hcju/proofwikiqa huggingface dataset:
Suppose $H \circ K$ is a subgroup of $G$. Let $h \circ k \in H \circ K$. Then $h \circ k$ is the inverse of some element $g$ of $H \circ K$. Thus we can write $g = h' \circ k'$ for some $h' \in H$ and $k' \in K$. So: {{begin-eqn}} {{eqn | l = h \circ k       | r = g^{-1}       | c = Inverse of Group Inverse: $g$ is the inverse of $h \circ k$ }} {{eqn | r = \paren {h' \circ k'}^{-1}       | c =  }} {{eqn | r = k'^{-1} \circ h'^{-1}       | c = Inverse of Group Product }} {{eqn | o = \in       | r = K \circ H       | c = as $K$ and $H$ are both groups }} {{end-eqn}} So by definition of subset: : $H \circ K \in K \circ H$ Now suppose $x \in K \circ H$. Then: {{begin-eqn}} {{eqn | l = x       | r = k \circ h       | c = for some $k \in K, h \in H$ }} {{eqn | r = \paren {\paren {k \circ h}^{-1} }^{-1}       | c = Inverse of Group Inverse }} {{eqn | r = \paren {h^{-1} \circ k^{-1} }^{-1}       | c = Inverse of Group Product }} {{eqn | o = \in       | r = H \circ K       | c = as $K$ and $H$ are both groups }} {{end-eqn}} So by definition of subset: :$K \circ H \in H \circ K$ The result follows by definition of set equality. {{qed}}
Suppose that $G$ is Eulerian. Then $G$ contains an Eulerian circuit, that is, a circuit that uses each vertex and passes through each edge exactly once. Since a circuit must be connected, $G$ is connected. Beginning at a vertex $v$, follow the Eulerian circuit through the graph. As the circuit passes through each vertex, it uses two edges: one going to the vertex and another leaving. Each edge is used exactly once, so each of the vertices must be even. Since the circuit must also end at $v$, it follows that $v$ is also even. {{qed}}
Suppose that an (undirected) graph $G$ is connected and its vertices all have even degree. If there is more than one vertex in $G$, then each vertex must have degree greater than $0$. Begin at a vertex $v$. From Graph with Even Vertices Partitions into Cycles, we know that $v$ will be on at least one cycle. Since $G$ is connected, there must be an edge $\set {v, v_1}$ for some vertex $v_1 \ne v$. Since $v_1$ has even degree greater than $0$, there is an edge $\set {v_1, v_2}$ where $v_2 \ne v_1$. These two edges make a trail from $v$ to $v_2$. Continue this trail, leaving each vertex on an edge that was not previously used, until returning to $v$. This is always possible, because $v$ is on a cycle. Call the circuit formed by this process $C_1$. If $C_1$ covers all the edges of $G$, then the proof is complete. Otherwise, remove all the edges that contribute to $C_1$ from $G$, leaving the graph $G_0$. The remaining vertices are still even, and since $G$ is connected there is some vertex $u$ in both $G_0$ and $C_1$. Repeat the same process as before, beginning at $u$. The new circuit, $C_2$, can be added to $C_1$ by starting at $v$, moving along $C_1$ to $u$, travelling around $C_2$ back to $u$ and then along the remainder of $C_1$ back to $v$. Repeat this process, adding each new circuit found to create a larger circuit. Since $G$ is finite, this process must end at some point, and the resulting circuit will be an Eulerian circuit. {{qed}}
