Q$, by De Morgan's Laws: Conjunction of Negations; * Instances of $\neg \neg P$ with $P$ by Double Negation Elimination. At any stage where a negation appears before a parenthesis, it will then appear before the statements inside the parenthesis. Thus the negation signs gradually move inwards or are eliminated. Eventually all remaining negation signs will appear next to simple statements. Hence the result. {{qed}} Category:Propositional Logic gzz56iayclxe1zjof26y760kwdu7lbx
A propositional variable is already trivially in conjunctive normal form (CNF). So we consider the general propositional formula $S$. First we convert to negation normal form (NNF). This is always possible, by Existence of Negation Normal Form of Statement. Now $S$ will be of the form: : $P_1 \land P_2 \land \cdots \land P_n$ where $P_1, P_2, \ldots, P_n$ are either: * Literals; * Statements of the form $\left({Q_1 \lor Q_2 \lor \ldots \lor Q_n}\right)$ If all the $Q_1, \ldots, Q_n$ are literals we have finished. Otherwise they will be of the form $Q_j = \left({R_1 \land R_2 \land \ldots \land R_m}\right)$ If the latter is the case, then use the Disjunction Distributes over Conjunction to convert: : $Q_1 \lor Q_2 \lor \ldots \lor \left({R_1 \land R_2 \land \ldots \land R_m}\right) \ldots \lor Q_n$ into: {{begin-eqn}} {{eqn | o =        | r = \left({Q_1 \lor Q_2 \lor \ldots \lor Q_n \lor R_1}\right)       | c =  }} {{eqn | o = \land       | r = \left({Q_1 \lor Q_2 \lor \ldots \lor Q_n \lor R_2}\right)       | c =  }} {{eqn | o = \land       | r = \ldots       | c =  }} {{eqn | o = \land       | r = \left({Q_1 \lor Q_2 \lor \ldots \lor Q_n \lor R_m}\right)       | c =  }} {{end-eqn}} It is taken for granted that Disjunction is Associative and Disjunction is Commutative. It can be seen then that each of the : $\left({Q_1 \lor Q_2 \lor \ldots \lor Q_n \lor R_k}\right)$ are terms in the CNF expression required. If any terms are still not in the correct format, then use the above operation until they are. {{qed}}
A propositional variable is already trivially in disjunctive normal form (DNF). So we consider the general propositional formula $S$. First we convert to negation normal form (NNF). This is always possible, by Existence of Negation Normal Form of Statement. Now $S$ will be of the form: : $P_1 \lor P_2 \lor \cdots \lor P_n$ where $P_1, P_2, \ldots, P_n$ are either: * Literals; * Statements of the form $\left({Q_1 \land Q_2 \land \ldots \land Q_n}\right)$ If all the $Q_1, \ldots, Q_n$ are literals we have finished. Otherwise they will be of the form $Q_j = \left({R_1 \lor R_2 \lor \ldots \lor R_m}\right)$ If the latter is the case, then use Conjunction Distributes over Disjunction to convert: : $Q_1 \land Q_2 \land \ldots \land \left({R_1 \lor R_2 \lor \ldots \lor R_m}\right) \ldots \land Q_n$ into: {{begin-eqn}} {{eqn | o =        | r = \left({Q_1 \land Q_2 \land \ldots \land Q_n \land R_1}\right)       | c =  }} {{eqn | o = \lor       | r = \left({Q_1 \land Q_2 \land \ldots \land Q_n \land R_2}\right)       | c =  }} {{eqn | o = \lor       | r = \ldots       | c =  }} {{eqn | o = \lor       | r = \left({Q_1 \land Q_2 \land \ldots \land Q_n \land R_m}\right)       | c =  }} {{end-eqn}} It is taken for granted that Conjunction is Associative and Conjunction is Commutative. It can be seen then that each of the : $\left({Q_1 \land Q_2 \land \ldots \land Q_n \land R_k}\right)$ are terms in the DNF expression required. If any terms are still not in the correct format, then use the above operation until they are. {{qed}}
