hcju/proofwikiqa huggingface dataset:
From Composite of Bijections is Bijection, $\psi \circ \phi$ is a bijection, as, by definition, an order isomorphism is also a bijection. By definition of composition of mappings, $\psi \circ \phi \left({x}\right) = \psi \left({\phi \left({x}\right)}\right)$. As $\phi$ is an order isomorphism, we have: :$\forall x_1, y_1 \in S_1: x_1 \preceq_1 y_1 \implies \phi \left({x_1}\right) \preceq_2 \phi \left({y_1}\right)$ As $\psi$ is an order isomorphism, we have: :$\forall x_2, y_2 \in S_2: x_2 \preceq_2 y_2 \implies \psi \left({x_2}\right) \preceq_3 \psi \left({y_2}\right)$ By setting $x_2 = \phi \left({x_1}\right), y_2 = \phi \left({y_1}\right)$, it follows that: :$\forall x_1, y_1 \in S_1: x_1 \preceq_1 y_1 \implies \psi \left({\phi \left({x_1}\right)}\right) \preceq_3 \psi \left({\phi \left({y_1}\right)}\right)$ Similarly we can show that: :$\forall x_3, y_3 \in S_3: x_3 \preceq_3 y_3 \implies \phi^{-1} \left({\psi^{-1} \left({x_3}\right)}\right) \preceq_1 \phi^{-1} \left({\psi^{-1} \left({y_3}\right)}\right)$ {{qed}}
{{Finish|Needs filling out.}} * Reflexive: Follows directly from Identity Mapping is Order Isomorphism. * Symmetric: Follows directly from Inverse of Order Isomorphism is Order Isomorphism. * Transitive: Follows directly from Composite of Order Isomorphisms is Order Isomorphism. {{qed}}
An ordered set is a relational structure where order isomorphism is a special case of relation isomorphism. The result follows directly from Relation Isomorphism is Equivalence Relation. {{qed}}
By definition, $\succeq$ is the inverse relation to $\preceq$. By Inverse Relation Properties, if a relation is reflexive, transitive and/or antisymmetric, then so is its inverse. The result follows. {{qed}}
From Subset Relation on Power Set is Partial Ordering, we have that $\subseteq$ is a partial ordering. We note in passing that for any set $S$: : From Supremum of Power Set, $\mathcal P \left({S}\right)$ has a supremum, that is, $S$ itself : From Infimum of Power Set, $\mathcal P \left({S}\right)$ has an infimum, that is, $\varnothing$. These are also the maximal and minimal elements of $\mathcal P \left({S}\right)$. Let $\mathbb S$ be a subset of $\mathcal P \left({S}\right)$. Then from Union is Smallest Superset: :$\left({\forall X \in \mathbb S: X \subseteq T}\right) \iff \bigcup \mathbb S \subseteq T$ and from Intersection is Largest Subset: :$\left({\forall X \in \mathbb S: T \subseteq X}\right) \iff T \subseteq \bigcap \mathbb S$ So $\bigcap \mathbb S$ is the infimum and $\bigcup \mathbb S$ is the supremum of $\left({\mathbb S, \subseteq}\right)$. Hence by definition $\mathcal P \left({S}\right)$ is a complete lattice. {{qed}}
Let $\left({S, \preceq}\right)$ be a totally ordered set. Then we have: : $\forall x, y \in S: x \preceq y \lor y \preceq x$ : $\forall x, y \in S: x \preceq y \implies \sup \left\{{x, y}\right\} = y \land \inf \left\{{x, y}\right\} = x$ : $\forall x, y \in S: y \preceq x \implies \sup \left\{{x, y}\right\} = x \land \inf \left\{{x, y}\right\} = y$ Thus the conditions for $\left({S, \preceq}\right)$ to be a lattice are fulfilled. {{qed}}
{{begin-eqn}} {{eqn | o =       | r = \forall a, b \in S: a \preceq b \lor b \preceq a       | c = {{Defof|Total Ordering}} }} {{eqn | o = \leadstoandfrom       | r = \forall a, b \in S: a \preceq b \lor a \succeq b       | c = {{Defof|Dual Ordering}} }} {{eqn | o = \leadstoandfrom       | r = \forall a, b \in S: \paren {a = b \lor a \prec b} \lor \paren {a = b \lor a \succ b}       | c = Strictly Precedes is Strict Ordering }} {{eqn | o = \leadstoandfrom       | r = \forall a, b \in S: a \prec b \lor a = b \lor a \succ b       | c = Rules of Commutation, Association and Idempotence }} {{end-eqn}} {{qed}}
