hcju/proofwikiqa huggingface dataset:
We need to prove that $\forall a, b, c, d \in G: \paren {a * b} * \paren {c * d} = \paren {a * c} * \paren {b * d}$. So: {{begin-eqn}} {{eqn | l = \paren {a * b} * \paren {c * d}       | r = \paren {a \circ b^{-1} } \circ \paren {c \circ d^{-1} }^{-1}       | c = Definition of $*$ }} {{eqn | r = \paren {a \circ b^{-1} } \circ \paren {\paren {d^{-1} }^{-1} \circ c^{-1} }       | c = Inverse of Group Product }} {{eqn | r = \paren {a \circ b^{-1} } \circ \paren {d \circ c^{-1} }       | c = Inverse of Group Inverse }} {{eqn | r = \paren {a \circ c^{-1} } \circ \paren {d \circ b^{-1} }       | c = {{Defof|Abelian Group}} }} {{eqn | r = \paren {a \circ c^{-1} } \circ \paren {\paren {d^{-1} }^{-1} \circ b^{-1} }       | c = Inverse of Group Inverse }} {{eqn | r = \paren {a \circ c^{-1} } \circ \paren {b \circ d^{-1} }^{-1}       | c = Inverse of Group Product }} {{eqn | r = \paren {a * c} * \paren {b * d}       | c = Definition of $*$ }} {{end-eqn}} {{qed}}
Let $\struct {S, *, \circ}$ be a semiring, all of whose elements of $S$ are cancellable for $*$. We expand the expression $\paren {a * b} \circ \paren {c * d}$ using the distributive law in two ways: {{begin-eqn}} {{eqn | l = \paren {a * b} \circ \paren {c * d}       | r = \paren {\paren {a * b} \circ c} * \paren {\paren {a * b} \circ d}       | c =  }} {{eqn | r = \paren {a \circ c} * \paren {b \circ c} * \paren {a \circ d} * \paren {b \circ d}       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \paren {a * b} \circ \paren {c * d}       | r = \paren {a \circ \paren {c * d} } * \paren {b \circ \paren {c * d} }       | c =  }} {{eqn | r = \paren {a \circ c} * \paren {a \circ d} * \paren {b \circ c} * \paren {b \circ d}       | c =  }} {{end-eqn}} So, by the fact that all elements of $\struct {S, *}$ are cancellable (and thus, a fortiori, are $a \circ c$ and $b \circ d$), we have: {{begin-eqn}} {{eqn | l = \paren {a \circ c} * \paren {b \circ c} * \paren {a \circ d} * \paren {b \circ d}       | r = \paren {a \circ c} * \paren {a \circ d} * \paren {b \circ c} * \paren {b \circ d}       | c =  }} {{eqn | ll= \leadsto       | l = \paren {b \circ c} * \paren {a \circ d}       | r = \paren {a \circ d} * \paren {b \circ c}       | c =  }} {{end-eqn}} As this is true for all $a, b, c, d \in S$, it is true in particular if $c = d = 1_S$. Thus it is clear that $b * a = a * b$, which is exactly to say that $*$ is commutative. Hence the result, by definition of additive semiring. {{Qed}} Category:Semirings gwveb67h30ny0u227ct67e0iioxn4vn
In a ring $\struct {R, +, \circ}$, $\struct {R, +}$ forms a group. From Group is not Empty, the group $\struct {R, +}$ contains at least the identity, so cannot be empty. So every ring $\struct {R, +, \circ}$ contains at least the identity for ring addition. {{qed}}
Because $\struct {R, +, \circ}$ is a ring, $\struct {R, +}$ is a group. Since $0_R$ is the identity in $\struct {R, +}$, we have $0_R + 0_R = 0_R$. From the Cancellation Laws, all group elements are cancellable, so every element of $\struct {R, +}$ is cancellable for $+$. Thus: {{begin-eqn}} {{eqn | l = x \circ \paren {0_R + 0_R}       | r = x \circ 0_R       | c = {{Defof|Ring Zero}} }} {{eqn | ll= \leadsto       | l = \paren {x \circ 0_R} + \paren {x \circ 0_R}       | r = x \circ 0_R       | c = {{Ring-axiom|D}} }} {{eqn | ll= \leadsto       | l = \paren {x \circ 0_R} + \paren {x \circ 0_R}       | r = \paren {x \circ 0_R} + 0_R       | c = {{Defof|Ring Zero}} }} {{eqn | ll= \leadsto       | l = x \circ 0_R       | r = 0_R       | c = Cancellation Laws }} {{end-eqn}} Next: {{begin-eqn}} {{eqn | l = \paren {0_R + 0_R} \circ x       | r = 0_R \circ x       | c = {{Defof|Ring Zero}} }} {{eqn | ll= \leadsto       | l = \paren {0_R \circ x} + \paren {0_R \circ x}       | r = 0_R \circ x       | c = {{Ring-axiom|D}} }} {{eqn | ll= \leadsto       | l = \paren {0_R \circ x} + \paren {0_R \circ x}       | r = 0_R + \paren {0_R \circ x}       | c = {{Defof|Ring Zero}} }} {{eqn | ll= \leadsto       | l = 0_R \circ x       | r = 0_R       | c = Cancellation Laws }} {{end-eqn}} {{Qed}}
