avewright/proofwiki-math huggingface dataset:
Theorem: Composite of Antisymmetric Relations is not necessarily Antisymmetric

Let $A$ be a set.
Let $\RR$ and $\SS$ be antisymmetric relations on $A$.
Then their composite $\RR \circ \SS$ is not necessarily also antisymmetric.

Proof:
;Proof by Counterexample:
Consider the ordering $\le$ on the natural numbers $\N$.
Consider its dual ordering $\ge$ also on $\N$.
Note that Dual Ordering is Ordering.
Both $\le$ and $\ge$ are {{afortiori}} antisymmetric relations.
We have:
:$1 \le 3$
:$3 \ge 2$
and similarly:
:$2 \le 3$
:$3 \ge 1$
Hence it follows that:
:$1 \le \circ \ge 2$
while at the same time:
:$2 \le \circ \ge 1$
and so while both $\le$ and $\ge$ are antisymmetric, their composite $\le \circ \ge$ is not.
Hence the result.
{{qed}}
Category:Antisymmetric Relations
Category:Composite Relations
gj4i28qi76npou7h7pt4g9vqkjj3www
Theorem: Composite of Orderings is not necessarily Ordering

Let $A$ be a set.
Let $\RR$ and $\SS$ be orderings on $A$.
Then their composite $\RR \circ \SS$ is not necessarily also an ordering on $A$.

Proof:
Let $\RR$ and $\SS$ be orderings as asserted.
Both $\RR$ and $\SS$ are {{afortiori}} both antisymmetric and transitive.
But we have:
:From Composite of Antisymmetric Relations is not necessarily Antisymmetric, it is not necessarily the case that $\RR \circ \SS$ is itself antisymmetric.
:From Composite of Transitive Relations is not necessarily Transitive, it is not necessarily the case that $\RR \circ \SS$ is itself transitive.
Hence it is not necessarily the case that $\RR \circ \SS$ is an ordering.
{{qed}}
Theorem: Composite of Transitive Relations is not necessarily Transitive

Let $A$ be a set.
Let $\RR$ and $\SS$ be transitive relations on $A$.
Then their composite $\RR \circ \SS$ is not necessarily also transitive.

Proof:
Recall the definition of composition of relations:
{{:Definition:Composition of Relations}}
;Proof by Counterexample
:240pxthumbright
Let $A = \set {a, b, c}$.
Let $\RR$ be defined as:
:$\RR = \set {\tuple {a, a}, \tuple {b, b}, \tuple {c, c}, \tuple {b, c} }$
Let $\SS$ be defined as:
:$\SS = \set {\tuple {a, a}, \tuple {b, b}, \tuple {c, c}, \tuple {a, b} }$
Both $\RR$ and $\SS$ can be seen to be transitive.
Then we have:
:$\RR \circ \SS = \set {\tuple {a, a}, \tuple {b, b}, \tuple {c, c}, \tuple {a, b}, \tuple {b, c} }$
We note that:
:$\tuple {a, b}, \tuple {b, c} \in \RR \circ \SS$
but:
:$\tuple {a, c} \notin \RR \circ \SS$
Hence, by definition, $\RR \circ \SS$ is not transitive.
{{qed}}
Category:Composite Relations
Category:Transitive Relations
g0vvswvx7mjomebuamnjkrft4bnlxrn
Theorem: Identity Mapping is Automorphism/Ordered Semigroups

Let $\struct {S, \circ, \preccurlyeq}$ be an ordered semigroup.
Then $I_S: \struct {S, \circ, \preccurlyeq} \to \struct {S, \circ, \preccurlyeq}$ is a ordered semigroup automorphism.

