hcju/proofwikiqa huggingface dataset:
Let $A \in \powerset S, B \in \powerset T$ be arbitrary. Then: {{begin-eqn}} {{eqn | l = \map {f^\to} {A \cap \map {f^\gets} B}       | o = \subseteq        | r = \map {f^\to} A \cap \map {f^\to} {\map {f^\gets} B}       | c = Image of Intersection under Mapping }} {{eqn | o = \subseteq        | r = \map {f^\to} A \cap B       | c = Subset of Codomain is Superset of Image of Preimage }} {{end-eqn}} Let $x \in \map {f^\to} A \cap B$. {{begin-eqn}} {{eqn | l = x       | o = \in       | r = \map {f^\to} A       | c = {{Defof|Set Intersection}} }} {{eqn | n = 1       | ll= \leadsto       | l = x       | r = \map f a       | c = for some $a \in A$ }} {{eqn-intertext|and:}} {{eqn | l = x       | o = \in       | r = B       | c = {{Defof|Set Intersection}} }} {{eqn | ll= \leadsto       | l = \map f a       | o = \in       | r = B       | c = from $(1)$ above }} {{eqn | ll= \leadsto       | l = a       | o = \in       | r = \map {f^\gets} B       | c = {{Defof|Inverse Image Mapping}} }} {{eqn | ll= \leadsto       | l = a       | o = \in       | r = A \cap \map {f^\gets} B       | c = {{Defof|Set Intersection}} }} {{eqn | ll= \leadsto       | l = \map f a       | o = \in       | r = \map {f^\to} {A \cap \map {f^\gets} B}       | c = {{Defof|Direct Image Mapping}} }} {{eqn | ll= \leadsto       | l = x       | o = \in       | r = \map {f^\to} {A \cap \map {f^\gets} B}       | c = as $x = \map f a$ }} {{eqn | ll= \leadsto       | l = \map {f^\to} A \cap B       | o = \subseteq       | r = \map {f^\to} {A \cap \map {f^\gets} B}       | c = {{Defof|Subset}} }} {{end-eqn}} Thus we have: :$\map {f^\to} {A \cap \map {f^\gets} B} \subseteq \map {f^\to} A \cap B$ and: :$\map {f^\to} A \cap B \subseteq \map {f^\to} {A \cap \map {f^\gets} B}$ and so: :$\forall A \in \powerset S, B \in \powerset T: \map {f^\to} {A \cap \map {f^\gets} B} = \map {f^\to} A \cap B$ by definition of set equality. {{qed}}
From Equation of Vertical Line, a vertical line in $\mathcal C$ through the point $\tuple {a, 0}$ on the $x$-axis has an equation $x = a$. A real function is by definition a mapping. Hence: :$\forall a_1, a_2 \in \Dom f: a_1 = a_2 \implies \map f {a_1} = \map f {a_2}$ where $\Dom f$ denotes the domain of $f$. Thus for each $a \in \Dom f$ there exists exactly one ordered pair $\tuple {a, y}$ such that $y = \map f a$. That is, there is exactly one point on $x = a$ which is also on the graph of $f$. The result follows. {{qed}}
