hcju/proofwikiqa huggingface dataset:
Let $N$ be divisible by $9$. Then: {{begin-eqn}} {{eqn | l = N       | o = \equiv       | r = 0 \pmod 9 }} {{eqn | ll= \leadstoandfrom       | l = a_0 + a_1 10 + a_2 10^2 + \cdots + a_n 10^n       | o = \equiv       | r = 0 \pmod 9 }} {{eqn | ll= \leadstoandfrom       | l = a_0 + a_1 1 + a_2 1^2 + \cdots + a_n 1^n       | o = \equiv       | r = 0 \pmod 9       | c = as $10 \equiv 1 \pmod 9$ }} {{eqn | ll= \leadstoandfrom       | l = a_0 + a_1 + \cdots + a_n       | o = \equiv       | r = 0 \pmod 9 }} {{end-eqn}} {{qed}}
This is a special case of Congruence of Sum of Digits to Base Less 1. {{qed}}
We have that: : $\sqrt 2$ is irrational. : $2$ is trivially rational, as $2 = \dfrac 2 1$. Consider the number $q = \sqrt 2^{\sqrt 2}$, which is irrational by the Gelfond-Schneider Theorem. Thus: : $q^{\sqrt 2} = \left({\sqrt 2^{\sqrt 2}}\right)^{\sqrt 2} = \sqrt 2 ^{\left({\sqrt 2}\right) \left({\sqrt 2}\right)} = \sqrt 2^2 = 2$ is rational. So $a = q = \sqrt 2^{\sqrt 2}$ and $b = \sqrt 2$ are the desired irrationals. {{qed}}
Given that $2$ is rational and $\sqrt 2$ is irrational, consider the number $q = \sqrt 2^{\sqrt 2}$. We consider the two cases. :$(1): \quad$ If $q$ is rational then $a = \sqrt 2$ and $b = \sqrt 2$ are the desired irrational numbers. :$(2): \quad$ If $q$ is irrational then $q^{\sqrt 2} = \paren {\sqrt 2^{\sqrt 2} }^{\sqrt 2} = \sqrt 2^{\paren {\sqrt 2} \paren {\sqrt 2} } = \sqrt 2^2 = 2$ is rational, so $a = q = \sqrt2^{\sqrt 2}$ and $b = \sqrt 2$ are the desired irrational numbers. {{qed}}
Consider the equation: :$\left({\sqrt 2}\right)^{\log_{\sqrt 2} 3} = 3$ We have that $\sqrt 2$ is irrational and $3$ is (trivially) rational. It remains to be proved $\log_{\sqrt 2} 3$ is irrational. We have: {{begin-eqn}} {{eqn | l = \log_{\sqrt 2} 3       | r = \frac{\log_2 3}{\log_2 \sqrt 2}       | c = Change of Base of Logarithm }} {{eqn | r = \frac{\log_2 3}{1/2}       | c = Logarithms of Powers: $\sqrt 2 = 2^{1/2}$ and $\log_2 2 = 1$ }} {{eqn | r = 2 \, \log_2 3       | c =  }} {{end-eqn}} From Irrationality of Logarithm, $\log_2 3$ is irrational. Therefore $2 \, \log_2 3$ is irrational. Thus there exist two irrational numbers $a = \sqrt 2$ and $b = \log_{\sqrt 2} 3$ such that $a^b$ is rational. {{qed}}
{{begin-eqn}} {{eqn | o =        | r = x \in A \cap \paren {B \cap C} }} {{eqn | o = \leadstoandfrom       | r = x \in A \land \paren {x \in B \land x \in C}       | c = {{Defof|Set Intersection}} }} {{eqn | o = \leadstoandfrom       | r = \paren {x \in A \land x \in B} \land x \in C       | c = Rule of Association: Conjunction }} {{eqn | o = \leadstoandfrom       | r = x \in \paren {A \cap B} \cap C       | c = {{Defof|Set Intersection}} }} {{end-eqn}} Therefore: :$x \in A \cap \paren {B \cap C}$ {{iff}} $x \in \paren {A \cap B} \cap C$ Thus it has been shown that: :$A \cap \paren {B \cap C}\ = \paren {A \cap B} \cap C$ {{qed}}