Let us refer to this group as $D_3$. Taking the group axioms in turn: === G0: Closure === From the Cayley table it is seen directly that $D_3$ is closed. {{qed|lemma}} === G1: Associativity === Composition of Mappings is Associative. {{qed|lemma}} === G2: Identity === The identity is $e = (A) (B) (C)$. {{qed|lemma}} === G3: Inverses === Each element can be seen to have an inverse: :$p^{-1} = q$ and so $q^{-1} = p$ :$r$, $s$ and $t$ are all self-inverse. {{qed|lemma}} No more need be done. $D_3$ is seen to be a group. {{qed}}
Let $\GF$ be a Galois field with $p$ elements: $\card \GF = p$. Let $A = \sqbrk {a_{i j} }_{n, n}$ be a matrix such that $\size A \ne 0$ and $a_{i j} \in \GF$. How many such matrices can be constructed? In order to avoid a zero determinant, the top row of the matrix, $\sequence {a_{1 j} }_{j \mathop = 1, \dotsc, n}$ must have at least one non-zero element. Therefore there are $p^n - 1$ possibilities for the top row: :the $p^n$ possible sequences of $n$ values from $\GF$, minus the one sequence $\paren {0, 0, \dotsc, 0}$. The only restriction on the second row is that it not be a multiple of the first. Therefore, there are the $p^n$ possible sequences again, minus the $p$ sequences which are multiples of the first row. Thus, continuing in this fashion, the $j^{th}$ row can be any of the $p^n$ possible sequences, minus the $p^{\paren {j - 1} }$ sequences which are linear combinations of previous rows. The number of possible matrices satisfying the conditions of $A$, then, is: :$\displaystyle \prod_{j \mathop = 1}^n \paren {p^n - p^{j - 1} }$ {{qed}} Category:Combinatorics Category:General Linear Group Category:Galois Fields 5dqxuyc8lf2t1unfa2x8z86s9hwle4c
The Cayley table of a finite group of order $n$ is a Latin square, from Group has Latin Square Property. For every $n \in \Z_{>0}$ there exists a cyclic group of order $n$. It follows that for every $n \in \Z_{>0}$ the Cayley table of the cyclic group of order $n$ is a Latin square of order $n$. {{qed}} Category:Latin Squares lcrxk2uq9h858urmw4nuqubls67htzj
From the definition, a Noetherian ring is also a commutative ring with unity. Let $f = a_n x^n + \cdots + a_1 x + a_0 \in A \sqbrk x$ be a polynomial over $x$. Let $I \subseteq A \sqbrk x$ be an ideal of $A \sqbrk x$. We will show that $I$ is finitely generated.  Let $f_1$ be an element of least degree in $I$, and let $\ideal {g_1, \ldots, g_r}$ denote the ideal generated by the polynomials $g_1, \dotsc, g_r$. For $i \ge 1$, if $\ideal {f_1, \ldots, f_i} \ne I$, then choose $f_{i + 1}$ to be an element of minimal degree in $I \setminus \ideal {f_1, \ldots, f_i}$. If $\ideal {f_1, \ldots, f_i} = I$ then stop choosing elements. Let $a_j$ be the leading coefficient of $f_j$. Since $A$ is Noetherian, the ideal $\ideal {a_1, a_2, \ldots} \subseteq A$ is generated by $a_1, a_2, \ldots, a_m$ for some $m \in \N$. We claim that $f_1, f_2, \ldots, f_m$ generate $I$. {{AimForCont}} not. Then our process chose an element $f_{m + 1}$, and $\displaystyle a_{m + 1} = \sum_{j \mathop = 1}^m u_j a_j$ for some $u_j \in A$. Since the degree of $f_{m + 1}$ is greater than or equal to the degree of $f_j$ for $j = 1, \ldots, m$, the polynomial: :$\displaystyle g = \sum_{j \mathop = 1}^m u_j f_j x^{\deg f_{m + 1} - \deg f_j} \in \ideal {f_1, \ldots, f_m}$ has the same leading coefficient and degree as $f_{m + 1}$. The difference $f_{m + 1} - g$ is not in $\ideal {f_1, \ldots, f_m}$ and has degree strictly less than $f_{m + 1}$, a contradiction of our choice of $f_{m + 1}$. Thus $I = \ideal {f_1, \ldots, f_m}$ is finitely generated, and the proof is complete. {{qed}} {{Namedfor|David Hilbert|cat = Hilbert}} Category:Noetherian Rings Category:Commutative Algebra 23a4xpjufa4jjg4kxbrowbcljw7efxy