Proof:
From Identity Mapping is Semigroup Automorphism:
:$I_S: \struct {S, \circ} \to \struct {S, \circ}$ is a semigroup automorphism.
From Identity Mapping is Order Isomorphism:
:$I_S: \struct {S, \preccurlyeq} \to \struct {S, \preccurlyeq}$ is an order isomorphism.
{{qed}}
Theorem: Inverse of Ordered Semigroup Isomorphism is Isomorphism

Let $\struct {S, \circ, \preceq}$ and $\struct {T, *, \preccurlyeq}$ be ordered semigroups.
Let $\phi: \struct {S, \circ, \preceq} \to \struct {T, *, \preccurlyeq}$ be a mapping.
Then:
:$\phi$ is an ordered semigroup isomorphism
{{iff}}:
:$\phi^{-1}: \struct {T, *, \preccurlyeq} \to \struct {S, \circ, \preceq}$ is also an ordered semigroup isomorphism.

Proof:
=== Sufficient Condition ===
Let $\phi$ be an ordered semigroup isomorphism.
Then by definition $\phi$ is a bijection.
Thus $\exists \phi^{-1}$ such that $\phi^{-1}$ is also a bijection from Bijection iff Inverse is Bijection.
That is:
:$\exists \phi^{-1}: \struct {T, *, \preccurlyeq} \to \struct {S, \circ, \preceq}$
From Inverse of Algebraic Structure Isomorphism is Isomorphism:
:$\phi^{-1}: \struct {T, *} \to \struct {S, \circ}$ is an isomorphism.
Indeed, from Isomorphism Preserves Semigroups, $\phi^{-1}$ is specifically a semigroup isomorphism.
Then from Inverse of Order Isomorphism is Order Isomorphism we have that:
:$\phi^{-1}: \struct {T, \preccurlyeq} \to \struct {S, \preceq}$ is an order isomorphism.
{{qed|lemma}}
Theorem: Composite of Ordered Semigroup Isomorphisms is Isomorphism

Let $\struct {S_1, \circ_1, \preccurlyeq_1}, \struct {S_2, \circ_2, \preccurlyeq_2}, \struct {S_3, \circ_3, \preccurlyeq_3}$ be ordered semigroups.
Let $\phi: S_1 \to S_2$ and $\psi: S_2 \to S_3$ be ordered semigroup isomorphisms.
Then the composite mapping $\psi \circ \phi: S_1 \to S_3$ is also an ordered semigroup isomorphism.

Proof:
From Composite of Isomorphisms in Algebraic Structure is Isomorphism, $\psi \circ \phi$ is an algebraic structure isomorphism.
From Isomorphism Preserves Semigroups, it follows that $\psi \circ \phi$ is a semigroup isomorphism from $\struct {S_1, \circ_1}$ to $\struct {S_3, \circ_3}$.
From Composite of Order Isomorphisms is Order Isomorphism, $\psi \circ \phi: \struct {S_1, \preceq_1} \to \struct {S_3, \preceq_3}$ is an order isomorphism.
Hence the result.
{{qed}}
Theorem: Integers Modulo m under Max Operation form Ordered Semigroup

Let $\Z_m$ denote the set of integers modulo $m$:
:$\Z_m = \set {0, 1, \ldots, m - 1}$
Let $\vee_m$ be the operation on $\Z_m$ defined as:
:$\forall a, b \in \Z_m: a \vee_m b = \max \set {a, b}$
Then the ordered algebraic structure $\struct {\Z_m, \vee_m, \le}$ is an ordered semigroup.

Proof:
Taking the semigroup axioms in turn:
Theorem: Composite of Connected Relation is not necessarily Connected

Let $A$ be a set.
Let $\RR$ and $\SS$ be connected relations on $A$.
Then their composite $\RR \circ \SS$ is not necessarily also connected.

