hcju/proofwikiqa huggingface dataset:
=== Necessary Condition === Let $\phi$ be an isomorphism. Then by definition $\phi$ is a bijective homomorphism. Thus by definition of bijection, $\phi$ is an injection. By definition of injection, there exists exactly one element $x$ of $G$ such that $\map \phi x = e_H$. From Epimorphism Preserves Identity, that element $x$ is $e_G$: :$\map \phi {e_G} = e_H$ Thus by definition of kernel: :$\map \ker \phi = \set {e_G}$ {{qed|lemma}} === Sufficient Condition === Let $K := \map \ker \phi = \set {e_G}$. From the Quotient Theorem for Epimorphisms: :$\mathcal R_\phi$ is compatible with $\oplus$ and thus from Kernel is Normal Subgroup of Domain: :$K \lhd G$ From Congruence Relation induces Normal Subgroup, $\mathcal R_\phi$ is the equivalence defined by $K$. Let $\mathcal R_K$ be the congruence modulo $K$ induced by $K$. Suppose $\map \phi x = \map \phi y$. Then: :$x \mathop {\mathcal R_K} y$ as $\mathcal R_\phi = \mathcal R_K$ from Congruence Modulo Subgroup is Equivalence Relation. Thus by Congruence Class Modulo Subgroup is Coset: :$x \oplus y^{-1} \in K$ Hence: :$x \oplus y^{-1} = e_G$ and so: :$x = y$ Thus $\phi$ is injective. By definition, an injective epimorphism is a isomorphism. {{qed}}
From Kernel is Trivial iff Monomorphism, $\phi$ is a monomorphism {{iff}} $K = \left\{{e_G}\right\}$. By definition, a group $G$ is an epimorphism is an isomorphism {{iff}} $G$ is also a monomorphism. Hence the result. {{qed}}
By Quotient Mapping on Structure is Canonical Epimorphism, both $q_K$ and $q_H$ are epimorphisms. From Composite of Group Epimorphisms is Epimorphism we have that $q_K \circ q_H: G \to \left({G / H}\right) / K$ is also an epimorphism. From Preimage of Normal Subgroup of Quotient Group under Quotient Mapping is Normal: :$L \lhd G$ By Quotient Theorem for Group Epimorphisms: :there exists a group isomorphism $\psi: G / L \to \left({G / H}\right) / K$ satisfying: ::$\psi \circ q_L = q_K \circ q_L$ Let $\phi = \psi^{-1}$. Then $\phi$ is a group isomorphism from $\left({G / H}\right) / K$ to $G / L$: :$\phi \circ q_k \circ q_H = \phi \circ \psi \circ q_L = q_L$ {{qed}}
Proof by Counterexample Consider the set: : $A := \left\{ {x \in \R: x > 1}\right\}$ Suppose $a$ were the smallest element of $A$. Then $a > 1$. But then: :$1 < \dfrac {a + 1} 2 < a$ and so $a$ has been shown not to be the smallest element of $A$ after all. Hence $A$ has no smallest element. Thus there exists a subset of $\R$ which has no smallest element. Thus $\R$ is not a well-ordered under $\leqslant$. {{qed}}
Let $c$ and $c'$ both be suprema of $T$ in $S$. From the definition of supremum, $c$ and $c'$ are upper bounds of $T$ in $S$. By that definition: :$c$ is an upper bound of $T$ in $S$ and $c'$ is a supremum of $T$ in $S$ implies that $c' \preceq c$ :$c'$ is an upper bound of $T$ in $S$ and $c$ is a supremum of $T$ in $S$ implies that $c \preceq c'$. So: :$c' \preceq c \land c \preceq c'$ and thus by the antisymmetry of the ordering $\preceq$: :$c = c'$ {{Qed}}
Let $c$ and $c'$ both be infima of $T$ in $S$. From the definition of infimum, $c$ and $c'$ are lower bounds of $T$ in $S$. By that definition: : $c$ is a lower bound of $T$ in $S$ and $c'$ is an infimum of $T$ in $S$ implies that $c \preceq c'$ : $c'$ is a lower bound of $T$ in $S$ and $c$ is an infimum of $T$ in $S$ implies that $c' \preceq c$. So: :$c' \preceq c \land c \preceq c'$ and thus by the antisymmetry of the ordering $\preceq$: :$c = c'$ {{Qed}}
