hcju/proofwikiqa huggingface dataset:
=== Sufficient Condition === Let $\FF$ converge. Then there is a point $x \in X$ such that $\FF$ converges to $x$. By Filter on Product Space Converges to Point iff Projections Converge to Projections of Point: :$\forall i \in I: \map {\pr_i} \FF$ converges to $x_i$ Thus :$\forall i \in I: \map {\pr_i} \FF$ converges. === Necessary Condition === Suppose that for each $i \in I$, $\map {\pr_i} \FF$ converges. For each $i \in I$, let $S_i$ be the set of points to which $\map {\pr_i} \FF$ converges. By our supposition, $S_i$ is non-empty for each $i \in I$. By the axiom of choice, there is a point $x \in X$ such that: :$\forall i \in I: \map {\pr_i} x \in S_i$ By Filter on Product Space Converges to Point iff Projections Converge to Projections of Point, $\FF$ converges to $x$. {{qed}} {{AoC}}
=== Sufficient Condition === Let $\FF$ converge. Then there is a point $x \in X$ such that $\FF$ converges to $x$. By Filter on Product Space Converges to Point iff Projections Converge to Projections of Point: :$\forall i \in I: \map {\pr_i} \FF$ converges to $x_i$. Thus, for each $i \in I$, $\map {\pr_i} \FF$ converges. === Necessary Condition === For each $i \in I$, let $\map {\pr_i} \FF$ converge. Since $X_i$ is a Hausdorff space for each $i \in I$, this implies that $\map {\pr_i} \FF$ converges to exactly one point for each $i \in I$. For each $i$, let $x_i$ be the point in $X_i$ to which $\map {\pr_i} \FF$ converges. Then by the definition of the product space, $x$ is an element of $X$. By Filter on Product Space Converges to Point iff Projections Converge to Projections of Point, $\FF$ converges to $x$. {{qed}}
First assume that $X$ is compact.  From Projection from Product Topology is Continuous, the projections $\pr_i : X \to X_i$ are continuous. It follows from Continuous Image of Compact Space is Compact that the $X_i$ are compact. Assume now that each $X_i$ is compact. By Equivalent Definitions of Compactness it is enough to show that every ultrafilter on $X$ converges. Thus let $\FF$ be an ultrafilter on $X$. From Image of Ultrafilter is Ultrafilter, for each $i \in I$, the image filter $\map {\pr_i} \FF$ is an ultrafilter on $X_i$. Each $X_i$ is compact by assumption. So by Equivalent Definitions of Compactness, each $\map {\pr_i} \FF$ converges. By Filter on Product Space Converges iff Projections Converge, $\FF$ converges. {{explain|Find the appropriate link for "converges" and explain why this proves the result.}} {{qed}} {{AoC|Filter on Product Space Converges iff Projections Converge}}
Let $f: \R \to \R$ be defined as: :$\forall x \in \R: f \left({x}\right) = x^2$ We have that $f$ is the pointwise product of the identity mapping with itself. By Product Rule for Continuous Functions and Identity Mapping is Continuous, $f$ is continuous. By Power Function is Unbounded Above: :$\exists q \in \R: f \left({q}\right) > x$ Then: :$0^2 = 0 \le x$ By the Intermediate Value Theorem: :$\exists y \in \R: 0 < y < q: y^2 = x$ {{qed}} Category:Real Numbers ltzo78jwbsk41ccg9u01ilvme4jfeo1
Let $a \in S$ be arbitrary. Then: {{begin-eqn}} {{eqn | l = a \vee a       | r = \sup \set {a, a}       | c = Definition of Join }} {{eqn | r = \sup \set a       | c = Axiom of Extension }} {{eqn | r = a       | c = Supremum of Singleton }} {{end-eqn}} Hence the result. {{qed}}
Since $a \preceq a$, $a$ is an upper bound of $\set a$. Let $b$ be another upper bound of $\set a$. Then necessarily $a \preceq b$. It follows that indeed: :$\sup \set a = a$ as desired. {{qed}}
