hcju/proofwikiqa huggingface dataset:
Suppose $T$ is an infinite particular point space. As $S$ is infinite, $\CC$ is also infinite. Let $x \in S, x \ne p$. Then any neighborhood of $x$ must contain $p$, by the nature of the particular point topology. But $p$ is contained in all elements of $\CC$. That is: :$\forall C \in \CC: p \in C$ So any neighborhood of $x$ intersects with all elements of the open cover $\CC$. As $\CC$ is infinite, it therefore has no open refinement which is locally finite. {{qed}}
=== Necessary Condition === When $U$ is path-connected, any $p \in U$ satisfies the assertion. Because $U$ is non-empty, such a $p$ exists. {{qed|lemma}} === Sufficient Condition === Suppose $p$ is such a point. For $q \in U$, denote $p \sim q$ when $p$ and $q$ are path-connected. To show that $U$ is path-connected, we have to show: :$\forall q, r \in U: q \sim r$ We already know that $p \sim q$ and $p \sim r$. From Path-Connectedness is Equivalence Relation, we find that hence $q \sim p$ by symmetry of $\sim$. Subsequently, we conclude $q \sim r$ from transitivity of $\sim$. Hence the result. {{qed}}
Let $C \subseteq S$ be a countable set. Since $S$ is uncountable, by Uncountable Set less Countable Set is Uncountable, so is $\relcomp S C$. Thus there exists some point $x \in \relcomp S C$ and $x \ne p$. By Clopen Points in Fort Space, $\set x \in \tau_p$.  By Empty Intersection iff Subset of Complement, we have $C \cap \set x = \O$. Therefore $C$ is not everywhere dense. Since $C$ is arbitrary, $T$ is not a separable space. {{qed}}
From Subset of Countably Infinite Set is Countable, $S \setminus \set p$ is countable. Let $f: \N \to S$ be an enumeration of $S \setminus \set p$. For brevity, let us write $s_n$ for $\map f n$. Now define, for $n \in \N$, $S_n = S \setminus \set {s_0, \ldots, s_{n - 1} }$. Note that $s_n \ne p$ for all $n \in \N$, so that $p \in S_n$ for all $n$. By Relative Complement of Relative Complement, $S \setminus S_n = \set {s_0, \ldots, s_{n - 1} }$ and so $S_n \in \tau_p$. Also, define $S'_n = \set {s_n}$. Since $s_n \ne p$, $S'_n$ is open in $\tau_p$ as well. From Finite Union of Countable Sets is Countable, $\BB := \set {S_n : n \in \N} \cup \set {S'_n : n \in \N}$ is countable. Let us verify that $\BB$ forms a basis for $\tau_p$. So let $U \subseteq S$ be open in $\tau_p$. If $p \notin U$ then $U = \displaystyle \bigcup \set {S'_n: s_n \in U}$ because the $s_n$ form an enumeration of $S \setminus \set p$. Now if $p \in U$, then by definition of the Fort topology, $S \setminus U$ must be finite. It follows that $A = \set {n \in \N: s_n \in S \setminus U}$ is also finite. From Subset of Naturals is Finite iff Bounded, there exists an $N \in \N$ such that $N - 1 > n$ for all $n \in A$. Therefore, $A \subseteq \set {0, \ldots, N - 1}$ and hence: :$S \setminus U \subseteq \set {s_0, \ldots, s_{N - 1} }$ From Set Complement inverts Subsets, also: :$S_N = S \setminus \set {s_0, \ldots, s_{N - 1} } \subseteq U$ It follows that $\BB$ is a basis for $\tau_p$. Since $\BB$ is countable, we conclude $T$ is second-countable. {{qed}}
We have that the Fort Space is Completely Normal. The result follows from tracing the relevant implications on Sequence of Implications of Separation Axioms. {{qed}}
We have: :Fort Space is Regular :Countable Fort Space is Second-Countable. The result follows from Urysohn's Metrization Theorem. {{qed}}