From its definition, the identity (here denoted by $e$) of a group $G$ commutes with all elements of $G$. So by definition of center: :$e \in \map Z {S_n}$ By definition of center: :$\map Z {S_n} = \set {\tau \in S_n: \forall \sigma \in S_n: \tau \sigma = \sigma \tau}$ Let $\pi, \rho \in S_n$ be permutations of $\N_n$. Let us choose an arbitrary $\pi \in S_n: \pi \ne e, \map \pi i = j, i \ne j$. Since $n \ge 3$, we can find $\rho \in S_n$ which interchanges $j$ and $k$ (where $k \ne i, j$) and fixes everything else. It follows that $\rho^{-1}$ does the same thing, and in particular both $\rho$ and $\rho^{-1}$ fix $i$. So: {{begin-eqn}} {{eqn | l = \map {\rho \pi \rho^{-1} } i       | r = \map {\rho \pi} i       | c =  }} {{eqn | r = \map \rho j       | c =  }} {{eqn | r = k       | c =  }} {{end-eqn}} So: :$\map {\rho \pi \rho^{-1} } i = k \ne j = \map \pi i$ From Conjugate of Commuting Elements, if $\rho$ and $\pi$ were to commute, $\rho \pi \rho^{-1} = \pi$. But they don't. Whatever $\pi \in S_n$ is, you can ''always'' find a $\rho$ such that $\rho \pi \rho^{-1} \ne \pi$. So no non-identity elements of $S_n$ commute with all elements of $S_n$. Hence: :$\map Z {S_n} = \set e$ {{Qed}}
Let us choose an arbitrary $\pi \in S_n: \pi \ne e, \pi \left({i}\right) = j, i \ne j$. Then $\pi \left({j}\right) \ne j$ since permutations are injective. Since $n \ge 3$, we can find $ k \ne j, k \ne \pi \left({j}\right)$ and $\rho \in S_n$ which interchanges $j$ and $k$ and fixes everything else. Let $\pi \left({j}\right) = m$. Then $m \ne j, m \ne k$ so $\rho$ fixes $m$. Then $ \rho \pi \left({j}\right) = \rho \left({m}\right) = m = \pi \left({j}\right)$ since $\rho$ fixes $m$. Now $k = \rho \left({j}\right)$ by definition of $\rho$. So $\pi \left({k}\right) = \pi \rho \left({j}\right)$. But $\pi \left({j}\right) \ne \pi \left({k}\right)$ since permutations are injective. Thus $\rho \pi \left({j}\right) \ne \pi \rho \left({j}\right)$. So arbitrary $\pi \ne e$ is not in the center since there exists a $\rho$ with which $\pi$ does not commute. Thus only $e$ is in the center, which, by definition, is trivial. {{Qed}}
First we note that, from Parity of Integer equals Parity of its Square, if an integer is even, its square root, if an integer, is also even. Thus it follows that: :$(1): \quad 2 \divides p^2 \implies 2 \divides p$ where $2 \divides p$ indicates that $2$ is a divisor of $p$. {{AimForCont}} that $\sqrt 2$ is rational. So: :$\sqrt 2 = \dfrac p q$ for some $p, q \in \Z$, and: :$\gcd \set {p, q} = 1$ where $\gcd$ denotes the greatest common divisor. Squaring both sides yields: :$2 = \dfrac {p^2} {q^2} \iff p^2 = 2 q^2$ Therefore from $(1)$: :$2 \divides p^2 \implies 2 \divides p$ That is, $p$ is an even integer. So $p = 2 k$ for some $k \in \Z$. Thus: :$2 q^2 = p^2 = \paren {2 k}^2 = 4 k^2 \implies q^2 = 2 k^2$ so by the same reasoning: :$2 \divides q^2 \implies 2 \divides q$ This contradicts our assumption that $\gcd \set {p, q} = 1$, since $2 \divides p, q$. Therefore, from Proof by Contradiction, $\sqrt 2$ cannot be rational. {{qed}}
{{AimForCont}} that $\sqrt 2$ is rational. Then $\sqrt 2 = \dfrac p q$ for some $p, q \in \Z_{>0}$ Consider the quantity $\paren {\sqrt 2 - 1}$: {{begin-eqn}} {{eqn | ll = 1       | l = <       | o = \sqrt 2       | r = <       | rr = 2       | c = Ordering of Squares in Reals }} {{eqn | lll = \leadsto       | ll = 0       | l = <       | o = \sqrt 2 - 1       | r = <       | rr = 1 }} {{end-eqn}} Now, observe that for any $n \in \Z_{>0}$: {{begin-eqn}} {{eqn | l = \paren {\sqrt 2 - 1}^n       | r = \sum_{k \mathop = 0}^n \binom n k \paren {\sqrt 2}^k \paren {-1}^{n-k}       | c = Binomial Theorem }} {{eqn | r = \sum_{\substack {0 \mathop \le k \mathop \le n \\ k \, \text{even} } } \binom n k 2^{k/2} \paren {-1}^{n - k} + \sqrt 2 \sum_{\substack {0 \mathop \le k \mathop \le n \\ k \, \text{odd} } } \binom n k 2^{\paren {k - 1}/2} \paren {-1}^{n-k} }} {{eqn | r = a_n + b_n \sqrt 2       | c = for some integers $a_n, b_n$ }} {{eqn | r = a_n + b_n \paren {\frac p q}       | c = recalling the assumption that $\sqrt 2 = \dfrac p q$ }} {{eqn | r = \frac {a_n q + b_n p} q }} {{eqn | o = \ge       | r = \frac 1 q       | c = as the numerator is an integer and $\sqrt 2 - 1 > 0$ }} {{end-eqn}} By Sequence of Powers of Number less than One: :$\displaystyle \lim_{n \mathop \to \infty} \paren {\sqrt 2 - 1}^n = 0$ where $\lim$ denotes limit. Recall the definition of $a_n$ and $b_n$. By Lower and Upper Bounds for Sequences: :$0 = \displaystyle \lim_{n \mathop \to \infty} \frac {a_n q + b_n p} q \ge \frac 1 q$ which is a contradiction. {{qed}}
{{AimForCont}} that $\sqrt 2$ is rational. Let $n$ be the smallest positive integer such that: :$\sqrt 2 = \dfrac m n$ for some $m \in \Z_{>0}$ Then: :$m = n \sqrt2 > n$ so: :$(1): \quad m - n > 0$ We also have: :$m = n \sqrt2 < 2 n$ so: :$m < 2 n$ and therefore: :$(2): \quad m - n < n$ Finally, we have {{begin-eqn}} {{eqn | l = m^2       | r = 2 n^2 }} {{eqn | ll= \leadsto       | l = m^2 - n m        | r = 2 n^2 - n m }} {{eqn | ll= \leadsto       | l = m \paren {m - n}        | r = \paren {2 n - m} n }} {{eqn | ll= \leadsto       | l = \dfrac {2 n - m} {m - n}        | r = \dfrac m n }} {{end-eqn}} It follows that: :$\dfrac {2 n - m} {m - n} = \sqrt2$ By $(1)$, the denominator of $\dfrac {2 n - m} {m - n}$ is positive. By $(2)$, the denominator of $\dfrac {2 n - m} {m - n}$ is less than $n$. We have thus written $\sqrt2$ as a fraction  with a smaller denominator than $n$, which is a contradiction. {{qed}}
Let $\O$ and $\O'$ both be empty sets. From Empty Set is Subset of All Sets, $\O \subseteq \O'$, because $\O$ is empty. Likewise, we have $\O' \subseteq \O$, since $\O'$ is empty. Together, by the definition of set equality, this implies that $\O = \O'$. Thus there is only one empty set. {{Qed}}
Let $A$ and $B$ both be empty sets. Thus: :$\forall x: \neg \paren {x \in A}$ and: :$\forall x: \neg \paren {x \in B}$ Hence: :$x \notin A \iff x \notin B$ and so: :$x \in A \iff x \in B$ vacuously. From Axiom of Extension: :$\forall x: \paren {x \in A \iff x \in B} \iff A = B$ Hence the result. {{Qed}}
This is apparent from inspection of the proof rules themselves. The rules concern only the broad structure of the propositional formulas involved, and this structure is unaffected by substitution. By performing the substitutions systematically throughout the given sequent, all applications of proof rules remain correct applications in the sequent. {{Handwaving}}
By hypothesis and substitution instance we have a proof, using primitive rules, of: : $P_1, P_2, \ldots, P_n \vdash Q$ By the Extended Rule of Implication, we have: : $\vdash P_1 \implies \left({P_2 \implies \left({P_3 \implies \left({\ldots \implies \left({P_n \implies Q}\right) \ldots }\right)}\right)}\right)$ {{Qed}}
From the Rule of Conjunction, we note the following. Any sequent: :$P_1, P_2 \vdash Q$ can be expressed as: :$P_1 \land p_2 \vdash Q$ Also, from the Rule of Simplification, any sequent: :$P_1 \land P_2 \vdash Q$ can be expressed as: :$P_1, P_2 \vdash Q$ Consider the expression: :$P_1, P_2, P_3, \ldots, P_{n - 1}, P_n \vdash Q$ By repeated application of the above, we can arrive at: :$P_1 \land \paren {P_2 \land \paren {P_3 \land \paren {\ldots \land \paren {P_{n - 1} \land P_n} \ldots} } } \vdash Q$ For convenience, the Rule of Substitution can be