Since $a \preceq a$, $a$ is a lower bound for $\set a$. Let $b$ be another lower bound for $\set a$. Then necessarily $b \preceq a$. It follows that indeed: :$\inf \set a = a$ as desired. {{qed}}
Let $a \in S$ be arbitrary. Then: {{begin-eqn}} {{eqn | l = a \wedge a       | r = \inf \set {a, a}       | c = {{Defof|Meet (Order Theory)}} }} {{eqn | r = \inf \set a       | c = Axiom of Extension }} {{eqn | r = a       | c = Infimum of Singleton }} {{end-eqn}} Hence the result. {{qed}}
Note that if one of the primes $p_i$ does not appear in the prime decompositions of either one of $a$ or $b$, then its corresponding index $k_i$ or $l_i$ will be zero. Let $d \divides a$. Then: :$d$ is of the form $p_1^{h_1} p_2^{h_2} \ldots p_r^{h_r}, \forall i: 1 \le i \le r, 0 \le h_i \le k_i$ :$d \divides a \iff \forall i: 1 \le i \le r, 0 \le h_i \le l_i$ So: :$d \divides a \land d \divides b \iff \forall i: 1 \le i \le r, 0 \le h_i \le \min \set {k_i, l_i}$ For $d$ to be at its greatest, we want the largest possible exponent for each of these primes. So for each $i \in \closedint 1 r$, $h_i$ needs to ''equal'' $\min \set {k_i, l_i}$. Hence the result: :$\gcd \set {a, b} = p_1^{\min \set {k_1, l_1} } p_2^{\min \set {k_2, l_2} } \ldots p_r^{\min \set {k_r, l_r} }$ {{Qed}}
{{begin-eqn}} {{eqn | l = \lcm \set {a, b}       | r = \frac {a b} {\gcd \set {a, b} }       | c = Product of GCD and LCM }} {{eqn | r = \frac {\paren {p_1^{k_1} p_2^{k_2} \cdots p_r^{k_r} } \paren {p_1^{l_1} p_2^{l_2} \ldots p_r^{l_r} } } {p_1^{\min \set {k_1, l_1} } p_2^{\min \set {k_2, l_2} } \cdots p_r^{\min \set {k_r, l_r} } }       | c = GCD from Prime Decomposition }} {{eqn | r = \frac {\paren {p_1^{k_1 + l_1} p_2^{k_2 + l_2} \cdots p_r^{k_r + l_r} } } {p_1^{\min \set {k_1, l_1} } p_2^{\min \set {k_2, l_2} } \cdots p_r^{\min \set {k_r, l_r} } }       | c =  }} {{eqn | r = p_1^{k_1 + l_1 - \min \set {k_1, l_1} } p_2^{k_2 + l_2 - \min \set {k_2, l_2} } \cdots p_r^{k_r + l_r - \min \set {k_r, l_r} }       | c =  }} {{eqn | r = p_1^{\max \set {k_1, l_1} } p_2^{\max \set {k_2, l_2} } \cdots p_r^{\max \set {k_r, l_r} }       | c = Sum Less Minimum is Maximum }} {{end-eqn}} {{Qed}}
Note that if one of the primes $p_i$ does not appear in the prime decompositions of either one of $a$ or $b$, then its corresponding index $k_i$ or $l_i$ will be zero. Let $a \divides m$. Then: :$m$ is of the form $p_1^{h_1} p_2^{h_2} \ldots p_r^{h_r}, \forall i: 1 \le i \le r, 0 \le k_i \le h_i$ :$a \divides l \iff \forall i: 1 \le i \le r, 0 \le l_i \le h_i$ So: :$a \divides m \land b \divides m \iff \forall i: 1 \le i \le r, 0 \le \max \set {k_i, l_i} \le h_i$ For $m$ to be at its smallest, we want the smallest possible exponent for each of these primes. So for each $i \in \closedint 1 r$, $h_i$ needs to ''equal'' $\max \set {k_i, l_i}$. Hence the result: :$\lcm \set {a, b} = p_1^{\max \set {k_1, l_1} } p_2^{\max \set {k_2, l_2} } \ldots p_r^{\max \set {k_r, l_r} }$ {{Qed}}
{{begin-eqn}} {{eqn | l = y       | o = >       | r = 1_R }} {{eqn | l = x \circ y       | o = >        | r = x \circ 1_R       | c = $x > 0_R$, Properties of Ordered Ring: $(6)$ }} {{eqn | l = x \circ y       | o = >       | r = x       | c = {{Defof|Unity of Ring}} }} {{end-eqn}} A similar argument shows that $y \circ x > x$. {{qed}} Category:Ordered Rings 3avgf1pkfjbsnrx11lyswl004xmyvda