From the definition, $T$ is a compact space if every open cover of $T$ has a finite subcover. Let $N \subset S$ be a countably infinite set such that $p \notin N$. Then $\relcomp S N$ is open in $T$ by the definition of the Fortissimo space. Let $q \in S$ such that $q \ne p$. As $p \in \relcomp S {\set q}$ it follows that $\set q$ is open in $T$. Then: :$\displaystyle S = \relcomp S N \cup \paren {\bigcup_{q \mathop \in N} \set q}$ Let $\CC$ be defined as: :$\CC := \set {\set q: q \in N} \cup \set {\relcomp S N}$ Then from the above, $\CC$ is an open cover of $S$. But all the sets of $\CC$ are pairwise disjoint. Therefore $\CC$ can have no finite subcover. Hence the space cannot be compact. {{qed}}
Let $T' = \struct {S \setminus \set p, \tau_p}$ be the subspace of $T$ induced by $S \setminus \set p$. From Discrete Subspace of Fortissimo Space, $T'$ is a discrete topological space. Let $H = \sequence {x_n}_{n \mathop \in \N}$ be a sequence of distinct terms of $S \setminus \set p$. That is: :$\forall n \in \N: x_n \ne p$ and that: :$n \ne m \implies x_n \ne x_m$ By definition, $H$ is a sequence of distinct terms of $S \setminus \set p$. From Convergence of Sequence in Discrete Space: Corollary, $H$ does not converge in $T'$. $H$ is by definition countable, so from Closed Sets of Fortissimo Space it follows that $H$ is closed. Thus by definition $S \setminus H$ is open. Also, as $p \notin H$ by definition of $H$, it follows that $P \in S \setminus H$. By definition, $S \setminus H$ is an open neighborhood of $p$. Suppose $H$ converges to the limit $p$ Then, by definition of convergence, for any open set $U \in \tau_p$ such that $p \in U$, there exists some $N \in \R$ such that $n > N \implies x_n \in U$. But $H \cap \paren {S \setminus H} = \O$, and so $H$ cannot converge to $p \in S \setminus H$. The result follows from the fact that every subsequence of $H$ is also a sequence of distinct terms of $S \setminus \set p$. {{qed}}
It suffices to show that $T$ has an infinite subset without limit points. Consider the set $S \setminus \set p$. Let $x \in S$. We have: {{begin-eqn}} {{eqn | l = \paren {S \setminus \paren {S \setminus \set p} } \cup \set x       | r = \set p \cup \set x       | c =  }} {{eqn | r = \set {p, x}       | c =  }} {{end-eqn}} By definition, $x$ is a limit point of $S \setminus \set p$ {{iff}} $\set {p, x}$ is not a neighborhood of $x$. By definition of Fortissimo space, $\set {p, x}$ is open in $T$. Hence it is a open neighborhood of $x$. Therefore $x$ is not a limit point of $S \setminus \set p$. Since $x$ is arbitrary, $S \setminus \set p$ has no limit points. Hence $T$ is not weakly countably compact. {{qed}}
Let $N_p$ be a neighborhood of $p$ such that $\relcomp S {N_p}$ is countable. Let $\psi: \relcomp S {N_p} \to \Z_{\ne 0}$ be a bijection between $\relcomp S {N_p}$ and the non-zero integers $\Z_{\ne 0}$. Let $\phi: S \to \set {0, 1}$ be the mapping defined as: :$\forall x \in S: \map \phi x = \begin {cases} 0 & : x \in N_p \\ \map \psi x & : x \notin N_p \end {cases}$ Then $\phi$ is a continuous mapping. However, $\phi$ is not bounded. Hence the result by definition of pseudocompact space. {{qed}}
Let $D = \set {0, 1}$. Let $\tuple {p, 0}$ belong to some infinite $A \subseteq S$. Then its twin $\tuple {p, 1}$ is a limit point of $A$. Hence the result by definition of weakly countably compact. {{qed}}