Let $H, K \in \mathbb G$. From Set of Subgroups forms Complete Lattice: :$\inf \set {H, K} = H \cap K$ {{qed}}
Recall that Set of Subgroups forms Complete Lattice. Let $L = \sup \left\{{H, K}\right\}$. Let either $H$ or $K$ be normal in $G$. Since $L$ contains $H$ and $K$, then $L$ contains $H \circ K$. The smallest subgroup of $G$ containing $H$ and $K$ is: : $\left \langle {H, K} \right \rangle$ the subgroup generated by $H$ and $K$. From Subset Product with Normal Subgroup as Generator: :$\left \langle {H, K} \right \rangle = H \circ K$ when either $H$ or $K$ is normal. The result follows. {{qed}}
Note that from Lexicographic Order is Ordering we have that $\struct {S_1 \times S_2, \preccurlyeq}$ is an ordered set. === Necessary Condition === Let $\struct {S_1, \preceq_1}$ and $\struct {S_2, \preceq_2}$ be totally ordered sets. Let $\tuple {x_1, x_2}, \tuple {y_1, y_2} \in S_1 \times S_2$. The following cases are to be examined: $(1): \quad x_1 \ne y_1$ As $\preceq_1$ is a total ordering it follows that either $x_1 \preceq_1 y_1$ or $y_1 \preceq_1 x_1$. Thus by definition of the lexicographic order on $S_1 \times S_2$: :either $\tuple {x_1, x_2} \preccurlyeq \tuple {y_1, y_2}$ or $\tuple {y_1, y_2} \preccurlyeq \tuple {x_1, x_2}$. $(2): \quad x_1 = y_1$ As $\preceq_2$ is a total ordering it follows that one of the following holds: :$x_2 \preceq_2 y_2$ in which case $\tuple {x_1, x_2} \preccurlyeq \tuple {y_1, y_2}$ :$y_2 \preceq_2 x_2$ in which case $\tuple {y_1, y_2} \preccurlyeq \tuple {x_1, x_2}$ :$x = y$ in which case $\tuple {x_1, x_2} = \tuple {y_1, y_2}$ by definition of the lexicographic order on $S_1 \times S_2$, In all cases $\tuple {x_1, x_2}$ is comparable with $\tuple {y_1, y_2}$. That is, $\preccurlyeq$ is a total ordering on $S_1 \times S_2$ {{qed|lemma}} === Sufficient Condition === Let $\left({S_1 \times S_2, \preccurlyeq}\right)$ be a totally ordered set. There are two cases to address: $(1): \quad$ Let $\tuple {x_1, x_2}, \tuple {y_1, y_2} \in S_1 \times S_2$ such that $x_1 = x_2$. As $\preccurlyeq$ is a total ordering it follows that one of the following holds: :$\tuple {x_1, x_2} \preccurlyeq \tuple {y_1, y_2}$ in which case $x_2 \preceq_2 y_2$ :$\tuple {y_1, y_2} \preccurlyeq \tuple {x_1, x_2}$ in which case $y_2 \preceq_2 x_2$ :$\tuple {x_1, x_2} = \tuple {y_1, y_2}$ in which case $x = y$ by definition of the lexicographic order on $S_1 \times S_2$. As the above holds for all $x_2, y_2 \in S_2$, it follows that $\preceq_2$ is a total ordering on $S_2$. $(2): \quad$ Let $\tuple {x_1, x_2}, \tuple {y_1, y_2} \in S_1 \times S_2$ such that $x_1 \ne x_2$. As $\preccurlyeq$ is a total ordering it follows that one of the following holds: :$\tuple {x_1, x_2} \preccurlyeq \tuple {y_1, y_2}$ in which case $x_1 \preceq_1 y_1$ :$\tuple {y_1, y_2} \preccurlyeq \tuple {x_1, x_2}$ in which case $y_1 \preceq_1 x_1$ by definition of the lexicographic order on $S_1 \times S_2$. As the above holds for all $x_1, y_1 \in S_1$, it follows that $\preceq_1$ is a total ordering on $S_1$. {{qed}}