From the Normal Form of Equation of Straight Line in Plane, a general straight line can be expressed in the form: :$x \cos \alpha + y \sin \alpha = p$ where: :$p$ is the length of a perpendicular $\mathcal P$ from $\mathcal L$ to the origin. :$\alpha$ is the angle made between $\mathcal P$ and the $x$-axis. As $\mathcal L$ is vertical, then by definition $\mathcal P$ is horizontal. By definition, the horizontal line through the origin is the $x$-axis itself. Thus $\alpha = 0$ and $p = a$ Hence the equation of $\mathcal L$ becomes: {{begin-eqn}} {{eqn | l = x \cos 0 + y \sin 0       | r = a       | c =  }} {{eqn | ll= \leadsto       | l = x       | r = a       | c = Sine of Zero is Zero, Cosine of Zero is One }} {{end-eqn}} Hence the result. {{qed}} Category:Equations of Straight Lines in Plane oee6f8ls5odc5y8hh8dzf1vgjk3lax1
From the Normal Form of Equation of Straight Line in Plane, a general straight line can be expressed in the form: :$x \cos \alpha + y \sin \alpha = p$ where: :$p$ is the length of a perpendicular $\mathcal P$ from $\mathcal L$ to the origin :$\alpha$ is the angle made between $\mathcal P$ and the $x$-axis. As $\mathcal L$ is horizontal, then by definition $\mathcal P$ is vertical. By definition, the vertical line through the origin is the $y$-axis itself. Thus: :$\alpha$ is a right angle, that is $\alpha = \dfrac \pi 2 = 90 \degrees$ :$p = b$ Hence the equation of $\mathcal L$ becomes: {{begin-eqn}} {{eqn | l = x \cos \dfrac \pi 2 + y \sin \dfrac \pi 2       | r = b       | c =  }} {{eqn | ll= \leadsto       | l = x \times 0 + y \times 1       | r = b       | c = Sine of Right Angle, Cosine of Right Angle }} {{eqn | ll= \leadsto       | l = y       | r = b       | c =  }} {{end-eqn}} Hence the result. {{qed}} Category:Equations of Straight Lines in Plane j9g6se6w41ekcnkgdentv3erwbwo7u9
By assumption:  :$\forall x \in S: \exists \map f x^{-1} : \map f x \circ \map f x^{-1} = \map f x^{-1} \circ \map f x  = 1_R$ Let $f^{-1} : S \to U_R$ be defined by: :$\forall x \in S : \map {f^{-1}} {x} = \map f x^{-1}$ Consider the mapping $f \circ’ f^{-1}$. For all $x \in S$:  {{begin-eqn}} {{eqn | l = \map {\paren {f \circ’ f^{-1} } } x       | r = \map f x \circ’ \map {f^{-1} } x }} {{eqn | r = \map f x \circ \map f x^{-1} }} {{eqn | r = 1_R }} {{end-eqn}} Hence $f \circ’ f^{-1} = f_{1_R}$. Similarly, $f^{-1} \circ’ f = f_{1_R}$. Hence $f$ is a unit of $\struct {R^S, +', \circ'}$ and the inverse of $f$ is the mapping $f^{-1}$.
Let $f^{-1}$ be the product inverse of $f$. Let $x \in R$. Then: {{begin-eqn}} {{eqn | l = 1_R        | r = \map {f_{1_R} } x       | c = Structure Induced by Ring with Unity Operations is Ring with Unity }} {{eqn | r = \map {\paren {f \circ' f^{-1} } } x       | c = {{Defof|Product Inverse|product inverse}} }} {{eqn | r = \map f x \circ \map {f^{-1} } x       | c = {{Defof|Pointwise Operation}} $\circ'$ }} {{end-eqn}} Similarly, $1_R = \map {f^{-1} } x \circ \map f x$ Hence $\map {f^{-1}} x$ is the product inverse of $\map f x$. So $\map f x$ is a unit in $R$ and $\map f x^{-1} = \map {f^{-1} } x$. The result follows.