used to substitute $R_1$ for: :$P_2 \land \paren {P_3 \land \paren {\ldots \land \paren {P_{n - 1} \land P_n} \ldots} }$ to get: :$P_1 \land R_1 \vdash Q$ From the Rule of Implication: :$\vdash \paren {P_1 \land R_1} \implies Q$ Using the Rule of Exportation, we then get: :$\vdash P_1 \implies \paren {R_1 \implies Q}$ Substituting back for $R_1$: :$\vdash P_1 \implies \paren {\paren {P_2 \land \paren {P_3 \land \paren {\ldots \land \paren {P_{n - 1} \land P_n} \ldots} } } \implies Q}$ Making a substitution of convenience again, substitute $R_2$ for: :$P_3 \land \paren {\ldots \land \paren {P_{n - 1} \land P_n} \ldots}$ Take the expression: :$\paren {P_2 \land R_2} \implies Q$ and express it as: :$P_2 \land R_2 \vdash Q$ and use the Rule of Exportation to get: :$P_2 \implies \paren {R_2 \implies Q}$ which, substituting back for $R_2$, gives: :$P_2 \implies \paren {\paren {P_3 \land \paren {\ldots \land \paren {P_{n - 1} \land P_n} \ldots} } \right) \implies Q}$ Similarly: :$\paren {P_3 \land \paren {\ldots \land \paren {P_{n - 1} \land P_n} \ldots} } \implies Q$ converts to: :$P_3 \implies \paren {\paren {\ldots \land \paren {P_{n - 1} \land P_n} \ldots} \implies Q}$ The pattern becomes apparent. Eventually: :$\paren {P_{n - 1} \land P_n} \implies Q$ which converts to: :$P_{n - 1} \implies \paren {P_n \implies Q}$ Substituting these back into our original expression: :$\vdash P_1 \implies \paren {P_2 \implies \paren {P_3 \implies \paren {\ldots \implies \paren {P_n \implies Q} \ldots} } }$ {{qed}}
The result follows directly from the truth table for the biconditional: $\begin{array}{|cc||ccc|} \hline p & q & p & \iff & q \\ \hline F & F & F & T & F \\ F & T & F & F & T \\ T & F & F & F & F \\ T & T & F & T & T \\ \hline \end{array}$ By inspection, it is seen that $\mathcal M \left({p \iff q}\right) = T$ precisely when $\mathcal M \left({p}\right) = \mathcal M \left({q}\right)$. {{qed}}
Let $v$ be an arbitrary interpretation. Then by definition of interderivable: :$\map v {p \iff q}$ {{iff}} $\map v p = \map v q$ Since $v$ is arbitrary, $\map v p = \map v q$ holds in all interpretations. That is: :$p \dashv \vdash q$ {{qed}}
{{begin-eqn}} {{eqn | l = x       | o = \in       | r = \paren {S \cup T}       | c =  }} {{eqn | ll= \leadstoandfrom       | l = x \in S       | o = \lor       | r = x \in T       | c = {{Defof|Set Union}} }} {{eqn | ll= \leadstoandfrom       | l = x \in T       | o = \lor       | r = x \in S       | c = Disjunction is Commutative }} {{eqn | ll= \leadstoandfrom       | l = x       | o = \in       | r = \paren {T \cup S}       | c = {{Defof|Set Union}} }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = x       | o = \in       | r = \paren {S \cap T} }} {{eqn | ll= \leadstoandfrom       | l = x \in S       | o = \land       | r = x \in T       | c = {{Defof|Set Intersection}} }} {{eqn | ll= \leadstoandfrom       | l = x \in T       | o = \land       | r = x \in S       | c = Conjunction is Commutative }} {{eqn | ll= \leadstoandfrom       | l = x       | o = \in       | r = \paren {T \cap S}       | c = {{Defof|Set Intersection}} }} {{end-eqn}} {{qed}}
