hcju/proofwikiqa huggingface dataset:
From Metric Induces Topology, a metric space induces a topological space. Then Equivalence of Definitions of Limit Point can be applied. {{qed}}
From Subset Relation is Ordering, $\subseteq$ is an ordering on $\mathcal F$. Let $F, F'$ be $F_\sigma$ sets of $T$. We have $F_\sigma$ Sets Closed under Union, so that $F \cup F' \in \mathcal F$. From Union is Smallest Superset and Subset of Union, it follows that $F \cup F'$ is the supremum of $F$ and $F'$. Similarly, we have $F_\sigma$ Sets Closed under Intersection, and so $F \cap F' \in \mathcal F$. From Intersection is Largest Subset and Intersection is Subset, it follows that $F \cap F'$ is the infimum of $F$ and $F'$. Thus any two elements of $\mathcal F$ are seen to have both a supremum and an infimum in $\mathcal F$. Hence $\left({\mathcal F, \subseteq}\right)$ is a lattice. {{qed}}
From Subset Relation is Ordering, $\subseteq$ is an ordering on $\mathcal G$. Let $G, G'$ be $G_\delta$ sets of $T$. We have $G_\delta$ Sets Closed under Union, so that $G \cup G' \in \mathcal G$. From Union is Smallest Superset and Subset of Union, it follows that $G \cup G'$ is the supremum of $G$ and $G'$. Similarly, we have $G_\delta$ Sets Closed under Intersection, and so $G \cap G' \in \mathcal G$. From Intersection is Largest Subset and Intersection is Subset, it follows that $G \cap G'$ is the infimum of $G$ and $G'$. Thus any two elements of $\mathcal G$ are seen to have both a supremum and an infimum in $\mathcal G$. Hence $\left({\mathcal G, \subseteq}\right)$ is a lattice. {{qed}}
Denote with $d$ the Euclidean metric on $\R^n$. Because $\sequence {x_i}_{i \in \N}$ is bounded, we find $v \in \R^n$ and $\epsilon \in \R_{>0}$ such that: :$\map d {v, x_i} < \epsilon$ for all $i \in \N$. Therefore, all $x_i$ are members of the closed $\epsilon$-ball $S = \map {B_\epsilon^-} v$. By Closed Ball in Euclidean Space is Compact, $S$ is compact. Thus $\left\langle{x_i}\right\rangle_{i \in \N}$ can be considered as a sequence in the compact metric space $\struct {S, d \restriction_{S \times S}}$. By Compact Subspace of Metric Space is Sequentially Compact in Itself, $\sequence {x_i}_{i \in \N}$ has a convergent subsequence in $S$. In particular, since $S$ is a metric subspace of $\R^n$, it follows that $\sequence {x_i}_{i \in \N}$ has a convergent subsequence in $\R^n$ as well. {{qed}}
Let the range of $\sequence {x_i}$ be $S$. By Closure of Bounded Subset of Metric Space is Bounded $\map \cl S$ is bounded in $\R^n$. By Topological Closure is Closed, $\map \cl S$ is closed in $\R^n$. By the Heine-Borel Theorem, $S$ is compact. The result follows from Compact Subspace of Metric Space is Sequentially Compact in Itself. {{qed}}
We have that all norms on $\R^n$ are equivalent. Choose the Euclidean norm $\norm {\, \cdot \,}_2$. Proof by induction will be used. === Basis for the induction === Let $n = 1$. Then the proof is given by Bolzano-Weierstrass theorem. === Induction hypothesis === Suppose, a bounded sequence $\sequence {\boldsymbol x_i}_{i \mathop \in \N}$ in $\R^n$ has a convergent subsequence $\sequence {\boldsymbol x_{i_k}}_{k \mathop \in \N}$. Then we have to show, that the same property holds for $\R^{n + 1}$. === Induction step === Let $\sequence {\mathbf x_n}_{n \mathop \in \N}$ be a bounded sequence in $\R^{n + 1}$. Split $\mathbf x_n$ into its first $n$ components and the last one: :$\mathbf x_n := \tuple {\boldsymbol \alpha_n, \beta_n}$ where $\boldsymbol \alpha_n \in \R^n$ and $\beta_n \in \R$ Then: {{begin-eqn}} {{eqn | l = \norm {\boldsymbol \alpha_n}_2       | o = \le       | r = \sqrt {\norm {\boldsymbol \alpha_n}_2^2 + \beta_n^2} }} {{eqn | r = \norm {\mathbf x_n}_2       | c = {{defof|Euclidean Norm}} }} {{end-eqn}} Hence, $\sequence {\boldsymbol \alpha_n}_{n \mathop \in \N}$ is a bounded sequence in $\R^n$. By induction hypothesis, it has a convergent subsequence $\sequence {\mathbf \alpha_{n_k} }_{k \mathop \in \N}$. Denote its limit by $\alpha$. Furthermore: {{begin-eqn}} {{eqn | l = \norm {\beta_n}_2       | o = \le       | r = \sqrt {\norm {\boldsymbol \alpha_n}_2^2 + \beta_n^2} }} {{eqn | r = \norm {\mathbf x_n}_2       | c = {{defof|Euclidean Norm}} }} {{end-eqn}} Therefore, $\sequence {\beta_n}_{n \mathop \in \N}$ is bounded. By Bolzano-Weierstrass theorem, $\sequence {\beta_n}_{n \mathop \in \N}$ has a convergent subsequence  $\sequence {\beta_{n_k}}_{k \mathop \in \N}$. Denote its limit by $\beta$. Then we have that: {{begin-eqn}} {{eqn | l = \lim_{k \to \infty} \mathbf x_{n_k}       | r = \lim_{k \to \infty} \tuple {\boldsymbol \alpha_{n_k}, \beta_{n_k} } }} {{eqn | r = \tuple {\boldsymbol \alpha, \beta}       | rr = \in \R^{n + 1} }} {{end-eqn}} Hence, the bounded sequence $\sequence {\mathbf x_n}_{n \mathop \in \N}$ has a convergent subsequence $\sequence {\mathbf x_{n_k}}_{k \mathop \in \N}$. {{qed}}
Let $\map f x = C e^x$. Then by Derivative of Exponential Function: :$\map {f'} x = \map f x$ From Exponential of Zero: :$\map f 0 = C$ Hence $C e^x$ is a solution of $y' = y$. Now suppose that a function $f$ satisfies $\map {f'} x = \map f x$.  Consider $\map h x = \map f x e^{-x}$. By the Product Rule for Derivatives: {{begin-eqn}} {{eqn | l = \map {h'} x       | r = \map {f'} x e^{-x} - \map f x e^{-x} }} {{eqn | r = \map f x e^{-x} - \map f x e^{-x} }} {{eqn | r = 0 }} {{end-eqn}} From Zero Derivative implies Constant Function, $h$ must be a constant function. Therefore, $\map h x = \map h 0 = \map f 0$. Recalling the definition of $h$, it follows that: :$\map f x = \map f 0 e^x$ {{qed}}
=== Existence === Differentiating $f \left({x}\right) = b e^{-A \left({x}\right)}$ {{WRT|Differentiation}} $x$: {{begin-eqn}} {{eqn | l = f' \left({x}\right)       | r = b e^{-A \left({x}\right)} \cdot \left({-A' \left({x}\right)}\right)       | c =  }} {{eqn | r = -b P \left({x}\right) e^{-A \left({x}\right)}       | c =  }} {{eqn | r = -P \left({x}\right) f \left({x}\right)       | c =  }} {{end-eqn}} So thedifferential equation becomes: :$f' \left({x}\right) + P \left({x}\right) f \left({x}\right) = -P \left({x}\right) f \left({x}\right) + P \left({x}\right) f \left({x}\right) = 0$ For the initial condition: {{begin-eqn}} {{eqn | l = f \left({a}\right)       | r = b e^{-A \left({a}\right)}       | c =  }} {{eqn | r = b e^{-\int_a^a P \left({x}\right) \ \mathrm d x}       | c =  }} {{eqn | r = b e^0       | c =  }} {{eqn | r = b       | c =  }} {{end-eqn}} Thus such a function exists satisfying the conditions. {{qed|lemma}} === Uniqueness === Suppose that $f$ is a function satisfying the initial conditions. Let $g \left({x}\right) = f \left({x}\right) e^{A \left({x}\right)}$. By the Product Rule: {{begin-eqn}} {{eqn | l = g' \left({x}\right)       | r = f' \left({x}\right) e^{A \left({x}\right)} + f \left({x}\right) e^{A \left({x}\right)}\cdot A' \left({x}\right)       | c =  }} {{eqn | r = e^{A \left({x}\right)} (f' \left({x}\right) + P \left({x}\right) f \left({x}\right))       | c =  }} {{eqn | r = 0       | c =  }} {{end-eqn}} So $g \left({x}\right)$ must be constant. Therefore: :$g \left({x}\right) = g \left({a}\right) = f \left({a}\right) e^{A \left({a}\right)} = f \left({a}\right) = b$ From this, we conclude that: :$f \left({x}\right) = g \left({x}\right) e^{-A \left({x}\right)} = b e^{-A \left({x}\right)}$ {{qed}}