Let $\epsilon > 0$, $x,y \in M_1$.  Let $K$ be a Lipschitz constant for $g$. First, suppose that $K \le 0$. Then: : $d_1 \left({x, y}\right) \le 0 d_2 \left({g \left({x}\right), g \left({y}\right)}\right)$ by the Lipschitz condition on $g$. So $d_1 \left({x, y}\right) \le 0 \implies d_1 \left({x, y}\right) = 0 \implies x = y$ for all $x$ and $y$ in $M_1$. Thus $g$ is a constant function, which is uniformly continuous. Next, suppose that $K > 0$. Take $\delta = \epsilon / K$. Then if $d_1 \left({x, y}\right) < \delta$, we have: :$K d_1 \left({x, y}\right) < \epsilon$ By the Lipschitz condition on $g$, we know that: :$d_2 \left({g \left({x}\right), g \left({y}\right)}\right) \le K d_1 \left({x, y}\right)$ These last two statements together imply $d_2 \left({g \left({x}\right), g \left({y}\right)}\right) < \epsilon$. Thus, $g$ is uniformly continuous on $M_1$. {{qed}} Category:Metric Spaces Category:Uniformly Continuous Mappings ea6xnew77ye1oii2vdcbsdzb65ni81j
Existence follows from Integers Modulo m under Addition form Cyclic Group. Uniqueness follows from Cyclic Groups of Same Order are Isomorphic. {{Qed}}
By Subgroup is Normal iff Left Cosets are Right Cosets, the set of left cosets for $N$ equals the set of right cosets. It follows that $G / N$ does not depend on whether left cosets are used to define it or right cosets. {{WLOG}}, we will work with the left cosets. By definition of quotient group, the elements of $G / N$ are the cosets of $N$ in $G$, where the group operation is defined as: :$\paren {a N} \paren {b N} = \paren {a b} N$ The operation has been shown in Coset Product is Well-Defined to be a well-defined operation. Now we need to demonstrate that $G / N$ is a group. === $\text G 0$: Closure === This follows from Coset Product is Well-Defined. As $a b \in G$, it follows that $\paren {a b} N$ is a left coset. Thus $G / N$ is closed. {{qed|lemma}} === $\text G 1$: Associativity === The associativity of coset product follows directly from Subset Product within Semigroup is Associative: {{begin-eqn}} {{eqn | l = a N \paren {b N c N}       | r = a N \paren {b c N}       | c =  }} {{eqn | r = a \paren {b c} N       | c =  }} {{eqn | r = \paren {a b} c N       | c =  }} {{eqn | r = \paren {a b} N c N       | c =  }} {{eqn | r = \paren {a N b N} c N       | c =  }} {{end-eqn}} Thus $G / N$ is associative. {{qed|lemma}} === $\text G 2$: Identity === The left coset $e N = N$ serves as the identity: {{begin-eqn}} {{eqn | ll= \forall x \in G:       | l = N \paren {x N}       | r = \paren {e N} \paren {x N}       | c = Coset by Identity }} {{eqn | r = \paren {e x} N       | c =  }} {{eqn | r = x N       | c =  }} {{end-eqn}} Similarly $\paren {x N} N = x N$. {{qed|lemma}} === $\text G 3$: Inverses === We have $\paren {x N}^{-1} = x^{-1} N$: {{begin-eqn}} {{eqn | l = \paren {x N} \paren {x^{-1} N}       | r = \paren {x x^{-1} } N       | c =  }} {{eqn | r = e N       | c =  }} {{eqn | r = N       | c =  }} {{end-eqn}} Similarly $\paren {x^{-1} N} \paren {x N} = N$. Thus $x^{-1} N$ is the inverse of $x N$. {{qed|lemma}} All the group axioms are seen to be fulfilled, and $G / N$ has been shown to be a group. {{qed}}