Sets have elements. Some of those elements may themselves be sets. So, given two sets $S$ and $T$, we can ask the question: Is $S$ an element of $T$? The answer will either be ''yes'' or ''no''. In particular, given any set $S$, we can ask the question: Is $S$ an element of $S$? Again, the answer will either be ''yes'' or ''no''. Thus, $\map P S = S \in S$ is a property on which we can use the comprehension principle to build this set: :$T = \set {S: S \in S}$ which is the set of all sets which contain themselves. Or we can apply the comprehension principle to build this set: :$R = \set {S: S \notin S}$ ($R$ for {{AuthorRef|Bertrand Russell|Russell}}, of course.) We ask the question: Is $R$ itself an element of $R$? There are two possible answers: ''yes'' or ''no''. If $R \in R$, then $R$ must satisfy the property that $R \notin R$, so from that contradiction we know that $R \in R$ does not hold. So the only other answer, $R \notin R$, must hold instead. But now we see that $R$ satisfies the conditions of the property that $R \in R$, so we can see that $R \notin R$ does not hold either. Thus we have generated a contradiction from the comprehension principle. {{qed}}
{{BeginTableau|\neg \left ({p \iff q}\right) \vdash \neg \left({p \implies q}\right) \lor \neg \left({q \implies p}\right)}} {{Premise|1|\neg \left ({p \iff q}\right)}} {{SequentIntro|2|1|\neg \left({\left ({p \implies q}\right) \land \left ({q \implies p}\right)}\right)|1   |Rule of Material Equivalence}} {{DeMorgan|3|1|\neg \left({p \implies q}\right) \lor \neg \left({q \implies p}\right)|2|Disjunction of Negations}} {{EndTableau}} {{qed|lemma}} {{BeginTableau|\neg \left({p \implies q}\right) \lor \neg \left({q \implies p}\right) \vdash \neg \left ({p \iff q}\right)}} {{Premise|1|\neg \left({p \implies q}\right) \lor \neg \left({q \implies p}\right)}} {{DeMorgan|2|1|\neg \left({\left ({p \implies q}\right) \land \left ({q \implies p}\right)}\right)|1|Disjunction of Negations}} {{SequentIntro|3|1|\neg \left ({p \iff q}\right)|2|Rule of Material Equivalence}} {{EndTableau}} {{qed}}
We apply the Method of Truth Tables. As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations. $\begin{array}{|cccc||ccccccccc|} \hline \neg & (p & \iff & q) & \neg & (p & \implies & q) & \lor & \neg & (q & \implies & p) \\ \hline F & F & T & F & F & F & T & F & F & F & F & T & F \\ T & F & F & T & F & F & T & T & T & T & T & F & F  \\ T & T & F & F & T & T & F & F & T & F & F & T & T  \\ F & T & T & T & F & T & T & T & F & F & T & T & T  \\ \hline \end{array}$ {{qed}}
{{BeginTableau|\neg \left ({p \iff q}\right) \vdash \left({p \lor q} \right) \land \neg \left({p \land q}\right)}} {{Premise|1|\neg \left ({p \iff q}\right)}} {{SequentIntro|2|1|\left({\neg p \land q}\right) \lor \left({p \land \neg q}\right)|1   |Non-Equivalence as Disjunction of Conjunctions: Formulation 1 }} {{Commutation|3|1|\left({p \land \neg q}\right) \lor \left({\neg p \land q}\right)|2|Disjunction}} {{Commutation|4|1|\left({p \land \neg q}\right) \lor \left({q \land \neg p}\right)|3|Conjunction}} {{SequentIntro|5|1   |\left ({\left({p \lor q}\right) \land \neg q}\right) \lor \left({\left({q \lor p}\right) \land \neg p}\right)   |4   |Conjunction of Disjunction with Negation is Conjunction with Negation }} {{Commutation|6|1   |\left ({\left({p \lor q}\right) \land \neg q}\right) \lor \left({\left({p \lor q}\right) \land \neg p}\right)   |5|Disjunction}} {{SequentIntro|7|1|\left({p \lor q}\right) \land \left({\neg q \lor \neg p}\right)|6   |Conjunction Distributes over Disjunction }} {{Commutation|8|1|\left({p \lor q}\right) \land \left({\neg p \lor \neg q}\right)|7|Disjunction}} {{DeMorgan|9|1|\left({p \lor q}\right) \land \neg \left({\neg \neg p \land \neg \neg