In the following, $\left({x_1, x_2}\right), \left({y_1, y_2}\right), \left({z_1, z_2}\right) \in S_1 \times S_2$. Checking in turn each of the criteria for an ordering: === Reflexivity === From Equality of Ordered Pairs: :$x_1 = x_2 \land y_1 = y_2 \iff \left({x_1, x_2}\right) = \left({y_1, y_2}\right)$ Thus: :$\left({x_1, x_2}\right) = \left({x_1, x_2}\right)$ and so: :$\left({x_1, x_2}\right) \preccurlyeq \left({x_1, x_2}\right)$ by definition of lexicographic order on $S_1 \times S_2$. So $\preccurlyeq$ has been shown to be reflexive. {{qed|lemma}} === Transitivity === Let: :$\left({x_1, x_2}\right) \preccurlyeq \left({y_1, y_2}\right)$ and: :$\left({y_1, y_2}\right) \preccurlyeq \left({z_1, z_2}\right)$ $(1): \quad$ Let $x_1 = y_1 = z_1$. Then by definition of lexicographic order on $S_1 \times S_2$: :$x_2 \preceq_2 y_2$ and: :$y_2 \preceq_2 z_2$ As $\preceq_2$ is an ordering, it is transitive. Thus: :$x_2 \preceq z_2$ and it follows by definition of lexicographic order on $S_1 \times S_2$ that: :$\left({x_1, x_2}\right) \preccurlyeq \left({z_1, z_2}\right)$ $(2): \quad$ Let $x_1 = y_1$ and $y_1 \ne z_1$. Then by definition of lexicographic order on $S_1 \times S_2$: :$y_1 \prec z_1$ But as $x_1 = y_1$: :$x_1 \prec z_1$ and so by definition of lexicographic order on $S_1 \times S_2$: :$\left({x_1, x_2}\right) \preccurlyeq \left({z_1, z_2}\right)$ $(3): \quad$ Let $x_1 \ne y_1$ and $y_1 = z_1$. Then by definition of lexicographic order on $S_1 \times S_2$: :$x_1 \prec y_1$ But as $y_1 = z_1$: :$x_1 \prec z_1$ and so by definition of lexicographic order on $S_1 \times S_2$: :$\left({x_1, x_2}\right) \preccurlyeq \left({z_1, z_2}\right)$ $(4): \quad$ Let $x_1 \ne y_1$ and $y_1 \ne z_1$. Then by definition of lexicographic order on $S_1 \times S_2$: :$x_1 \prec y_1$ and: :$y_1 \prec z_1$ As $\preceq_1$ is an ordering, it is transitive. :$x_1 \prec z_1$ So by definition of lexicographic order on $S_1 \times S_2$: :$\left({x_1, x_2}\right) \preccurlyeq \left({z_1, z_2}\right)$ Thus in all cases it can be seen that: :$\left({x_1, x_2}\right) \preccurlyeq \left({z_1, z_2}\right)$ So $\preccurlyeq$ has been shown to be transitive. {{qed|lemma}} === Antisymmetry === Suppose that: :$\left({x_1, x_2}\right) \preccurlyeq \left({y_1, y_2}\right)$ and: :$\left({y_1, y_2}\right) \preccurlyeq \left({x_1, x_2}\right)$ Suppose $x_1 \ne y_1$. Then by definition of lexicographic order on $S_1 \times S_2$: :$x_1 \preceq_1 y_1$ and: :$y_1 \preceq_1 x_1$ But $\preceq_1$ is an ordering and so $x_1 = y_1$. From that contradiction it follows that $x_1 = y_1$. Then by definition of lexicographic order on $S_1 \times S_2$: :$x_2 \preceq_2 y_2$ and: :$y_2 \preceq_2 x_2$ As $\preceq_2$ is an ordering, it is antisymmetric. Therefore: :$x_2 = y_2$ and so: :$\left({x_1, x_2}\right) = \left({y_1, y_2}\right)$ So $\preccurlyeq$ has been shown to be antisymmetric. {{qed|lemma}} $\preccurlyeq$ has been shown to be reflexive, transitive and antisymmetric. Hence by definition it is an ordering. {{qed}} Category:Lexicographic Order 20j34o6ba1mr2xw3oudisfdaldqfog1