=== Existence === If $P$ and $Q$ are continuous, then they are integrable, and we may use the Fundamental Theorem of Calculus on definite integrals involving these functions. {{begin-eqn}} {{eqn|l = f'(x)      |r = -bP(x)e^{-A(x)} - P(x)e^{-A(x)}\int_a^x Q(t) e^{A(t)}dt + e^{-A(x)} Q(x) e^{A(x)}      |c = by the Product Rule  }} {{eqn|l =       |r = -P(x) (be^{-A(x)} + e^{-A(x)}\int_a^x Q(t) e^{A(t)}dt) + Q(x)      |c = by factoring }} {{eqn|l =      |r = -P(x) f(x) + Q(x)      |c = by Substitution }} {{end-eqn}} Therefore, $f'(x) + P(x) f(x) = Q(x)$.  For the initial condition, {{begin-eqn}} {{eqn|l = f(a)      |r = be^{-A(a)} + e^{-A(a)} \int_a^a Q(t) e^{A(t)}dt      |c = by Substitution of $a$ into $f(x)$ }} {{eqn|l =       |r = be^0 + e^0 \cdot 0      |c = by Integral on Zero Interval }} {{eqn|l =      |r = b      |c = Simplification using arithmetic }} {{end-eqn}} === Uniqueness === Suppose $f$ is a solution to the differential equation and satisfies the initial condition.  Let $g(x) = f(x) e^{A(x)}$. {{begin-eqn}} {{eqn|l = g'(x)      |r = f'(x) e^{A(x)} + P(x) f(x) e^{A(x)}      |c = by the Product Rule }} {{eqn|l =       |r = e^{A(x)} Q(x)      |c = by Substitution }} {{end-eqn}} Moreover, $g(a) = b$.  By the Second Fundamental Theorem of Calculus, :$\displaystyle g(x) = \int_a^x Q(t) e^{A(t)} dt + b$ Furthermore, $\displaystyle f(x) = g(x) e^{-A(x)}$.  Therefore, we can conclude that :$\displaystyle f(x) = be^{-A(x)} + e^{-A(x)}\int_a^x Q(t) e^{A(t)} dt$ {{qed}}
Because $B \ne \varnothing$ it follows that $B$ has an element $x$. If $x$ is the smallest element of $B$ then the theorem holds. Otherwise, there is an element $y \in B$ such that $x \npreceq y$. By Well-Ordering is Total Ordering, $y \prec x$. Thus $S_x$, the $\preceq$-initial segment of $x$, is not empty. By the hypothesis, $S_x$ is a set. By Intersection is Subset, $B \cap S_x \subseteq S_x$ Thus by the subset axiom, $B \cap S_x$ is a set. Thus by the definition of proper well-ordering, $B \cap S_x$ has a smallest element $a$. Let $z$ be any element of $B$. Suppose for the sake of contradiction that $z \prec a$. Then by Extended Transitivity, $z \prec x$, so $z \in B \cap S_x$. But this contradicts the fact that $a$ is the smallest element of $B \cap S_x$. Thus $z \nprec a$. Since $\preceq$ is a total ordering, $a \preceq z$. As this holds for all $z \in B$, $a$ is the smallest element of $B$. {{qed}}
{{AimForCont}} that $A \not \subseteq B$. Then: :$A \setminus B \ne 0$. By Proper Well-Ordering Determines Smallest Elements, $A \setminus B$ must have some $\prec$-minimal element. Thus: :$\displaystyle \exists x \in \left({ A \setminus B }\right): \left({ A \setminus B }\right) \cap \prec^{-1} \left({ x }\right) = \varnothing$ implies that: :$A \cap \prec^{-1} \left({ x }\right) \subseteq B$ Hence this fulfils the hypothesis for $(1)$. We have that $x \in A$. so by $(1)$: : $x \in B$ But this contradicts the fact that $x \in \left({A \setminus B}\right)$. Thus by Proof by Contradiction: :$A \subseteq B$. It follows by definition of set equality that: :$A = B$ {{qed}}
We have that $\Epsilon$ creates a well-ordering on any ordinal. Also, the initial segments of $x$ are sets. Therefore from Proper Well-Ordering Determines Smallest Elements: : $B$ has an $\Epsilon$-minimal element.