q}\right)|8|Disjunction}} {{DoubleNegElimination|10|1|\left({p \lor q}\right) \land \neg \left({p \land q}\right)|9}} {{EndTableau}} {{BeginTableau|\left({p \lor q} \right) \land \neg \left({p \land q}\right) \vdash \neg \left ({p \iff q}\right)}} {{Premise|1|\left({p \lor q}\right) \land \neg \left({p \land q}\right)}} {{DeMorgan|2|1|\left({p \lor q}\right) \land \left({\neg p \lor \neg q}\right)|1|Disjunction of Negations}} {{Commutation|3|1|\left({p \lor q}\right) \land \left({\neg p \lor \neg q}\right)|2|Disjunction}} {{SequentIntro|4|1   |\left ({\left({p \lor q}\right) \land \neg q}\right) \lor \left({\left({p \lor q}\right) \land \neg p}\right)   |3   |Conjunction Distributes over Disjunction }} {{SequentIntro|5|1|\left({p \land \neg q}\right) \lor \left({q \land \neg p}\right)|4|Conjunction of Disjunction with Negation is Conjunction with Negation}} {{Commutation|6|1|\left({q \land \neg p}\right) \lor \left({p \land \neg q}\right)|5|Disjunction}} {{Commutation|7|1|\left({\neg p \land q}\right) \lor \left({p \land \neg q}\right)|6|Conjunction}} {{SequentIntro|8|1|\neg \left ({p \iff q}\right)|6|Non-Equivalence as Disjunction of Conjunctions: Formulation 1}} {{EndTableau}} {{qed}}
We apply the Method of Truth Tables. As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations. $\begin{array}{|cccc||cccccccc|} \hline \neg & (p & \iff & q) & (p & \lor & q) & \land & \neg & (p & \land & q) \\ \hline F & F & T & F & F & F & F & F & T & F & F & F \\ T & F & F & T & F & T & T & T & T & F & F & T  \\ T & T & F & F & T & T & F & T & T & T & F & F  \\ F & T & T & T & T & T & T & F & F & T & T & T  \\ \hline \end{array}$ {{qed}}
Let $a x^2 + b x + c = 0$. Then: {{begin-eqn}} {{eqn | l = 4 a^2 x^2 + 4 a b x + 4 a c       | r = 0       | c = multiplying through by $4 a$ }} {{eqn | ll= \leadsto       | l = \paren {2 a x + b}^2 - b^2 + 4 a c       | r = 0       | c = Completing the Square }} {{eqn | ll= \leadsto       | l = \paren {2 a x + b}^2       | r = b^2 - 4 a c }} {{eqn | ll= \leadsto       | l = x       | r = \frac {-b \pm \sqrt {b^2 - 4 a c} } {2 a} }} {{end-eqn}} {{qed}}
Let $H$ be a subgroup of $G$. Suppose $H$ is normal in $G$. Then $\forall x \in G, a \in H: \exists b \in H: x a = b x$. Thus, $x a x^{-1} = b \in H$ implying $x H x^{-1} \subseteq H$. Conversely, suppose $\forall x \in G: x H x^{-1} \subseteq H$. Then for $g \in G$, we have $g H g^{-1} \subseteq H$, which implies $g H \subseteq H g$. Also, for $g^{-1} \in G$, we have $g^{-1} H (g^{-1})^{-1} = g^{-1} H g \subseteq H$ which implies $H g \subseteq g H$. Therefore, $g H = H g$ meaning $H \lhd G$. {{qed}} Category:Normal Subgroups htkkjglfl8j117gvu9foubxcy7ss1b0
=== Necessary Condition === {{:Equality of Ordered Pairs/Necessary Condition}}{{qed|lemma}} === Sufficient Condition === {{:Equality of Ordered Pairs/Sufficient Condition}}{{qed}}
Suppose $e_1$ and $e_2$ are both identity elements of $\struct {S, \circ}$. Then by the definition of identity element: :$\forall s \in S: s \circ e_1 = s = e_2 \circ s$ Then: :$e_1 = e_2 \circ e_1 = e_2$ So: :$e_1 = e_2$ and there is only one identity element after all. {{qed}}
Let $e_S$ be an identity of $\left({S, \circ}\right)$. Then by definition, $e_S$ is both a left identity and a right identity. By More than one Left Identity then no Right Identity, if there is more than one of either, there cannot be one of the other. So there can be only one of each. By Left and Right Identity are the Same, they are one and the same thing. {{qed}}