Because $\struct {S, \circ}$ is a monoid, it is a fortiori also a semigroup. From Powers of Commuting Elements of Semigroup Commute: :$\forall m, n \in \N_{>0}: \paren {\circ^m a} \circ \paren {\circ^n b} = \paren {\circ^n b} \circ \paren {\circ^m a}$ That is: :$\forall m, n \in \N_{>0}: a^m \circ b^n = b^n \circ a^m$ It remains to be shown that the result holds for the cases where $m = 0$ and $n = 0$. Let $n \in \N$: {{begin-eqn}} {{eqn | l = a^m \circ b^0       | r = a^m \circ e       | c = Definition of $b^0$ }} {{eqn | r = a^m       | c = {{Defof|Identity Element}} }} {{eqn | r = e \circ a^m       | c = {{Defof|Identity Element}} }} {{eqn | r = b^0 \circ a^m       | c = Definition of $b^0$ }} {{end-eqn}} Similarly, let $m \in \N$: {{begin-eqn}} {{eqn | l = a^0 \circ b^n       | r = e \circ b^n       | c = Definition of $a^0$ }} {{eqn | r = b^n       | c = {{Defof|Identity Element}} }} {{eqn | r = b^n \circ e       | c = {{Defof|Identity Element}} }} {{eqn | r = b^n \circ a^0       | c = Definition of $a^0$ }} {{end-eqn}} and: {{begin-eqn}} {{eqn | l = a^0 \circ b^0       | r = e \circ e       | c = Definition of $a^0$ and $b^0$ }} {{eqn | r = b^0 \circ a^0       | c = Definition of $a^0$ and $b^0$ }} {{end-eqn}} Thus: :$a^m \circ b^n = b^n \circ a^m$ holds for $n = 0$ and $m = 0$. Thus: :$\forall m, n \in \N: a^m \circ b^n = b^n \circ a^m$ {{qed}}
Because $\struct {S, \circ}$ is a monoid, it is a fortiori also a semigroup. From Power of Product of Commuting Elements in Semigroup equals Product of Powers: :$\forall n \in \N_{>0}: \circ^n \paren {a \circ b} = \paren {\circ^n a} \circ \paren {\circ^n b}$ That is: :$\forall n \in \N_{>0}: \paren {a \circ b}^n = a^n \circ b^n$ It remains to be shown that the result holds for the cases where $n = 0$. Thus: {{begin-eqn}} {{eqn | l = \paren {a \circ b}\^0       | r = \circ^0 \paren {a \circ b}       | c = Definition of $\circ$ }} {{eqn | r = e       | c = Definition of $\circ^0$ }} {{eqn | r = e \circ e       | c = {{Defof|Identity Element}} }} {{eqn | r = \paren {\circ^0 a} \circ \paren {\circ^0 b}       | c = Definition of $\circ^0$ }} {{eqn | r = a^0 \circ b^0       | c = Definition of $\circ$ }} {{end-eqn}} Thus: :$\paren {a \circ b}^n = a^n \circ b^n$ holds for $n = 0$. Thus: :$\forall n \in \N: \paren {a \circ b}^n = a^n \circ b^n$ {{qed}}
Because $\struct {S, \circ}$ is a monoid, it is a fortiori a semigroup. Hence, from Index Laws for Semigroup: Product of Indices: :$\forall m, n \in \N_{>0}: \circ^{n m} a = \circ^m \paren {\circ^n a} = \circ^n \paren {\circ^m a}$ That is: :$\forall m, n \in \N_{>0}: a^{n m} = \paren {a^n}^m = \paren {a^m}^n$ It remains to be shown that the result holds for the cases where $m = 0$ and $n = 0$. {{begin-eqn}} {{eqn | l = n \times 0       | r = 0       | c =  }} {{eqn | r = 0 \times  m       | c =  }} {{eqn | ll= \leadsto       | l = \circ^{n \times 0} a       | r = e       | c =  }} {{eqn | ll= \leadsto       | r = \circ^{0 \times m} a       | c =  }} {{end-eqn}} so the condition holds for either $n = 0$ or $m = 0$. Finally, we also have: :$\circ^n \paren {\circ^0 a} = e = \circ^0 \paren {\circ^m a}$ :$\circ^0 \paren {\circ^n a}\= e = \circ^m \paren {\circ^0 a}$ {{qed}}