From Equation of Horizontal Line, a horizontal line in $\mathcal C$ through the point $\tuple {0, b}$ on the $y$-axis has an equation $y = b$. By hypothesis, $f$ is a surjection. Hence: :$\forall b \in \R: \exists a \in \R: b = \map f a$ Thus for each $b \in \R$ there exists at least one ordered pair $\tuple {a, b}$ such that $b = \map f a$. That is, there exists at least one point on $y = b$ which is also on the graph of $f$. The result follows. {{qed}}
From Equation of Horizontal Line, a horizontal line in $\mathcal C$ through the point $\tuple {0, b}$ on the $y$-axis has an equation $y = b$. By hypothesis, $f$ is a injection. Hence: :$\forall a_1, a_2 \in \Dom f: \map f {a_1} = \map f {a_2} \implies a_1 = a_2$ where $\Dom f$ denotes the domain of $f$. Thus for each $b \in \Cdm f$ there exists exactly one ordered pair $\tuple {a, b}$ such that $b = \map f a$. That is, there is exactly one point on $y = b$ which is also on the graph of $f$. The result follows. {{qed}}
By definition, a bijection is a mapping which is both an injection and a surjection. Let $\mathcal L$ be a horizontal line through a point $b$ in the codomain of $f$. From Graph of Real Surjection in Coordinate Plane intersects Every Horizontal Line: :$\mathcal L$ intersects the graph of $f$ on at least one point $P = \tuple {a, b}$ where $b = \map f a$. From Graph of Real Injection in Coordinate Plane intersects Horizontal Line at most Once: :$\mathcal L$ intersects the graph of $f$ on at most one point $P = \tuple {a, b}$ where $b = \map f a$. The result follows. {{qed}}
Let $X, Y, Z$ be sets defined as: {{begin-eqn}} {{eqn | l = X       | r = \set {a, b, c}       | c =  }} {{eqn | l = Y       | r = \set {1, 2, 3, 4}       | c =  }} {{eqn | l = Z       | r = \set {z, y, z}       | c =  }} {{end-eqn}} Let $f: X \to Y$ be defined in two-row notation as: :$\dbinom {a \ b \ c } {1 \ 2 \ 3}$ which is seen by inspection to be an injection. Let $g: Y \to Z$ be defined in two-row notation as: :$\dbinom {1 \ 2 \ 3 \ 4} {x \ y \ y \ z}$ which is seen by inspection to be a surjection. The composition $f \circ g$ is seen to be: :$\dbinom {a \ b \ c} {x \ y \ y}$ which is: :not an injection (both $b$ and $c$ map to $y$) :not a surjection (nothing maps to $z$). Hence the result. {{qed}} Category:Injections Category:Surjections Category:Composite Mappings pr38md0rxug63sezves2glotq8qwr01
We note that both $g \circ f$ and $h \circ g$ are both injections and surjections by definition of bijection. First it is shown that $g$ is a bijection. We are given that: :$g \circ f$ is a bijection. From Injection if Composite is Injection it follows that $f$ is an injection. From Surjection if Composite is Surjection it follows that $g$ is a surjection. Similarly, we are given that: :$h \circ g$ is a bijection. From Injection if Composite is Injection it follows that $g$ is an injection. From Surjection if Composite is Surjection it follows that $h$ is a surjection. Thus we have that $g$ is a bijection. {{qed|lemma}} Next it is shown that $f$ is a bijection. {{AimForCont}} $f$ is not a surjection. Then: :$\exists y \in B: \neg \exists x \in A: \map f x = y$ But then $g$ is a surjection and so: :$\forall z \in C: \exists y \in B: z = \map g y$ and so: :$\exists z \in C: \neg \exists x \in A: \map g {\map f x} = z$ and so $g \circ f$ is not a surjection. This contradicts our assertion that $g \circ f$ is a bijection and so a surjection. Hence by Proof by Contradiction $f$ must be a surjection