Let $a, b, c \in G$ and let $a^{-1}$ be the inverse of $a$. Suppose $b a = c a$. Then: {{begin-eqn}} {{eqn | l = \paren {b a} a^{-1}       | r = \paren {c a} a^{-1}       | c =  }} {{eqn | ll= \leadsto       | l = b \paren {a a^{-1} }       | r = c \paren {a a^{-1} }       | c = {{Defof|Associative Operation}} }} {{eqn | ll= \leadsto       | l = b e       | r = c e       | c = {{Defof|Inverse Element}} }} {{eqn | ll= \leadsto       | l = b       | r = c       | c = {{Defof|Identity Element}} }} {{end-eqn}} Thus, the '''right cancellation law''' holds. The proof of the '''left cancellation law''' is analogous. {{qed}}
From its definition, a group is a monoid, all of whose elements have inverses and thus are invertible. From Invertible Element of Monoid is Cancellable, it follows that all its elements are therefore cancellable. {{qed}}
Suppose $x = b a = c a$. By Group has Latin Square Property, there exists exactly one $y \in G$ such that $x = y a$. That is, $x = b a = c a \implies b = c$. Similarly, suppose $x = a b = a c$. Again by Group has Latin Square Property, there exists exactly one $y \in G$ such that $x = a y$. That is, $a b = a c \implies b = c$. {{qed}}
Recall that Center of Group is Abelian Subgroup. Since $g x = x g$ for each $g \in G$ and $x \in \map Z G$: :$g \map Z G = \map Z G g$ Thus: :$\map Z G \lhd G$ {{qed}}
We have: :$\forall a \in G: x \in \map Z G^a \iff a x a^{-1} = x a a^{-1} = x \in \map Z G$ Therefore: :$\forall a \in G: \map Z G^a = \map Z G$ and $\map Z G$ is a normal subgroup of $G$. {{qed}}
First, note that from the definition of composition of relations, the following must be the case before the above expression is even to be defined: :$\Dom {\RR_2} = \Cdm {\RR_1}$ :$\Dom {\RR_3} = \Cdm {\RR_2}$ The two composite relations can be seen to have the same domain, thus: {{begin-eqn}} {{eqn | l = \Dom {\paren {\RR_3 \circ \RR_2} \circ \RR_1}       | r = \Dom {\RR_1}       | c = Domain of Composite Relation }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \Dom {\RR_3 \circ \paren {\RR_2 \circ \RR_1} }       | r = \Dom {\RR_2 \circ \RR_1}       | c = Domain of Composite Relation }} {{eqn | r = \Dom {\RR_1}       | c = Domain of Composite Relation }} {{end-eqn}} ... and also the same codomain, thus: {{begin-eqn}} {{eqn | l = \Cdm {\paren {\RR_3 \circ \RR_2} \circ \RR_1}       | r = \Cdm {\RR_3 \circ \RR_2}       | c = Codomain of Composite Relation }} {{eqn | r = \Cdm {\RR_3}       | c = Codomain of Composite Relation }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \Cdm {\RR_3 \circ \paren {\RR_2 \circ \RR_1} }       | r = \Cdm {\RR_3}       | c = Codomain of Composite Relation }} {{end-eqn}} So they are equal {{iff}} they have the same value at each point in their common domain, which this shows: {{begin-eqn}} {{eqn | ll= \forall x \in \Dom {\RR_1}:       | l = \map {\paren {\paren {\RR_3 \circ \RR_2} \circ \RR_1} } x       | r = \map {\paren {\RR_3 \circ \RR_2} } {\map {\RR_1} x}       | c = {{Defof|Composition of Relations}} }} {{eqn | r = \map {\RR_3} {\map {\RR_2} {\map {\RR_1} x} }       | c = {{Defof|Composition of Relations}} }} {{eqn | r = \map {\RR_3} {\map {\paren {\RR_2 \circ \RR_1} } x}       | c = {{Defof|Composition of Relations}} }} {{eqn | r = \map {\paren {\RR_3 \circ \paren {\RR_2 \circ \RR_1} } } x       | c = {{Defof|Composition of Relations}} }} {{end-eqn}} {{qed}}