The proof proceeds by the Principle of Mathematical Induction on $n$, the cardinality of $B$. Let $S$ be the set of $n \in \N$ such that every proper subset of any set with $m$ elements is finite and has (strictly) fewer than $n$ elements. === Basis for the Induction === The case $n = 0$ is verified as follows: Suppose $\card B = 0$. From Cardinality of Empty Set, $B = \O$.  There are no sets $A$ such that $A \subsetneq \O$. Therefore $0 \in S$ by vacuous truth. This is the basis for the induction. === Induction Hypothesis === Fix $k \in \N$ with $k \ge 0$. Suppose that $k \in S$. That is: :every proper subset of any set with $k$ elements is finite and has (strictly) fewer than $k$ elements. This is our induction hypothesis. It remains to be proved that $k + 1 \in S$: :every proper subset of any set with $k + 1$ elements is finite and has (strictly) fewer than $k + 1$ elements. === Induction Step === This is our induction step: Suppose we have any set $B$ such that $\card B = k + 1$. Let $A \subsetneqq B$. From the definition of $\subsetneqq$: :$\exists b \in B: b \notin A$ Therefore: :$A \subseteq B \setminus \set b$ From Cardinality Less One: :$\card {B - \set b} = k$ Suppose $A = B - \set b$. Then $\card A = k < k + 1$ and so has fewer than $k + 1$ elements and is finite by definition. Otherwise: :$A \subsetneqq B - \set b$ So by the induction hypothesis: :$A$ is finite and has fewer than $k$ elements. But as $k < k + 1$ it also follows that $A$ is finite and has fewer than $k + 1$ elements. Thus $A \in S$. The result follows by the Principle of Mathematical Induction. {{qed}}
=== Sufficient Condition === This is shown in Convergent Sequence in Hausdorff Space has Unique Limit. Note that it does not require first-countability. {{qed|lemma}} === Necessary Condition === We prove the contrapositive. Suppose $T$ is not Hausdorff. Then by definition: :$\exists x, y \in S, x \ne y: \forall U, V \in \tau: x \in U, y \in V: U \cap V \ne \O$ By first-countability, $x$ and $y$ has countable local basis. Let $\set {X_n: n \in \N}$ and $\set {Y_n: n \in \N}$ be countable local bases for $x$ and $y$ respectively. Let $U_n = \displaystyle \bigcap_{k \mathop = 1}^n X_n$ and $V_n = \displaystyle \bigcap_{k \mathop = 1}^n Y_n$. We have for all $n \in \N$: :$x \in U_n, y \in V_n$ :$U_m \subseteq U_n, V_m \subseteq V_n$ for all $m \ge n$ We have that: :$\forall n \in \N: U_n \cap V_n \ne \O$ hence we can pick any $z_n \in U_n \cap V_n$. We show that $\sequence {z_n}$ converges to both $x$ and $y$. Let $A$ be an open set such that $x \in A$. By definition of a local basis: :$\exists N \in \N: X_N \subseteq A$ We also have: :$\forall n \ge N: U_n \subseteq U_N \subseteq X_N \subseteq A$ Therefore for any $n \ge N$: :$z_n \in U_n \cap V_n \subseteq U_n \subseteq A$ By definition of convergence, $\sequence {z_n}$ converges to $x$. Similarly $\sequence {z_n}$ converges to $y$. Hence $\sequence {z_n}$ does not have unique limit point. {{qed}} Category:First-Countable Spaces Category:Hausdorff Spaces hjqs2yokgfjt4z0o78tp2r10yrx7q1e
Let $x$ and $y$ be '''topologically distinguishable'''. Then either: :$\exists U \in \tau: x \in U \subseteq N_x \subseteq X: y \notin N_x$ or: :$\exists V \in \tau: y \in V \subseteq N_y \subseteq X: x \notin N_y$ Suppose $x = y$. Then: :$\neg \left({\exists U \in \tau: x \in U \subseteq N_x \subseteq X: y \notin N_x}\right)$ and: :$\neg \left({\exists V \in \tau: y \in V \subseteq N_y \subseteq X: x \notin N_y}\right)$ Hence the result by Proof by Contradiction: :$x \ne y$ and so by Singleton Equality: :$\left\{{x}\right\} \ne \left\{{y}\right\}$ {{qed}} Category:Separation Axioms fibtb9jvxk7u2j8s651wkn5xs3chb5l