Suppose that an (undirected) graph $G$ is connected and its vertices all have even degree. From Graph with Even Vertices Partitions into Cycles, we can split $G$ into a number of cycles $\mathbb S = C_1, C_2, \ldots, C_k$. Start at any vertex $v$ on cycle $C_1$ and traverse its edges until we encounter a vertex of another cycle of $\mathbb S$, $C_2$ say. The edges of $C_2$ are then traversed, and then the traversal of $C_1$ is resumed when it is returned to. As $C_1$ is traversed, the journey is interrupted so as to traverse any other cycles of $\mathbb S$ in the same way that $C_2$ is traversed. Eventually the beginning of $C_1$ is reached, which is vertex $v$. Thus there exists a circuit which includes $C_1$ and at least one other cycle (unless $C_1$ is the only cycle), as $G$ is connected. If this circuit contains all the cycles $C_1, C_2, \ldots, C_k$, the required Eulerian circuit has been found. If not, then the circuit just generated is traversed. Because $G$ is connected, other cycles in $\mathbb S$ will be encountered. These, again, are traversed as they are encountered. This process is continued till all the cycles have been included in the circuit. At this stage, the required Eulerian circuit has been found. Hence $G$ is Eulerian. {{qed}}
=== Proof 1 === {{:Characteristics of Eulerian Graph/Sufficient Condition/Proof 1}} === Proof 2 === {{:Characteristics of Eulerian Graph/Sufficient Condition/Proof 2}}
{{ProofWanted|will be very similar to Characterization of Measures}} Category:Measure Theory e35wgep67cz8j4ef317n2xgmesxmbo2
For each $i \in I$, for each $\family {S_{j_i} }_{j_i \mathop \in J_i} \in \powerset X^{J_i} \cap \DD_i$, that: :$\map {\phi_i} {\family {S_{j_i} }_{j_i \mathop \in J_i} } \in \powerset X$ follows directly from the fact that $\powerset X$ is the codomain of $\phi_i$. Hence $\powerset X$ is a magma of sets for $\family {\phi_i}: i \in I$ on $X$. {{qed}} Category:Magmas of Sets mew9sx26augs5d0w4fk4edell2c17cf
=== Uniqueness === Suppose that both $\mathcal S$ and $\mathcal T$ are magmas of sets for $\Phi$ generated by $\mathcal G$. Applying condition $(2)$ for these twice, we obtain: :$\mathcal S \subseteq \mathcal T$ :$\mathcal T \subseteq \mathcal S$ By definition of set equality: : $\mathcal S = \mathcal T$ {{qed|lemma}} === Existence === Define $\mathcal S$ by: :$\mathcal S := \displaystyle \bigcap \left\{{\mathcal T: \mathcal G \subseteq \mathcal T}\right\}$ where $\mathcal T$ ranges over the magmas of sets for $\Phi$ on $X$. From Power Set is Magma of Sets, there intersection is at least one such $\mathcal T$. From Intersection of Magmas of Sets is Magma of Sets, $\mathcal S$ is a magma of sets for $\Phi$. By Intersection Preserves Subsets: General Result: Corollary, $\mathcal G \subseteq \mathcal S$. By Intersection is Subset: General Result, $\mathcal S \subseteq \mathcal T$ for every other magma of sets $\mathcal T$ containing $\mathcal G$. Thus $\mathcal S$ is a magma of sets for $\Phi$ generated by $\mathcal G$. {{qed}} Category:Magmas of Sets bn428z8bwb6ktuupoy65ga4qlc8p00r
{{explain|the notation in the below}} For each $i \in I$ and $j \in J$, we have: :$\map {\phi_i} {\sequence {S_{j, j_i} }_{j_i \mathop \in J_i} } \in \mathcal S_j$ Thus, if for each $j \in J$, one has: :$(1): \quad \sequence {S_{j_i} }_{j_i \mathop \in J_i} \in \mathcal S_j^{J_i}$ it follows from definition of set intersection that: :$\map {\phi_i} {\sequence {S_{j_i} }_{j_i \mathop \in J_i} } \in \mathcal S$ The condition $(1)$, for each $j \in J$, comes down to: :$\sequence {S_{j_i} }_{j_i \mathop \in J_i} \in \mathcal S^{J_i}$ It follows that $\mathcal S$ satisfies the requirement on $\phi_i$ to be a magma of sets, for each $i \in I$. That is to say, $\mathcal S$ is a magma of sets for $\Phi$. {{qed}} Category:Magmas of Sets 2ijx4tjhyv5ztlfxnwuxk7fpvvuevhz
