hcju/proofwikiqa huggingface dataset:
That $\beta$ is an equivalence relation is proved in Equivalence Relation on Integers Modulo 5 induced by Squaring. Let: {{begin-eqn}} {{eqn | l = x, x'       | o = \in       | r = \eqclass x \beta }} {{eqn | l = y, y'       | o = \in       | r = \eqclass y \beta }} {{end-eqn}} We have: {{begin-eqn}} {{eqn | l = x^2       | o = \equiv       | r = \paren {x'}^2       | rr= \pmod 5       | c =  }} {{eqn | l = y^2       | o = \equiv       | r = \paren {y'}^2       | rr= \pmod 5       | c =  }} {{eqn | ll= \leadsto       | l = \paren {x y}^2       | o = \equiv       | r = \paren {x' y'}^2       | rr= \pmod 5       | c =  }} {{end-eqn}} Hence the result. {{qed}}
Let $\sequence {x_n}$ be the constant sequence:  :$\forall n, x_n = \lambda$ Given $\epsilon \gt 0$, then: :$\forall n \ge 1: \norm {x_n - \lambda} = \norm {\lambda - \lambda} = \norm {0} = 0 < \epsilon$ The result follows. {{qed}} Category:Sequences Category:Normed Division Rings h5vohsh8cy0tm21orkru2mv48ys7ish
By Cauchy Sequence is Bounded: :$\exists M \in \R_{>0}: \forall n, \norm {x_n} \le M$ Given $\epsilon > 0$. Since $\sequence {x_n}$ converges to $0$ then:  :$\exists N \in \N: \forall n > N, \norm {x_n} < \dfrac \epsilon M$ Hence: {{begin-eqn}} {{eqn | l = \norm {x_n y_n - 0}       | r = \norm {x_n y_n} }} {{eqn | r = \norm {x_n} \norm {y_n}       | c = {{NormAxiom|2}} }} {{eqn | o = <       | r = \norm {x_n} M       | c = $\sequence {y_n}$ is bounded by $M$ }} {{eqn | o = <       | r = \paren { \dfrac \epsilon M } M       | c = $\sequence {x_n}$ converges to $0$ }} {{eqn | r = \epsilon }} {{end-eqn}} Similarly, $\norm {y_n x_n - 0} < \epsilon$ The result follows from convergence in normed division rings. {{qed}}
Given $\epsilon > 0$: By the definition of a Cauchy sequence then: :$\exists N': \forall n, m > N', \norm {y_n - y_m} < \epsilon$ Hence $\forall n, m > \paren {N' + N}$: {{begin-eqn}} {{eqn | l = \norm {x_n - x_m }       | r = \norm {y_{n - N} - y_{m - N} }       | c = $n, m > N$ }} {{eqn | o = <       | r = \epsilon       | c = $n - N, m - N > N'$ }} {{end-eqn}} The result follows. {{qed}}
Let $\epsilon > 0$ be given. By the definition of a convergent sequence in $R$ with limit $l$: :$\exists N': \forall n > N', \norm {y_n - l} < \epsilon$ Hence: {{begin-eqn}} {{eqn | lo= \forall n > \paren {N' + N}:       | l = \norm {x_n - l}       | r = \norm {y_{n - N} - l}       | c = $n > N$ }} {{eqn | o = <       | r = \epsilon       | c = $n - N > N'$ }} {{end-eqn}} The result follows. {{qed}}
=== Maximal Left Ideal implies Quotient Ring is Division Ring === {{:Maximal Left and Right Ideal iff Quotient Ring is Division Ring/Maximal Left Ideal implies Quotient Ring is Division Ring}} {{qed|lemma}} === Quotient Ring is Division Ring implies Maximal Left Ideal === {{:Maximal Left and Right Ideal iff Quotient Ring is Division Ring/Quotient Ring is Division Ring implies Maximal Left Ideal}} {{qed|lemma}} === Maximal Right Ideal implies Quotient Ring is Division Ring === The proof for Maximal Right Ideal implies Quotient Ring is Division Ring is similar to the proof for Maximal Left Ideal implies Quotient Ring is Division Ring with product orders reversed. It can be found here. {{qed|lemma}} === Quotient Ring is Division Ring implies Maximal Right Ideal === The proof for Quotient Ring is Division Ring implies Maximal Right Ideal is similar to the proof for Quotient Ring is Division Ring implies Maximal Left Ideal with product orders reversed. It can be found here. {{qed}}