We have: {{begin-eqn}} {{eqn | l = \paren {x + \paren {-x} } \circ y       | r = 0_R \circ y       | c = {{Defof|Ring Zero}} }} {{eqn | r = 0_R       | c = Ring Product with Zero }} {{eqn | ll= \leadstoandfrom       | l = \paren {x \circ y} + \paren {\paren {-x} \circ y}       | r = 0_R       | c = {{Ring-axiom|D}} }} {{end-eqn}} So from {{GroupAxiom|3}} as applied to $\struct {R, +}$: :$\paren {-x} \circ y = -\paren {x \circ y}$ The proof that $x \circ \paren {-y} = -\paren {x \circ y}$ follows identical lines. {{qed}}
Let $\struct {R, +, \circ}$ be a ring. If $\struct {R, \circ}$ has an identity, then it is a monoid. From Identity of Monoid is Unique, it follows that such an identity is unique. {{qed}}
The single element of the null ring serves as an identity for both of the operations. So, in this particular ring, the unity and the zero are the same element. A non-null ring contains some non-zero element $a$. Since $1_R \circ a = a \ne 0_R = 0_R \circ a$, then $1_R \ne 0_R$. So if a ring is non-null, its unity cannot be zero. So a ring is null {{iff}} its unity is also its zero. {{qed}}
Let $\left({R, +, \circ}\right)$ be a ring with unity $1_R$. From Identity is Self-Inverse: :$1_R^{-1} = 1_R \in R \implies 1_R \in U_R$ {{qed}} Category:Rings with Unity 1kgq0vcsz8xqaswu08wx5qsyj7zb6a3
From Unity is Unit: :$1_R \in U_R$ It remains to be shown that $-1_R \in U_R$. From {{Ring-axiom|A}}, $\struct {R, +}$ is an abelian group. Therefore: :$1_R \in R \implies -1_R \in R$. From Product of Ring Negatives: :$-1_R \circ -1_R = 1_R \circ 1_R = 1_R$ Thus $-1_R$ has a ring product inverse (itself) and therefore $-1_R \in U_R$. We have: {{begin-eqn}} {{eqn | l = 1_R \circ 1_R       | r = 1_R       | c =  }} {{eqn | l = 1_R \circ -1_R       | r = -1_R       | c =  }} {{eqn | l = -1_R \circ 1_R       | r = -1_R       | c =  }} {{eqn | l = -1_R \circ -1_R       | r = 1_R       | c =  }} {{end-eqn}} This exhausts all the ways we can form a ring product between $1_R$ and $-1_R$. That is: :$\forall x, y \in \set {1_R, -1_R}: x \circ y \in \set {1_R, -1_R}$ Hence the result from the Two-Step Subgroup Test. {{qed}} Category:Rings with Unity Category:Subgroups 4xfqsv3mwxeuhx44xgisxjrxwdhk8k6
Let the unity of $\left({R, +, \circ}\right)$ be $1_R$. {{begin-eqn}} {{eqn | l = \left({- \left({z^{-1} }\right)}\right) \circ \left({- z}\right)       | r = z^{-1} \circ z       | c = Product of Ring Negatives }} {{eqn | r = 1_R       | c = Inverse under $\circ$ }} {{eqn | r = z \circ z^{-1}       | c = Inverse under $\circ$ }} {{eqn | r = \left({- z}\right) \circ \left({- \left({z^{-1} }\right)}\right)       | c = Product of Ring Negatives }} {{end-eqn}} Thus: : $\left({- z}\right)^{-1} = - \left({z^{-1}}\right)$ {{qed}}
{{begin-eqn}} {{eqn | n=1       | l=-\left({x \circ z^{-1} }\right)       | r=\left({- x}\right) \circ z^{-1}       | c=Product with Ring Negative }} {{eqn | r=x \circ \left({- \left({z^{-1} }\right)}\right)       | c=Product with Ring Negative }} {{eqn | r=x \circ \left({\left({- z}\right)^{-1} }\right)       | c=Negative of Product Inverse }} {{end-eqn}} {{qed|lemma}} {{begin-eqn}} {{eqn | n=2       | l=-\left({z^{-1} \circ x}\right)       | r=z^{-1} \circ \left({- x}\right)       | c=Product with Ring Negative }} {{eqn | r=\left({-\left({z^{-1} }\right)}\right) \circ x       | c=Product with Ring Negative }} {{eqn | r=\left({\left({- z}\right)^{-1} }\right) \circ x       | c=Negative of Product Inverse }} {{end-eqn}} {{qed}}