So $f$ is both an injection and a surjection, and so a bijection. {{qed|lemma}} Next it is shown that $h$ is a bijection. {{AimForCont}} $h$ is not an injection. Then: :$\exists y_1, y_2 \in C: \map h {y_1} = \map h {y_2}$ But then $g$ is a bijection and so: :$\exists x_1, x_2 \in B: y_1 = \map g {x_1} \ne \map g {x_2} = y_2$ and so: :$\exists x_1, x_2 \in B: \map h {\map g {x_1} } = \map h {\map g {x_2} }$ and so $h \circ g$ is not an injection. This contradicts our assertion that $h \circ g$ is a bijection and so an injection. Hence by Proof by Contradiction $h$ must be an injection So $h$ is both an injection and a surjection, and so a bijection. {{qed|lemma}} Hence the result. {{qed}}
We note that $g \circ f$ is both an injection and a surjection by definition of bijection. We are given that: :$g \circ f$ is a bijection. From Injection if Composite is Injection it follows that $f$ is an injection. From Surjection if Composite is Surjection it follows that $g$ is a surjection. {{qed|lemma}} First suppose that $g$ is a bijection. We already have that $f$ is an injection. {{AimForCont}} $f$ is not a surjection. Then: :$\exists y \in B: \neg \exists x \in A: \map f x = y$ But then $g$ is a surjection and so: :$\forall z \in C: \exists y \in B: z = \map g y$ and so: :$\exists z \in C: \neg \exists x \in A: \map g {\map f x} = z$ and so $g \circ f$ is not a surjection. This contradicts our assertion that $g \circ f$ is a bijection and so a surjection. Hence by Proof by Contradiction $f$ must be a surjection So $f$ is both an injection and a surjection, and so a bijection. {{qed|lemma}} Now suppose that $f$ is a bijection. {{AimForCont}} $g$ is not an injection. Then: :$\exists y_1, y_2 \in B: \map g {y_1} = \map g {y_2}$ But then $f$ is a bijection and so: :$\exists x_1, x_2 \in A: y_1 = \map f {x_1} \ne \map f {x_2} = y_2$ and so: :$\exists x_1, x_2 \in A: \map g {\map f {x_1} } = \map g {\map f {x_2} }$ and so $g \circ f$ is not an injection. This contradicts our assertion that $g \circ f$ is a bijection and so an injection. Hence by Proof by Contradiction $g$ must be an injection. So $g$ is both an injection and a surjection, and so a bijection. {{qed|lemma}} Hence the result. {{qed}} Category:Bijections Category:Composite Mappings r59vy38dyxt59ge73vn9psa1a8qfaos
First note that from Composition of Mappings is Associative: :$\paren {h \circ g} \circ f = h \circ \paren {g \circ f}$ and so on. However, while there is no need to use parenthesis to establish the order of composition of mappings, in the following the technique will be used in order to clarify what is being done. We have that $\paren {h \circ g} \circ f$ and $\paren {f \circ h} \circ g$ are injections. From Injection if Composite is Injection it follows that: :$f$ is an injection :$g$ is an injection. Similarly, because $h \circ \paren {g \circ f}$ and $f \circ \paren {h \circ g}$ are injections: :$g \circ f$ is an injection :$h \circ g$ is an injection. We have that $g \circ \paren {f \circ h}$ is a surjection. From Surjection if Composite is Surjection it follows that $g$ is a surjection. Similarly, because $\paren {g \circ f} \circ h$ is a surjection: :$g \circ f$ is a surjection. Thus we have deduced that: :$g$ is a bijection and: :$g \circ f$ is a bijection. From Mapping Composed with Bijection forming Bijection is Bijection it follows that: :$f$ is a bijection. Thus from Inverse of Bijection is Bijection we have that $f^{-1}: B \to A$ and $g^{-1}: C \to B$ are both bijections. Similarly, $\paren {g \circ f}^{-1}: C \to A$ is a bijection, which from Inverse of Composite Bijection is the same as $f^{-1} \circ g^{-1}$. {{finish|Can we now prove that $\paren {g \circ f}^{-1} {{=}} h?$}}