Let $D = \set {0, 1}$. Let $\CC$ be an open cover of $T \times D$. Then $\exists A \times B \in \CC$ such that $\tuple {p, 0} \in A \times B$. We must have $\relcomp S A$ is countable and $B = D$. Hence $\relcomp S A \times D$, a product of countable sets, must be countable. So $A \times D$, together with an open neighborhood of each of the elements of $\relcomp S A \times D$, is a countable subcover of $\CC$. Hence the result by definition of Lindelöf space. {{qed}}
{{proof wanted|Use the same argument as Fortissimo Space is not Sigma-Compact}}
{{proof wanted|Use the same argument as Fortissimo Space is not Pseudocompact}}
From Meager Sets in Arens-Fort Space, we have that $A \subseteq S$ is meager in $T$ {{iff}} $A = \set {\tuple {0, 0} }$. So as $\set {\tuple{0, 0} } \ne S \subseteq S$, it follows that $T$ is non-meager. {{qed}}
Let $\map {S_m} V$ denote the set: :$\map {S_m} V := \set {n: \tuple {m, n} \notin V}$ where $V \subseteq \Z_{\ge 0} \times \Z_{\ge 0}$ (the same set $S_m$ used in the definition of the Arens-Fort space). Let $U = \set {\tuple {n, m}: \exists k: m = 2 k} \setminus \set {\tuple {0, 0} }$. From the definition of the Arens-Fort space, $U$ is open in $T$ since $\left({0, 0}\right) \notin U$. We have that: :$\tuple {0, 0} \in \relcomp S U$ and: :$\map {S_m} U$ is infinite for any $m \in \N$. Thus by definition of the Arens-Fort space, $\relcomp S U$ is not open in $T$. So $U$ is not closed in $T$. Let $U^-$ denote the closure of $U$. We have that: :$\tuple {0, 0} \notin \relcomp S {U \cup \set {\tuple {0, 0} } }$ Therefore by definition of the Arens-Fort space, $\relcomp S {U \cup \set {\tuple {0, 0} } }$ is open in $T$. Therefore $U \cup \set {\tuple {0, 0} }$ is closed in $T$. From Set Closure is Smallest Closed Set in Topological Space, it follows that $U^- = U \cup \set {\tuple {0, 0} }$. We have that: :$\tuple {0, 0} \in U \cup \set {\tuple {0, 0} }$ and: :$\map {S_m} {U \cup \set {\tuple {0, 0} } }$ is infinite for any $m \in \N$. Therefore by definition of the Arens-Fort space, $U \cup \set {\tuple {0, 0} }$ is not open. Thus we have created an open set $U$ whose closure $U^-$ is not itself open. Thus by definition the  Arens-Fort space is not extremally disconnected. {{qed}}
We have: :Modified Fort Space is Scattered :Modified Fort Space is $T_1$ We also have: :Scattered $T_1$ Space is Totally Disconnected Hence the result. {{qed}}
We have: :Modified Fort Space is Totally Disconnected :Totally Disconnected and Locally Connected Space is Discrete We also have: :Modified Fort Space is not $T_2$ :Discrete Space satisfies all Separation Properties Hence modified Fort space is not the discrete space, and the result follows. {{qed}}
By Isolated Points in Subsets of Modified Fort Space: :$\exists x \in H: x$ is isolated By Point in Topological Space is Open iff Isolated, $\set x$ is open in $T$. By Modified Fort Space is $T_1$ and definition of $T_1$ space, $\set x$ is closed in $T$. Therefore $\relcomp S {\set x}$ is open in $T$. Then we have: {{begin-eqn}} {{eqn | l = H       | o = \subseteq       | r = S }} {{eqn | r = \set x \cup \relcomp S {\set x}       | c = Union with Relative Complement }} {{eqn | l = H \cap \set x \cap \relcomp S {\set x}       | o = \subseteq       | r = \set x \cap \relcomp S {\set x}       | c = Intersection is Subset }} {{eqn | r = \O       | c = Intersection with Relative Complement is Empty }} {{eqn | ll = \leadsto       | l = H \cap \set x \cap \relcomp S {\set x}       | r = \O       | c = Subset of Empty Set }} {{eqn | l = H \cap \set x       | r = \set x }} {{eqn | o = \ne       | r = \O }} {{eqn | l = H \cap \relcomp S {\set x}       | r = H \setminus \set x       | c = Set Difference as Intersection with Relative Complement }} {{eqn | o = \ne       | r = \O       | c = $H$ has more than one point }} {{end-eqn}} This shows that $H$ is disconnected. {{qed}}