Follows directly from Product of Negative with Product Inverse and the definition of division product. {{qed}}
First we demonstrate the operation has the specified property: {{begin-eqn}} {{eqn | l = \frac a b + \frac c d       | r = a \circ b^{-1} + c \circ d^{-1}       | c = {{Defof|Division Product}} }} {{eqn | r = a \circ b^{-1} \circ d \circ d^{-1} + c \circ d^{-1} \circ b \circ b^{-1}       | c = {{Defof|Inverse Element}} and {{Defof|Identity Element}} under $\circ$ }} {{eqn | r = \paren {a \circ d} \circ \paren {d^{-1} \circ b^{-1} } + \paren {b \circ c} \circ \paren {d^{-1} \circ b^{-1} }       | c = {{Defof|Commutative Operation}} }} {{eqn | r = \paren {a \circ d + b \circ c} \circ \paren {b \circ d}^{-1}       | c = {{Defof|Distributive Operation}} $\circ$ over $+$ }} {{eqn | r = \frac {a \circ d + b \circ c} {b \circ d}       | c = {{Defof|Division Product}} }} {{end-eqn}} Notice that this works only if $\struct {R, +, \circ}$ is commutative. Now we show that $+$ is well-defined. Let $a, c, a', c' \in D, b, d, b', d' \in D^*$ such that $\dfrac a b = \dfrac {a'} {b'}$ and $\dfrac c d = \dfrac {c'} {d'}$. Then: {{begin-eqn}} {{eqn | l = a \circ b'       | r = \paren {a \circ b^{-1} } \circ b \circ b'       | c =  }} {{eqn | r = \paren {a' \circ \paren {b'}^{-1} } \circ b \circ b'       | c = as $a / b = a' / b'$ }} {{eqn | r = a' \circ b       | c =  }} {{end-eqn}} Similarly, $c \circ d' = c' \circ d$. Hence: {{begin-eqn}} {{eqn | l = \paren {\paren {a' \circ d'} + \paren {c' \circ b'} } \circ b \circ d       | r = a' \circ b \circ d' \circ d + c' \circ d \circ b' \circ b       | c =  }} {{eqn | r = a \circ b' \circ d' \circ d + c \circ d' \circ b' \circ b       | c =  }} {{eqn | r = \paren {\paren {a \circ d} + \paren {c \circ b} }  \circ b' \circ d'       | c =  }} {{end-eqn}} Thus: {{begin-eqn}} {{eqn | l = \frac {\paren {a' \circ d'} + \paren {c' \circ b'} } {b' \circ d'}       | r = \paren {\paren {\paren {a' \circ d'} + \paren {c' \circ b'} } \circ b \circ d} \circ \paren {b^{-1} \circ d^{-1} \circ \paren {b'}^{-1} \circ \paren {d'}^{-1} }       | c =  }} {{eqn | r = \paren {\paren {\paren {a \circ d} + \paren {c \circ b} } \circ b' \circ d'} \circ \paren {b^{-1} \circ d^{-1} \circ \paren {b'}^{-1} \circ \paren {d'}^{-1} }       | c =  }} {{eqn | r = \paren {\paren {a \circ d} + \paren {c \circ b} } \circ b^{-1} \circ d^{-1}       | c =  }} {{eqn | r = \frac {\paren {a \circ d} + \paren {c \circ b} } {b \circ d}       | c =  }} {{end-eqn}} showing that $+$ is indeed well-defined. {{Qed}}
{{begin-eqn}} {{eqn | l = \frac a c       | r = \frac b d       | c =  }} {{eqn | ll= \leadstoandfrom       | l = a \circ c^{-1}       | r = b \circ d^{-1}       | c = {{Defof|Division Product}} }} {{eqn | ll= \leadstoandfrom       | l = a \circ c^{-1} \circ c \circ d       | r = b \circ d^{-1} \circ c \circ d       | c = {{Defof|Cancellable Element}} of $U_R$ }} {{eqn | ll= \leadstoandfrom       | l = \paren {a \circ d} \circ \paren {c^{-1} \circ c}       | r = \paren {b \circ c} \circ \paren {d^{-1} \circ d}       | c = {{Defof|Commutative Operation}} }} {{eqn | ll= \leadstoandfrom       | l = a \circ d       | r = b \circ c       | c = {{Defof|Identity Element}} and {{Defof|Inverse Element}} }} {{end-eqn}} {{qed}} {{refactor|Construct a proof based on the below}} Alternatively, a proof can be built using Addition of Division Products.