It will suffice to define partial mappings such that the axiom for a magma of sets crystallises into the axioms for a $\sigma$-algebra. Let $X$ be any set, and let $\mathcal P \left({X}\right)$ be its power set. Define: :$\phi_1: \mathcal P \left({X}\right) \to \mathcal P \left({X}\right): \phi_1 \left({S}\right) := X$ :$\phi_2: \mathcal P \left({X}\right) \to \mathcal P \left({X}\right): \phi_2 \left({S}\right) := X \setminus S$ :$\phi_3: \mathcal P \left({X}\right)^\N \to \mathcal P \left({X}\right): \phi_3 \left({\left({S_n}\right)_{n \in \N}}\right) := \displaystyle \bigcup_{n \mathop \in \N} S_n$ It is blatantly obvious that $\phi_1, \phi_2$ and $\phi_3$ capture the axioms for a $\sigma$-algebra. {{finish|Bladibladiblabla, i.e. show that they are mappings, and that the MoS property translates into the $\sigma$-axiom}} {{qed}} Category:Sigma-Algebras Category:Magmas of Sets 3zk2z9nu7yzyliyrmdcll0h7rgpcdob
The proof proceeds by transfinite induction on $y$. === Basis for the Induction === {{begin-eqn}} {{eqn | l = \left({x \cdot \varnothing}\right)       | r = \varnothing       | c = Definition of Ordinal Multiplication }} {{eqn | l = \varnothing       | o = \in       | r = \operatorname{On}       | c = Empty Set is Ordinal }} {{eqn | ll= \implies       | l = \left({x \cdot \varnothing}\right)       | o = \in       | r = \operatorname{On}       | c = Substitutivity of Equality }} {{end-eqn}} This proves the basis for the induction. === Induction Step === {{begin-eqn}} {{eqn | l = \left({x \cdot y}\right)       | o = \in       | r = \operatorname{On}       | c =  }} {{eqn | ll= \implies       | l = \left({\left({x \cdot y}\right) + y}\right)       | o = \in       | r = \operatorname{On}       | c = Ordinal Addition is Closed }} {{eqn | l = \left({x \cdot y^+}\right)       | r = \left({\left({x \cdot y}\right) + y}\right)       | c = Definition of Ordinal Multiplication }} {{eqn | ll= \implies       | l = \left({x \cdot y^+}\right)       | o = \in       | r = \operatorname{On}       | c = Substitutivity of Equality }} {{end-eqn}} This proves the induction step. === Limit Case === {{begin-eqn}} {{eqn | lo= \forall z \in y:       | l = \left({x \cdot z}\right)       | o = \in       | r = \operatorname{On}       | c = Hypothesis }} {{eqn | ll= \implies       | l = \bigcup_{z \mathop \in y} \left({x \cdot z}\right)       | o = \in       | r = \operatorname{On}       | c = Union of Ordinals is Ordinal }} {{eqn | ll=        | l = \left({x \cdot y}\right)       | r = \bigcup_{z \mathop \in y} \left({x \cdot z}\right)       | c = Definition of Ordinal Multiplication }} {{eqn | ll= \implies       | l = \left({x \cdot y}\right)       | o = \in       | r = \operatorname{On}       | c = Substitutivity of Equality }} {{end-eqn}} This proves the limit case. {{qed}}
Let $g: \N \to \N$ be defined by: :$\map g n = m + n$ Applying the Principle of Recursive Definition to $0$ and $g$, we get the following function $f: \N \to \N$: :$\map f n = \begin{cases} 0 & : n = 0 \\ m + \map f k & : n = k + 1 \end{cases}$ which is seen to be equivalent to $m \times n$ for all $m,n \in \N$. Hence $m \times n \in \N$. {{qed}}