The proof proceeds by induction: If $n = 1$, then $\circ^n x = x$. So: : $\circ^n x \ge x$ Suppose that $\circ^n x \ge x$. Then since $x > 1$: : $\circ^n x > 1$ By Product of Positive Element and Element Greater than One, $x \circ \left({\circ^n x}\right) > x$, so: :$\circ^{n + 1} x \ge x$ {{qed}} Category:Ordered Rings jp4f4cxys96x080ks18yg4zni6n7txu
=== Necessary Condition === Suppose that each element of $S$ has a strict successor in $S$. If $x$ is any element of $S$, then $x$ has a strict successor. Therefore, $x$ is not an upper bound of $S$. {{qed|lemma}} === Sufficient Condition === Suppose $S$ has no upper bound in $S$. Let $x \in S$. Since $S$ has no upper bound, $x$ is not an upper bound of $S$. Thus for some $p \in S$, $p \not\le x$. By the definition of a directed set, there is a $y \in S$ such that $p \le y$ and $x \le y$. Since $p \not\le x$, it follows that $x \ne y$. Hence, since we already know that $x \le y$, in fact $x < y$. Thus $y$ is a strict successor of $x$. {{qed}} Category:Order Theory 64s3c8t5iwrvm98ky606f08v4dr0q00
For all $r \in \R$, we have $r = r + 0 i \in C$. Thus the set of real numbers $\R$ can be considered a subset of $\C$. As the Real Numbers are Uncountable, it follows from Sufficient Conditions for Uncountability, proposition $(1)$, that $\C$ is uncountably infinite. {{qed}} Category:Complex Numbers Category:Set Theory 136ianqjlmsm2mh37zvydw8vz6bb46p
Suppose that $s = \inf \bigcup \mathbb T \in S$. By Set is Subset of Union, $T \subseteq \bigcup \mathbb T$ for all $T \in \mathbb T$. Hence by Infimum of Subset: :$\forall T \in \mathbb T: s \preceq \inf T$ Suppose now that $a \in S$ satisfies: :$\forall T \in \mathbb T: a \preceq \inf T$ Then by transitivity of $\preceq$: :$\forall t \in T: a \preceq t$ Since this holds for any $T \in \mathbb T$, also: :$\forall t \in \bigcup \mathbb T: a \preceq t$ Hence $a \preceq s$, by definition of infimum. That is, $s = \inf \left\{{\inf T: T \in \mathbb T}\right\}$. {{qed|lemma}} Suppose now that $r = \inf \left\{{\inf T: T \in \mathbb T}\right\} \in S$. By definition of infimum, for all $T \in \mathbb T$ and $t \in T$: :$\inf T \preceq t$ By transitivity of $\preceq$: :$\forall T \in \mathbb T: \forall t \in T: r \preceq t$ Hence for all $t \in \bigcup \mathbb T$: :$r \preceq t$ Suppose that $a \in S$ satisfies: :$\forall t \in \bigcup \mathbb T: a \preceq t$ In particular, for any $T \in \mathbb T$, since $T \subseteq \bigcup \mathbb T$: :$a \preceq \inf T$ and therefore by definition of infimum, also: :$a \preceq r$ That is, $r = \inf \bigcup \mathbb T$. {{qed}} Category:Order Theory fq5xoc97v99rszpr7xohflxeoe0z8e5
For $\struct {S, \vee, \preceq}$ to be an ordered structure is equivalent to, for all $a, b, c \in S$: :$a \preceq b \implies a \vee c \preceq b \vee c$ :$a \preceq b \implies c \vee a \preceq c \vee b$ Since Join is Commutative, it suffices to prove the first of these implications. By definition of join: :$a \vee c = \sup \set {a, c}$ where $\sup$ denotes supremum. By Join Succeeds Operands: :$b \preceq b \vee c$ :$c \preceq b \vee c$ Now also $a \preceq b$, and by transitivity of $\preceq$ we find that: :$a \preceq b \vee c$ Thus $b \vee c$ is an upper bound for $\set {a, c}$. Hence: :$a \vee c \preceq b \vee c$ by definition of supremum. {{qed}}