{{handwaving}} By the definitional abbreviation for the conditional: :$\mathbf A \implies \mathbf B =_{\text{def}} \neg \mathbf A \lor \mathbf B$ the Rule of Addition can be written as: : $\neg q \lor \left({p \lor q}\right)$ This evaluates as follows: :$\begin{array}{|cc|c|ccc|} \hline \neg & q & \lor & (p & \lor & q) \\ \hline 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 1 \\ 2 & 2 & 0 & 0 & 0 & 2 \\ 1 & 0 & 0 & 1 & 0 & 0 \\ 0 & 1 & 0 & 1 & 1 & 1 \\ 2 & 2 & 0 & 1 & 2 & 2 \\ 1 & 0 & 0 & 2 & 0 & 0 \\ 0 & 1 & 0 & 2 & 2 & 1 \\ 2 & 2 & 0 & 2 & 0 & 2 \\ \hline \end{array}$ {{qed}}
{{handwaving}} By the definitional abbreviation for the conditional: :$\mathbf A \implies \mathbf B =_{\text{def}} \neg \mathbf A \lor \mathbf B$ the Rule of Commutation can be written as: :$\neg \left({p \lor q}\right) \lor \left({q \lor p}\right)$ This evaluates as follows: :$\begin{array}{|cccc|c|ccc|} \hline \neg & (p & \lor & q) & \lor & (q & \lor & p) \\ \hline 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 1 & 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 2 & 0 & 2 & 0 & 0 \\ 1 & 1 & 0 & 0 & 0 & 0 & 0 & 1 \\ 0 & 1 & 1 & 1 & 0 & 1 & 1 & 1 \\ 2 & 1 & 2 & 2 & 0 & 2 & 2 & 1 \\ 1 & 2 & 0 & 0 & 0 & 0 & 0 & 2 \\ 2 & 2 & 2 & 1 & 0 & 1 & 2 & 2 \\ 1 & 2 & 0 & 2 & 0 & 2 & 0 & 2 \\ \hline \end{array}$ {{qed}}
{{handwaving}} By the definitional abbreviation for the conditional: :$\mathbf A \implies \mathbf B =_{\text{def}} \neg \mathbf A \lor \mathbf B$ the Factor Principle can be written as: :$\neg \left({\neg p \lor q}\right) \lor \left({\neg \left({r \lor p}\right) \lor \left ({r \lor q}\right)}\right)$ This evaluates as follows: :$\begin{array}{|ccccc|c|cccccccc|} \hline \neg & (\neg & p & \lor & q) & \lor & (\neg & (r & \lor & p) & \lor & (r & \lor & q)) \\ \hline 1 & 1 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 1 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ 2 & 1 & 0 & 2 & 2 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 2 \\ 1 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 1 & 0 & 1 & 0 & 1 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 0 & 1 \\ 1 & 0 & 1 & 0 & 2 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 0 & 2 \\ 1 & 2 & 2 & 0 & 0 & 0 & 1 & 0 & 0 & 2 & 0 & 0 & 0 & 0 \\ 2 & 2 & 2 & 2 & 1 & 0 & 1 & 0 & 0 & 2 & 0 & 0 & 0 & 1 \\ 1 & 2 & 2 & 0 & 2 & 0 & 1 & 0 & 0 & 2 & 0 & 0 & 0 & 2 \\ 1 & 1 & 0 & 0 & 0 & 0 & 1 & 1 & 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 1 & 0 & 1 & 1 & 0 & 1 & 1 & 0 & 0 & 1 & 1 & 1 & 1 \\ 2 & 1 & 0 & 2 & 2 & 0 & 1 & 1 & 0 & 0 & 2 & 1 & 2 & 2 \\ 1 & 0 & 1 & 0 & 0 & 0 & 0 & 1 & 1 & 1 & 0 & 1 & 0 & 0 \\ 1 & 0 & 1 & 0 & 1 & 0 & 0 & 1 & 1 & 1 & 0 & 1 & 1 & 1 \\ 1 & 0 & 1 & 0 & 2 & 0 & 0 & 1 & 1 & 1 & 0 & 1 & 2 & 2 \\ 1 & 2 & 2 & 0 & 0 & 0 & 2 & 1 & 2 & 2 & 0 & 1 & 0 & 0 \\ 2 & 2 & 2 & 2 & 1 & 0 & 2 & 1 & 2 & 2 & 2 & 1 & 1 & 1 \\ 1 & 2 & 2 & 0 & 2 & 0 & 2 & 1 & 2 & 2 & 0 & 1 & 2 & 2 \\ 1 & 1 & 0 & 0 & 0 & 0 & 1 & 2 & 0 & 0 & 0 & 2 & 0 & 0 \\ 0 & 1 & 0 & 1 & 1 & 0 & 1 & 2 & 0 & 0 & 2 & 2 & 2 & 1 \\ 2 & 1 & 0 & 2 & 2 & 0 & 1 & 2 & 0 & 0 & 0 & 2 & 0 & 2 \\ 1 & 0 & 1 & 0 & 0 & 0 & 2 & 2 & 2 & 1 & 0 & 2 & 0 & 0 \\ 1 & 0 & 1 & 0 & 1 & 0 & 2 & 2 & 2 & 1 & 0 & 2 & 2 & 1 \\ 1 & 0 & 1 & 0 & 2 & 0 & 2 & 2 & 2 & 1 & 0 & 2 & 0 & 2 \\ 1 & 2 & 2 & 0 & 0 & 0 & 1 & 2 & 0 & 2 & 0 & 2 & 0 & 0 \\ 2 & 2 & 2 & 2 & 1 & 0 & 1 & 2 & 0 & 2 & 2 & 2 & 2 & 1 \\ 1 & 2 & 2 & 0 & 2 & 0 & 1 & 2 & 0 & 2 & 0 & 2 & 0 & 2 \\ \hline \end{array}$ {{qed}}
