hcju/proofwikiqa huggingface dataset:
{{BeginTableau|\left({p \dashv \vdash q}\right) \vdash p \iff q}} {{Premise|1|p \dashv \vdash q}} {{SequentIntro|2|1|\left ({p \vdash q}\right) \land \left ({q \vdash p}\right)|1   |Definition of Interderivable}} {{Assumption|3|p}} {{Simplification|4|1, 3|p \vdash q|2|1}} {{Implication|5|1|p \implies q|3|4}} {{Assumption|6|q}} {{Simplification|7|1, 6|q \vdash p|2|2}} {{Implication|8|1|q \implies p|6|7}} {{BiconditionalIntro|9|1|p \iff q|5|8}} {{EndTableau}} {{Qed}} Category:Equivalences are Interderivable s1qqzy938u5n0099qdwkh50zrgg4tkc
Let $\mathbb S \subseteq \powerset S$. Suppose that $\forall X \in \mathbb S: X \subseteq T$. Consider any $\displaystyle x \in \bigcup \mathbb S$. By definition of set union, it follows that: :$\exists X \in \mathbb S: x \in X$ But as $X \subseteq T$ it follows that $x \in T$. Thus it follows that: :$\displaystyle \bigcup \mathbb S \subseteq T$ So: :$\displaystyle \paren {\forall X \in \mathbb S: X \subseteq T} \implies \bigcup \mathbb S \subseteq T$ {{qed}} Category:Set Union Category:Subsets b4h728rptgi1xeujekujfg47szo7nag
{{BeginTableau|\vdash \paren {p \land \neg q} \implies \paren {\neg \paren {p \implies q} } }} {{Assumption|1|p \land \neg q}} {{SequentIntro|2|1|\neg \paren {p \implies q}|1|Conjunction with Negative Equivalent to Negation of Implication: Formulation 1}} {{Implication|3||\paren {p \land \neg q} \implies \paren {\neg \paren {p \implies q} }|1|2}} {{EndTableau}} {{qed}}
{{BeginTableau|\vdash \left({\neg \left({p \implies q}\right)}\right) \implies \left({p \land \neg q}\right)}} {{Assumption|1|\neg \left({p \implies q}\right)}} {{SequentIntro|2|1|p \land \neg q|1|Conjunction with Negative Equivalent to Negation of Implication: Formulation 1}} {{Implication|3||\left({\neg \left({p \implies q}\right)}\right) \implies \left({p \land \neg q}\right)|1|2}} {{EndTableau}} {{qed}} {{LEM|Conjunction with Negative Equivalent to Negation of Implication/Formulation 1/Reverse Implication}} Category:Conjunction with Negative Equivalent to Negation of Implication sbryqpjb0wps953pi6kun8dd03dtd07
{{BeginTableau|\vdash \paren {p \land \neg q} \iff \paren {\neg \paren {p \implies q} } }} {{Assumption|1|p \land \neg q}} {{SequentIntro|2|1|\neg \paren {p \implies q}|1|Conjunction with Negative Equivalent to Negation of Implication: Formulation 1}} {{Implication|3||\paren {p \land \neg q} \implies \paren {\neg \paren {p \implies q} }|1|2}} {{Assumption|4|\neg \paren {p \implies q} }} {{SequentIntro|5|4|p \land \neg q|4|Conjunction with Negative Equivalent to Negation of Implication: Formulation 1}} {{Implication|6||\paren {\neg \paren {p \implies q} } \implies \paren {p \land \neg q}|4|5}} {{BiconditionalIntro|7||\paren {p \land \neg q} \iff \paren {\neg \paren {p \implies q} }|3|6}} {{EndTableau|qed}}
Let $a \in A$. Note that there is at least one point in $A$, as $A$ is non-empty. If $x \in A$, then there is a line segment joining $a$ and $x$. By definition of star convex set, it follows that $A$ is star convex, and $a$ is a star center. {{qed}} Category:Vector Spaces 2fqyb2qiefdy7qshallf7xpvdwu1sxh