Let $a, b, c \in S$. Let $a \preceq b$. By the definition of join semilattice: : $a \vee b = b$ Thus: : $\left({a \vee b}\right) \vee c = b \vee c$ Since $\vee$ is associative, commutative, and idempotent: :$\left({a \vee c}\right) \vee \left({b \vee c}\right) = b \vee c$ Therefore, $a \vee c \preceq b \vee c$. From Join is Commutative, we conclude that: : $c \vee a \preceq c \vee b$ {{qed}}
For $\left({S, \wedge, \preceq}\right)$ to be an ordered structure is equivalent to, for all $a,b,c \in S$: :$a \preceq b \implies a \wedge c \preceq b \wedge c$ :$a \preceq b \implies c \wedge a \preceq c \wedge b$ Since Meet is Commutative, it suffices to prove the first of these implications. By definition of meet: :$b \wedge c = \inf \left\{{b, c}\right\}$ where $\inf$ denotes infimum. By Meet Precedes Operands: :$a \wedge c \preceq a$ :$a \wedge c \preceq c$ Now also $a \preceq b$, and by transitivity of $\preceq$ we find that: :$a \wedge c \preceq b$ Thus $a \wedge c$ is a lower bound for $\left\{{b, c}\right\}$. Hence: :$a \wedge c \preceq b \wedge c$ by definition of infimum. {{qed}}
By definition of join semilattice, $\vee$ is closed. The other three defining properties for a semilattice follow respectively from: :Join is Commutative :Join is Associative :Join is Idempotent Hence $\struct {S, \vee}$ is a semilattice. {{qed}}
By definition of meet semilattice, $\wedge$ is closed. The other three defining properties for a semilattice follow respectively from: :Meet is Commutative :Meet is Associative :Meet is Idempotent Hence $\struct {S, \wedge}$ is a semilattice. {{qed}}
Let us verify that $\preceq$ satisfies the three conditions for an ordering. === Reflexivity === Since $\circ$ is idempotent, for all $a \in S$: :$a \circ a = a$ hence $a \preceq a$. Thus $\preceq$ is reflexive. {{qed|lemma}} === Antisymmetry === Suppose that $a \preceq b$ and $b \preceq a$. Then from the first relation: :$a \circ b = b$ and from the second: :$b \circ a = a$ Since $\circ$ is commutative, it follows that $a = b$. Hence $\preceq$ is antisymmetric. {{qed|lemma}} === Transitivity === Suppose that $a \preceq b$ and $b \preceq c$. Then: {{begin-eqn}} {{eqn | l = a \circ c       | r = a \circ \paren {b \circ c}       | c = Since $b \preceq c$ }} {{eqn | r = \paren {a \circ b} \circ c       | c = $\circ$ is associative }} {{eqn | r = b \circ c       | c = Since $a \preceq b$ }} {{eqn | r = c       | c = Since $b \preceq c$ }} {{end-eqn}} Hence, $a \preceq c$. Thus $\preceq$ is transitive. {{qed|lemma}} Having verified all three conditions, it follows that $\preceq$ is an ordering. {{qed}}
We apply the Method of Truth Tables to the propositions in turn. As can be seen for all boolean interpretations by inspection, where the truth values under the main connective on the {{LHS}} is $T$, that under the one on the {{RHS}} is also $T$: $\begin{array}{|ccccccc||ccc|} \hline (p & \implies & q) & \land & (q & \implies & r) & p & \implies & r \\ \hline F & T & F & T & F & T & F & F & T & F \\ F & T & F & T & F & T & T & F & T & T \\ F & T & T & F & T & F & F & F & T & F \\ F & T & T & T & T & T & T & F & T & T \\ T & F & F & F & F & T & F & T & F & F \\ T & F & F & F & F & T & T & T & T & T \\ T & T & T & F & T & F & F & T & F & F \\ T & T & T & T & T & T & T & T & T & T \\ \hline \end{array}$ Hence the result. {{qed}}
{{BeginTableau|p \implies q, q \implies r, p \vdash r}} {{Premise|1|p \implies q}} {{Premise|2|q \implies r}} {{Premise|3|p}} {{SequentIntro|4|1, 2|p \implies r|1, 2|Hypothetical Syllogism: Formulation 1}} {{ModusPonens|5|1, 2, 3|r|4|3}} {{EndTableau}} {{qed}}