Let $\RR \subseteq S \times T$ be reflexive. Then: :$\forall x \in S: \tuple {x, x} \in \RR$ By the definition of complement of $\RR$: :$\tuple {x, y} \in \RR \implies \tuple {x, y} \notin \relcomp {S \times S} \RR$ The same applies to $\tuple {x, x}$, and thus: :$\forall x \in S: \tuple {x, x} \notin \relcomp {S \times S} \RR$ Thus $\relcomp {S \times S} \RR$ is antireflexive. Similarly, by definition: :$\forall x \in S: \tuple {x, x} \notin \RR \implies \neg \tuple {x, x} \notin \relcomp {S \times S} \RR$ By Double Negation it follows that: :$\tuple {x, x} \in \relcomp {S \times S} \RR$ The converses of these results follow from the fact that: :$\relcomp {S \times S} {\relcomp {S \times S} \RR} = \RR$ by Relative Complement of Relative Complement. {{Qed}} Category:Reflexive Relations Category:Relative Complement iqnuxo8bs5jeattlg5gkrvi05m9rgsj
Let $\mathcal R \subseteq S \times S$ be symmetric. Then from Symmetry of Relations is Symmetric: :$\tuple {x, y} \in \mathcal R \iff \tuple {y, x} \in \mathcal R$ {{AimForCont}} $\relcomp {S \times S} {\mathcal R} \subseteq S \times S$ is not symmetric. Then: :$\exists \tuple {x, y} \in \relcomp {S \times S} {\mathcal R}: \tuple {y, x} \notin \relcomp {S \times S} {\mathcal R}$ But then by definition of complement of $\mathcal R$: :$\tuple {y, x} \in \mathcal R$ As $\mathcal R$ is symmetric it follows that: :$\tuple {x, y} \in \mathcal R$ This contradicts the premise that $\tuple {x, y} \in \relcomp {S \times S} {\mathcal R}$. Hence by Proof by Contradiction it follows that $\relcomp {S \times S} {\mathcal R}$ is symmetric. {{qed}} The converse follows similarly. Category:Symmetric Relations Category:Relative Complement og8lpnl3stud43136bscyl1muu27udf
Let $\left({S, \preceq_1}\right)$ and $\left({T, \preceq_2}\right)$ be ordered sets. Let $\phi: \left({S, \preceq_1}\right) \to \left({T, \preceq_2}\right)$ be strictly increasing. From Strictly Precedes is Strict Ordering: :$x \preceq_1 y \implies x = y \lor x \prec_1 y$ So: {{begin-eqn}} {{eqn | l = x       | r = y       | c =  }} {{eqn | ll= \implies       | l = \phi \left({x}\right)       | r = \phi \left({y}\right)       | c = Definition of Mapping }} {{eqn | ll= \implies       | l = \phi \left({x}\right)       | o = \preceq_2       | r = \phi \left({y}\right)       | c = as $\preceq_2$, being an ordering, is reflexive }} {{end-eqn}} This leaves us with: {{begin-eqn}} {{eqn | l = x       | o = \prec_1       | r = y       | c =  }} {{eqn | ll= \implies       | l = \phi \left({x}\right)       | o = \prec_2       | r = \phi \left({y}\right)       | c = Definition of Strictly Increasing }} {{eqn | ll= \implies       | l = \phi \left({x}\right)       | o = \preceq_2       | r = \phi \left({y}\right)       | c = Strictly Precedes is Strict Ordering }} {{end-eqn}} {{Qed}}