By definition of coproduct: :for all sets $X$ and mappings $f_1: S_1 \to X$ and $f_1: S_1 \to X$ ::there exists a unique mapping $h: C \to X$ such that: :::$h \circ i_1 = f_1$ :::$h \circ i_2 = f_2$ Let $X := S_1$ and $f_1: S_1 \to X = I_{S_1}$ where $I_{S_1}$ denotes the identity mapping. We have: :$h \circ i_1 = I_{S_1}$ From Identity Mapping is Injection, $I_{S_1}$ is an injection. From Injection if Composite is Injection it follows that $i_1$ is an injection. By setting $X := S_2$ and $f_2 := I_{S_2}$ it follows in the same way that $i_2$ is also an injection. Hence the result. {{qed}}
For $\struct {S_1 \sqcup S_2, i_1, i_2}$ to be a coproduct, it is necessary that: :for all sets $X$ and all mappings $f_1: S_1 \to X$ and $f_2: S_2 \to X$ ::there exists a unique mapping $h: S_1 \sqcup S_2 \to X$ such that: :::$h \circ i_1 = f_1$ :::$h \circ i_2 = f_2$ Let $h$ be the mapping defined as: :$\forall \tuple {s, t} \in S_1 \sqcup S_2: \map h {s, t} = \begin {cases} \map {f_1} s & : t = 1 \\ \map {f_2} s & : t = 2 \end {cases}$ Then: {{begin-eqn}} {{eqn | l = \map {h \circ i_1} {s_1}       | r = \map h {s_1, 1}       | c =  }} {{eqn | r = \map {f_1} {s_1}       | c =  }} {{eqn-intertext|and:}} {{eqn | l = \map {h \circ i_2} {s_2}       | r = \map h {s_2, 2}       | c =  }} {{eqn | r = \map {f_2} {s_2}       | c =  }} {{end-eqn}} The existence of $h$ is apparent. It remains to prove that $h$ is unique. Let $h_1: S_1 \sqcup S_2 \to X$ and $h_2: S_1 \sqcup S_2 \to X$ both be mappings with the properties of $h$. Let $\tuple {s, t} \in S_1 \sqcup S_2$ be arbitrary. Then: {{begin-eqn}} {{eqn | l = \map {h_1 \circ i_1} {s_1}       | r = \map {f_1} {s_1}       | c =  }} {{eqn | r = \map {h_2 \circ i_1} {s_1}       | c = as $h_1$ and $h_2$ have the same properties }} {{eqn | ll= \leadsto       | l = \map {h_1} {s_1, 1}       | r = \map {h_2} {s_1, 1}       | c = for all $s_1 \in S_1$ }} {{end-eqn}} and: {{begin-eqn}} {{eqn | l = \map {h_1 \circ i_2} {s_2}       | r = \map {f_2} {s_2}       | c =  }} {{eqn | r = \map {h_2 \circ i_2} {s_2}       | c = as $h_1$ and $h_2$ have the same properties }} {{eqn | ll= \leadsto       | l = \map {h_2} {s_2, 2}       | r = \map {h_2} {s_2, 2}       | c = for all $s_2 \in S_2$ }} {{end-eqn}} Hence $h_1 = h_2$ and so $h$ is unique Hence $\struct {S_1 \sqcup S_2, i_1, i_2}$ is a coproduct of $S_1$ and $S_2$. {{qed}}
Let $X$ be an arbitrary set. Let $f_1: S_1 \to X$ and $f_2: S_2 \to X$ be arbitrary mappings. Let $h_C: C \to X$ be the unique mapping such that: :$h_C \circ i_1 = f_1$ :$h_C \circ i_2 = f_2$ Let $h_D: D \to X$ be the unique mapping such that: :$h_D \circ j_1 = f_1$ :$h_D \circ j_2 = f_2$ The existence and uniqueness of $h_C$ and $h_D$ follow from the fact of $\struct {C, i_1, i_2}$ and $\struct {D, j_1, j_2}$ both being coproducts of $S_1$ and $S_2$. We have that: :$h_C \circ i_1 = f_1 = h_D \circ j_1$ :$h_C \circ i_2 = f_2 = h_D \circ j_2$ {{ProofWanted}}