Let $A$ be both closed and open in $T$. Let $S = A \mid B$ be a separation of $S$ into the open sets $A$ and $B$. Then by definition $B$ is also both closed and open in $T$. Let $a \in A$. Suppose $b \in B$. As $A$ is open it must contain all but a finite number of points of $S$. Thus $A$ must itself be infinite. But then $B$ can not be open, as $b \in B$ and there is more than a finite number of points in $A$. So if $a \in A$ then $b \in A$ as well. {{qed}}
We have: :Totally Separated Space is Completely Hausdorff and Urysohn :Completely Hausdorff Space is Hausdorff Space But we have: :Modified Fort Space is not Hausdorff The result follows from Modus Tollendo Tollens. {{qed}}
We have Modified Fort Space is not $T_3$. We also have Zero Dimensional Space is $T_3$. The result follows by Modus Tollendo Tollens. {{qed}}
We have that a modified Fort space is $T_1$. We also have that a dense-in-itself subset of a $T_1$ space is infinite. But from Isolated Points in Subsets of Modified Fort Space, we have that any subset of $T$ with more than two points has at least one isolated point. So any dense-in-itself subset of $T$ must have an isolated point. Hence the result, by definition of scattered space. {{qed}}
Let $H \subseteq S$ contain more than two points. From {{Defof|Modified Fort Space}}, we can write $S = N \cup \set a \cup \set b$, where $N$ is an infinite set. Suppose $H \cap N \ne \O$. Let $x \in H \cap N$. Then: :$\set x \subset N$ Therefore: :$\set x \in \tau_{a, b}$ This shows that $x$ is isolated. Suppose $H \cap N = \O$. By Empty Intersection iff Subset of Complement: :$H \subseteq \relcomp S N = \set {a, b}$ Because $H$ contains at least $2$ elements: :$H = \set {a, b}$ Because $a \in \relcomp S {\set b}$ and $\relcomp S {\relcomp S {\set b} } = \set b$ is a finite set: :$\relcomp S {\set b} \in \tau_{a, b}$ Thus: :$H \cap \relcomp S {\set b} = \set a$ This shows that $a$ is isolated. Therefore $H$ must contain an isolated point. {{qed}}
Let $\sequence {x_n}_{n \mathop \in \N}$ be an infinite sequence in $T$. Suppose $\sequence {x_n}$ takes an infinite number of distinct values in $S$. Then there is an infinite subsequence $\sequence {x_{n_r} }_{r \mathop \in \N}$ with distinct terms. Let $U$ be a neighborhood of $a$. Then $S \setminus U$ is a finite set by definition. Thus there exists $N \in \N$ such that $\forall r > N: x_{n_r} \in U$. Thus $\sequence {x_{n_r} }$ converges to $a$. Otherwise $\sequence {x_n}$ only takes a finite number of distinct values. Then, since $\sequence {x_n}$ is infinite, there exists $x \in S$ such that: :$\forall N \in \N: \exists n > N: x = x_n$ This implies that we can take a subsequence of $\sequence {x_n}$ which is constant, and which converges to that constant. We can conclude then that, by definition, $T$ is a sequentially compact space. {{qed}}