{{begin-eqn}} {{eqn | l = \left({x \cdot \varnothing}\right)       | r = \varnothing       | c = Definition of Ordinal Multiplication }} {{end-eqn}} For $\left({\varnothing \cdot x}\right) = \varnothing$, the proof shall proceed by Transfinite Induction on $x$. === Basis for the Induction === {{begin-eqn}} {{eqn | l = \left({\varnothing \cdot \varnothing}\right)       | r = \varnothing       | c = Definition of Ordinal Multiplication }} {{end-eqn}} This proves the basis for the induction. === Induction Step === {{begin-eqn}} {{eqn | l = \left({\varnothing \cdot x}\right)       | r = \varnothing       | c = Inductive Hypothesis }} {{eqn | ll= \implies       | l = \left({\left({\varnothing \cdot x}\right) + \varnothing}\right)       | r = \varnothing       | c = Definition of Ordinal Addition }} {{eqn | l = \left({\varnothing \cdot x^+}\right)       | r = \left({\left({\varnothing \cdot x}\right) + \varnothing}\right)       | c = Definition of Ordinal Multiplication }} {{eqn | ll= \implies       | l = \left({\varnothing \cdot x^+}\right)       | r = \varnothing       | c = Equality is Transitive }} {{end-eqn}} This proves the induction step. === Limit Case === {{begin-eqn}} {{eqn | lo= \forall y \in x:       | l = \left({\varnothing \cdot y}\right)       | r = \varnothing       | c = Hypothesis }} {{eqn | ll= \implies       | l = \bigcup_{y \mathop \in x} \left({\varnothing \cdot y}\right)       | r = \varnothing       | c = Indexed Union Equality }} {{eqn | l = \left({\varnothing \cdot x}\right)       | r = \bigcup_{y \mathop \in x} \left({\varnothing \cdot y}\right)       | c = Definition of Ordinal Multiplication }} {{eqn | ll= \implies       | l = \left({\varnothing \cdot x}\right)       | r = \varnothing       | c = Equality is Transitive }} {{end-eqn}} This proves the limit case. {{qed}}
{{begin-eqn}} {{eqn|l = \left({x \cdot 1}\right)      |r = \left({x \cdot \varnothing^+}\right)      |c = Definition of One ($1$) }} {{eqn|r = \left({\left({x \cdot \varnothing}\right) + x}\right)      |c = Definition of Ordinal Multiplication }} {{eqn|r = \left({\varnothing + x}\right)      |c = Definition of Ordinal Multiplication }} {{eqn|r = x      |c = Ordinal Addition by Zero }} {{end-eqn}} {{qed|lemma}} The proof of the other equality shall proceed by Transfinite Induction. === Basis for the Induction === {{begin-eqn}} {{eqn|l = \left({1 \cdot \varnothing}\right)      |r = \varnothing      |c = Definition of Ordinal Multiplication }} {{end-eqn}} This proves the basis for the induction. === Induction Step === {{begin-eqn}} {{eqn|l = \left({1 \cdot x}\right)      |r = x      |c = Inductive Hypothesis }} {{eqn|ll = \implies      |l = \left({\left({1 \cdot x}\right) + 1}\right)      |r = x^+      |c = Ordinal Addition by One }} {{eqn|l = \left({1 \cdot x^+}\right)      |r = \left({\left({1 \cdot x}\right) + 1}\right)      |c = Definition of Ordinal Multiplication }} {{eqn|ll = \implies      |l = \left({1 \cdot x^+}\right)      |r = x^+      |c = Equality is Transitive }} {{end-eqn}} This proves the induction step. === Limit Case === {{begin-eqn}} {{eqn|ll = \forall y \in x:      |l = \left({1 \cdot y}\right)      |r = y      |c = Hypothesis }} {{eqn|ll = \implies      |l = \bigcup_{y \mathop \in x} \left({1 \cdot y}\right)      |r = \bigcup_{y \mathop \in x} y      |c = Indexed Union Equality }} {{eqn|ll = \implies      |l = \left({1 \cdot x}\right)      |r = \bigcup_{y \mathop \in x} y      |c = Definition of Ordinal Multiplication }} {{eqn|ll = \implies      |l = \left({1 \cdot x}\right)      |r = x      |c = Union of Limit Ordinal }} {{end-eqn}} This proves the limit case. {{qed}}