By definition of strict total ordering, $\prec$ is: : $(1): \quad$ a relation which is transitive and antireflexive  : $(2): \quad$ a relation which is connected. By Reflexive Reduction of Transitive Antisymmetric Relation is Strict Ordering: : $\prec \restriction_T$ is a strict ordering. It follows from Restriction of Connected Relation is Connected that: :$\prec \restriction_T$ is connected. Thus $\prec \restriction_T$ is a strict ordering which is connected. So by definition $\prec \restriction_T$ is a strict total ordering of $T$. {{qed}}
Consider $\N$ defined as a Peano structure $\struct {P, 0, s}$. The result follows from Principle of Recursive Definition for Peano Structure. {{qed}}
Consider $\N$ defined as the von Neumann construction of the natural numbers. The result follows from Principle of Recursive Definition for Minimal Infinite Successor Set. {{qed}}
Consider $\N$, defined as a naturally ordered semigroup $\struct {S, \circ, \preceq}$. Let the mapping $f$ be defined as: :$\map f x = \begin{cases} a & : x = 0 \\ \map s {\map f n} & : x = n \circ 1 \end{cases}$ if $\map f n$ is defined. Let $S' = \set {n \in S: \map f n \text{ is defined} }$. Then: :$0 \in S'$ and: :$n + 1 \in S'$ so by Principle of Mathematical Induction for Naturally Ordered Semigroup: :$S' = S$ Thus the domain of $f$ is $S$. Consequently, $f$ is a mapping from $S$ into $T$ which satisfies: :$\map f 0 = a$ and: :$\map f {n \circ 1} = \map s {\map f n}$ for all $n \in S$. {{qed}} === Objections === $(1):$ In the above argument, $S'$ is not precisely defined. In order for a set to be defined, that definition should be able to be expressed solely in terms of logic and definitions from set theory. In this case, the expression "is defined" does not meet that criterion. $(2):$ The mapping $f$ is not defined properly. A mapping needs to be defined as a set of ordered pairs which needs to satisfy a condition. In the above, it is indeed specified that :$\tuple {0, a} \in f$ and: :$\tuple {n \circ 1, \map s x} \in f$ whenever $\tuple {n, x} \in f$ Thus it appears either that: :$f$ itself is used to define $f$ or else that: :$f$ itself changes during the process in which it is being defined. Neither of these possibilities can be accepted. $(3):$ The only property of $\struct {S, \circ, \preceq}$ used in the argument is that it satisfies Principle of Mathematical Induction for Naturally Ordered Semigroup. However, consider the commutative semigroup $\struct {D, +}$ which has elements $0$ and $1$, such that: :$D$ is the only subset of $D$: ::containing $0$ :and ::containing $x + 1$ whenever it contains $x$. Then: :for any set $T$ :any mapping $g: T \to T$ :any element $a \in T$ there exists a mapping $f: D \to T$ such that: :$\map f y = \begin{cases} a & : y = 0 \\ \map s {\map f x} & : y = x + 1 \end{cases}$ But consider the additive group of integers modulo $2$: :$\struct {\Z_2, +_2}$ which is indeed a commutative semigroup containing $0$ and $1$ which satisfies the hypothesis. But if $g: \N \to \N$ is the mapping defined as: :$\map g n = n + 1$ there is no mapping $f: \Z_2 \to \N$ which satisfies: :$\map f y = \begin{cases} 0 & : y = 0 \\ \map s {\map f x} & : y = x +_2 1 \end{cases}$ for all $x \in \Z_2$. Hence the argument is invalid.