The day of the week on which the $13$th falls is directy dependent upon the day of the week that the $1$st of the month falls. From Months that Start on the Same Day of the Week, the months can be grouped into equivalence classes according to which day the month starts: For a non-leap year, the set of equivalence classes is: :$\set {\set {\text {January}, \text {October} }, \set {\text {February}, \text {March}, \text {November} }, \set {\text {April}, \text {July} }, \set {\text {May} }, \set {\text {June} }, \set {\text {August} }, \set {\text {September}, \text {December} } }$ For a leap year, the set of equivalence classes is: :$\set {\set {\text {January}, \text {April}, \text {July} }, \set {\text {February}, \text {August} }, \set {\text {March}, \text {November} }, \set {\text {May} }, \set {\text {June} }, \set {\text {September}, \text {December} }, \set {\text {October} } }$ It can be seen that none of these equivalence classes has fewer than $1$ day, and none has more than $3$. Hence the result. {{qed}}
That $\alpha$ is an equivalence relation is proved in Equivalence Relation on Natural Numbers such that Quotient is Power of Two. Let $x$ be a natural number whose $\alpha$-equivalence class is $\eqclass x \alpha$. If $x$ is odd, then $\eqclass x \alpha$ contains that odd number $x$ We have that $x$ is of the form $x = 2^r y$ for some $r \in \Z$ where $y$ is an odd number. Thus by definition $y \in \eqclass x \alpha$ is an odd number which is contained in $\eqclass x \alpha$. Thus it has been shown that an arbitrary $\eqclass x \alpha$ contains at least one odd number. It remains to be demonstrated that $\eqclass x \alpha$ contains no more than $1$ odd number. {{AimForCont}} $\eqclass n \alpha$ contains the odd numbers $p$ and $q$ such that $p \ne q$. Then: :$p = 2^n q$ for some $n \in \Z$ If $n = 0$ then $p = q$ which contradicts $q \ne p$. Thus $n \ne 0$. Let $n > 0$. Then: :$p = r q$ for $r \in \Z$ such that $r = 2^n$. Thus $p$ is has $2$ as a divisor. This contradicts the supposition that $p$ is odd. Let $n < 0$. Then: :$p = \dfrac q {2^m}$ where $m = -n$ and so $m > 0$. Thus: :$q = r p$ for $r \in \Z$ such that $r = 2^m$.   Thus $q$ has $2$ as a divisor. This contradicts the supposition that $q$ is odd. Thus all cases of $n$ lead to a contradiction. It follows that no $\alpha$-equivalence class can contain more than one odd number. {{qed}}
That $\alpha$ is an equivalence relation is proved in Equivalence Relation on Natural Numbers such that Quotient is Power of Two. We are given that $c \mathrel \alpha d$. {{WLOG}}, suppose $c < d$. If $d < c$ then the same argument holds, mutatis mutandis. By definition of $\alpha$, we have that: :$c = 2^n d$ for some $n \in \Z$. As $c < d$ it follows that $2^n > 1$. That is, $n > 0$ and so $2^n \in \Z$. So by definition $c$ is a divisor of $d$. The result follows. {{qed}}
Let $\alpha$ denote the relation defined on the natural numbers $\N$ by: :$\forall x, y \in \N: x \mathrel \alpha y \iff \exists n \in \Z: x = 2^n y$ From Equivalence Relation on Natural Numbers such that Quotient is Power of Two, $\alpha$ is an equivalence relation. From Equivalence Class under $\alpha$ Contains $1$ Odd Number, each odd integer between $1$ and $2 n$ is in its own equivalence class. Each even integer $m$ between $1$ and $2 n$ is equal to an odd integer $p$ strictly less than $m$ multiplied by a power of $2$. Hence each such $m$ is an element of one of the equivalence classes of one of the $n$ odd integers between $1$ and $2 n$. Thus each element of $S$ is an element of one of exactly $n$ equivalence classes of $\alpha$. But there are $n + 1$ elements of $S$. So by the Pigeonhole Principle, at least $2$ elements of $S$ are in the same equivalence class. Thus let $a, b \in S: a \mathrel \alpha b$. From One of Pair of Equivalent Elements is Divisor of the Other, either $a \divides b$ or $b \divides a$. Hence the result. {{qed}}