{{begin-eqn}} {{eqn | l = x \in \bigcup A       | o = \leadsto       | r = \exists y \in A: x \in y       | c =  }} {{eqn | o = \leadsto       | r = \exists y \subseteq \bigcup A: x \subseteq y       | c =  }} {{eqn | o = \leadsto       | r = x \subseteq \bigcup A       | c =  }} {{end-eqn}} From this, we conclude that $\displaystyle \bigcup A$ is a transitive class. From Class is Transitive iff Union is Subset, it follows that: :$\displaystyle \bigcup A \subseteq A \subseteq \On$ By Subset of Well-Ordered Set is Well-Ordered, $A$ is also well-ordered by $\Epsilon$. Thus by Alternative Definition of Ordinal, $\bigcup A$ is an ordinal.
First, we must show that $\displaystyle \bigcup A$ is an upper bound. Take any member $a \in A$. Then by Subset of Union: : $\displaystyle a \subseteq \bigcup A$ By Ordering on Ordinal is Subset Relation: : $a \le A$ By generalizing for all $a \in A$: :$\displaystyle \forall x \in A: x \le \bigcup A$ Similarly, suppose now that $x$ is an upper bound of $A$. We shall denote $<$ for ordering on the ordinal numbers. By Ordering on Ordinal is Subset Relation and Transitive Set is Proper Subset of Ordinal iff Element of Ordinal, $<$ is the same as both $\in$ and $\subsetneq$. Then: {{begin-eqn}} {{eqn | l = z \in \bigcup A       | o = \implies       | r = \exists y: \left({z \in y \land y \in A}\right)       | c = Definition of union  }} {{eqn | o = \implies       | r = \exists y: \left({z \in y \land y < x}\right)       | c = by hypothesis (since $y \in A$, $y < x$) }} {{eqn | o = \implies       | r = z \in x       | c = by transitivity of $\in$: see Alternative Definition of Ordinal }} {{end-eqn}} Thus, by definition of subset: : $\displaystyle \bigcup A \subseteq x$ Therefore: :$\displaystyle \forall y \in A: y \le x \implies \bigcup A \le x$ {{qed}}
$x$ is an ordinal and so by definition is also a set. :$x \in \left({x \cup \left\{ {x}\right\} }\right) \land x \subset \left({x \cup \left\{ {x}\right\} }\right)$ so by applying the definition of a successor set: :$x \in x^+ \land x \subset x^+$ {{qed}}
{{AimForCont}} such an ordinal $y$ exists. Then, by Ordering on Ordinal is Subset Relation: :$x \in y$ and from Transitive Set is Proper Subset of Ordinal iff Element of Ordinal: :$y \in x^+$ Applying the definition of a successor set, we have: :$y \in x \lor y = x$ But this creates a membership loop, in that: :$x \in y \in x \lor x \in x$ By No Membership Loops, we have created a contradiction. The result follows from Proof by Contradiction. {{qed}}
Let $\N$ be defined as the von Neumann construction $\omega$. By definition of the ordering on von Neumann construction: :$m \le n \iff m \subseteq n$ From Von Neumann Construction of Natural Numbers is Minimally Inductive, $\omega$ is minimally inductive class under the successor mapping. The result follows from the Sandwich Principle: :$\forall m, n \in \omega: m \subseteq n \lor n \subseteq m$ {{qed}}
For this proof, we shall use $\prec$, $\in$, and $\subset$ interchangeably. We are justified in doing this because of Ordering on Ordinal is Subset Relation and Transitive Set is Proper Subset of Ordinal iff Element of Ordinal. {{begin-eqn}} {{eqn | l = x       | o = \in       | r = a       | c = By Hypothesis }} {{eqn | ll= \implies       | l = x       | o = \subseteq       | r = \bigcup A       | c =  }} {{eqn | ll= \implies       | l = x       | o = \preceq       | r = \bigcup A       | c =  }} {{eqn | o = \prec       | r = \left({\bigcup A}\right)^+       | c = Ordinal is Less than Successor }} {{eqn | ll= \implies       | l = x       | o = \prec       | r = \left({\bigcup A}\right)^+       | c =  }} {{end-eqn}} {{qed}}