=== Forward implication === If $\mathcal F$ converges to $p$, then it contains ''every'' neighborhood of $p$, and the set of neighborhoods of $p$ is trivially a neighborhood sub-basis at $p$. {{qed|lemma}} === Reverse Implication === Let $S_p$ be a neighborhood sub-basis at $p$. Let $S_p \subseteq \mathcal F$. Let $N$ be a neighborhood of $p$. Then by the definition of neighborhood sub-basis, there is a finite $T_N \subseteq S_p$ such that: :$\bigcap T_N \subseteq N$. Since a filter is closed under finite intersections, $\bigcap T_N \in \mathcal F$. Then $\bigcap T_N \in \mathcal F$ and $\bigcap T_N \subseteq N$, so by the definition of a filter, $N \in \mathcal F$. Since $\mathcal F$ contains every neighborhood of $p$, it converges to $p$.  {{qed}} Category:Filter Theory Category:Topological Bases 50pbzj0ow5lfvxhhnf75aln5p0i6353
{{BeginTableau|\vdash \paren {\paren {p \implies r} \land \paren {q \implies r} } \iff \paren {\paren {p \lor q} \implies r} }} {{TheoremIntro|1|\paren {\paren {p \implies r} \land \paren {q \implies r} } \implies \paren {\paren {p \lor q} \implies r}|Proof by Cases: Forward Implication}} {{TheoremIntro|2|\paren {\paren {p \lor q} \implies r} \implies \paren {\paren {p \implies r} \land \paren {q \implies r} }|Proof by Cases: Reverse Implication}} {{BiconditionalIntro|3||\paren {\paren {p \implies r} \land \paren {q \implies r} } \iff \paren {\paren {p \lor q} \implies r}|1|2}} {{EndTableau}} {{qed}}
{{BeginTableau|\paren {p \implies q} \iff \paren {\neg p \lor q} }} {{Assumption|1|p \implies q}} {{SequentIntro|2|1|\neg p \lor q|1|Rule of Material Implication: Formulation 1}} {{Implication|3||\paren {p \implies q} \implies \paren {\neg p \lor q}|1|2}} {{Assumption|4|\neg p \lor q}} {{SequentIntro|5|4|p \implies q|4|Rule of Material Implication: Formulation 1}} {{Implication|6||\paren {\neg p \lor q} \implies \paren {p \implies q}|4|5}} {{BiconditionalIntro|7||\paren {p \implies q} \iff \paren {\neg p \lor q}|3|6}} {{EndTableau|qed}} {{LEM|Rule of Material Implication/Formulation 1/Forward Implication|3}}
{{BeginTableau|\left({p \implies q}\right) \implies \left({\neg p \lor q}\right)}} {{Assumption|1|p \implies q}} {{SequentIntro|2|1|\neg p \lor q|1|Rule of Material Implication: Formulation 1}} {{Implication|3||\left({p \implies q}\right) \implies \left({\neg p \lor q}\right)|1|2}} {{EndTableau}} {{qed}} {{LEM|Rule of Material Implication/Formulation 1/Forward Implication|3}}
{{BeginTableau|\left({\neg p \lor q}\right) \implies \left({p \implies q}\right)}} {{Assumption|1|\neg p \lor q}} {{SequentIntro|2|1|p \implies q|1|Rule of Material Implication: Formulation 1}} {{Implication|3||\left({\neg p \lor q}\right) \implies \left({p \implies q}\right)|1|2}} {{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. $\begin{array}{|ccccc||ccccccc|} \hline p & \implies & (q & \land & r) & (p & \implies & q) & \land & (p & \implies & r) \\ \hline F & T & F & F & F & F & T & F & T & F & T & F \\ F & T & F & F & T & F & T & F & T & F & T & T \\ F & T & T & F & F & F & T & T & T & F & T & F \\ F & T & T & T & T & F & T & T & T & F & T & T \\ T & F & F & F & F & T & F & F & F & T & F & F \\ T & F & F & F & T & T & F & F & F & T & T & T \\ T & F & T & F & F & T & T & T & F & T & F & F \\ T & T & T & T & T & T & T & T & T & T & T & T \\ \hline \end{array}$ {{qed}}
We apply the Method of Truth Tables. As can be seen by inspection, the truth values under the main connective is true for all boolean interpretations. $\begin{array}{|ccccc|c|ccccccc|} \hline (p & \implies & (q & \land & r)) & \iff & ((p & \implies & q) & \land & (p & \implies & r)) \\ \hline F & T & F & F & F & T & F & T & F & T & F & T & F \\ F & T & F & F & T & T & F & T & F & T & F & T & T \\ F & T & T & F & F & T & F & T & T & T & F & T & F \\ F & T & T & T & T & T & F & T & T & T & F & T & T \\ T & F & F & F & F & T & T & F & F & F & T & F & F \\ T & F & F & F & T & T & T & F & F & F & T & T & T \\ T & F & T & F & F & T & T & T & T & F & T & F & F \\ T & T & T & T & T & T & T & T & T & T & T & T & T \\ \hline \end{array}$ {{qed}}