Proof:
Recall the definition of composition of relations:
{{:Definition:Composition of Relations}}
;Proof by Counterexample
:240pxthumbright
Let $A = \set {a, b, c}$.
Let $\RR$ be defined as:
:$\RR = \set {\tuple {a, b}, \tuple {b, c}, \tuple {c, a} }$
Let $\SS$ be defined as $\RR^{-1}$, that is, the inverse of $\RR$:
:$\SS = \set {\tuple {b, a}, \tuple {c, b}, \tuple {a, c} }$
Both $\RR$ and $\SS$ can be seen to be connected.
Then we have:
:$\RR \circ \SS = \set {\tuple {a, a}, \tuple {b, b}, \tuple {c, c} }$
and it is immediately apparent that $\RR \circ \SS$ is not connected.
{{qed}}
Category:Composite Relations
Category:Connected Relations
5kiy490yyghidztrvrs82gxqjec3rwq
Theorem: Composite of Total Relations is Total

Let $A$ be a set.
Let $\RR$ and $\SS$ be total relations on $A$.
Then their composite $\RR \circ \SS$ is also total.

Proof:
Recall the definition of composition of relations:
{{:Definition:Composition of Relations}}
Hence in this particular context:
:$\RR \circ \SS := \set {\tuple {x, z} \in A \times A: \exists y \in A: \tuple {x, y} \in \SS \land \tuple {y, z} \in \RR}$
Let $x$ and $y$ in $A$ be arbitrary.
Because $\SS$ is total, either $x \mathrel \SS y$ or $y \mathrel \SS x$.
{{WLOG}}, suppose $x \mathrel \SS y$.
From Relation is Connected and Reflexive iff Total, $\RR$ is reflexive.
That is:
:$y \mathrel \RR y$
Thus we have:
:$\exists y \in A: \tuple {x, y} \in \SS \land \tuple {y, y} \in \RR$
and so by definition of composition of relations:
:$\tuple {x, y} \in \RR \circ \SS$
Similarly for $y \mathrel \SS x$:
:$\exists x \in A: \tuple {y, x} \in \SS \land \tuple {x, x} \in \RR$
giving:
:$\tuple {y, x} \in \RR \circ \SS$
So, by definition, $\RR \circ \SS$ is total.
{{qed}}
Category:Composite Relations
Category:Total Relations
jwup2yg6uv9vggl0xg2vjjfbdw5t42n
Theorem: Dual of Ordered Semigroup is Ordered Semigroup

Let $\struct {S, \circ, \preccurlyeq}$ be an ordered semigroup.
Then its dual $\struct {S, \circ, \succcurlyeq}$ is also an ordered semigroup.

Proof:
From Dual Ordering is Ordering, we have that $\struct {S, \succcurlyeq}$ is an ordered set.
We also note from the definition that $\struct {S, \circ}$ is a semigroup.
It remains to be demonstrated that $\succcurlyeq$ is compatible with $\circ$.
Recall that $\struct {S, \circ, \preccurlyeq}$ is an ordered semigroup.
Hence {{afortiori}} $\preccurlyeq$ is compatible with $\circ$.
Let $x, y \in S$ be arbitrary such that $x \succcurlyeq y$.
We have:
{{begin-eqn}}
{{eqn | l = x
      | o = \succcurlyeq
      | r = y
      | c = 
}}
{{eqn | ll= \leadsto
      | l = y
      | o = \preccurlyeq
      | r = x
      | c = {{Defof|Dual Ordering}}
}}
{{eqn | ll= \leadsto
      | l = \paren {y \circ z}
      | o = \preccurlyeq
      | r = \paren {x \circ z}
      | c = {{Defof|Relation Compatible with Operation}} 
}}
{{eqn | ll= \leadsto
      | l = \paren {x \circ z}
      | o = \succcurlyeq
      | r = \paren {y \circ z}
      | c = 
}}
{{end-eqn}}
and similarly:
{{begin-eqn}}
{{eqn | l = x
      | o = \succcurlyeq
      | r = y
      | c = 
}}
{{eqn | ll= \leadsto
      | l = y
      | o = \preccurlyeq
      | r = x
      | c = {{Defof|Dual Ordering}}
}}
{{eqn | ll= \leadsto
      | l = \paren {z \circ y}
      | o = \preccurlyeq
      | r = \paren {z \circ x}
      | c = {{Defof|Relation Compatible with Operation}} 
}}
{{eqn | ll= \leadsto
      | l = \paren {z \circ x}
      | o = \succcurlyeq
      | r = \paren {z \circ y}
      | c = 
}}
{{end-eqn}}
Hence the result.
{{qed}}
Theorem: Inversion Mapping is Isomorphism from Ordered Abelian Group to its Dual