Let $\left({S, \preceq_1}\right)$ and $\left({T, \preceq_2}\right)$ be ordered sets. Let $\phi: \left({S, \preceq_1}\right) \to \left({T, \preceq_2}\right)$ be strictly decreasing. From Strictly Precedes is Strict Ordering: :$x \preceq_1 y \iff x = y \lor x \prec_1 y$ So: {{begin-eqn}} {{eqn | l = x       | r = y       | c =  }} {{eqn | ll= \implies       | l = \phi \left({y}\right)       | r = \phi \left({x}\right)       | c = Definition of mapping }} {{eqn | ll= \implies       | l = \phi \left({y}\right)       | o = \preceq_2       | r = \phi \left({x}\right)       | c = as $\preceq_2$, being an ordering, is reflexive }} {{end-eqn}} This leaves us with: {{begin-eqn}} {{eqn | l = x       | o = \prec_1       | r = y       | c =  }} {{eqn | ll= \implies       | l = \phi \left({y}\right)       | o = \prec_2       | r = \phi \left({x}\right)       | c = Definition of Strictly Decreasing }} {{eqn | ll= \implies       | l = \phi \left({y}\right)       | o = \preceq_2       | r = \phi \left({x}\right)       | c = Strictly Precedes is Strict Ordering }} {{end-eqn}} {{Qed}}
If $\phi$ is strictly monotone, then it is either strictly increasing or strictly decreasing. If $\phi$ is strictly increasing, then by Strictly Increasing Mapping is Increasing, $\phi$ is increasing. If $\phi$ is strictly decreasing, then by Strictly Decreasing Mapping is Decreasing, $\phi$ is decreasing. Thus $\phi$ is monotone by definition. {{qed}}
Let $\phi$ be an order embedding from $\struct {S, \preceq_1}$ into $\struct {T, \preceq_2}$. Then $\phi$ is an injection into $\struct {T, \preceq_2}$ by definition. From Restriction of Mapping to Image is Surjection, a mapping from a set to the image of that mapping is a surjection. Thus the surjective restriction of $\phi$ onto $S'$ is an order embedding which is also a surjection. Hence the result from Order Isomorphism is Surjective Order Embedding. {{qed}}
{{begin-eqn}} {{eqn | l = x, y       | o = \in       | r = S       | c =  }} {{eqn | lo= \land       | l = x       | o = \ne       | r = y       | c =  }} {{eqn | ll= \leadsto       | l = x       | o = \prec_1       | r = y }} {{eqn | lo= \lor       | l = y       | o = \prec_1       | r = x       | c = Trichotomy Law }} {{eqn | ll= \leadsto       | l = \map \phi x       | o = \prec_2       | r = \map \phi y }} {{eqn | lo= \lor       | l = \map \phi y       | o = \prec_2       | r = \map \phi x       | c = $\phi$ is strictly monotone }} {{eqn | ll= \leadsto       | l = \map \phi x       | o = \ne       | r = \map \phi y       | c = Definition of $\prec_2$ }} {{end-eqn}} Hence the result. {{qed}}
=== Forward Implication === Let $\phi$ be an order embedding. Let $x, y \in S$ with $x \prec_1 y$, where $\prec_1$ is the strict predecessor relation. Then: {{begin-eqn}} {{eqn | o =       | r = x \prec_1 y \land x \ne y       | c = Definition of Strictly Precedes }} {{eqn | o = \leadsto       | r = \map \phi x \preceq_2 \map \phi y       | c = {{Defof|Order Embedding}} }} {{eqn | o = \leadsto       | r = \map \phi x \prec_2 \map \phi y       | c = $\phi$ is injective }}  {{end-eqn}} So by definition, $\phi$ is strictly increasing. {{qed|lemma}} {{transclude:Mapping from Totally Ordered Set is Order Embedding iff Strictly Increasing/Reverse Implication   |section  = proofs   |title    = Reverse Implication   |header   = 3   |increase = 2   |link     = true }}
Let $\struct {S, \preceq}$ be a finite totally ordered set. From Condition for Well-Foundedness, $\struct {S, \preceq}$ is well-founded {{iff}} there is no infinite sequence $\sequence {a_n}$ of elements of $S$ such that $\forall n \in \N: a_{n + 1} \prec a_n$. If it were the case that $\struct {S, \preceq}$ had such an infinite sequence, then at least some of the terms would be repeated in that sequence. So there would be, for example: :$s_i \preceq s_j \preceq s_k \preceq s_i$ and $\preceq$ would therefore not be transitive and so not a totally ordered set. So $\struct {S, \preceq}$ is well-founded. The result follows from the definition of well-ordered set. {{qed}}