{{begin-eqn}} {{eqn | l = \frac a c \circ \frac b d       | r = a \circ c^{-1} \circ b \circ d^{-1}       | c = {{Defof|Division Product}} }} {{eqn | r = \paren {a \circ b} \circ \paren {d^{-1} \circ c^{-1} }       | c = {{Defof|Commutative Operation}} }} {{eqn | r = \paren {a \circ b} \circ \paren {c \circ d}^{-1}       | c = Inverse of Product }} {{eqn | r = \frac {a \circ b} {c \circ d}       | c = {{Defof|Division Product}} }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \frac {1_R} {\paren {a / b} }       | r = 1_R / \paren {a \circ b^{-1} }       | c = {{Defof|Division Product}} }} {{eqn | r = 1_R \circ \paren {a \circ b^{-1} }^{-1}       | c = {{Defof|Division Product}} }} {{eqn | r = \paren {a \circ b^{-1} }^{-1}       | c = {{Defof|Identity Element}} of $\circ$ }} {{eqn | r = \paren {\frac a b}^{-1}       | c = {{Defof|Division Product}} }} {{eqn | r = \paren {a \circ b^{-1} }^{-1}       | c = {{Defof|Division Product}} }} {{eqn | r = \paren {b^{-1} }^{-1} \circ a^{-1}       | c = Inverse of Group Product }} {{eqn | r = b \circ a^{-1}       | c = Inverse of Group Inverse }} {{eqn | r = \frac b a       | c = {{Defof|Division Product}} }} {{end-eqn}} {{qed}}
Follows directly from the definition of proper zero divisor. If $y \ne 0_R$ and $x \circ y = 0_R$ and $x \in R^*$ (which is has to be if it's a proper zero divisor), then all the criteria of being a zero divisor are fulfilled by $y$. {{qed}} Category:Ring Theory otdlkr5hues1586vxgj40zmmh38byr4
Let $\struct {R, +, \circ}$ be a ring with unity whose unity is $1_R$ and whose zero is $0_R$. Let $x$ be a unit of $\struct {R, +, \circ}$. {{AimForCont}} $x$ is such that: :$x \circ y = 0_R, y \ne 0_R$ Then: {{begin-eqn}} {{eqn | l = \paren {x^{-1} \circ x} \circ y       | r = 0_R       | c =  }} {{eqn | ll= \leadsto       | l = 1_R \circ y       | r = 0_R       | c = {{Defof|Inverse Element}} }} {{eqn | ll= \leadsto       | l = 1_R       | r = 0_R       | c = as $y \ne 0_R$ }} {{end-eqn}} From this contradiction it follows that $x$ cannot have such a property. Thus by Proof by Contradiction $x$ is not a zero divisor. {{qed}} Category:Rings with Unity jjh16erxnmb4k3yno1on1yjw5d2unfe
Let $\struct {R, +, \circ}$ be a ring. Let $x \divides 0_R$ in $R$. Then: {{begin-eqn}} {{eqn | lo= \exists y \in R, y \ne 0:       | l = x \circ y       | r = 0_R       | c = {{Defof|Zero Divisor of Ring}} }} {{eqn | ll= \leadsto       | lo= \forall z \in R:       | l = z \circ \paren {x \circ y}       | r = z \circ 0_R }} {{eqn | r = 0_R       | c = Property of Ring Zero }} {{eqn | ll= \leadsto       | lo= \forall z \in R:       | l = \paren {z \circ x} \circ y       | r = 0_R       | c = Associativity of $\circ$ }} {{end-eqn}} So $z \circ x \divides 0_R$ in $R$. The same thing happens if we form the product $\paren {x \circ y} \circ z$. {{Qed}} Category:Ring Theory gid9x63l0fomedgnaji3f26ntjao24n
{{begin-eqn}} {{eqn | l = \paren {x \circ y}       | o = \divides       | r = 0_R }} {{eqn | ll= \leadsto       | l = \exists z \divides 0_R \in R: \paren {x \circ y} \circ z       | r = 0_R, x \ne 0_R, y \ne 0_R       | c = {{Defof|Zero Divisor of Ring}} }} {{eqn | ll= \leadsto       | l = x \circ \paren {y \circ z}       | r = 0_R       | c = {{Ring-axiom|M1}} }} {{eqn | ll= \leadsto       | l = x \divides 0_R       | o = \lor       | r = \paren {y \circ z} \divides 0_R       | c = {{Defof|Zero Divisor of Ring}} }} {{eqn | ll= \leadsto       | l = x \divides 0_R       | o = \lor       | r = y \divides 0_R       | c = Zero Product with Proper Zero Divisor is with Zero Divisor applies, as $z \divides 0_R$ }} {{end-eqn}} {{qed}} Category:Ring Theory 3dh13ct8ok76uxfnchcs2qkr33s2on2