Let $\struct {G, \circ, \preccurlyeq}$ be an ordered Abelian group.
Let $\struct {G, \circ, \succcurlyeq}$ be the dual of $\struct {G, \circ, \preccurlyeq}$.
Let $\phi: \struct {G, \circ, \preccurlyeq} \to \struct {G, \circ, \succcurlyeq}$ be the inversion mapping from $\struct {G, \circ, \preccurlyeq}$ to $\struct {G, \circ, \succcurlyeq}$ defined as:
:$\forall x \in G: \map \phi x = x^{-1}$
where $x^{-1}$ denotes the inverse of $x$ in $\struct {G, \circ}$.
Then $\phi$ is an isomorphism.

Proof:
$\struct {G, \circ, \preccurlyeq}$ is {{afortiori}} an ordered semigroup.
Hence from Dual of Ordered Semigroup is Ordered Semigroup, $\struct {G, \circ, \succcurlyeq}$ is also an ordered semigroup.
That is, $\succcurlyeq$ is compatible with $\circ$.
From Inversion Mapping is Permutation, {{afortiori}} the inversion mapping is a bijection.
From Inversion Mapping is Automorphism iff Group is Abelian, $\phi: \struct {G, \circ} \to \struct {G, \circ}$ is an automorphism and so {{afortiori}} an isomorphism.
It remains to be shown that  $\phi: \struct {G, \preccurlyeq} \to \struct {G, \succcurlyeq}$ is an order isomorphism.
We note that from Inversion Mapping is Involution that:
:$\phi = \phi^{-1}$
Let $x, y \in G$ be arbitrary.
We have:
{{begin-eqn}}
{{eqn | l = x
      | o = \preccurlyeq
      | r = y
      | c = 
}}
{{eqn | ll= \leadsto
      | l = y
      | o = \succcurlyeq
      | r = x
      | c = {{Defof|Dual Ordering}}
}}
{{eqn | ll= \leadsto
      | l = x^{-1}
      | o = \succcurlyeq
      | r = y^{-1}
      | c = Inversion Mapping Reverses Ordering in Ordered Group
}}
{{eqn | ll= \leadsto
      | l = \map \phi x
      | o = \succcurlyeq
      | r = \map \phi y
      | c = 
}}
{{end-eqn}}
Then we have:
{{begin-eqn}}
{{eqn | l = x
      | o = \succcurlyeq
      | r = y
      | c = 
}}
{{eqn | ll= \leadsto
      | l = y
      | o = \preccurlyeq
      | r = x
      | c = {{Defof|Dual Ordering}}
}}
{{eqn | ll= \leadsto
      | l = x^{-1}
      | o = \preccurlyeq
      | r = y^{-1}
      | c = Inversion Mapping Reverses Ordering in Ordered Group
}}
{{eqn | ll= \leadsto
      | l = \map \phi x
      | o = \preccurlyeq
      | r = \map \phi y
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \map {\phi^{-1} } x
      | o = \preccurlyeq
      | r = \map {\phi^{-1} } y
      | c = as $\phi = \phi^{-1}$
}}
{{end-eqn}}
Hence we have shown that $\phi$ is an order-preserving mapping in both directions.
Hence the result.
{{qed}}
Theorem: Hensel's Lemma/P-adic Integers/Lemma 1

:There exists a unique $p$-adic expansion $\ds \sum_{n \mathop = 0}^\infty d_n p^n$:
::$\ds \forall k : a_k = \sum_{n \mathop = 0}^k d_n p^n$ satisfies:
{{begin-eqn}}
{{eqn | n = 1
      | l = \map F {a_k}
      | o = \equiv
      | r = 0
      | rr= \pmod {p^{k+1}\Z_p}
}}
{{eqn | n = 2
      | l = a_k
      | o = \equiv
      | r = \alpha_0
      | rr= \pmod {p\Z_p}
}}
{{end-eqn}}