=== Necessary Condition === Let $J$ be a left ideal of $\struct {R, +, \circ}$. Then conditions $(1)$ to $(3)$ hold by virtue of the ring axioms and $J$ being a left ideal. {{qed|lemma}} === Sufficient Condition === Suppose conditions $(1)$ to $(3)$ hold. Conditions $(1)$ and $(2)$ satisfy the criteria for the One-Step Subgroup Test, thus $J$ is a subgroup of $\struct {R, +}$. As $(3)$ defines the condition for $J$ to be a left ideal, the result follows. {{qed}} Category:Ideal Theory bw7k1b76pdqxqegok553egp44mhlhau
=== Necessary Condition === Let $J$ be a right ideal of $\struct {R, +, \circ}$. Then conditions $(1)$ to $(3)$ hold by virtue of the ring axioms and $J$ being a right ideal. {{qed|lemma}} === Sufficient Condition === Suppose conditions $(1)$ to $(3)$ hold. Conditions $(1)$ and $(2)$ satisfy the criteria for the One-Step Subgroup Test, thus $J$ is a subgroup of $\struct {R, +}$. As $(3)$ defines the condition for $J$ to be a right ideal, the result follows. {{qed}} Category:Ideal Theory 8p0999gzey1h19m7itieqbjnx1yv3x0
Since $J \subset R$, it follows from Quotient Ring of Ring with Unity is Ring with Unity that $R / J$ is a ring with unity. We now need to prove that every non-zero element of $\struct {R / J, +,  \circ}$ has an inverse for $\circ$ in $R / J$. By Left Inverse for All is Right Inverse it is sufficient to show that $\struct {R / J, +,  \circ}$ has a left inverse for every non-zero element. Let $x + J \in R / J$ not be the zero element of $R / J$. That is, $x + J \ne J$ and $x \notin J$. Take $K \subseteq R$ such that $K = \set {j + r \circ x: j \in J, r \in R}$. That is, $K$ is the subset of $R$ which can be expressed as a sum of an element of $J$ and a left product in $R$ of $x$. Now $0_R \in K$ as $0_R \in J$ and $0_R \in R$, giving $0_R + 0_R \circ x = 0_R$. So: :$(1): \quad K \ne \O$ Now let $g, h \in K$. That is: : $g = j_1 + r_1 \circ x, h = j_2 + r_2 \circ x$ Then: {{begin-eqn}} {{eqn | l = g + \paren{-h }       | r = \paren{j_1 + r_1 \circ x } - \paren{j_2 + r_2  \circ x } }} {{eqn | r = \paren{j_1 - j_2 } + \paren{r_1 \circ x  - r_2  \circ x } }} {{eqn | r = \paren{j_1 - j_2 } + \paren{r_1  - r_2 } \circ x }} {{eqn | r = K       | o = \in       | c = $j_1 - j_2 \in J$ from Test for Ideal and $r_1 - r_2 \in R$ }} {{end-eqn}} So we have: :$(2) \quad \forall g, h \in K, g-h \in K$ Now consider $g = \paren {j_1 + r_1 \circ x} \in K, y \in R$. Then: {{begin-eqn}} {{eqn | l = y \circ g       | r = y \circ \paren {j_1 + r_1 \circ x} }} {{eqn | r = \paren {y \circ j_1 } + y  \circ \paren { r_1 \circ x } }} {{eqn | r = \paren {y \circ j_1 } + \paren {y \circ r_1} \circ x }} {{eqn | r = K       | o = \in       | c = $\paren{y \circ j_1} \in J$, as $J$ is an ideal, while $y \circ r_1 \in R$ }} {{end-eqn}} Thus: :$(3) \quad y \circ g \in K$ So Test for Left Ideal can be applied to statements $(1)$ to $(3)$, and it is seen that $K$ is a left ideal of $R$. Now: {{begin-eqn}} {{eqn | l = j       | o = \in       | r = J       | c =  }} {{eqn | ll= \leadsto       | l = j + 0_R \circ x       | o = \in       | r = K       | c =  }} {{eqn | ll= \leadsto       | l = j       | o = \in       | r = K       | c =  }} {{eqn | ll= \leadsto       | l = J       | o = \subseteq       | r = K       | c =  }} {{end-eqn}} As $x = 0_R + 1_R \circ x$, it follows that $x \in K$ also. As $x \notin J$, it follows that $K$ is a left ideal such that $J \subset K \subseteq R$. As $J$ is a maximal left ideal, it follows that $K = R$. Thus $1_R \in K$ and thus: :$\exists j_0 \in J, s \in R: 1_R = j_0 + s \circ x$ So: :$1_R + \paren {- s \circ x} = j_0 \in J$ Hence: :$1_R + J = s \circ x + J = \paren {s + J} \circ \paren {x + J}$ So in the ring with unity $\struct {R / J, +,  \circ}$, the left inverse of $x + J$ is $s + J$. The result follows.
Since $J \subset R$, it follows from Quotient Ring of Ring with Unity is Ring with Unity that $R / J$ is a ring with unity. We now need to prove that every non-zero element of $\struct {R / J, +,  \circ}$ has an inverse for $\circ$ in $R / J$. By Right Inverse for All is Left Inverse it is sufficient to show that $\struct {R / J, +,  \circ}$ has a right inverse for every non-zero element. Let $x + J \in R / J$ not be the zero element of $R / J$. That is, $x + J \ne J$ and $x \notin J$. Take $K \subseteq R$ such that $K = \set {j + x \circ r: j \in J, r \in R}$. That is, $K$ is the subset of $R$ which can be expressed as a sum of an element of $J$ and a right product in $R$ of $x$. Now $0_R \in K$ as $0_R \in J$ and $0_R \in R$, giving $0_R + x \circ 0_R = 0_R$. So: :$(1): \quad K \ne \O$ Now let $g, h \in K$. That is: :$g = j_1 + x \circ r_1, h = j_2 + x \circ r_2$ Then: {{begin-eqn}} {{eqn | l = g + \paren{-h }       | r = \paren{j_1 + x \circ r_1 } - \paren{j_2 + x \circ r_2 } }} {{eqn | r = \paren{j_1 - j_2 } + \paren{x \circ r_1  - x \circ r_2 } }} {{eqn | r = \paren{j_1 - j_2 } + x \circ \paren{r_1  - r_2 } }} {{eqn | r = K       | o = \in       | c = $j_1 - j_2 \in J$ from Test for Ideal and $r_1 - r_2 \in R$ }} {{end-eqn}} So we have: :$(2) \quad \forall g, h \in K, g - h \in K$ Now consider $g = \paren {j_1 + x \circ r_1} \in K, y \in R$. Then: {{begin-eqn}} {{eqn | l = g \circ y       | r = \paren {j_1 + x \circ r_1} \circ y }} {{eqn | r = \paren {j_1 \circ y} + \paren { x \circ r_1} \circ y }} {{eqn | r = \paren {j_1 \circ y} + x \circ \paren {r_1 \circ y}  }} {{eqn | r = K       | o = \in       | c = $\paren {j_1 \circ y} \in J$, as $J$ is an ideal, while $r_1 \circ y \in R$ }} {{end-eqn}} Thus: : $(3) \quad g \circ y \in K$ So Test for Right Ideal can be applied to statements $(1)$ to $(3)$, and it is seen that $K$ is a right ideal of $R$. Now: {{begin-eqn}} {{eqn | l = j       | o = \in       | r = J       | c =  }} {{eqn | ll= \leadsto       | l = j + x \circ 0_R       | o = \in       | r = K       | c =  }} {{eqn | ll= \leadsto       | l = j       | o = \in       | r = K       | c =  }} {{eqn | ll= \leadsto       | l = J       | o = \subseteq       | r = K       | c =  }} {{end-eqn}} Since $x = 0_R + x \circ 1_R$, then $x \in K$ too. Since $x \notin J$, then $K$ is a right ideal such that $J \subset K \subseteq R$. Since $J$ is a maximal right ideal, then $K = R$. Thus $1_R \in K$ and thus: :$\exists j_0 \in J, s \in R: 1_R = j_0 + x \circ s$ So: :$1_R + \paren {- x \circ s} = j_0 \in J$ Hence: :$1_R + J = x \circ s + J = \paren {x + J} \circ \paren {s + J}$ So in the ring with unity $\struct {R / J, +,  \circ}$, the right inverse of $x + J$ is $s + J$. The result follows.