Let $\struct {S, \preceq}$ be a well-ordered set. Let $T \subseteq S$. Let $X \subseteq T$. By Subset Relation is Transitive, $X \subseteq S$. By the definition of a well-ordered set, $X$ has a smallest element. It follows by definition that $T$ is well-ordered. Hence the result. {{Qed}}
{{begin-eqn}} {{eqn | l = x \circ \left({y \circ z}\right)       | r = \left({x \circ y}\right) \circ z       | c = $\circ$ is associative }} {{eqn | r = \left({y \circ x}\right) \circ z       | c = $x$ commutes with $y$ }} {{eqn | r = y \circ \left({x \circ z}\right)       | c = $\circ$ is associative }} {{eqn | r = y \circ \left({z \circ x}\right)       | c = $x$ commutes with $z$ }} {{eqn | r = \left({y \circ z}\right) \circ x       | c = $\circ$ is associative }} {{end-eqn}} {{qed}}
Let $\circ$ be an associative operation on $S$. === Necessary Condition === Suppose $\circ$ is anticommutative. Let $a \circ a = x$ for some $a \in S$. Then: {{begin-eqn}} {{eqn | l = \left({a \circ a}\right) \circ a       | r = a \circ \left({a \circ a}\right)       | c = $\circ$ is Associative }} {{eqn | ll= \implies       | l = a \circ x       | r = x \circ a       | c =  }} {{eqn | ll= \implies       | l = x       | r = a       | c = $\circ$ is Anticommutative }} {{eqn | ll= \implies       | l = a \circ a       | r = a       | c =  }} {{end-eqn}} So $\circ$ being associative and anticommutative implies that $\circ$ is idempotent. Now, let $a \circ b \circ a = x$ for some $a, b$. {{begin-eqn}} {{eqn | l = \left({a \circ a}\right) \circ b \circ a       | r = a \circ b \circ a       | c = Idempotence of $\circ$, proved above }} {{eqn | r = a \circ b \circ \left({a \circ a}\right)       | c = Idempotence of $\circ$ }} {{eqn | ll= \implies       | l = a \circ \left({a \circ b \circ a}\right)       | r = \left({a \circ b \circ a}\right) \circ a       | c = $\circ$ is Associative }} {{eqn | ll= \implies       | l = a \circ x       | r = x \circ a       | c =  }} {{eqn | ll= \implies       | l = x       | r = a       | c = $\circ$ is Anticommutative }} {{eqn | ll= \implies       | l = a \circ b \circ a       | r = a       | c =  }} {{end-eqn}} So $\circ$ being associative and anticommutative implies, via the fact (also proved) that $\circ$ is idempotent, that $a \circ b \circ a = a$. {{qed|lemma}} === Sufficient Condition === Now, suppose $\circ$ (which we take to be associative) is idempotent and: :$\forall a, b \in S: a \circ b \circ a = a$ It remains to be shown that $\circ$ is anticommutative. Suppose $a \circ b = b \circ a$. Then: {{begin-eqn}} {{eqn | l = a \circ b       | r = b \circ a       | c =  }} {{eqn | ll= \implies       | l = a \circ b \circ a       | r = a \circ a \circ b = a \circ b = a       | c =  }} {{eqn | l = a \circ b \circ a       | r = b \circ a \circ a = b \circ a = a       | c =  }} {{eqn | l = b \circ a \circ b       | r = b \circ b \circ a = b \circ a = b       | c =  }} {{eqn | l = b \circ a \circ b       | r = a \circ b \circ b = a \circ b = b       | c =  }} {{eqn | ll= \implies       | l = a       | r = a \circ b = b       | c =  }} {{eqn | l = a       | r = b \circ a = b       | c =  }} {{eqn | ll= \implies       | l = a       | r = b       | c =  }} {{end-eqn}} So, $\circ$ being idempotent and $a \circ b \circ a = a$ implies that $\circ$ is anticommutative. {{Qed}}