=== Sufficient Condition === If $z \circ x = 0_R$ or $x \circ z = 0_R$ for some $x \in R^*$, then $z$ can not be cancellable for $\circ$ since $z \circ 0_R = 0_R = 0_R \circ z$. === Necessary Condition === Let $z \circ x = z \circ y$ where $x \ne y$. Then: {{begin-eqn}} {{eqn | l = z \circ \paren {x + \paren {-y} }       | r = z \circ x + z \circ \paren {-y}       | c = $\circ$ distributes over $+$ }} {{eqn | r = z \circ x + \paren {-z \circ y}       | c = Product with Ring Negative }} {{eqn | r = 0_R       | c = as $z \circ x = z \circ y$ }} {{end-eqn}} But $x \ne y$, so $x + \paren {-y} \ne 0$. Thus $z$ is a zero divisor. Similarly if $x \circ z = y \circ z$ where $x \ne y$. {{Qed}}
=== Necessary Condition === Let $\struct {R, +, \circ}$ be a non-null ring with no zero divisors. The set $R^* = R \setminus \set {0_R} \ne \O$ as $\struct {R, +, \circ}$ is non-null. All elements of $R$ are not zero divisors, and therefore are cancellable. $\struct {R, +, \circ}$ is closed under $\circ$, from the fact that there are no zero divisors, and also that $\struct {R, \circ}$ is also closed. From Restriction of Associative Operation is Associative, ring product is associative on $\struct {R^*, +, \circ}$, as it is associative on $\struct {R, +, \circ}$. Thus $\struct {R^*, \circ}$ is a semigroup. {{qed|lemma}} === Sufficient Condition === Let $\struct {R^*, \circ}$ be a semigroup. Then: :$\neg \exists x, y \in R^*: x \circ y \notin R^*$ Thus: :$\neg \exists x, y \in R^*: x \circ y = 0_R$ and the result follows. {{qed}}
We have $0_R \circ 0_R = 0_R$, so that sorts out one element. Let $R^*$ be the ring $R$ without the zero: $R^* = R \setminus \left\{{0_R}\right\}$. By Ring Element is Zero Divisor iff not Cancellable, all elements of $R^*$ that are not zero divisors are cancellable. Therefore all elements of $R^*$ are cancellable. Suppose $x \circ x = x$ where $x \ne 0_R$. Then $x \circ x = x = x \circ 1_R$. As $x$ is cancellable, the result follows. {{Qed}}
A division ring by definition is a ring with unity, and therefore not null. A division ring by definition has no zero divisors, so $\struct {R^*, \circ}$  is a semigroup. $1_R \in \struct {R^*, \circ}$ and so the identity of $\circ$ is in $\struct {R^*, \circ}$. By the definition of a division ring, each element of $\struct {R^*, \circ}$ is a unit, and therefore has a unique inverse in $\struct {R^*, \circ}$. Thus $\struct {R^*, \circ}$ is a semigroup with an identity and inverses and so is a group. {{qed}} Category:Division Rings Category:Group Theory 7qcq9nzppvyldguhqwc62adejiwizur