We apply the Method of Truth Tables to the propositions in turn. As can be seen for all boolean interpretations by inspection, where the truth values under the main connective on the {{LHS}} (that is, the rightmost $\land$) is $T$, that under the instance of $r$ on the {{RHS}} is also $T$: $\begin{array}{|ccccccccc||c|} \hline ((p & \implies & q) & \land & (q & \implies & r)) & \land & p & r \\ \hline F & T & F & T & F & T & F & F & F & F \\ F & T & F & T & F & T & T & F & F & T \\ F & T & T & F & T & F & F & F & F & F \\ F & T & T & T & T & T & T & F & F & T \\ T & F & F & F & F & T & F & F & T & F \\ T & F & F & F & F & T & T & F & T & T \\ T & T & T & F & T & F & F & F & T & F \\ T & T & T & T & T & T & T & T & T & T \\ \hline \end{array}$ Hence the result. {{qed}}
Let $x, y, z \in S$. Let $x \mathop {\mathcal Q} y$. Then for some $\mathcal R \in \mathcal F$: :$x \mathop{\mathcal R} y$. Since $\mathcal R$ is a relation compatible with $\circ$: :$\left({x \circ z}\right) \mathop {\mathcal R} \left({y \circ z}\right)$ Since $\mathcal R \in \mathcal F$: :$\mathcal R \subseteq \bigcup \mathcal F = \mathcal Q$ Thus  :$\left({x \circ z}\right) \mathop {\mathcal Q} \left({y \circ z}\right)$ We have shown that: :$\forall x, y, z \in S: x \mathop {\mathcal Q} y \implies \left({x \circ z}\right) \mathop {\mathcal Q} \left({y \circ z}\right)$ A similar argument shows that: :$\forall x, y, z \in S: x \mathop {\mathcal Q} y \implies \left({z \circ x}\right) \mathop {\mathcal Q} \left({z \circ y}\right)$ So $Q$ is a relation compatible with $\circ$. {{qed}} Category:Compatible Relations 9hi6f9d88lb8x6c8dddtz5pm5zxj44r
Let $x, y, z \in G$. Suppose that $\neg \left({\left({x \circ z}\right) \mathop {\mathcal Q} \left({y \circ z}\right)}\right)$. Then by the definition of $\mathcal Q$: :$\left({x \circ z}\right) \mathop {\mathcal R} \left({y \circ z}\right)$ Since $\mathcal R$ is compatible with $\circ$: :$\left({x \circ z}\right) \circ z^{-1} \mathop{\mathcal R} \left({y \circ z}\right) \circ z^{-1}$. By Group Axiom $(G1)$: Associativity and the Group Axiom $(G3)$: Inverses: :$x \mathop {\mathcal R} y$ so by the definition of $\mathcal Q$: $\neg \left({x \mathop {\mathcal Q} y}\right)$ By the Rule of Transposition: :$\forall x, y, z \in G: x \mathop {\mathcal Q} y \implies \left({x \circ z}\right) \mathop {\mathcal Q} \left({y \circ z}\right)$ A similar argument shows that: :$\forall x, y, z \in G: x \mathop {\mathcal Q} y \implies \left({z \circ x}\right) \mathop {\mathcal Q} \left({z \circ y}\right)$ Thus, by definition, $\mathcal Q$ is a relation compatible with $\circ$. {{qed}} Category:Compatible Relations Category:Group Theory h36d75mihbt885114501519gb9llmu8
Since $\mathcal R$ is compatible with $\circ$: :$\forall a,b,c \in S: a \mathrel{\mathcal R} b \implies a \circ c \mathrel{\mathcal R} b \circ c$ In particular, letting $a = x$, $b = y$, and $c = z$, we see that: :$x \mathrel{\mathcal R} y \implies x \circ z \mathrel{\mathcal R} y \circ z$ On the other hand, letting $a = x \circ z$, $b = y \circ z$, and $c = z^{-1}$, we see that: :$x \circ z \mathrel{\mathcal R} y \circ z \implies \left({x \circ z}\right) \circ z^{-1} \mathrel{\mathcal R} \left({y \circ z}\right) \circ z^{-1}$ By the associativity of $\circ$ and the definition of inverse, this reduces to: :$x \circ z \mathrel{\mathcal R} y \circ z \implies x \mathrel{\mathcal R} y$ We have thus shown that: :$x \mathrel{\mathcal R} y \iff x \circ z \mathrel{\mathcal R} y \circ z$ A precisely similar argument shows that: :$x \mathrel{\mathcal R} y \iff z \circ x \mathrel{\mathcal R} z \circ y$ so $\mathcal R$ is strongly compatible with $\circ$. {{qed}} Category:Compatible Relations Category:Group Theory 0n37g0lfrnlig9tbsl2hbz1cgpxioqg