Let $\circ$ be both associative and anticommutative. Then from Associative Idempotent Anticommutative: : $\forall x, z \in S: x \circ z \circ x = x$ and $\circ$ is idempotent. Consider $x \circ y \circ z \circ x \circ z$. We have: {{begin-eqn}} {{eqn | l = x \circ y \circ z \circ x \circ z       | r = x \circ \left({y \circ z}\right) \circ x \circ z       | c =  }} {{eqn | r = x \circ z       | c =  }} {{end-eqn}} Also: {{begin-eqn}} {{eqn | l = x \circ y \circ z \circ x \circ z       | r = x \circ y \circ \left({z \circ x \circ z}\right)       | c =  }} {{eqn | r = x \circ y \circ z       | c =  }} {{end-eqn}} Hence the result. {{qed}}
{{begin-eqn}} {{eqn | l = x \sqbrk c y       | r = c       | c = {{Defof|Constant Operation}} }} {{eqn | l = y \sqbrk c x       | r = c       | c = {{Defof|Constant Operation}} }} {{eqn | ll= \leadsto       | l = x \sqbrk c y       | r = y \sqbrk c x       | c =  }} {{end-eqn}} Hence the result by definition of commutativity. {{Qed}} Category:Constant Operation Category:Commutativity rgvaxlfl8v7zwybh4pmlxf52y5rnnxb
{{begin-eqn}} {{eqn | l = \left({x \left[{c}\right] y}\right) \left[{c}\right] z       | r = c \left[{c}\right] z       | c = Definition of Constant Operation }} {{eqn | r = c       | c = Definition of Constant Operation }} {{end-eqn}} {{begin-eqn}} {{eqn | l = x \left[{c}\right] \left({y \left[{c}\right] z}\right)       | r = x \left[{c}\right] c       | c = Definition of Constant Operation }} {{eqn | r = c       | c = Definition of Constant Operation }} {{end-eqn}} {{qed}} Category:Constant Operation jph7k2jvg81vrd0hlsz5994dno3s6nk
To simplify our notation: : Let $\max \left({x, y}\right)$ be (temporarily) denoted $x \overline \wedge y$ : Let $\min \left({x, y}\right)$ be (temporarily) denoted $x \underline \vee y$. There are three cases to consider: :$(1): \quad x \le y$ :$(2): \quad y \le x$ :$(3): \quad x = y$ $(1): \quad$ Let $x \le y$. Then: {{begin-eqn}} {{eqn | l=x \overline \wedge y       | r=y       | rr==y \overline \wedge x }} {{eqn | l=x \underline \vee y       | r=x       | rr==y \underline \vee x }} {{end-eqn}} $(2): \quad$ Let $y \le x$. Then: {{begin-eqn}} {{eqn | l=x \overline \wedge y       | r=x       | rr==y \overline \wedge x }} {{eqn | l=x \underline \vee y       | r=y       | rr==y \underline \vee x }} {{end-eqn}} $(3): \quad$ Let $x = y$. Then: {{begin-eqn}} {{eqn | l=x \overline \wedge y       | r=x = y       | rr==y \overline \wedge x }} {{eqn | l=x \underline \vee y       | r=x = y       | rr==y \underline \vee x }} {{end-eqn}} Thus $\overline \wedge$, i.e. $\max$, and $\underline \vee$, i.e. $\min$, have been shown to be commutative in all cases. {{qed}} Category:Max and Min Operations 49tfykvm4ap31r0om7omi80kwm1vsjx
Follows immediately from the definitions of Max and Min. {{Qed}} Category:Max and Min Operations Category:Idempotence 98gysw9vcuahm70qrfopw6qzd1kffdn
This follows from set equality and Equality of Ordered Pairs. Category:Abstract Algebra Category:Equality lx4ikvvqj4hwbu3b5xw5bd8y3kckaqx