=== Necessary Condition === If $S$ is a subring of $\struct {R, +, \circ}$, the conditions hold by virtue of the ring axioms as applied to $S$. === Sufficient Condition === Conversely, suppose the conditions hold. We check that the ring axioms hold for $S$. :$(1): \quad$ '''{{Ring-axiom|A}}:''' By $1$ and $2$ above, it follows from: :::: the One-Step Subgroup Test ::::Subgroup of Abelian Group is Abelian :::that $\struct {S, +}$ is an abelian subgroup of $\struct {R, +}$, and therefore an abelian group. :$(2): \quad$ '''{{Ring-axiom|M0}}:''' From $3$, $\struct {S, \circ}$ is closed. :$(3): \quad$ '''{{Ring-axiom|M1}}:''' From Restriction of Associative Operation is Associative, $\circ$ is associative on $R$, therefore also associative on $S$. :$(4): \quad$ '''{{Ring-axiom|D}}:''' From Restriction of Operation Distributivity, $\circ$ distributes over $+$ for the whole of $R$, therefore for $S$ also. So $\struct {S, +, \circ}$ is a ring, and therefore a subring of $\struct {R, +, \circ}$. {{qed}}
By Idempotent Elements of Ring with No Proper Zero Divisors, it follows that the unity of a subdomain is the unity of the integral domain it's a subdomain of. {{qed}}
Certainly $0_R \in \map {C_R} S$ as $0_R$ commutes (trivially) with all elements of $R$. Suppose $x, y \in \map {C_R} S$. Then: {{begin-eqn}} {{eqn | l = \forall s \in S: s \circ \paren {x + \paren {-y} }       | r = s \circ x + s \circ \paren {-y}       | c = Distributivity of $\circ$ over $+$ }} {{eqn | r = x \circ s + \paren {-y} \circ s       | c = $x$ and $y$ are in $\map {C_R} S$ }} {{eqn | r = \paren {x + \paren {-y} } \circ s       | c = Distributivity of $\circ$ over $+$ }} {{end-eqn}} So: :$x + \paren {-y} \in \map {C_R} s$ Suppose $x, y \in \map {C_R} S$ again. Then from Element Commutes with Product of Commuting Elements: :$x \circ y \in \map {C_R} S$ Thus all the conditions are fulfilled for Subring Test, and $\map {C_R} S$ is a subring of $R$. {{qed}}
Follows directly from the definition of center and Centralizer of Ring Subset is Subring. {{qed}}
From Null Ring and Ring Itself Subrings, $\struct {R, +, \circ}$ is a subring of $\struct {R, +, \circ}$. Also: :$\forall x, y \in \struct {R, +, \circ}: x \circ y \in R$ thus fulfilling the condition for $R$ to be an ideal of $R$. {{qed}}
Let $u \in J$, where $u \in U_R$. Also by definition, we have $u^{-1} \in U_R$. Let $x \in R$. {{begin-eqn}} {{eqn | o =        | r = x \in R       | c =  }} {{eqn | o = \leadsto       | r = x \circ u^{-1} \in R       | c = as $R$ is closed }} {{eqn | o = \leadsto       | r = \paren {x \circ u^{-1} } \circ u \in J       | c = {{Defof|Ideal of Ring}} }} {{eqn | o = \leadsto       | r = x \in J       | c = Ring properties: $u \circ u^{-1} = 1_R$ }} {{end-eqn}} Thus $R \subseteq J$. As $J \subseteq R$ by definition, it follows that $J = R$. {{Qed}}
=== Necessary Condition === Let $J$ be an ideal of $\struct {R, +, \circ}$. Then conditions $(1)$ to $(3)$ hold by virtue of the ring axioms and $J$ being an ideal. {{qed|lemma}} === Sufficient Condition === Suppose conditions $(1)$ to $(3)$ hold. As $r \in J \implies r \in R$, if $(3)$ holds for $J$, then $J$ is closed under $\circ$ and condition $(3)$ of Subring Test holds. Thus, $J$ is a subring of $R$. As $(3)$ defines the condition for $J$, being a subring, to be an ideal, the result holds. So $J$ is an ideal of $\struct {R, +, \circ}$. {{qed}}