Let $x, y, z \in S$. Suppose that $x \mathrel \QQ y$. Then for each $\RR \in \mathscr F$: :$x \mathrel \RR y$ Then since $\RR$ is a relation compatible with $\circ$: :$\paren {x \circ z} \mathrel \RR \paren {y \circ z}$ Since this holds for each $\RR \in \mathscr F$: :$\paren {x \circ z} \mathrel \QQ \paren {y \circ z}$ We have shown that: :$\forall x, y, z \in S: x \mathrel \QQ y \implies \paren {x \circ z} \mathrel \QQ \paren {y \circ z}$ A similar argument shows that: :$\forall x, y, z \in S: x \mathrel \QQ y \implies \paren {z \circ x} \mathrel \QQ \paren {z \circ y}$ so $Q$ is a relation compatible with $\circ$. {{qed}} Category:Compatible Relations 6ahue29b035401wxcwed9oazy028n3m
By the definition of an ordered group, $\preceq$ is a relation compatible with $\circ$. Thus by Properties of Relation Compatible with Group Operation/CRG1, we obtain the first two results: :$(\operatorname{OG}1.1):\quad x \preceq y \iff x \circ z \preceq y \circ z$ :$(\operatorname{OG}1.2):\quad x \preceq y \iff z \circ x \preceq z \circ y$ By Reflexive Reduction of Relation Compatible with Group Operation is Compatible, $\prec$ is compatible with $\circ$. Thus by again Properties of Relation Compatible with Group Operation/CRG1, we obtain the remaining results: :$(\operatorname{OG}1.1'):\quad x \prec y \iff x \circ z \prec y \circ z$ :$(\operatorname{OG}1.2'):\quad x \prec y \iff z \circ x \prec z \circ y$ and so the theorem is established. {{qed}} Category:Ordered Groups d7o5mhc9vz9d79jgkpk5orj2dbvhol0
=== $(1)$ implies $(2)$ === Let $\struct {S, \preceq}$ be an ordered set. Let $\struct {S, \succeq}$ be its dual. By assumption, $\Sigma$ is true for $\struct {S, \succeq}$. By Local Duality, this implies $\Sigma^*$ is true for $\struct {S, \preceq}$. Since $\struct {S, \preceq}$ was arbitrary, the result follows. {{qed}} === $(2)$ implies $(1)$ === From Dual of Dual Statement (Order Theory): :$\paren {\Sigma^*}^* = \Sigma$ The result thus follows from applying the other implication to $\Sigma^*$. {{qed}}
=== $(1)$ implies $(2)$ === By assumption, $\Sigma$ is true for $\struct {S, \preceq}$. By Dual of Dual Ordering, the dual statement $\Sigma^*$ applied to $\struct {S, \succeq}$ is the same as $\Sigma$ applied to $\struct {S, \preceq}$. Hence $\Sigma^*$ is true for $\struct {S, \succeq}$. {{qed|lemma}} === $(2)$ implies $(1)$ === From Dual of Dual Statement (Order Theory), $\paren {\Sigma^*}^* = \Sigma$. From Dual of Dual Ordering, $\struct {S, \preceq}$ is the dual of $\struct {S, \succeq}$. The result thus follows from applying the other implication to $\Sigma^*$ and $\struct {S, \succeq}$. {{qed}}
We apply the Method of Truth Tables to the proposition. As can be seen for all boolean interpretations by inspection, where the truth value under the main connective on the {{LHS}} is $T$, that under the one on the {{RHS}} is also $T$: $\begin{array}{|cccccc||cc|} \hline (p & \implies & q) & \land & \neg & q & \neg & p \\ \hline F & T & F & T & T & F & T & F \\ F & T & T & F & F & T & T & F \\ T & F & F & F & T & F & F & T \\ T & T & T & F & F & T & F & T \\ \hline \end{array}$ Hence the result. {{qed}} Note that the two formulas are not equivalent, as the relevant columns do not match exactly.