Proof:
The Second Principle of Recursive Definition is used to construct the sequence
$\sequence {d_n}$.
Let $T$ be the set of $p$-adic digits.
For each $k \in \N_{>0}$, let:
:$\ds S_k = \set{\tuple{b_0, b_1, \ldots, b_{k-1}} \subseteq T^k : \map F {\sum_{n \mathop = 0}^{k-1} b_n p^n } \equiv 0 \pmod{p^k\Z_p} \quad \text{and} \quad \sum_{n \mathop = 0}^{k-1} b_n p^n \equiv \alpha_0 \pmod{p\Z_p}}$
Let $d_0$ be the first $p$-adic digit of the canonical expansion of $\alpha_0$.
Theorem: Simple Order Product of Pair of Ordered Semigroups is Ordered Semigroup

Let $\struct {S_1, \circ_1, \preccurlyeq_1}$ and $\struct {S_2, \circ_2, \preccurlyeq_2}$ be ordered semigroups.
Let $\struct {S_1 \times S_2, \odot} := \struct {S_1, \circ_1} \times \struct {S_2, \circ_2}$ denote the '''external direct product''' of $\struct {S_1, \circ_1}$ and $\struct {S_2, \circ_2}$.
Let $\struct {S_1 \times S_2, \preccurlyeq_s} := \struct {S_1, \preccurlyeq_1} \otimes^s \struct {S_2, \preccurlyeq_2}$ denote the '''simple (order) product''' of $\struct {S_1, \preccurlyeq_1}$ and $\struct {S_2, \preccurlyeq_2}$.
Then $\struct {S_1 \times S_2, \odot, \preccurlyeq_s}$ is also an ordered semigroup.

Proof:
From Simple Order Product of Pair of Ordered Sets is Ordered Set, $\struct {S_1 \times S_2, \otimes^s}$ is an ordered set.
From External Direct Product of Semigroups, $\struct {S_1 \times S_2, \odot}$ is a semigroup.
It remains to be shown that $\preccurlyeq_s$ is compatible with $\odot$.
Let $\tuple {x_1, x_2}, \tuple {y_1, y_2} \in S_1 \times S_2$ be arbitrary such that $\tuple {x_1, x_2} \preccurlyeq_s \tuple {y_1, y_2}$.
{{begin-eqn}}
{{eqn | q = 
      | l = \tuple {x_1, x_2}
      | o = \preccurlyeq_s
      | r = \tuple {y_1, y_2}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = x_1
      | o = \preccurlyeq_1
      | r = y_1
      | c = {{Defof|Simple Order Product}}
}}
{{eqn | lo= \land
      | l = x_2
      | o = \preccurlyeq_2
      | r = y_2
      | c = 
}}
{{eqn | ll= \leadsto
      | q = \forall \tuple {z_1, z_2} \in S_1 \times S_2
      | l = x_1 \circ_1 z_1
      | o = \preccurlyeq_1
      | r = y_1 \circ_1 z_1
      | c = {{Defof|Relation Compatible with Operation}}
}}
{{eqn | lo= \land
      | l = x_2 \circ_2 z_2
      | o = \preccurlyeq_2
      | r = y_2 \circ_2 z_2
      | c = 
}}
{{eqn | lo= \land
      | l = z_1 \circ_1 x_1
      | o = \preccurlyeq_1
      | r = z_1 \circ_1 y_1
      | c = 
}}
{{eqn | lo= \land
      | l = z_2 \circ_2 x_2
      | o = \preccurlyeq_2
      | r = z_2 \circ_2 y_2
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \tuple {x_1 \circ_1 z_1, x_2 \circ_2 z_2}
      | o = \preccurlyeq_s
      | r = \tuple {y_1 \circ_1 z_1, y_2 \circ_2 z_2}
      | c = {{Defof|Simple Order Product}}
}}
{{eqn | lo= \land
      | l = \tuple {z_1 \circ_1 x_1, z_2 \circ_2 x_2}
      | o = \preccurlyeq_s
      | r = \tuple {z_1 \circ_1 y_1, z_2 \circ_2 y_2}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \tuple {x_1, x_2} \odot \tuple {z_1, z_2}
      | o = \preccurlyeq_s
      | r = \tuple {y_1, y_2} \odot \tuple {z_1, z_2}
      | c = {{Defof|External Direct Product}}
}}
{{eqn | lo= \land
      | l = \tuple {z_1, z_2} \odot \tuple {x_1, x_2}
      | o = \preccurlyeq_s
      | r = \tuple {z_1, z_2} \odot \tuple {y_1, y_2}
      | c = 
}}
{{end-eqn}}
and the proof is complete.
{{qed}}
Theorem: Convergent Series can be Added Term by Term