=== Forward Implication === Let $S$ be a non-empty subset of $Y$. By Compact Subspace of Linearly Ordered Space: Lemma, $S$ has a supremum $k$ in $Y$. {{explain|supremum WRT $\preceq \restriction_Y$}} {{AimForCont}} that $S$ has an upper bound $b$ in $X$ such that $b \prec k$. Let: :$\AA = \set {s^\preceq: s \in S} \cup \set {b^\succeq}$ where: :$s^\preceq$ denotes the lower closure of $s$ in $S$ :$b^\succeq$ denotes the upper closure of $s$ in $S$. Then $\AA$ is an open cover of $Y$. But $\AA$ has no finite subcover, contradicting the fact that $Y$ is compact. {{finish|above lines need proof}} {{explain|Pronounce proper incantations about duality.}} A similar argument proves the corresponding statement for infima, so $(2)$ holds. {{qed|lemma}} === Reverse Implication === {{:Compact Subspace of Linearly Ordered Space/Reverse Implication}}
Let $x, y \in X$ with $x \ne y$. Since $\le$ is a total ordering, either $x \prec y$ or $y \prec x$. {{WLOG}}, assume that $x \prec y$. If there is a $z \in X$ such that $x \prec z \prec y$, then ${\downarrow}z$ and ${\uparrow}z$ separate $x$ and $y$. {{explain|We are replacing the ${\downarrow}z$ and ${\uparrow}z$ notation with notation as defined in Definition:Upper Closure and Definition:Lower Closure -- make sure the correct one is used here}} Otherwise, by Mind the Gap, ${\downarrow}y$ and $\uparrow x$ separate $x$ and $y$. Since any two distinct points can be separated by neighborhoods, $\left({X, \tau}\right)$ is a Hausdorff space. {{qed}} Category:Order Topology e5qhmlhfjbqqdy2rxg2mmxjakz97rs6
First it is demonstrated that $\left({Y, \preceq}\right)$ is a complete lattice. Let $\phi: Y \to \left[{0 \,.\,.\, 1}\right]$ be defined as: :$\phi \left({y}\right) = \begin{cases} y & : y \in \left[{0 \,.\,.\, 1}\right) \\ 1 & : y = 4 \end{cases}$ Then $\phi$ is a order isomorphism. {{explain|The above needs to be proved.}} {{qed|lemma}} We have that $\left[{0 \,.\,.\, 1}\right]$ is a complete lattice. {{explain|This is probably around here somewhere.}} Next is is shown that $\left({Y, \preceq}\right)$ is closed in $X$. Let $x \in X \setminus Y$. Then: : $x \in \left({2 \,.\,.\, 3}\right)$ Thus: :$x \in \left({\dfrac {x+2} 2 \,.\,.\, \dfrac {x+3} 2}\right) \in \tau'$ Since the complement of $Y$ is open, $Y$ is closed. Finally it is shown that $\left({Y, \tau'}\right)$ is not compact. Let: : $\mathcal A = \left\{ {x^\preceq: x \in \left[{{0}\,.\,.\,{1}}\right) }\right\} \cup \left\{ {\left({\dfrac 5 2}\right)^\succeq}\right\}$ where: : $x^\preceq$ denotes the lower closure of $x$ : $x^\succeq$ denotes the upper closure of $x$ Then $\mathcal A$ is an open cover of $Y$ with no finite subcover. {{explain|Prove the above statement.}} {{qed}} {{finish|The remaining parts of this proof need to be completed.}} Category:Order Topology i5nrywsoteeccup06eje27v86t8vy83
{{BeginTableau|\vdash \neg p \implies \left({p \implies q}\right)}} {{Assumption|1|\neg p}} {{SequentIntro|2|1|p \implies q|1|False Statement implies Every Statement: Formulation 1}} {{Implication|3||\neg p \implies \left({p \implies q}\right)|1|2}} {{EndTableau}} {{qed}}