=== Proof of Ideal === First we establish that $\ideal a$ is an ideal of $R$, by verifying the conditions of Test for Ideal. $\ideal a \ne \O$, as $1_R \circ a = a \in \ideal a$. 	 Let $x, y \in \ideal a$. Then: {{begin-eqn}} {{eqn | ll= \exists r, s \in R:       | l = x       | r = r \circ a, y = s \circ a       | c =  }} {{eqn | ll= \leadsto       | l = x + \paren {-y}       | r = r \circ a + \paren {-s \circ a}       | c =  }} {{eqn | r = \paren {r + \paren {-s} } \circ a       | c =  }} {{eqn | ll= \leadsto       | l = x + \paren {-y}       | o = \in       | r = \ideal a       | c =  }} {{end-eqn}} Let $s \in \ideal a, x \in R$. {{begin-eqn}} {{eqn | l = s       | o = \in       | r = \ideal a, x \in R       | c =  }} {{eqn | ll= \leadsto       | l = \exists r \in R: s       | r = r \circ a       | c =  }} {{eqn | ll= \leadsto       | l = x \circ s       | r = x \circ r \circ a       | c =  }} {{eqn | o = \in       | r = \ideal a       | c =  }} {{end-eqn}} and similarly $s \circ x \in \ideal a$. Thus by Test for Ideal, $\ideal a$ is an ideal of $R$. {{qed|lemma}} === Proof that Principal Ideal is Smallest === Let $J$ be an ideal of $R$ such that $a \in J$. By the definition of an ideal: :$\forall r, s \in R: r \circ a \circ s \in J$ Also, $J$ is a group under $+$. So every element of $\ideal a$ is in $J$. Thus $\ideal a \subseteq J$. {{qed}}
Proof by induction: For all $n \in \N_{>0}$, let $\map P n$ be the proposition: :$\tuple {a_1, a_2, \ldots, a_n} = \tuple {b_1, b_2, \ldots, b_n} \iff \forall i: 1 \le i \le n: a_i = b_i$ $\map P 1$ is true, as this just says $\tuple {a_1} = \tuple {b_1} \iff a_1 = b_1$ which is trivial. === Basis for the Induction === $\map P 2$ is the case: :$\tuple {a_1, a_2} = \tuple {b_1, b_2} \iff a_1 = b_1, a_2 = b_2$ which has been proved in Equality of Ordered Pairs. This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true. So this is our induction hypothesis: :$\tuple {a_1, a_2, \ldots, a_k} = \tuple {b_1, b_2, \ldots, b_k} \iff \forall i: 1 \le i \le n: a_i = b_i$ Then we need to show: :$\tuple {a_1, a_2, \ldots, a_{k + 1} } = \tuple {b_1, b_2, \ldots, b_{k + 1} } \iff \forall i: 1 \le i \le n: a_i = b_i$ === Induction Step === This is our induction step: {{begin-eqn}} {{eqn | l = \tuple {a_1, a_2, \ldots, a_{k + 1} }       | r = \tuple {b_1, b_2, \ldots, b_{k + 1} }       | c =  }} {{eqn | ll= \leadstoandfrom       | l = \tuple {a_1, \tuple {a_2, \ldots, a_{k + 1} } }       | r = \tuple {b_1, \tuple {b_2, \ldots, b_{k + 1} } }       | c =  }} {{eqn | ll= \leadstoandfrom       | l = a_1       | r = b_1       | c =  }} {{eqn | ll= \land       | l = \tuple {a_2, \ldots, a_{k + 1} }       | r = \tuple {b_2, \ldots, b_{k + 1} }       | c = Basis for the Induction }} {{end-eqn}} But from the induction hypothesis we have that: :$\tuple {a_2, \ldots, a_{k + 1} } = \tuple {b_2, \ldots, b_{k + 1} } \iff \forall i: 1 \le 2 \le k + 1: a_i = b_i$ So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\forall n \in \N_{>0}: \tuple {a_1, a_2, \ldots, a_n} = \tuple {b_1, b_2, \ldots, b_n} \iff \forall i: 1 \le i \le n: a_i = b_i$ {{qed}}
We note that by Set is Subset of Itself, $S \subseteq S$ and so $S \in \powerset S$ from the definition of the power set. From Union with Superset is Superset, we have: :$A \subseteq S \iff A \cup S = S = S \cup A$. By definition of power set: :$A \subseteq S \iff A \in \powerset S$ So: :$\forall A \in \powerset S: A \cup S = S = S \cup A$ Thus we see that $S$ acts as the zero. {{qed}}
From Empty Set is Element of Power Set: :$\O \in \powerset S$ From Intersection with Empty Set: :$\forall A \subseteq S: A \cap \O = \O = \O \cap A$ By definition of power set: :$A \subseteq S \iff A \in \powerset S$ So: :$\forall A \in \powerset S: A \cap \O = \O = \O \cap A$ Thus we see that $\O$ acts as the zero element for $\struct {\powerset S, \cap}$. {{qed}}