Let $\ds \sum_{n \mathop = 1}^\infty a_n$ and $\ds \sum_{n \mathop = 1}^\infty b_n$ be two convergent series converging to $A$ and $B$ respectively.
Then:
:$\ds \sum_{n \mathop = 1}^\infty \paren {a_n + b_n} = A + B$

Proof:
{{begin-eqn}}
{{eqn | l = A + B
      | r = \sum_{n \mathop = 1}^\infty a_n + \sum_{n \mathop = 1}^\infty b_n
}}
{{eqn | r = \lim_{N \mathop \to \infty} \sum_{n \mathop = 1}^N + \lim_{N \mathop \to \infty} \sum_{n \mathop = 1}^N b_n
      | c = {{Defof|Convergent Series of Numbers}}
}}
{{eqn | r = \lim_{N \mathop \to \infty} \sum_{n \mathop = 1}^N a_n + \sum_{n \mathop = 1}^N b_n
      | c = Sum Rule for Real Sequences
}}
{{eqn | r = \lim_{N \mathop \to \infty} \sum_{n \mathop = 1}^N \paren {a_n + b_n}
      | c = rearranging a finite sum
}}
{{eqn | r = \sum_{n \mathop = 1}^\infty \paren {a_n + b_n}
      | c = {{Defof|Convergent Series of Numbers}}
}}
{{end-eqn}}
{{qed}}
Category:Series
Category:Convergence
r9v2as9qfuef9asrpcw97bkjf26cd52
Theorem: Reflexive Reduction of Relation Compatible with Cancellable Operation is Compatible

Let $\struct {S, \circ}$ be an algebraic structure such that $\circ$ is a cancellable operation.
Let $\RR$ be a relation on $S$ which is compatible with $\circ$.
Let $\RR^\ne$ be the reflexive reduction of $\RR$.
Then $\RR^\ne$ is compatible with $\circ$.

Proof:
{{AimForCont}} $\RR^\ne$ is not compatible with $\circ$.
Let $x, y \in S$ such that:
:$x \mathrel \RR y$
but:
:$x \mathrel {\RR^\ne} y$
Then by definition of reflexive reduction:
:$x \ne y$
Then as $\RR^\ne$ is not compatible with $\circ$:
:$\exists z \in S: \lnot \paren {z \circ x \mathrel {\RR^\ne} z \circ y}$
But as $\RR$ is compatible with $\circ$:
:$z \circ x \mathrel \RR z \circ y$
That means:
:$z \circ x = z \circ y$
As $\circ$ is cancellable this leads to:
:$x = y$
which is a contradiction.
Hence $\RR^\ne$ must be compatible with $\circ$.
{{qed}}
Category:Cancellability
Category:Compatible Relations
Category:Reflexive Reductions
k7hqoweuchqa97fp86egplczvackgip
Theorem: Lexicographically Ordered Pair of Ordered Semigroups with Cancellable Elements/Converse

If $\circ_1$ is not a cancellable operation, then it may not necessarily be the case that $\preccurlyeq_l$ is compatible with $\odot$.