=== Sufficient Condition === Let $Y$ be a compact subspace of $T$. From: : Order Topology is Hausdorff : Compact Subspace of Hausdorff Space is Closed it follows that $Y$ is closed in $T$. From Compact Subspace of Linearly Ordered Space: Lemma, $\left({Y, \preceq {\restriction_Y} }\right)$ is a complete lattice. {{finish|Not much to do, just need to show that a complete lattice in this context is bounded.}} Hence $Y$ is closed and bounded in $T$. === Necessary Condition === Let $Y$ be a closed and bounded subspace of $T$. Let $S$ be a non-empty subset of $Y$. Since $Y$ is bounded and $S \subseteq Y$, $S$ is bounded. Since $X$ is Dedekind complete, $S$ has a supremum and infimum in $X$. We will show that $\sup S, \inf S \in Y$. Suppose for the sake of contradiction that $\sup S \notin Y$. By Closed Set in Linearly Ordered Space, $b$ is not a supremum of $S$, a contradiction. Thus $\sup S \in Y$. A similar argument shows that $\inf S \in Y$. Thus by Compact Subspace of Linearly Ordered Space, $Y$ is compact. {{qed}} Category:Linearly Ordered Spaces Category:Dedekind Complete Sets Category:Heine-Borel Theorem 2ezmkf6n61wq6ysih05rnyoss72km89
The forward implication follows from Heine-Borel Theorem: Dedekind-Complete Space. Suppose that $X$ is not Dedekind complete. Then $X$ has a non-empty subset $S$ with an upper bound $b$ in $X$ but no supremum in $X$. Let $a \in S$ and let $Y = {\bar \downarrow} S \cap {\bar \uparrow} a$. $Y$ is nonempty and bounded below by $a$ and above by $b$. === $Y$ is closed in $X$ === Let $x \in X \setminus Y$. Then $x \prec a$ or $x$ strictly succeeds every element of $S$. If $x \prec a$, then $x \in {\downarrow} a \subseteq X \setminus Y$. If $x$ strictly succeeds each element of $S$, then it is an upper bound of $S$. Since $S$ has no supremum in $X$, it has an upper bound $p \prec x$. Then $x \in {\uparrow p} \subseteq X \setminus Y$. {{explain|What does ${\downarrow} a$ and ${\uparrow p}$ mean?}} ===  $Y$ is not compact === Let $\AA = \set { {\downarrow} s: s \in  S}$. Then $\AA$ is an open cover of $Y$ with no finite subcover. {{qed}} Category:Order Topology Category:Compact Spaces 2ze14h1qxh81mb7rrsz45wixi8rqibh
=== Lemma === {{:Jordan Polygon Theorem/Lemma 1}}{{qed|lemma}} We show that $\R^2 \setminus \partial P$ is not path-connected. Find any $q_1 \in R^2 \setminus \partial P$ and $\theta \in \R$ such that the ray $\LL_\theta = \set {q_1 + s \map g \theta: s \in \R_{\ge 0} }$ has exactly one crossing of $\partial P$. Find any $q_2 \in \LL_\theta$ that lies on the ray after the crossing, so the ray $\set {q_2 + s \map g \theta: s \in \R_{\ge 0} }$ does not intersect $\partial P$. Then $\map {\mathrm {par} } {q_1} = 1 \ne 0 = \map {\mathrm {par} } {q_2}$. From Jordan Polygon Parity Lemma, it follows that $q_1$ and $q_2$ cannot be connected by a path. {{qed|lemma}} As $R^2 \setminus \partial P$ is not path-connected, it follows from the Lemma that $R^2 \setminus \partial P$ is a union of exactly two disjoint path-connected sets, which we denote as $U_1$ and $U_2$. Let $q \in \R^2 \setminus \partial P$, and let $\map d {q, \partial P}$ be the Euclidean distance between $q$ and $\partial P$. From Distance between Closed Sets in Euclidean Space, it follows that $\map d {q, \partial P} > 0$. When we put $\epsilon = \map d {q, \partial P} / 2$, we have $\map {B_\epsilon} q \subseteq \R^2 \setminus \partial P$. As Open Ball is Convex Set, it follows that $\map {B_\epsilon} q$ is path-connected, so $\map {B_\epsilon} q$ is a