Let $K_I$ denote the class of all nonlimit ordinals. Aiming for a contradiction, suppose $\omega$ is not a limit ordinal. Every element of $\omega$ is a nonlimit ordinal. So, if $\omega$ is also a nonlimit ordinal, $\omega + 1 \subseteq K_I$. By the definition of $\omega$: :$\omega \in \omega$ which violates No Membership Loops and is thus contradictory. It follows from Proof by Contradiction that $\omega$ is a limit ordinal. {{Qed}}
Let $F$ be a function whose domain is $\omega$. By the axiom of replacement, the range of $F$ is a set. Let the range of $F$ be denoted $\map \WW F$. Then: {{begin-eqn}} {{eqn | o =        | r = \exists x \in \map \WW F: \paren {\map \WW F \cap x} = \O       | c = Axiom of Foundation }} {{eqn | o = \leadsto       | r = \exists x: \paren {\paren {\map \WW F \cap x} = \O \land \exists n \in \omega: x = \map F n}       | c = as $x \in \map \WW F$, $x = \map F n$ for some $n \in \omega$ }} {{eqn | o = \leadsto       | r = \exists n \in \omega: \paren {\map \WW F \cap \map F n} = \O       | c = Logical manipulation eliminating $x$ }} {{end-eqn}} But: :$\map F {n^+} \in \map \WW F$ So: :$\map F {n^+} \notin \map F n$ {{qed}}
Let $\phi_1$ denote the mapping creating the order isomorphism between $\left({\in, A}\right)$ and $\left({\prec, S}\right)$. Let $\phi_2$ denote the mapping creating the order isomorphism between $\left({\in, B}\right)$ and $\left({\prec, S}\right)$. Then $\phi_2^{-1}: S \to B$ is an order isomorphism by Inverse of Order Isomorphism is Order Isomorphism. $\phi_1 \circ \phi_2^{-1}: A \to B$ is an order isomorphism by Composite of Order Isomorphisms is Order Isomorphism. So $\left({\in, A}\right)$ and $\left({\in, B}\right)$ are order isomorphic ordinals. By Isomorphic Ordinals are Equal: :$A = B$ {{qed}}
Proof by transfinite induction: Suppose that: : $\forall x \in \alpha: f \left({x}\right) = g \left({x}\right)$ for some arbitrary ordinal $\alpha < y$. Then $\alpha < z$. {{explain|Find the link to the result proving this.}} Hence: {{begin-eqn}} {{eqn | l = \forall x \in \alpha: f \left({x}\right) = g \left({x}\right)       | o = \implies       | r = \left({f \restriction \alpha}\right) = \left({g \restriction \alpha}\right)       | c = Equality of Restrictions }} {{eqn | o = \implies       | r = G \left({f \restriction \alpha}\right) = G \left({g \restriction \alpha}\right)       | c = Substitution }} {{eqn | o = \implies       | r = f \left({\alpha}\right) = g \left({\alpha}\right)       | c = By hypothesis }} {{end-eqn}} So applying induction: : $\forall \alpha < y: f \left({\alpha}\right) = g \left({\alpha}\right)$ {{qed}}