Proof:
Recall from Lexicographically Ordered Pair of Ordered Semigroups with Cancellable Elements that if $\circ_1$ is cancellable, then $\preccurlyeq_l$ is compatible with $\odot$.
Let $\Z_2$ denote the set of integers modulo $2$:
:$\Z_2 = \set {0, 1}$
Let $\vee_2$ be the operation on $\Z_2$ defined as:
:$\forall a, b \in \Z_2: a \vee_2 b = \max \set {a, b}$
From Integers Modulo m under Max Operation form Ordered Semigroup, the ordered algebraic structure $\struct {\Z_2, \vee_2, \le}$ is an ordered semigroup.
Let $\struct {\Z_2 \times \Z_2, \odot, \le_l}$ be the external direct product of $\struct {\Z_2, \vee_2, \le}$ with itself, such that:
:$\odot$ is the operation defined as:
::$\forall \tuple {x_1, y_1}, \tuple {x_2, y_2} \in \Z_2 \times \Z_2: \tuple {x_1, y_1} \odot \tuple {x_2, y_2} := \tuple {x_1 \vee_2 x_2, y_1 \vee_2, y_2}$
:$\le_l$ is the lexicographic ordering defined as:
::$\tuple {x_1, x_2} \le_l \tuple {y_1, y_2} \iff \paren {x_1 < y_1} \lor \paren {x_1 = y_1 \land x_2 \le y_2}$
From Lexicographic Order is Ordering we have that $\struct {S_1 \times S_2, \preccurlyeq_l}$ is an ordered set.
From External Direct Product of Semigroups, $\struct {S_1 \times S_2, \odot}$ is a semigroup.
However, we note that $\vee_2$ is not cancellable:
:$0 \vee_2 1 = 1 \vee_2 1$
but:
:$0 \ne 1$
We are now to demonstrate that $\le_l$ is ''not'' compatible with $\odot$.
Indeed:
{{begin-eqn}}
{{eqn | l = \tuple {0, 1}
      | o = \le_l
      | r = \tuple {1, 0}
      | c = 
}}
{{eqn | l = \tuple {0, 1} \odot \tuple {1, 0}
      | r = \tuple {1, 1}
      | c = 
}}
{{eqn | l = \tuple {1, 0} \odot \tuple {1, 0}
      | r = \tuple {1, 0}
      | c = 
}}
{{end-eqn}}
Thus we have that:
{{begin-eqn}}
{{eqn | l = \tuple {1, 0} \odot \tuple {1, 0}
      | o = \le_l
      | r = \tuple {0, 1} \odot \tuple {1, 0}
      | c = 
}}
{{end-eqn}}
Hence and so.
{{qed}}
Theorem: Structure Induced by Semigroup Operation is Semigroup

Let $\struct {T, \circ}$ be a semigroup.
Let $S$ be a set.
Let $\struct {T^S, \oplus}$ be the structure on $T^S$ induced by $\circ$.
Then $\struct {T^S, \oplus}$ is a semigroup.

Proof:
Taking the semigroup axioms in turn:
Theorem: Structure Induced on Set of All Mappings to Ordered Semigroup is Ordered Semigroup

Let $S$ be a set.
Let $\struct {T, \odot, \preccurlyeq}$ be an ordered semigroup.
Let $\struct {T^S, \otimes, \preccurlyeq}$ denote the algebraic structure on $T^S$ induced by $\odot$.
Then $\struct {T^S, \otimes, \preccurlyeq}$ is an ordered semigroup.