subset of either $U_1$ or $U_2$. Then, both $U_1$ and $U_2$ are open. From Path-Connected Space is Connected, it follows that $U_1$ and $U_2$ are connected. Then, $\R^2 \setminus \partial P$ is a union of two components. {{qed|lemma}} From Boundary of Polygon is Jordan Curve, it follows that $\partial P$ is equal to the image of a Jordan curve $\gamma: \closedint 0 1 \to \R^2$. From Continuous Image of Compact Space is Compact/Corollary 2, it follows that $\partial P$ is bounded. That is, there exist $a \in \R^2$ and $R \in \R_{>0}$ such that $\partial P \subseteq \map {B_R} a$. If $x_1, x_2 \in \R^2 \setminus \map {B_R} a$, $x_1$ to $x_2$ can be joined by a path in $\R^2 \setminus \map {B_R} a$ following: :the circumference of the two circles with center $a$ and radii $\map d {a, x_1}$ and $\map d {a, x_2}$ :a line segment joining the two circumferences. Then $\R^2 \setminus \map {B_R} a$ is path-connected, so $\R^2 \setminus \map {B_R} a$ is a subset of one of the components of $\R^2 \setminus \partial P$, say $U_1$. As $\R^2 \setminus \map {B_R} a \subseteq U_1$, it follows that $U_1$ is unbounded, so $U_1$ is the exterior of $\gamma$. Then $U_2 \subseteq \map {B_R} a$, so $U_2$ is bounded, which implies that $U_2$ is the interior of $\gamma$. {{qed}} {{namedfor|Marie Ennemond Camille Jordan|cat = Jordan}}
From Boundary of Polygon is Jordan Curve, it follows that $\partial P$ is equal to the image of a Jordan curve. From the Jordan Polygon Theorem, it follows that $\operatorname{Int} \left({P}\right)$ and $\operatorname{Ext} \left({P}\right)$ are path-connected. Then, Jordan Polygon Parity Lemma shows that $N \left({q}\right) = \operatorname{par} \left({q}\right)$, where $\operatorname{par} \left({q}\right)$ denotes the parity of $q$. From Jordan Polygon Theorem, it follows that $\operatorname{Ext} \left({P}\right)$ is unbounded, while $\operatorname{Int} \left({P}\right)$ is bounded. As $\partial P$ is the image of a Jordan curve, it follows from Continuous Image of Compact Space is Compact/Corollary 2 that $\partial P$ is also bounded. Then, there exists $R \in \R_{>0}$ such that $\operatorname{Int} \left({P}\right) \cup \partial P \subseteq B_R \left({\mathbf 0}\right)$. If $q_0 \in \R^2 \setminus B_R \left({\mathbf 0}\right)$, then $q_0 \in \operatorname{Ext} \left({P}\right)$. Then, the ray $\left\{ {q_0 + s q_0: s \in \R_{\ge 0} }\right\} \subseteq \R^2 \setminus B_R \left({\mathbf 0}\right)$, so there are zero crossings between the ray and $\partial P$. From Jordan Polygon Parity Lemma, it follows that $\operatorname{par} \left({q_0}\right) = 0$. As $\operatorname{Ext} \left({P}\right)$ is path-connected, it follows from the lemma that for all $q \in \operatorname{Ext} \left({P}\right)$, we have $\operatorname{par} \left({q}\right) = 0$. If $q_1 \in \R^2 \setminus \partial P$ with $\operatorname{par} \left({q}\right) = 1$, it follows that $q_1 \notin \operatorname{Ext} \left({P}\right)$. As $\R^2 \setminus \partial P = \operatorname{Int} \left({P}\right) \cup \operatorname{Ext} \left({P}\right)$, it follows that $q_1 \in \operatorname{Int} \left({P}\right)$. Again, Jordan Polygon Parity Lemma shows that for all $q \in \operatorname{Int} \left({P}\right)$, we have $\operatorname{par} \left({q}\right) = 1$. So if instead $q_0 \in \R^2 \setminus \partial P$ with $\operatorname{par} \left({q}\right) = 0$, the only possibility is that $q_0 \in \operatorname{Ext} \left({P}\right)$. {{qed}}