Checking in turn each of the criteria for equivalence: === Reflexivity === We have that for all $x \in \Z$: :$x^2 \equiv x^2 \pmod n$ It follows by definition of $\RR_n$ that: :$x \mathrel {\RR_n} x$ Thus $\RR_n$ is seen to be reflexive. {{qed|lemma}} === Symmetry === {{begin-eqn}} {{eqn | l = x       | o = \RR_n       | r = y       | c =  }} {{eqn | ll= \leadsto       | l = x^2       | o = \equiv        | r = y^2       | rr= \pmod n       | c =  }} {{eqn | ll= \leadsto       | l = y^2       | o = \equiv        | r = x^2       | rr= \pmod n       | c =  }} {{eqn | ll= \leadsto       | l = y       | o = \RR_n       | r = x       | c =  }} {{end-eqn}} Thus $\RR_n$ is seen to be symmetric. {{qed|lemma}} === Transitivity === Let: :$x \mathrel {\RR_n} y$ and $y \mathrel {\RR_n} z$ for $x, y, z \in \Z$. Then by definition: {{begin-eqn}} {{eqn | l = x^2       | o = \equiv        | r = y^2       | rr= \pmod n       | c =  }} {{eqn | l = y^2       | o = \equiv        | r = z^2       | rr= \pmod n       | c =  }} {{eqn | ll= \leadsto       | l = x^2       | o = \equiv        | r = z^2       | rr= \pmod n       | c =  }} {{eqn | ll= \leadsto       | l = x       | o = \RR_n       | r = z       | c =  }} {{end-eqn}} Thus $\RR_n$ is seen to be transitive. {{qed|lemma}} $\RR_n$ has been shown to be reflexive, symmetric and transitive. Hence by definition it is an equivalence relation. {{qed}} g1l8x8wgatc81977erysfjynifph6ux
By definition of a left ideal then $\circ$ is well-defined. === $(\text M 1)$ : Scalar Multiplication (Left) Distributes over Module Addition === Follows directly from {{Ring-axiom|D}} {{qed|lemma}} === $(\text M 2)$ : Scalar Multiplication (Right) Distributes over Scalar Addition === Follows directly from {{Ring-axiom|D}} {{qed|lemma}} === $(\text M 3)$ : Associativity of Scalar Multiplication === Follows directly from {{Ring-axiom|M1}} {{qed}}
By definition of a right ideal then $\circ$ is well-defined. === $(\text {RM} 1)$ : Scalar Multiplication (Right) Distributes over Module Addition === Follows directly from {{Ring-axiom|D}} {{qed|lemma}} === $(\text {RM} 2)$ : Scalar Multiplication (Left) Distributes over Scalar Addition === Follows directly from {{Ring-axiom|D}} {{qed|lemma}} === $(\text {RM} 3)$ : Associativity of Scalar Multiplication === Follows directly from {{Ring-axiom|M1}} {{qed}}
By definition of the opposite ring: :$\forall x, y \in R: x * y = y \times x$. By definition of the ring $R$, $\struct {R, +}$ is an abelian group. To complete the proof, it remains to be shown that $\struct {R, *}$ is a semigroup. That is, it remains to show that $\struct{R, *}$ is associative. Let $a, b, c \in R$. {{begin-eqn}} {{eqn | l = a * \paren {b * c}       | r = \paren{b * c} \times a       | c = Definition of $*$ }} {{eqn | r = \paren {c \times b} \times a       | c = Definition of $*$ }} {{eqn | r = c \times \paren {b \times a}       | c = {{Ring-axiom|M1}} }} {{eqn | r = c \times \paren {a * b}       | c = Definition of $*$ }} {{eqn | r = \paren {a * b} * c       | c = Definition of $*$ }} {{end-eqn}} The result follows. {{qed}}
By definition of the opposite ring: :$\forall x, y \in S: x * y = y \times x$ :$\forall x, y \in S: x \circ y = y * x$ Hence for all $x,y \in S$: :$x \circ y = y * x = x \times y$ The result follows. {{qed}} Category:Ring Theory k14myncy6bkp8xbic8ndvx15pkuigex