Proof:
From Structure Induced by Semigroup Operation is Semigroup, $\struct {T^S, \otimes}$ is a semigroup
From Ordered Set of All Mappings is Ordered Set, $\struct {T^S, \preccurlyeq}$ is an ordered set.
It remains to be demonstrated that $\preccurlyeq$ is compatible with $\odot$.
Let $f, g \in T^S$ such that $f \preccurlyeq g$.
That is:
:$\forall x \in S: \map f x \preccurlyeq \map g x$
We are given that $\struct {T, \odot, \preccurlyeq}$ be an ordered semigroup.
Hence {{afortiori}} $\preccurlyeq$ is compatible with $\odot$.
Thus:
{{begin-eqn}}
{{eqn | q = \forall h \in T^S
      | l = \map h x \odot \map f x
      | o = \preccurlyeq
      | r = \map h x \odot \map g x
      | c = {{Defof|Relation Compatible with Operation}}
}}
{{eqn | lo= \land
      | l = \map f x \odot \map h x
      | o = \preccurlyeq
      | r = \map g x \odot \map h x
      | c = 
}}
{{eqn | ll= \leadsto
      | l = f \otimes h
      | o = \preccurlyeq
      | r = g \otimes h
      | c = {{Defof|Pointwise Operation}}
}}
{{eqn | lo= \land
      | l = h \otimes f
      | o = \preccurlyeq
      | r = h \otimes g
      | c = 
}}
{{end-eqn}}
Hence the result.
{{qed}}
Theorem: Power Set with Intersection and Subset Relation is Ordered Semigroup

Let $S$ be a set and let $\powerset S$ be its power set.
Let $\struct {\powerset S, \cap, \subseteq}$ be the ordered structure formed from the set intersection operation and subset relation.
Then $\struct {\powerset S, \cap, \subseteq}$ is an ordered semigroup.

Proof:
From Power Set with Intersection is Commutative Monoid, $\struct {\powerset S, \cap}$ is {{afortiori}} a semigroup.
From Subset Relation is Ordering, $\struct {\powerset S, \subseteq}$ is an ordered set.
It remains to be shown that $\subseteq$ is compatible with $\cap$.
Let $A, B \subseteq S$ be arbitrary such that $A \subseteq B$.
Thus:
{{begin-eqn}}
{{eqn | l = A
      | o = \subseteq
      | r = B
      | c = 
}}
{{eqn | ll= \leadsto
      | q = \forall T \subseteq S
      | l = A \cap T
      | o = \subseteq
      | r = B \cap T
      | c = Set Intersection Preserves Subsets: Corollary
}}
{{eqn | ll= \leadsto
      | q = \forall T \subseteq S
      | l = T \cap A
      | o = \subseteq
      | r = T \cap B
      | c = Intersection is Commutative
}}
{{end-eqn}}
Hence the result.
{{qed}}
Theorem: Power Set with Intersection and Superset Relation is Ordered Semigroup

Let $S$ be a set and let $\powerset S$ be its power set.
Let $\struct {\powerset S, \cap, \supseteq}$ be the ordered structure formed from the set intersection operation and superset relation.
Then $\struct {\powerset S, \cap, \supseteq}$ is an ordered semigroup.

Proof:
From Power Set with Intersection is Commutative Monoid, $\struct {\powerset S, \cap}$ is {{afortiori}} a semigroup.
From Subset Relation is Ordering, $\struct {\powerset S, \subseteq}$ is an ordered set.
We have that $\supseteq$ is the dual to $\subseteq$.
Hence $\struct {\powerset S, \supseteq}$ is an ordered set.
It remains to be shown that $\supseteq$ is compatible with $\cap$.
Let $A, B \subseteq S$ be arbitrary such that $A \supseteq B$.
Thus:
{{begin-eqn}}
{{eqn | l = A
      | o = \supseteq
      | r = B
      | c = 
}}
{{eqn | ll= \leadsto
      | q = \forall T \subseteq S
      | l = A \cap T
      | o = \supseteq
      | r = B \cap T
      | c = Set Intersection Preserves Subsets: Corollary
}}
{{eqn | ll= \leadsto
      | q = \forall T \subseteq S
      | l = T \cap A
      | o = \supseteq
      | r = T \cap B
      | c = Intersection is Commutative
}}
{{end-eqn}}
Hence the result.
{{qed}}