{{begin-eqn}} {{eqn | l = x + 1       | r = x + \O^+       | c = {{Defof|One (Ordinal)|(ordinal) $1$}} }} {{eqn | r = \left({x + \O}\right)^+       | c = {{Defof|Ordinal Addition}} }} {{eqn | r = x^+       | c = Ordinal Addition by Zero }} {{end-eqn}} {{qed}} Category:Ordinal Arithmetic 5xcm4zlomfuadm9rooebhnadrjawc3x
Since $C$ is closed, $S \setminus C$ is open. By definition of Borel $\sigma$-algebra, $S \setminus C \in \mathcal B \left({\tau}\right)$. By axiom $(2)$ for $\sigma$-algebras: :$S \setminus \left({S \setminus C}\right) \in \mathcal B \left({\tau}\right)$ and this set equals $C$ by Set Difference with Set Difference since $C \subseteq S$. The result follows by definition of measurable set. {{qed}} Category:Sigma-Algebras Category:Closed Sets lxvmduxwzoyeybd13pyy1vftqq1uv3f
Let $d \divides n$. We can write $n$ as $n = d p_1 p_2 p_3 \cdots p_r$, where $p_1, p_2, \ldots, p_r$ are all the primes (not necessarily distinct) which divide $\dfrac n d$. Thus, repeatedly using Euler Phi Function of Product with Prime: {{begin-eqn}} {{eqn | l = \map \phi d       | o = \divides       | r = \map \phi {d p_1}       | c =  }} {{eqn | ll= \leadsto       | l = \map \phi {d p_1}       | o = \divides       | r = \map \phi {d p_1 p_2}       | c =  }} {{eqn | ll= \leadsto       | l = \map \phi {d p_1 p_2}       | o = \divides       | r = \map \phi {d p_1 p_2 p_3}       | c =  }} {{eqn | ll= \leadsto       | o = \cdots       | c =  }} {{eqn | ll= \leadsto       | l = \map \phi {d p_1 p_2 \cdots p_{r - 1} }       | o = \divides       | r = \map \phi {d p_1 p_2 \cdots p_{r - 1} p_r}       | c =  }} {{end-eqn}} As the last expression is $\map \phi n$, the result follows from Divisor Relation on Positive Integers is Partial Ordering. {{qed}} Category:Euler Phi Function gaf16afs9orzxul0phku8ayz4qe8tpf
From Commutative $B$-Algebra Induces Abelian Group we have that there exists an abelian group $\left({G, \circ}\right)$ such that: :$\forall a, b \in G: a \circ b^{-1} = a * b$  where $a * b$ is defined by the binary operation in $\left({G, *}\right)$. From Abelian Group Induces Entropic Structure, we have directly that $\left({G, *}\right)$ is an entropic structure. {{qed}} Category:B-Algebras Category:Entropic Structures budcghna3fdih36g7m93wk9ed3u1wzj
From B-Algebra Induces Group, the algebraic structure $\left({X, *}\right)$ is a group such that: :$\forall x \in X: 0 \circ x$ is the inverse element of $x$ under $*$. It remains to show that $*$ is a commutative operation. Let $x, y \in X$: {{begin-eqn}} {{eqn | l=x * y       | r=x \circ \left({0 \circ y}\right)       | c=by definition of $*$ }} {{eqn | r=y \circ \left({0 \circ x}\right)       | c=by definition of commutative $B$-algebra }} {{eqn | r=y * x       | c=by definition of $*$ }} {{end-eqn}} Hence the result. {{qed}}
=== Sufficient Condition === The proof of the sufficient condition shall proceed by Transfinite Induction on $y$. === Basis for the Induction === Both $x < 0$ and $\left({ x \cdot z }\right) < \left({ 0 \cdot z }\right)$ are contradictory, so the iff statement holds for the condition that $y = 0$. This proves the basis for the induction. === Induction Step === Suppose the biconditional statement holds for $y$.  