By definition of immediate successor element, this is equivalent to the statement: :$\forall n \in \Z: n + 1$ is the immediate successor to $n$ We have that Integers form Ordered Integral Domain. From One Succeeds Zero in Well-Ordered Integral Domain: :$\not \exists r \in \Z: 0 < r < 1$ From Properties of Ordered Ring: :$a < b \implies n + a < n + b$ Putting $a = 0, b = 1, m = n + r$: :$\not \exists m \in \Z: n + 0 < m < n + 1$ Hence the result. {{qed}}
{{begin-eqn}} {{eqn | l = 0 \times x       | r = \paren {0 + 0} \times x       | c = Real Number Axioms: $\R A 3$ }} {{eqn | r = 0 \times x + 0 \times x       | c = Real Number Axioms: $\R D$ }} {{eqn | ll= \leadsto       | l = 0 \times x       | r = 0       | c = Real Addition Identity is Zero: Corollary }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = -0 + 0       | r = 0       | c = Real Number Axioms: $\R A 4$ }} {{eqn | ll= \leadsto       | l = -0       | r = 0       | c = Real Addition Identity is Zero: Corollary }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = 0       | r = \paren {-x} + x       | c = Real Number Axioms: $\R A 4$: Inverse for Addition }} {{eqn | ll= \leadsto       | l = -\paren {-x} + 0       | r = -\paren {-x} + \paren {-x} + x       | c = adding $-\paren {-x}$ to both sides }} {{eqn | ll= \leadsto       | l = -\paren {-x} + 0       | r = \paren {-\paren {-x} + \paren {-x} } + x       | c = Real Number Axioms: $\R A 1$: Associativity }} {{eqn | ll= \leadsto       | l = -\paren {-x} + 0       | r = 0 + x       | c = Real Number Axioms: $\R A 4$: Inverse for Addition }} {{eqn | ll= \leadsto       | l = -\paren {-x}       | r = x       | c = Real Number Axioms: $\R A 3$: Identity for Addition }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = x \times \paren {\paren {-y} + y}       | r = x \times 0       | c = Real Number Axioms: $\R A 4$: Inverse for Addition }} {{eqn | r = 0       | c = Real Zero is Zero Element }} {{eqn | ll= \leadsto       | l = \paren {x \times \paren {-y} } + \paren {x \times y}       | r = 0       | c = Real Number Axioms: $\R D$: Distributivity }} {{eqn | ll= \leadsto       | l = \paren {\paren {x \times \paren {-y} } + \paren {x \times y} } + \paren {-\paren {x \times y} }       | r = 0 + \paren {-\paren {x \times y} }       | c = adding $-\paren {x \times y}$ to both sides }} {{eqn | ll= \leadsto       | l = \paren {x \times \paren {-y} } + \paren {\paren {x \times y} + \paren {-\paren {x \times y} } }       | r = 0 + \paren {-\paren {x \times y} }       | c = Real Number Axioms: $\R A 1$: Associativity }} {{eqn | ll= \leadsto       | l = \paren {x \times \paren {-y} } + 0       | r = 0 + \paren {-\paren {x \times y} }       | c = Real Number Axioms: $\R A 4$: Inverse for Addition }} {{eqn | ll= \leadsto       | l = x \times \paren {-y}       | r = -\paren {x \times y}       | c = Real Number Axioms: $\R A 3$: Identity for Addition }} {{end-eqn}} Similarly: {{begin-eqn}} {{eqn | l = \paren {x + \paren {-x} } \times y       | r = 0 \times y       | c = Real Number Axioms: $\R A 4$: Inverse for Addition }} {{eqn | r = 0       | c = Real Zero is Zero Element }} {{eqn | ll= \leadsto       | l = \paren {x \times y} + \paren {\paren {-x} \times y}       | r = 0       | c = Real Number Axioms: $\R D$: Distributivity }} {{eqn | ll= \leadsto       | l = -\paren {x \times y} + \paren {\paren {x \times y} + \paren {\paren {-x} \times y} }       | r = -\left({x \times y}\right) + 0       | c = adding $-\left({x \times y}\right)$ to both sides }} {{eqn | ll= \leadsto       | l = \paren {-\paren {x \times y} + \paren {x \times y} } + \paren {\paren {-x} \times y}       | r = -\paren {x \times y} + 0       | c = Real Number Axioms: $\R A 1$: Associativity }} {{eqn | ll= \leadsto       | l = 0 + \paren {\paren {-x} \times y}       | r = -\paren {x \times y} + 0       | c = Real Number Axioms: $\R A 4$: Inverse for Addition }} {{eqn | ll= \leadsto       | l = \paren {-x} \times y       | r = -\paren {x \times y}       | c = Real Number Axioms: $\R A 3$: Identity for Addition }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = -\paren {x + y}       | r = \paren {-1} \times \paren {x + y}       | c = Multiplication by Negative Real Number: Corollary }} {{eqn | r = \paren {\paren {-1} \times x} + \paren {\paren {-1} \times y}       | c = Real Number Axioms: $\R D$: Distributivity }} {{eqn | r = \paren {-x} + \paren {-y}       | c = Multiplication by Negative Real Number: Corollary }} {{eqn | r = -x - y       | c = {{Defof|Real Subtraction}} }} {{end-eqn}} {{qed}}