=== Necessary Condition === Suppose that $C$ be closed. Let $S$ be a non-empty subset of $C$. Let $b \in X \setminus C$. We will show that $b$ is not a supremum of $S$. If $b$ is not an upper bound of $S$, then by definition $b$ cannot be a supremum of $S$. Suppose, then, that $b$ is an upper bound of $S$. Since $C$ is closed and $b \notin C$, there must be an open interval or open ray $U$ containing $b$ that is disjoint from $Y$. Since $b$ is an upper bound for $S$ and $S$ is not empty, $U$ cannot be a downward-pointing ray. Thus $U$ is either an open interval $\openint a q$ or an upward-pointing open ray ${\uparrow}a$. Then $a \prec b$. Since $b \in U$, and $b$ is an upper bound of $S$, no element strictly succeeding all elements of $U$ can be in $S$. By the above and the fact that $S \subseteq C$, ${\uparrow}a \cap S = \O$, so $a$ is an upper bound of $S$. Since $a \prec b$, $b$ is not a supremum of $S$. A similar argument shows that $b$ is not an infimum. {{qed|lemma}} === Sufficient Condition === Suppose that no nonempty subset of $C$ has a supremum or infimum relative to $X$ that lies in $X \setminus C$. Let $p \in X \setminus C$. ==== Case 1: $p$ is an upper bound of $C$ ==== Since $C$ is a non-empty subset of $C$, it does not have a supremum in $X \setminus C$. Thus $p$ is not a supremum of $C$. Therefore, $C$ has an upper bound $a \in X$ such that $a \prec p$. Thus ${\uparrow_X}a$ contains $p$ and is disjoint from $C$, so $p$ is not an adherent point of $C$. ==== Case 2: $p$ is a lower bound of $C$ ==== The same approach used for case 1 proves $p$ is not an adherent point of $C$. ==== Case 3: $p$ is neither an upper nor a lower bound of $C$ ==== In this case, $C \cap {\downarrow_X} p$ and $C \cap {\uparrow_X} p$ are nonempty, and their union is $C$. Thus $p$ is not a supremum of $C \cap {\downarrow_X} p$ and is not an infimum of $C \cap {\uparrow_X}p$. Thus there are $a, b \in X$ such that $a \prec p \prec b$, $a$ is an upper bound of $C \cap {\downarrow_X}p$, and $b$ is a lower bound of $C \cap {\downarrow_X} p$. Then $\openint a b$ contains $p$ and is disjoint from $C$, so $p$ is not an adherent point of $C$. Since $C$ contains all of its adherent points, it is closed. {{qed}}
{{BeginTableau|p \implies \paren {q \lor r} \vdash \paren {p \implies q} \lor  \paren {p \implies r} }} {{Assumption|1|p \implies \paren {q \lor r} }} {{Assumption|2|p}} {{ModusPonens|3|1, 2|q \lor r|1|2}} {{IdentityLaw|4|2|p|2}} {{Assumption|5|q}} {{Implication|6|5|p \implies q|4|5}} {{Addition|7|5|\paren {p \implies q} \lor \paren {p \implies r}|6|1}} {{Assumption|8|r}} {{SequentIntro|9|8|p \implies r|8|True Statement is implied by Every Statement}} {{Addition|10|8|\paren {p \implies q} \lor \paren {p \implies r}|9|2}} {{ProofByCases|11|1|\paren {p \implies q} \lor \paren {p \implies r}|3|2|7|8|10}} {{EndTableau}} {{qed}} Category:Implication is Left Distributive over Disjunction iuw2ti0tdligmxfmhge8tid9aanq9iv
{{BeginTableau|\vdash \left({p \implies q}\right) \implies \left({\left({r \lor p}\right) \implies \left ({r \lor q}\right)}\right)}} {{Assumption|1|p \implies q}} {{SequentIntro|2|1|\left({\left({r \lor p}\right) \implies \left ({r \lor q}\right)}\right)|1   |Factor Principles: Disjunction on Left: Formulation 1}} {{Implication|3|1|\left({p \implies q}\right) \implies \left({\left({r \lor p}\right) \implies \left ({r \lor q}\right)}\right)|1|2}} {{EndTableau}} {{qed}}