{{begin-eqn}} {{eqn | l = \map \tan {-x}       | r = \frac {\map \sin {-x} } {\map \cos {-x} }       | c = Tangent is Sine divided by Cosine }} {{eqn | r = \frac {-\sin x} {\cos x}       | c = Sine Function is Odd; Cosine Function is Even }} {{eqn | r = -\tan x       | c = Tangent is Sine divided by Cosine }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \map \tan {a + b}       | r = \frac {\map \sin {a + b} } {\map \cos {a + b} }       | c = Tangent is Sine divided by Cosine }} {{eqn | r = \frac {\sin a \cos b + \cos a \sin b} {\cos a \cos b - \sin a \sin b}       | c = Sine of Sum and Cosine of Sum }} {{eqn | r = \frac {\frac {\sin a} {\cos a} + \frac {\sin b} {\cos b} } {1 - \frac {\sin a \sin b} {\cos a \cos b} }       | c = dividing top and bottom by $\cos a \cos b$ }} {{eqn | r = \frac {\tan a + \tan b} {1 - \tan a \tan b}       | c = Tangent is Sine divided by Cosine }} {{end-eqn}} {{qed}}
This proof assumes the definition of $\ln$ as : :$\ln x = \displaystyle \int_1^x \frac 1 t \ \mathrm dt$ As $\ln$ is defined as a definite integral, the result follows from the corollary to the first fundamental theorem of calculus. {{qed}}
This proof assumes the definition of $\ln$ as the inverse of the exponential function. As the Exponential Function is Differentiable, the result follows from the differentiability of inverse functions. {{qed}}
By definition, there exists a countable basis $\BB$ for $\tau$. Using the axiom of countable choice, we can obtain a choice function $\phi$ for $\BB \setminus \set \O$. Define: :$H = \set {\map \phi B: B \in \BB \setminus \set \O}$ By Image of Countable Set under Mapping is Countable, it follows that $H$ is countable. It suffices to show that $H$ is everywhere dense in $T$. Let $x \in U \in \tau$. By Equivalence of Definitions of Analytic Basis, there exists a $B \in \BB$ such that $x \in B \subseteq U$. Then $\map \phi B \in U$, and so $H \cap U$ is non-empty. Hence, $x$ is an adherent point of $H$. By Equivalence of Definitions of Adherent Point, it follows that $x \in H^-$, where $H^-$ denotes the closure of $H$. Therefore, $H^- = S$, and so $H$ is everywhere dense in $T$ by definition. {{qed}} {{ACC}}
From Countable Basis of Real Number Line we have that $\struct {\R, \tau_d}$ has a countable basis. The result follows directly from the definition of a second-countable space. {{qed}}
Let $\CC$ be the set of subsets of $\R$ defined as: :$\CC = \set {\openint n {n + 2}: n \in \Z}$ Then $\CC$ is an open cover of $\R$ which is countable. However, there is no finite subcover for $\R$ of $\CC$. Hence the result, by definition of countably compact. {{qed}}
We have that a Real Number Line satisfies all Separation Axioms. Specifically, $\struct {\R, \tau_d}$ is a Hausdorff space. Consider $\CC$ the set of subsets of $\R$ defined as: :$\CC = \set {\closedint n {n + 1}: n \in \Z}$ where $\closedint n {n + 1}$ is the closed real interval between successive integers. By the Heine-Borel Theorem, each element of $\CC$ is compact. Every element of $\R$ is contained in at least one of the elements of $\CC$. Hence, by definition, $\R$ is locally compact. {{qed}}
We have that a Real Number Line satisfies all Separation Axioms. Specifically, $\struct {\R, \tau_d}$ is a Hausdorff space. Consider $\CC$ the set of subsets of $\R$ defined as: :$\CC = \set {\closedint n {n + 1}: n \in \Z}$ where $\closedint n {n + 1}$ is the closed real interval between successive integers. By the Heine-Borel Theorem, each element of $\CC$ is compact. $\CC$ itself is countable, as there is a (trivial) one-to-one-correspondence between $\CC$ and $\Z$. Every element of $\R$ is contained in at least one of the elements of $\CC$. Thus $\R$ is the union of $\CC$. Hence, by definition, $\R$ is $\sigma$-compact. {{qed}}