{{begin-eqn}} {{eqn | o =        | r = T \subseteq S       | c =  }} {{eqn | ll= \leadsto       | o =        | r = \forall a, b, c \in T: a, b, c \in S       | c = {{Defof|Subset}} }} {{eqn | ll= \leadsto       | o =        | r = a \mathop {\circ {\restriction_T} } \paren {b \mathop {\circ {\restriction_T} } c}       | c =  }} {{eqn | r = a \circ \paren {b \circ c}       | c =  }} {{eqn | r = \paren {a \circ b} \circ c       | c = as $\circ$ is associative }} {{eqn | r = \paren {a \mathop {\circ {\restriction_T} } b} \mathop {\circ {\restriction_T} } c       | c =  }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = T       | o = \subseteq       | r = S       | c =  }} {{eqn | ll= \leadsto       | l = \forall a, b \in T       | o = :       | r = a, b \in S       | c = {{Defof|Subset}} }} {{eqn | ll= \leadsto       | l = a \mathop {\circ {\restriction_T} } b       | r = a \circ b       | c =  }} {{eqn | r = b \circ a       | c = $\circ$ is commutative }} {{eqn | r = b \mathop {\circ {\restriction_T} } a       | c =  }} {{end-eqn}} {{qed}}
By Singleton of Element is Subset: : $x \in S \iff \set x \subseteq S$ By the definition of idempotence: : $x \circ x = x \in \set x$ Thus $\set x$ is a subset of $S$ which is closed under $\circ$. By the definition of submagma, the result follows. {{qed}}
By definition: : $T \circ T = \left\{{x = a \circ b: a, b \in T}\right\}$ === Necessary Condition === Let $\left({T, \circ}\right)$ be a magma. Then $T$ is closed. That is: :$\forall x, y \in T: x \circ y \in T$ Thus: : $x \circ y \in T \circ T \implies x \circ y \in T$ {{qed|lemma}} === Sufficient Condition === Let $T \circ T \subseteq T$. Then: : $x \circ y \in T \circ T \implies x \circ y \in T$ That is, $T$ is closed. Therefore $\left({T, \circ}\right)$ is a magma by definition. {{qed}}
Let $X, Y, Z \in \powerset S$. Then: {{begin-eqn}} {{eqn | l = X \circ_\PP \paren {Y \circ_\PP Z}       | r = \set {x \circ \paren {y \circ z}: x \in X, y \in Y, z \in Z}       | c =  }} {{eqn | r = \set {\paren {x \circ y} \circ z: x \in X, y \in Y, z \in Z}       | c =  }} {{eqn | r = \paren {X \circ_\PP Y} \circ_\PP Z       | c =  }} {{end-eqn}} demonstrating that $\circ_\PP$ is associative on $\powerset S$. {{qed}}
Let $\struct {S, \circ}$ be a magma in which $\circ$ is commutative. Let $X, Y \in \powerset S$. Then: {{begin-eqn}} {{eqn | l = X \circ_\mathcal P Y       | r = \set {x \circ y: x \in X, y \in Y} }} {{eqn | l = Y \circ_\mathcal P X       | r = \set {y \circ x: x \in X, y \in Y} }} {{end-eqn}} from which it follows that $\circ_\mathcal P$ is commutative on $\powerset S$. {{qed}}
Let $x \in X, z \in Z$. Then: :$x \circ z \in X \circ Z$ and $z \circ x \in Z \circ X$ Now: :$Y \circ Z = \set {y \circ z: y \in Y, z \in Z}$ :$Z \circ Y = \set {z \circ y: y \in Y, z \in Z}$ But by the definition of a subset: :$x \in X \implies x \in Y$ Thus: :$x \circ z \in Y \circ Z$ and $z \circ x \in Z \circ Y$ and the result follows. {{Qed}}
Let $x \in T$ be cancellable in $S$. Then by definition, $x \in T$ is left cancellable in $S$. It follows from Left Cancellable Element is Left Cancellable in Subset that $x \in T$ is left cancellable in $T$. Again by definition, $x \in T$ is right cancellable in $S$. It follows from Right Cancellable Element is Right Cancellable in Subset that $x \in T$ is right cancellable in $T$. Thus $x$ is also both left cancellable and right cancellable in $T$. So by definition, $x$ is cancellable in $T$. {{qed}}
From Restriction of Associative Operation is Associative, if $\circ$ is associative on $\struct {S, \circ}$, then it will also be associative on $\struct {T, \circ}$. Thus we do not need to check for associativity in $\struct {T, \circ}$, as that has been inherited from its extension $\struct {S, \circ}$. So, once we have established that $T \subseteq S$, all we need to do is to check for $\circ$ to be a closed operation. {{Qed}}