{{BeginTableau|\vdash \left({p \implies q}\right) \implies \left({\left({p \lor r}\right) \implies \left ({q \lor r}\right)}\right)}} {{Assumption|1|p \implies q}} {{SequentIntro|2|1|\left({p \lor r}\right) \implies \left ({q \lor r}\right)|1   |Factor Principles: Disjunction on Right: Formulation 1}} {{Implication|3|1|\left({p \implies q}\right) \implies \left({\left({p \lor r}\right) \implies \left ({q \lor r}\right)}\right)|1|2}} {{EndTableau}} {{qed}}
{{BeginTableau|\vdash \paren {p \implies q} \lor \paren {q \implies r} }} {{ExcludedMiddle|1|q \lor \neg q}} {{Assumption|2|q}} {{SequentIntro|3|2|p \implies q|2|True Statement is implied by Every Statement}} {{Addition|4|2|\paren {p \implies q} \lor \paren {q \implies r}|3|1}} {{Assumption|5|\neg q}} {{SequentIntro|6|5|q \implies r|5|False Statement implies Every Statement}} {{Addition|7|5|\paren {p \implies q} \lor \paren {q \implies r}|6|2}} {{ProofByCases|8||\paren {p \implies q} \lor \paren {q \implies r}|1|2|4|5|7}} {{EndTableau|qed}} {{LEM}}
We apply the Method of Truth Tables to the proposition $\vdash p \lor \neg p$. As can be seen by inspection, the truth value of the main connective, that is $\lor$, is $T$ for each boolean interpretation for $p$. $\begin{array}{|cccc|} \hline p & \lor & \neg & p \\ \hline F & T & T & F \\ T & T & F & T \\ \hline \end{array}$ {{qed}}
{{BeginTableau|\paren {p \land q} \implies r \vdash \paren {p \implies r} \lor \paren {q \implies r} }} {{Premise        |  1 | \paren {p \land q} \implies r}} {{SequentIntro   |  2 | 1 | \neg \paren {p \lor q} \lor r | 1 | Rule of Material Implication}} {{SequentIntro   |  3 | 1 | \neg p \lor \neg q \lor r | 2 | De Morgan's Laws: Disjunction of Negations}} {{Addition       |  4 | 1 | r \lor \neg p \lor \neg q \lor r | 3 | 2}} {{Commutation    |  5 | 1 | \neg p \lor r \lor \neg q \lor r | 4 | Disjunction}} {{SequentIntro   |  6 | 1 | \paren {p \implies r} \lor \neg q \lor r | 5 | Rule of Material Implication}} {{SequentIntro   |  7 | 1 | \paren {p \implies r} \lor \paren {q \implies r} | 6 | Rule of Material Implication}} {{EndTableau}} {{qed}} Category:Principle of Composition lrbsgl53wcyrryvmzb11mbkzydx5lg1
{{BeginTableau|\left({\left({p \implies r}\right) \lor  \left({q \implies r}\right)}\right) \iff \left({\left({p \land q}\right) \implies r}\right)}} {{Assumption|1|\left({p \implies r}\right) \lor  \left({q \implies r}\right)}} {{SequentIntro|2|1|\left({p \land q}\right) \implies r|1|Principle of Composition/Formulation 1}} {{Implication|3||\left({\left({p \implies r}\right) \lor  \left({q \implies r}\right)}\right) \implies \left({\left({p \land q}\right) \implies r}\right)|1|2}} {{Assumption|4|\left({p \implies r}\right) \lor  \left({q \implies r}\right)}} {{SequentIntro|5|4|\left({p \land q}\right) \implies r|4|Principle of Composition/Formulation 1}} {{Implication|6||\left({\left({p \land q}\right) \implies r}\right) \implies \left({\left({p \implies r}\right) \lor  \left({q \implies r}\right)}\right)|4|5}} {{BiconditionalIntro|7||\left({\left({p \implies r}\right) \lor  \left({q \implies r}\right)}\right) \iff \left({\left({p \land q}\right) \implies r}\right)|3|6}} {{EndTableau}} {{qed}}
From the definition of topological group, $\phi$ is continuous. By Inversion Mapping is Involution, $\phi$ is an involution. By Continuous Involution is Homeomorphism, $\phi$ is a homeomorphism. {{qed}} Category:Inversion Mappings Category:Homeomorphisms Category:Topological Groups rjcmxcmvc5l269f81xlhd2baxldobxe