From Injection iff Left Inverse, $f$ is an injection {{iff}} $f$ has a left inverse mapping. From Surjection iff Right Inverse, $f$ is a surjection {{iff}} $f$ has a right inverse mapping. Putting these together, it follows that: : $f$ is both an injection and a surjection {{iff}}: : $f$ has both a left inverse and a right inverse.
This is demonstrated in Mapping is Injection and Surjection iff Inverse is Mapping.
Recall the definition of the inverse of $f$: $f^{-1} \subseteq T \times S$ is the relation defined as: :$f^{-1} = \set {\tuple {t, s}: t = \map f s}$ Let $f: S \to T$ be a mapping such that: ::$(1): \quad f$ is an injection ::$(2): \quad f$ is a surjection. By Inverse of Injection is Many-to-One Relation, $f^{-1}$ is many-to-one. From Inverse of Surjection is Relation both Left-Total and Right-Total $\map {f^{-1} } y$ is left-total. Thus $f^{-1}$ is: :many-to-one and :left-total. Hence, by definition, $f^{-1}$ is a mapping. {{qed}}
Let $f: S \to T$ be a mapping such that: ::$(1): \quad f$ is an injection ::$(2): \quad f$ is a surjection. Let $t \in T$. Then as $f$ is a surjection: :$\exists s \in S: t = \map f s$ As $f$ is an injection, there is only one $s \in S$ such that $t = \map f s$. Define $\map g t = s$. As $t \in T$ is arbitrary, it follows that: :$\forall t \in T: \exists s \in S: \map g t = s$ such that $s$ is unique for a given $t$. That is, $g: T \to S$ is a mapping. By the definition of $g$: :$(1): \quad \forall t \in T: \map f {\map g t} = t$ Let $s \in S$. Let: :$(2): \quad s' = \map g {\map f s}$ Then: {{begin-eqn}} {{eqn | l = \map f {s'}       | r = \map f {\map g {\map f s} }       | c =  }} {{eqn | r = \map f s       | c = from $(1)$ }} {{eqn | ll= \leadsto       | l = s       | r = s'       | c = as $f$ is an injection }} {{eqn | r = \map g {\map f s}       | c = from $(2)$ }} {{end-eqn}} Thus $f: S \to T$ and $g: T \to S$ are inverse mappings of each other. {{qed}}
This is divided into two parts: === Inverse is Mapping implies Mapping is Injection === {{:Inverse is Mapping implies Mapping is Injection}} === Inverse is Mapping implies Mapping is Surjection === {{:Inverse is Mapping implies Mapping is Surjection}} Category:Mapping is Injection and Surjection iff Inverse is Mapping asmhfla8yh05d10a2qzv6kjevi6p702
Let $K$ be a right ideal of $R$ such that $J \subsetneq K \subset R$. Let $x \in K \setminus  J$. As $x \notin J$ then $x + J \ne J$, the zero in $R / J$. As $R / J$ is a division ring then $x + J \in R / J$ has an inverse, say $s + J$. That is: :$1_R + J = \paren {x + J} \circ \paren {s + J} = \paren {x \circ s} + J$ By Left Cosets are Equal iff Product with Inverse in Subgroup then: :$1_R - x \circ s \in J \subsetneq K$ By the definition of a right ideal: :$x \in K$ and $s \in R \implies x \circ s \in K$ :$1_R - x \circ s \in K$ and $x \circ s \in K \implies \paren {1_R - x \circ s} + \paren {x \circ s} = 1_R \in K$ :$1_R \in K \implies \forall y \in R, 1_R \circ y = y \in K$ Hence $K = R$ The result follows.