By Diagonal Relation is Universally Compatible, $\Delta_G$ is compatible with $\circ$. By Complement of Relation Compatible with Group is Compatible , $\Delta_G^c$ is also compatible with $\circ$. {{qed}} Category:Compatible Relations n07d4bm10j5n1oe24z8vocmgjasgc45
By Diagonal Relation is Universally Compatible, the diagonal relation is compatible with $\circ$. Then by Union of Relations Compatible with Operation is Compatible, $\mathcal R^=$ is compatible with $\circ$. {{qed}}
By definition of reflexive reduction, for all $a, b \in S$: :$a \mathrel {\RR^\ne} b$ {{iff}} $a \mathrel \RR b$ but $a \ne b$. By definition of the diagonal relation $\Delta_S$: :$a \ne b$ {{iff}} $\tuple {a, b} \notin \Delta_S$ Thus, considered as subsets of $S \times S$, we have: :$\RR^\ne = \RR \setminus \Delta_S$ By Diagonal Relation is Universally Compatible, $\Delta_S$ is compatible with $\circ$. Thus by Set Difference of Relations Compatible with Group Operation is Compatible, $\RR^\ne$ is compatible with $\circ$. {{qed}} Category:Compatible Relations Category:Reflexive Reductions 5bruh6rzeh2dor1v9o7f8hwoz6mv5cv
{{BeginTableau|\vdash p \lor \neg p}} {{TheoremIntro|1|\neg\neg (p \lor \neg p)|Negation of Excluded Middle is False/Form 2}} {{DoubleNegElimination|2||p \lor \neg p|1}} {{EndTableau}} {{Qed}}
Denote with $\preceq'$ the dual of $\succeq$. By definition of dual ordering, we thus have for all $a, b \in S$: :$a \preceq b$ iff $b \succeq a$ :$b \succeq a$ iff $a \preceq' b$ Hence $a \preceq b$ iff $a \preceq' b$. The result follows from Equality of Relations. {{qed}}
By definition, the dual statement $\Sigma^*$ is formed by replacing the ordering $\preceq$ with its dual $\succeq$. By Dual of Dual Ordering, applying this operation twice results in the original sentence $\Sigma$ again. {{qed}} Category:Order Theory sgenggbsvmbmw7wypp9g6hrj4lesfpe
=== Double Negation Introduction === {{:Double Negation/Double Negation Introduction/Sequent Form/Formulation 1/Proof}} === Double Negation Elimination === {{:Double Negation/Double Negation Elimination/Sequent Form/Formulation 1/Proof}}
We apply the Method of Truth Tables to the proposition. As can be seen by inspection, appropriate truth values match for both boolean interpretations. $\begin{array}{|c||ccc|} \hline p & \neg & \neg & p \\ \hline F & F & T & F \\ T & T & F & T \\ \hline \end{array}$ Hence $p \dashv \vdash \neg \neg p$. {{qed}}
=== Proof of Forward Implication === {{:Rule of Transposition/Formulation 1/Forward Implication/Proof}} === Proof of Reverse Implication === {{:Rule of Transposition/Formulation 1/Reverse Implication/Proof}}
We apply the Method of Truth Tables to the proposition. As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations. $\begin{array}{|ccc||ccccc|} \hline p & \implies & q & \neg & q & \implies & \neg & p \\ \hline F & T & F & T & F & T & T & F \\ F & T & T & F & T & T & T & F \\ T & F & F & T & F & F & F & T \\ T & T & T & F & T & T & F & T \\ \hline \end{array}$ {{qed}}