Then: {{begin-eqn}} {{eqn | l = x < y^+       | o = \implies       | r = x < y \lor x = y       | c = Definition:Successor Set }} {{eqn | l = x < y \land z > 0       | o = \implies       | r = \left({ z \cdot x }\right) < \left({ z \cdot y }\right)       | c = Inductive Hypothesis }} {{eqn | l = x = y       | o = \implies       | r = \left({ z \cdot x }\right) = \left({ z \cdot y }\right)       | c = Leibniz's law }} {{eqn | l = z > 0       | o = \implies       | r = \left({ z \cdot y }\right) < \left({ \left({ z \cdot y }\right) + z }\right)       | c = Membership is Left Compatible with Ordinal Addition }} {{eqn | o = \implies       | r = \left({ z \cdot y }\right) < \left({ z \cdot y^+ }\right)       | c = Definition:Ordinal Multiplication }} {{eqn | o = \implies       | r = \left({ z \cdot x }\right) < \left({ z \cdot y^+ }\right)       | c = Transitivity of $<$ and Leibniz's law }} {{end-eqn}} In either case, $\left({ z \cdot x }\right) < \left({ z \cdot y^+ }\right)$ This proves the induction step. === Limit Case === Suppose $y$ is a limit ordinal: {{begin-eqn}} {{eqn | o =        | r = \forall w \in y: \left({ \left({ x < w \land z > 0 }\right) \iff \left({ z \cdot x }\right) < \left({ z \cdot w }\right) }\right)       | c = Hypothesis }} {{eqn | o = \implies       | r = \left({ \exists w \in y: \left({ x < w \land z > 0 }\right) \iff \exists w \in y: \left({ z \cdot x }\right) < \left({ z \cdot w }\right) }\right)       | c = Predicate Logic Manipulation }} {{eqn | o = \implies       | r = \left({ \left({ x < \bigcup y \land z > 0 }\right) \iff \left({ z \cdot x }\right) < \bigcup_{w \mathop \in y} \left({ z \cdot w }\right) }\right)       | c = Definition of Set Union }} {{eqn | o = \implies       | r = \left({ \left({ x < y \land z > 0 }\right) \iff \left({ z \cdot x }\right) < \bigcup_{w \mathop \in y} \left({ z \cdot w }\right) }\right)       | c = Union of Limit Ordinal }} {{eqn | o = \implies       | r = \left({ \left({ x < y \land z > 0 }\right) \iff \left({ z \cdot x }\right) < \left({ z \cdot y }\right) }\right)       | c = Definition of Ordinal Multiplication }} {{end-eqn}} This proves the limit case. === Necessary Condition === Conversely, suppose $\left({ z \cdot x }\right) < \left({ z \cdot y^+ }\right)$. Then $z \ne 0$ because if it were equal, both sides of the inequality would be $0$. So $z > 0$. Furthermore: :$y < x \implies \left({ z \cdot y }\right) < \left({ z \cdot x }\right)$ :$y = x \implies \left({ z \cdot y }\right) = \left({ z \cdot x }\right)$ So if $\left({ z \cdot x }\right) < \left({ z \cdot y }\right)$, then $y \ne x$ and $y \not < x$, so $x < y$ by Ordinal Membership is Trichotomy. {{qed}}
For the proof, $<$, $\in$, and $\subsetneq$ will be used interchangeably. This is justified by Transitive Set is Proper Subset of Ordinal iff Element of Ordinal. Since $z \ne 0$ and $z \not < 0$, $0 < z$ by Ordinal Membership is Trichotomy. Note that: {{begin-eqn}} {{eqn | l = x < y \land z > 0       | o = \implies |r = \left({ z \cdot x }\right) < \left({ z \cdot y }\right)       | c = Membership is Left Compatible with Ordinal Multiplication }} {{eqn | l = y < x \land z > 0       | o = \implies |r = \left({ z \cdot y }\right) < \left({ z \cdot x }\right)       | c = Membership is Left Compatible with Ordinal Multiplication }} {{end-eqn}} In addition: {{begin-eqn}} {{eqn | l = \left({ z \cdot x }\right) = \left({ z \cdot y }\right)       | o = \implies |r = \left({ z \cdot x }\right) \not < \left({ z \cdot y }\right) \land \left({ z \cdot y }\right) \not < \left({ z \cdot x }\right)       | c = No Membership Loops }} {{end-eqn}} This contradicts the consequents of the first two equations, so: {{begin-eqn}} {{eqn | o = \implies       | r = x \not < y \land y \not < x       | c = Rule of Transposition, Praeclarum Theorema }} {{eqn | o = \implies       | r = x = y       | c = Ordinal Membership is Trichotomy }} {{end-eqn}} {{qed}}