=== Left Cancellable === {{:Left Cancellable iff Left Regular Representation Injective}} === Right Cancellable === {{:Right Cancellable iff Right Regular Representation Injective}}
Let $\struct {S, \circ}$ be an algebraic structure with more than one left identity. Take any two of these, and call them $e_{L_1}$ and $e_{L_2}$, where $e_{L_1} \ne e_{L_2}$. Suppose $\struct {S, \circ}$ has a right identity. Call this right identity $e_R$. Then, by the behaviour of $e_R$, $e_{L_1}$ and $e_{L_2}$: : $e_{L_1} = e_{L_1} \circ e_R = e_R$ : $e_{L_2} = e_{L_2} \circ e_R = e_R$ So $e_{L_1} = e_R = e_{L_2}$, which contradicts the supposition that $e_{L_1}$ and $e_{L_2}$ are different. Therefore, in an algebraic structure with more than one left identity, there can be no right identity. {{qed}}
From Left Operation is Associative, $\leftarrow$ is associative. By the nature of the left operation, $\struct {S, \leftarrow}$ is closed: :$\forall x, y \in S: x \leftarrow y = x \in S$ whatever $S$ may be. So $\struct {S, \leftarrow}$ is a semigroup. From the definition of left operation: :$\forall x, y \in S: x \leftarrow y = x$ from which it is apparent that all elements of $S$ are right identities. From More than one Right Identity then no Left Identity, it also follows that there is no left identity. {{qed}}
From Right Operation is Associative, $\rightarrow$ is associative. By the nature of the right operation, $\struct {S, \rightarrow}$ is closed: :$\forall x, y \in S: x \rightarrow y = y \in S$ whatever $S$ may be. So $\struct {S, \rightarrow}$ is a semigroup. From the definition of right operation: :$\forall x, y \in S: x \rightarrow y = y$ from which it is apparent that all elements of $S$ are left identities. {{qed}} From More than one Left Identity then no Right Identity, it also follows that there is no right identity.
Let $\struct {S, \circ}$ be an algebraic structure such that: :$\exists e_L \in S: \forall x \in S: e_L \circ x = x$ :$\exists e_R \in S: \forall x \in S: x \circ e_R = x$ Then $e_L = e_L \circ e_R = e_R$ by both the above, hence the result. The uniqueness of the left and right identity is a direct result of Identity is Unique. {{Qed}}
Suppose that: :$\forall a \in S: \exists x, y \in S: s \circ x = a = y \circ s$. Since $s \in S$, it follows that: :$\exists v, w \in S: s \circ v = s = w \circ s$. Let $a \in S$. Then: :$\exists x, y \in S: s \circ x = a = y \circ s$. Thus: {{begin-eqn}} {{eqn | l = a       | r = s \circ x       | c =  }} {{eqn | ll= \leadsto       | l = w \circ a       | r = w \circ s \circ x       | c =  }} {{eqn | r = s \circ x       | c =  }} {{eqn | r = a       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = a       | r = y \circ s       | c =  }} {{eqn | ll= \leadsto       | l = a \circ v       | r = y \circ s \circ v       | c =  }} {{eqn | r = y \circ s       | c =  }} {{eqn | r = a       | c =  }} {{end-eqn}} Hence: :$w \circ a = a$ and $a \circ v = a$ for any $a \in S$. In particular: :Letting $a = v$ in the first of these gives $w \circ v = v$ :Letting $a = w$ in the second gives $w \circ v = w$. Thus $v = w \circ v = w$ is the identity element in $S$. {{qed}}
Let $\left({S, \circ}\right)$ be a monoid whose identity is $e$. Let $x, y \in S$ such that $x \circ e = y \circ e$ Then, by the definition of the identity: :$x = x \circ e = y \circ e = y$ ... thus $x = y$ and the result is proved. {{qed}} Category:Monoids eatawhcnkyf0usqdi6ji6dxzxuyafou