{{NotZFC}} We shall prove this using the first principle of transfinite induction. Assume $y$ is an ordinal such that $y \subseteq A$. This implies $\forall x < y: x \in A$. By definition, $y$ is a limit ordinal, $y = \varnothing$, or $\exists x: y = x^+$. If $y$ is a limit ordinal, then, since: :$\forall x < y: x \in A$ it follows that $y \in A$. If $y = \varnothing$, then $y \in A$ by hypothesis. Finally, assume $y = z^+$ for some ordinal $z$. Since $z < z^+ = y$ and $\forall x < y: x \in A$, it follows that $z \in A$. By hypothesis, we conclude $y = z^+ \in A$. In any case, $y \in A$. Therefore, for all ordinals $y$, if $y \subseteq A$, then $y \in A$. By the first principle of transfinite induction, $\operatorname{On} \subseteq A$. {{qed}} Category:Ordinals naaka7r4oqapyyn0kugeenl5yh7jmro
{{begin-eqn}} {{eqn | l = \map F \O       | r = \map G {F \restriction \O}       | c = Hypothesis }} {{eqn | r = \map G \O       | c = Restriction of $\O$ }} {{eqn | r = a       | c = Definition of $G$ }} {{end-eqn}} {{qed|lemma}} {{begin-eqn}} {{eqn | l = \map F {\beta^+}       | r = \map G {F \restriction \beta^+}       | c = Hypothesis }} {{eqn | r = \map H {\map {F \restriction \beta^+} {\bigcup \beta^+} }       | c = Definition of $G$ }} {{eqn | r = \map H {\map F \beta}       | c = Union of successor set is the original set by Union of Ordinals is Least Upper Bound }} {{end-eqn}} {{qed|lemma}} {{begin-eqn}} {{eqn | l = \map F \beta       | r = \map G {F \restriction \beta}       | c = Hypothesis }} {{eqn | r = \bigcup \Img {F \restriction \beta}       | c = Definition of $G$ }} {{eqn | r = \bigcup_{\gamma \mathop \in \beta} \map F \gamma       | c =  }} {{end-eqn}} {{qed|lemma}} We can proceed in the fourth part by Transfinite Induction. === Basis for the Induction === {{begin-eqn}} {{eqn | l = \map F \O       | r = a       | c = First part }} {{eqn | r = \map A \O       | c = Hypothesis }} {{end-eqn}} This proves the basis for the induction. {{qed|lemma}} === Induction Step === {{begin-eqn}} {{eqn | l = \map F \beta       | r = \map A \beta       | c =  }} {{eqn | ll= \leadsto       | l = \map H {\map F \beta}       | r = \map H {\map A \beta}       | c = Substitutivity of equality }} {{eqn | ll= \leadsto       | l = \map F {\beta^+}       | r = \map A {\beta^+}       | c = Second part }} {{end-eqn}} This proves the induction step. {{qed|lemma}} === Limit Case === {{begin-eqn}} {{eqn | lo= \forall \gamma \in \beta:       | l = \map F \gamma       | r = \map A \gamma       | c =  }} {{eqn | ll= \leadsto       | l = \bigcup_{\gamma \mathop \in \beta} \map F \gamma       | r = \bigcup_{\gamma \mathop \in \beta} \map A \gamma       | c = Substitutivity of equality }} {{eqn | ll= \leadsto       | l = \map F \beta       | r = \map A \beta       | c = Third part }} {{end-eqn}} This proves the limit case. {{qed}}
Construct $K$ and $F$ as in the First Principle of Transfinite Recursion. Set $f = \left({F \restriction x}\right)$. Then since $x \subseteq \operatorname{Dom} F$, the domain of $f$ is $x$. {{begin-eqn}} {{eqn | l = \forall y \in x: f \left({y}\right)       | r = F \left({y}\right)       | c = from the fact that $f$ is a restriction }} {{eqn | r = G \left({F \restriction y}\right)       | c = Transfinite Recursion: Theorem 1 }} {{eqn | r = G \left({f \restriction y}\right)       | c =  }} {{end-eqn}} Thus such a mapping $f$ exists. Suppose there are two mappings $f$ and $g$ that satisfy these conditions. We will use the first principle of transfinite induction to show that $f = g$. Take $y \in x$. Suppose $\forall z \in y: f \left({z}\right) = g \left({z}\right)$. Then: :$ \left({f \restriction y}\right) =  \left({g \restriction y}\right)$ and so: {{begin-eqn}} {{eqn | l = f \left({y}\right)       | r = G \left({f \restriction y}\right)       | c = Assumption }} {{eqn | r = G \left({g \restriction y}\right)       | c = Substitutivity of Equality }} {{eqn | r = g \left({y}\right)       | c = Hypothesis }} {{end-eqn}} So $f \left({y}\right) = g \left({y}\right)$ for all $y \in x$ by transfinite induction. Since $x$ is the domain of $f$ and $g$, it follows that $f = g$. Thus the mapping $f$ is unique. {{qed}}
Proof by induction: For all $n \in \N_{> 0}$, let $P \left({n}\right)$ be the proposition: :$\left({a_1 \circ a_2 \circ \cdots \circ a_n}\right)^{-1} = a_n^{-1} \circ \cdots \circ a_2^{-1} \circ a_1^{-1}$ $P \left({1}\right)$ is (trivially) true, as this just says: : $\left({a_1}\right)^{-1} = a_1^{-1}$ === Basis for the Induction === $P \left({2}\right)$ is the case: :$\left({a_1 \circ a_2}\right)^{-1} = a_2^{-1} \circ a_1^{-1}$ which has been proved in Inverse of Product in Monoid. This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 2$, then it logically follows that $P \left({k+1}\right)$ is true. So this is our induction hypothesis: :$\left({a_1 \circ a_2 \circ \cdots \circ a_k}\right)^{-1} = a_k^{-1} \circ \cdots \circ a_2^{-1} \circ a_1^{-1}$ Then we need to show: :$\left({a_1 \circ a_2 \circ \cdots \circ a_{k+1}}\right)^{-1} = a_{k+1}^{-1} \circ \cdots \circ a_2^{-1} \circ a_1^{-1}$ === Induction Step === This is our induction step: {{begin-eqn}} {{eqn | l = \left({a_1 \circ a_2 \circ \cdots \circ a_k \circ a_{k+1} }\right)^{-1}       | r = \left({\left({a_1 \circ a_2 \circ \cdots \circ a_k}\right) \circ a_{k+1} }\right)^{-1}       | c = General Associativity Theorem }} {{eqn | r = a_{k+1}^{-1} \circ \left({a_1 \circ a_2 \circ \cdots \circ a_k}\right)^{-1}       | c = Basis for the Induction }} {{eqn | r = a_{k+1}^{-1} \circ \left({a_k^{-1} \circ \cdots \circ a_2^{-1} \circ a_1^{-1} }\right)       | c = Induction Hypothesis }} {{eqn | r = a_{k+1}^{-1} \circ a_k^{-1} \circ \cdots \circ a_2^{-1} \circ a_1^{-1}       | c = General Associativity Theorem }} {{end-eqn}} So $P \left({k}\right) \implies P \left({k+1}\right)$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\forall n \in \N_{> 0}: \left({a_1 \circ a_2 \circ \cdots \circ a_n}\right)^{-1} = a_n^{-1} \circ \cdots \circ a_2^{-1} \circ a_1^{-1}$ {{qed}} Category:Monoids Category:Inverse Elements Category:Inverse of Product 1cvstn53zogxege6zrvmhavbsluy7uc
We have that a group is a monoid, all of whose elements are invertible. The result follows from Inverse of Product in Monoid. {{Qed}}
{{begin-eqn}} {{eqn | l = \paren {a \circ b} \circ \paren {b^{-1} \circ a^{-1} }       | r = \paren {\paren {a \circ b} \circ b^{-1} } \circ a^{-1}       | c = {{GroupAxiom|1}} }} {{eqn | r = \paren {a \circ \paren {b \circ b^{-1} } } \circ a^{-1}       | c = {{GroupAxiom|1}} }} {{eqn | r = \paren {a \circ e} \circ a^{-1}       | c = {{GroupAxiom|3}} }} {{eqn | r = a \circ a^{-1}       | c = {{GroupAxiom|2}} }} {{eqn | r = e       | c = {{GroupAxiom|3}} }} {{end-eqn}} The result follows from Group Product Identity therefore Inverses: :$\paren {a \circ b} \circ \paren {b^{-1} \circ a^{-1} } = e \implies \paren {a \circ b}^{-1} = b^{-1} \circ a^{-1}$ {{Qed}}