The proof shall proceed by Transfinite Induction on $z$. === Basis for the Induction === By definition of ordinal multiplication: :$\left({x \cdot 0}\right) = 0$ so the statement simply reduces to: :$0 \le 0$ This proves the basis for the induction. === Induction Step === {{begin-eqn}} {{eqn | l = x       | o = \le       | r = y }} {{eqn | ll= \implies       | l = \left({x \cdot z}\right)       | o = \le       | r = \left({y \cdot z}\right)       | c = Inductive Hypothesis }} {{eqn | ll= \implies       | l = \left({\left({x \cdot z}\right) + x}\right)       | o = \le       | r = \left({\left({y \cdot z}\right) + y}\right)       | c = Membership is Left Compatible with Ordinal Addition and Subset is Right Compatible with Ordinal Addition }} {{eqn | ll= \implies       | l = \left({x \cdot z^+}\right)       | o = \le       | r = \left({y \cdot z^+}\right)       | c = Definition of Ordinal Multiplication }} {{end-eqn}} This proves the induction step. === Limit Case === {{begin-eqn}} {{eqn | l = x       | o = \le       | r = y }} {{eqn | ll= \implies       | lo= \forall w \in z:       | l = \left({x \cdot w}\right)       | o = \le       | r = \left({y \cdot w}\right)       | c = Inductive Hypothesis }} {{eqn | ll= \implies       | l = \bigcup_{w \mathop \in z} \left({x \cdot w}\right)       | o = \le       | r = \bigcup_{w \mathop \in z} \left({y \cdot w}\right)       | c = Indexed Union Subset }} {{eqn | ll= \implies       | l = \left({x \cdot z}\right)       | o = \le       | r = \left({y \cdot z}\right)       | c = Definition of Ordinal Multiplication }} {{end-eqn}} This proves the limit case. {{qed}}
=== Necessary Condition === Suppose that $\paren {x \cdot y} = 0$ and that $x \ne 0$. By Ordinal Multiplication by Zero: : $\paren {x \cdot 0} = 0$ Therefore: : $\paren {x \cdot y} = \paren {x \cdot 0}$ Thus, by Ordinal Multiplication is Left Cancellable, we have that $y = 0$. {{qed|lemma}} === Sufficient Condition === If $x = 0$, then by Leibniz's Rule: :$\paren {x \cdot y} = \paren {0 \cdot y}$ If $y = 0$, then also by Leibniz's Rule: :$\paren {x \cdot y} = \paren {x \cdot 0}$ By Ordinal Multiplication by Zero, both $\paren {0 \cdot y} = 0$ and $\paren {x \cdot 0} = 0$. {{qed}}
The proof proceeds by induction on $n$, the length of $\left \langle {a_k} \right \rangle_{1 \mathop \le k \mathop \le n}$. === Basis for the Induction === The case $n = 1$ states that: :$b \circ a_1 = a_1 \circ b$ That is, $b$ commutes with $a_1$, which holds by hypothesis. This constitutes the basis for the induction. === Induction Hypothesis === Fix $n \in \N$ with $n \ge 1$. Assume that the theorem holds for sequences of length $n$. This is our induction hypothesis. === Induction Step === This is our induction step: Let $\left\langle{a_k}\right\rangle_{1 \mathop \le k \mathop \le n+1}$ be a sequence of length $n + 1$ in $S$. Suppose that $b$ commutes with $a_k$ for all $k \in \left[{0 \,.\,.\, n+1}\right]$. Then (suppressing brackets as $\circ$ is associative): {{begin-eqn}} {{eqn | l = b \circ a_1 \circ \cdots \circ a_n \circ a_{n+1}       | r = a_1 \circ \cdots \circ a_n \circ b \circ a_{n+1}       | c = Induction Hypothesis }} {{eqn | r = a_1 \circ \cdots a_n \circ a_{n+1} \circ b       | c = $a_{n+1}$ commutes with $b$ }} {{end-eqn}} We conclude that the theorem holds for sequences of length $n+1$. The result follows by the Principle of Mathematical Induction. {{qed}}