We have: :Real Number Line is Metric Space :Closed Set in Metric Space is $G_\delta$ Hence the result. {{qed}}
Let $\CC$ be an open cover for $\R$. Then $\CC$ covers each of the closed real intervals $\closedint n {n + 1}$ for all $n \in \Z$. By the Heine-Borel Theorem, each of $\closedint n {n + 1}$ is compact. So, for each of these intervals $\closedint n {n + 1}$, it follows that $\CC$ can be reduced to a sequence $\sequence {G_i^{\paren n} }$ of finite subcovers. Then each of $G_i^{\paren n} \cap \openint {n - 1} {n + 2}$ forms a refinement of $\CC$ which is locally finite. Hence, by definition, $\R$ is paracompact. {{qed}}
To show that $\map d {x, y}$ is a quasimetric, we need to show that $d: \R \times \R \to \R$ satisfies the following conditions for all $x, y, z \in \R$: :$(\text M 1): \quad \map d {x, x} = 0$ :$(\text M 2): \quad \map d {x, y} + \map d {y, z} \ge \map d {x, z}$ :$(\text M 4): \quad x \ne y \implies \map d {x, y} > 0$ === Proof of $\text M 1$ === :$\map d {x, x} = x - x = 0$ So $(\text M 1)$ is shown to hold. {{qed|lemma}} === Proof of $\text M 2$ === Suppose $x \le y \le z$. Then: {{begin-eqn}} {{eqn | l = \map d {x, y} + \map d {y, z}       | r = \paren {y - x} + \paren {z - y}       | c =  }} {{eqn | r = \paren {z - x}       | c =  }} {{eqn | r = \map d {x, z}       | c =  }} {{end-eqn}} Equality holds, hence so does the inequality. {{finish|Other cases still to do.}} === Proof of $\text M 4$ === Suppose $x \ne y$. There are two possibilities: :$(1): \quad y \ge x$, in which case: ::$\map d {x, y} = y - x > 0$ :$(2): \quad y < x$, in which case: ::$\map d {x, y} = 2 \paren {x - y} > 0$ So $(\text M 4)$ is shown to hold. {{qed|lemma}} All criteria $(\text M 1)$, $(\text M 2)$ and $(\text M 4)$ are shown to hold, and so $\struct {\R, \tau_d}$ is shown to be a  quasimetric space. {{qed}}
{{BeginTableau|p \uparrow p \vdash \neg p}} {{Premise|1|p \uparrow p}} {{SequentIntro|2|1|\neg \left({p \land p}\right)|1|Definition of Logical NAND}} {{Idempotence|3|1|\neg p|2|Conjunction}} {{EndTableau}} {{BeginTableau|\neg p \vdash p \uparrow p}} {{Premise|1|\neg p}} {{Idempotence|2|1|\neg \left({p \land p}\right)|1|Conjunction}} {{SequentIntro|3|1|p \uparrow p|2|Definition of Logical NAND}} {{EndTableau}} {{qed}}
We apply the Method of Truth Tables: :$\begin{array}{|ccc||cc|} \hline p & \uparrow & p & \neg & p \\ \hline F & T & F & T & F \\ T & F & T & F & T \\ \hline \end{array}$ As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations. {{qed}}
{{BeginTableau|p \uparrow q \vdash q \uparrow p}} {{Premise|1|p \uparrow q}} {{SequentIntro|2|1|\neg \left({p \land q}\right)|1||Definition of Logical NAND}} {{Commutation|3|1|\neg \left({q \land p}\right)|2|Conjunction}} {{SequentIntro|4|1|q \uparrow p|3||Definition of Logical NAND}} {{EndTableau}} {{qed|lemma}} {{BeginTableau|q \uparrow p \vdash p \uparrow q}} {{Premise|1|q \uparrow p}} {{SequentIntro|2|1|\neg \left({q \land p}\right)|1||Definition of Logical NAND}} {{Commutation|3|1|\neg \left({p \land q}\right)|2|Conjunction}} {{SequentIntro|4|1|p \uparrow q|3||Definition of Logical NAND}} {{EndTableau}} {{qed}}