From Epimorphism Preserves Groups, we have that if $\struct {R_1, +_1}$ is a group then so is $\struct {R_2, +_2}$. From Epimorphism Preserves Semigroups, we have that if $\struct {R_1, \circ_1}$ is a semigroup then so is $\struct {R_2, \circ_2}$. From Epimorphism Preserves Distributivity, we have that if $\circ_1$ distributes over $+_1$ then $\circ_2$ distributes over $+_2$. So it follows from the definition of a ring that if $\struct {R_1, +_1, \circ_1}$ is a ring then so is $\struct {R_2, +_2, \circ_2}$. {{qed}}
From the definition of a ring, both $\left({R_1, +_1}\right)$ and $\left({R_2, +_2}\right)$ are abelian groups. The result follows from the definitions of ring homomorphism and group homomorphism. {{Qed}} Category:Ring Homomorphisms Category:Group Homomorphisms 0vdhqve9vlm1czgtl0xvsx794hmny89
By definition, an integral domain is a ring. So, from Ring Product with Zero: :$\forall x \in D: 0_D = x \circ 0_D$. The result follows from the definition of divisor. {{qed}}
The element $1_D$ is the unity of $\struct {D, +, \circ}$, and so: :$1_D \in D: x = 1_D \circ x$ Similarly, from Product of Ring Negatives: :$-1_D \in D: x = \paren {-1_D} \circ \paren {-x}$ The result follows from the definition of divisor. {{qed}}
This is a special case of Unit of Integral Domain divides all Elements, as Unity is Unit. Furthermore, from Unity and Negative form Subgroup of Units we also have that $-1_D$ is a unit of $D$. Hence the result. {{qed}}
Follows directly from the definition of divisor: :$\forall x \in D: \exists 1_D \in D: x = 1_D \circ x$ {{qed}} Category:Integral Domains 8xrney68i9349jnuparynsqc0h08i4k
Let $x \in D, u \in U_D$ such that $x \divides u$. By definition: :$\exists t \in D: u = t \circ x$ Thus: :$1_D = u^{-1} \circ u = u^{-1} \circ t \circ x$ Also, as $D$ is an integral domain and hence a commutative ring, we have: :$u^{-1} \circ t \circ x = 1_D = x \circ u^{-1} \circ t$ It follows by definition that $x$ is a unit, as it has an inverse, namely $u^{-1} \circ t$. {{qed}}
Checking in turn each of the criteria for equivalence: === Reflexive === Clearly $x \divides x$ as $x = 1_D \circ x$, so $x \sim x$. Thus $\sim$ is reflexive. {{qed|lemma}} === Symmetric === By the definition: : $x \sim y \iff x \divides y \land y \divides x \iff y \sim x$ Thus $\sim$ is symmetric. {{qed|lemma}} === Transitive === {{begin-eqn}} {{eqn | o =        | r = x \sim y \land y \sim z       | c =  }} {{eqn | o = \leadsto       | r = \paren {x \divides y \land y \divides x} \land \paren {y \divides z \land z \divides y}       | c = {{Defof|Associate/Integral Domain|index = 1|Associate}} }} {{eqn | o = \leadsto       | r = \paren {x \divides y \land y \divides z} \land \paren {z \divides y \land y \divides x}       | c = Rule of Association and Rule of Commutation }} {{eqn | o = \leadsto       | r = x \divides z \land z \divides x       | c = Divisor Relation in Integral Domain is Transitive }} {{eqn | o = \leadsto       | r = x \sim z       | c = {{Defof|Associate/Integral Domain|index = 1|Associate}} }} {{end-eqn}} Thus $\sim$ is transitive. {{qed|lemma}} All criteria are fulfilled for $\sim$ to be an equivalence relation. Hence the result. {{Qed}}
Let $x \divides y \land y \divides z$. Then from the definition of divisor, we have: :$x \divides y \iff \exists s \in D: y = s \circ x$ :$y \divides z \iff \exists t \in D: z = t \circ y$ Then: :$z = t \circ \paren {s \circ x} = \paren {t \circ s} \circ x$ Thus: :$\exists \paren {t \circ s} \in D: z = \paren {t \circ s} \circ x$ and the result follows. {{Qed}}