==== Necessary Condition ==== Let $f^{-1}: T \to S$ be a mapping. Then by definition: :$\forall y \in T: \exists x \in S: \tuple {y, x} \in f^{-1}$ Thus for all $y \in T$ there exists '''at least''' one $x \in S$ such that $\tuple {y, x} \in f^{-1}$. Also by definition of mapping: :$\tuple {x_1, y} \in f^{-1} \land \tuple {x_2, y} \in f^{-1} \implies x_1 = x_2$ Thus for all $y \in T$ there exists '''at most''' one $x \in S$ such that $\tuple {y, x} \in f^{-1}$. Hence it has been demonstrated that $y \in T$ there exists a unique $x \in S$ such that $\tuple {y, x} \in f^{-1}$. {{qed|lemma}} ==== Sufficient Condition ==== Let $f$ be such that for all $y \in T$ there exists a unique $x \in S$ such that $\tuple {y, x} \in f^{-1}$. Then by definition $f^{-1}$ is a mapping.
Let $K$ be a left ideal of $R$ such that $J \subsetneq K \subset R$. Let $x \in K \setminus J$. As $x \notin J$, then $x + J \ne J$, the zero in $R / J$. As $R / J$ is a division ring, $x + J \in R / J$ has an inverse, say $s + J$. That is: :$1_R + J =  \paren {s + J} \circ \paren {x + J} = \paren {s \circ x} + J$ By Left Cosets are Equal iff Product with Inverse in Subgroup then: :$1_R - s \circ x \in J \subsetneq K$ By the definition of a left ideal then: :$x \in K$ and $s \in R \implies s \circ x \in K$ :$1_R - s \circ x \in K$ and $s \circ x \in K \implies \paren {1_R - s \circ x} + \paren {s \circ x } = 1_R \in K$ :$1_R \in K \implies \forall y \in R, y \circ 1_R = y \in K$ Hence $K = R$ The result follows.
==== Necessary Condition ==== Let $f: S \to T$ be a mapping such that $f^{-1}: T \to S$ is also a mapping. Then as $f$ is a mapping: :for all $x \in S$ there exists a unique $y \in T$ such that $\tuple {x, y} \in f$. Similarly, as $f^{-1}$ is also a mapping: :for all $y \in T$ there exists a unique $x \in S$ such that $\tuple {y, x} \in f^{-1}$. {{qed|lemma}} ==== Sufficient Condition ==== Let $f \subseteq S \times T$ be a relation such that: :$(1): \quad$ for each $x \in S$ there exists one and only one $y \in T$ such that $\tuple {x, y} \in f$ :$(2): \quad$ for each $y \in T$ there exists one and only one $x \in S$ such that $\tuple {x, y} \in f$. Then by definition: : from $(1)$, $f$ is a mapping. : from $(2)$, $f^{-1}$ is a mapping.
By Null Sequences form Maximal Left and Right Ideal then $\NN$ is an ideal of the ring $\CC$ that is also a maximal left ideal. By Maximal Left and Right Ideal iff Quotient Ring is Division Ring then the quotient ring $\CC / \NN$ is a division ring {{qed}}
Let $f: S \to T$ be a bijection by definition 1. Then by definition: :$f$ is an injection :$f$ is a surjection By definition of injection: :every element of $T$ is the image of '''at most''' $1$ element of $S$. By definition of surjection: :every element of $T$ is the image of '''at least''' $1$ element of $S$. So: :for each $y \in T$ there exists one and only one $x \in S$ such that $\tuple {x, y} \in f$. Thus $f$ is a bijection by definition 4. {{qed|lemma}} Let $f: S \to T$ be a bijection by definition 4. Then by definition: :for each $y \in T$ there exists one and only one $x \in S$ such that $\tuple {x, y} \in f$. But: :every element of $T$ is the image of '''at most''' $1$ element of $S$ defines an injection and: :every element of $T$ is the image of '''at least''' $1$ element of $S$ defines a surjection. From Injection iff Left Inverse, $f$ is an injection {{iff}} $f$ has a left inverse mapping. From Surjection iff Right Inverse, $f$ is a surjection {{iff}} $f$ has a right inverse mapping. Putting these together, it follows that: :$f$ is an injection :$f$ is a surjection Thus $f$ is a bijection by definition 1.