We apply the Method of Truth Tables: :$\begin{array}{|ccc||ccc|} \hline p & \uparrow & q & q & \uparrow & p \\ \hline F & T & F & F & T & F \\ F & T & T & T & T & F \\ T & T & F & F & T & T \\ T & F & T & T & F & T \\ \hline \end{array}$ As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations. {{qed}}
{{BeginTableau|\neg \paren {\paren {p \uparrow q} \uparrow r \implies p \uparrow \paren {q \uparrow r} } }} {{Assumption|1|\paren {p \uparrow q} \uparrow r \implies p \uparrow \paren {q \uparrow r} }} {{Assumption|2|p \land \neg r}} {{Simplification|3|2|p|2|1}} {{Simplification|4|2|\neg r|2|2}} {{Addition|5|2|\paren {\neg q} \lor \paren {\neg r}|4|2}} {{DeMorgan|6|2|\neg \paren {q \land r}|5|Disjunction of Negations}} {{SequentIntro|7|2|q \uparrow r|6   |Definition of Logical NAND}} {{Conjunction|8|2|p \land \paren {q \uparrow r}|3|7}} {{DoubleNegIntro|9|2|\neg \neg \paren {p \land \paren {q \uparrow r} }|8}} {{SequentIntro|10|2|\neg \paren {p \uparrow \paren {q \uparrow r} }|9   |Definition of Logical NAND}} {{Addition|11|2|\paren {\neg \paren {p \uparrow q} } \lor \paren {\neg r}|4|2}} {{DeMorgan|12|2|\neg \paren {\paren {p \uparrow q} \land r}|11|Disjunction of Negations}} {{SequentIntro|13|2|\paren {p \uparrow q} \uparrow r|12   |Definition of Logical NAND}} {{DoubleNegIntro|14|2|\neg \neg \paren {\paren {p \uparrow q} \uparrow r}|13}} {{Conjunction|15|2|\neg \neg \paren {\paren {p \uparrow q} \uparrow r} \land \neg \paren {p \uparrow \paren {q \uparrow r} }|13|10}} {{DeMorgan|16|2|\neg \paren {\neg \paren {\paren {p \uparrow q} \uparrow r} \lor \paren {p \uparrow \paren {q \uparrow r} } }|15|Conjunction of Negations}} {{SequentIntro|17|2   |\neg \paren {\paren {p \uparrow q} \uparrow r \implies p \uparrow \paren {q \uparrow r} }   |16   |Disjunction and Implication }} {{NonContradiction|18|1, 2|1|17}} {{EndTableau}} Taking $p = \top$, $r = \bot$, we find $\vdash p \land \neg r$, and conclude our initial assumption was false. {{qed}}
We apply the Method of Truth Tables: :$\begin{array}{|ccccc||ccccc|} \hline p & \uparrow & (q & \uparrow & r) & (p & \uparrow & q) & \uparrow & r \\ \hline F & T & F & T & F & F & T & F & T & F \\ F & T & F & T & T & F & T & F & F & T \\ F & T & T & T & F & F & T & T & T & F \\ F & T & T & F & T & F & T & T & F & T \\ T & F & F & T & F & T & T & F & T & F \\ T & F & F & T & T & T & T & F & F & T \\ T & F & T & T & F & T & F & T & T & F \\ T & T & T & F & T & T & F & T & T & T \\ \hline \end{array}$ As can be seen by inspection, the truth values under the main connectives do not match for all boolean interpretations. {{qed}}
{{BeginTableau|p \downarrow p \vdash \neg p}} {{Premise|1|p \downarrow p}} {{SequentIntro|2|1|\neg \left({p \lor p}\right)|1|Definition of Logical NOR}} {{Idempotence|3|1|\neg p|2|Disjunction}} {{EndTableau}} {{BeginTableau|\neg p \vdash p \downarrow p}} {{Premise|1|\neg p}} {{Idempotence|2|1|\neg \left({p \lor p}\right)|1|Disjunction}} {{SequentIntro|3|1|p \downarrow p|2|Definition of Logical NOR}} {{EndTableau}} {{qed}}
Apply the Method of Truth Tables: :$\begin{array}{|ccc||cc|} \hline p & \downarrow & p & \neg & p \\ \hline F & T & F & T & F \\ T & F & T & F & T \\ \hline \end{array}$ As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations. {{qed}}
Apply the Method of Truth Tables: :$\begin{array}{|ccc||ccc|} \hline p & \downarrow & q & q & \downarrow & p \\ \hline F & T & F & F & T & F \\ F & F & T & T & F & F \\ T & F & F & F & F & T \\ T & F & T & T & F & T \\ \hline \end{array}$ As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations. {{qed}}