Denote with $\mathscr H_2 - (A1)$ the proof system resulting from $\mathscr H_2$ by removing axiom $(A1)$. Consider $\mathscr C_2$, Instance 2 of constructed semantics. We will prove that: * $\mathscr H_2 - (A1)$ is sound for $\mathscr C_2$; * Axiom $(A1)$ is not a tautology in $\mathscr C_2$ which leads to the conclusion that $(A1)$ is not a theorem of $\mathscr H_2 - (A1)$. === Soundness of $\mathscr H_2 - (A1)$ for $\mathscr C_2$ === Starting with the axioms: {{begin-axiom}} {{axiom|n = A2        |lc = Rule of Addition        |m = q \implies (p \lor q)        |rc = Proof of Tautology }} {{axiom|n = A3        |lc = Rule of Commutation        |m = (p \lor q) \implies (q \lor p)        |rc = Proof of Tautology }} {{axiom|n = A4        |lc = Factor Principle        |m = (q \implies r) \implies \left({ (p \lor q) \implies (p \lor r)}\right)        |rc = Proof of Tautology }} {{end-axiom}} Next it needs to be shown that the rules of inference of $\mathscr H_2$ preserve $\mathscr C_2$-tautologies. ==== Rule $RST \, 1$: Rule of Uniform Substitution ==== By definition, any WFF is assigned a value $0$, $1$ or $2$. Thus, in applying Rule $RST \, 1$, we are introducing $0$, $1$ or $2$ in the position of a propositional variable. But all possibilities of assignments of $0$s, $1$s and $2$s to such propositional variables were shown not to affect the resulting value $0$ of the axioms. Hence Rule $RST \, 1$ preserves $\mathscr C_2$-tautologies. ==== Rule $RST \, 2$: Rule of Substitution by Definition ==== Because the definition of $\mathscr C_2$ was given in terms of Rule $RST \, 2$, it cannot affect any of its results. ==== Rule $RST \, 3$: Rule of Detachment ==== Suppose $\mathbf A$ and $\mathbf A \implies \mathbf B$ both take value $0$. Then using Rule $RST \, 2$, definition $(2)$, we get: :$\neg \mathbf A \lor \mathbf B$ taking value $0$ by assumption. But $\neg \mathbf A$ takes value $1$ by definition of $\neg$. So from the definition of $\lor$ it must be that $\mathbf B$ takes value $0$. Hence Rule $RST \, 3$ also produces only WFFs of value $0$. ==== Rule $RST \, 4$: Rule of Adjunction ==== Suppose $\mathbf A$ and $\mathbf B$ take value $0$. Then: {{begin-eqn}} {{eqn|l = \mathbf A \land \mathbf B      |r = 0 \land 0 }} {{eqn|r = \neg ( \neg 0 \lor \neg 0 )      |c = Rule $RST \, 2 \, (1)$ }} {{eqn|r = \neg ( 1 \lor 1 ) }} {{eqn|r = \neg 1 }} {{eqn|r = 0 }} {{end-eqn}} proving that Rule $RST \, 4$ also produces only $0$s from $0$s. Hence $\mathscr H_2 - (A1)$ is sound for $\mathscr C_2$. === $(A1)$ is not a $\mathscr C_2$-tautology === Recall axiom $(A1)$, the Rule of Idempotence: :$(p \lor p) \implies p$ Under $\mathscr C_2$, we apply a single definitional abbreviation and have the following: :$\begin{array}{|cccc|c|c|} \hline \neg & (p & \lor & p) & \lor & p \\ \hline 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 1 & 1 & 0 & 1 \\ 1 & 2 & 0 & 2 & 2 & 2 \\ \hline \end{array}$ Hence according to the definition of $\mathscr C_2$, $(A1)$ is not a tautology. Therefore $(A1)$ is independent from $(A2)$, $(A3)$, $(A4)$. {{qed}}