hcju/proofwikiqa huggingface dataset:
=== Necessary Condition === Suppose that $\sequence {x_n}$ converges to $l$ relative to $d$. When $\map {B_\epsilon} l$ denotes the open $\epsilon$-ball of $l$, this means: :$\forall \epsilon \in \R_{>0}: \exists N \in \R: n > N \implies x_n \in \map {B_\epsilon} l$ Let $U \in \tau$ with $l \in U$. By definition of induced topology, there exists $\epsilon_0 \in \R_{>0}$ such that: :$\map {B_{\epsilon_0} } l \subseteq U$ Then there exists $N_0 \in \R$ such that for all $n > N_0$: :$x_n \in \map {B_{\epsilon_0} } l \subseteq U$ Hence, $\sequence {x_n}$ converges to $l$ in the induced topology $\tau$. {{qed|lemma}} === Sufficient Condition === Suppose that $\sequence {x_n}$ converges to $l$ in $\tau$. Let $\epsilon \in \R_{>0}$. Then $l \in \map {B_\epsilon} l$, and $\map {B_\epsilon} l \in \tau$. Then there exists $N \in \N$ such that for all $n > N$, we have $x_n \in \map {B_\epsilon} l$. Hence, $\sequence {x_n}$ converges to $l$ relative to $d$. {{qed}} Category:Metric Spaces 4gt7042v97zx44p62ujy1c5l8nh60wp
{{begin-eqn}} {{eqn | o =       | r = \lim_{n \mathop \to \infty} \map {d_2} {\map \phi {x_n}, \map \phi p} }} {{eqn | r = \lim_{n \mathop \to \infty} \map {d_1} {x_n, p}       | c = {{Defof|Isometry (Metric Spaces)|index = 1}} }} {{eqn | r = 0       | c = {{Defof|Convergent Sequence|subdef = Metric Space|index = 3}} }} {{end-eqn}} Hence, by definition, $\sequence {\map \phi {x_n}}$ converges to $\map \phi p$. {{qed}}
=== Necessary Condition === Let $\struct {S, \preceq}$ be Dedekind complete. Let $A \subseteq S$ be non-empty and bounded below. Let $B \subseteq S$ be set of all lower bounds for $A$. Then every element of $A$ is an upper bound for $B$. Therefore, $B$ is non-empty and bounded above. By the definition of Dedekind completeness, $B$ admits a supremum $x \in S$. By the definition of supremum, it follows that every element of $A$ succeeds $x$. That is, $x$ is a lower bound for $A$. If $y \in S$ is a lower bound for $A$, then $y \in B$, and so $y \preceq x$. Hence, $x$ is the infimum of $A$. {{qed|lemma}} === Sufficient Condition === Follows directly from the Necessary Condition and Dual of Dual Ordering. {{qed}}
{{BeginTableau|\vdash \left({\left({p \iff q}\right) \land \left({q \iff r}\right)}\right) \implies \left({p \iff r}\right)}} {{Assumption|1|\left({p \iff q}\right) \land \left({q \iff r}\right)}} {{Simplification|2|1|p \iff q|1|1}} {{Simplification|3|1|q \iff r|1|2}} {{SequentIntro|4|1|p \iff r|2, 3|Biconditional is Transitive: Formulation 1}} {{Implication|5||\left({\left({p \iff q}\right) \land \left({q \iff r}\right)}\right) \implies \left({p \iff r}\right)|1|4}} {{EndTableau}} {{qed}}
{{BeginTableau|\left({p \land q}\right) \lor \left({\neg p \land \neg q}\right) \vdash p \iff q}} {{Premise|1|\left({p \land q}\right) \lor \left({\neg p \land \neg q}\right)}} {{Assumption|2|p \land q}} {{Assumption|3|p}} {{Simplification|4|2|q|2|2}} {{Implication|5|2|p \implies q|3|4}} {{Assumption|6|q}} {{Simplification|7|2|p|2|1}} {{Implication|8|2|q \implies p|6|7}} {{BiconditionalIntro|9|2|p \iff q|5|8}} {{Assumption|10|\neg p \land \neg q}} {{Assumption|11|\neg p}} {{Simplification|12|10|\neg q|10|2}} {{Implication|13|10|\neg p \implies \neg q|11|12}} {{SequentIntro|14|10|q \implies p|13|Rule of Transposition}} {{Assumption|15|\neg q}} {{Simplification|16|10|\neg p|10|1}} {{Implication|17|10|\neg q \implies \neg p|15|16}} {{SequentIntro|18|10|p \implies q|17|Rule of Transposition}} {{BiconditionalIntro|19|10|p \iff q|18|14}} {{ProofByCases|20|1|p \iff q|1|2|9|10|19}} {{EndTableau}} {{qed}} Category:Biconditional as Disjunction of Conjunctions kpkzs6086yvdup4ehfnuooz56hz7783
We apply the Method of Truth Tables. As can be seen by inspection, in all cases the truth values under the main connectives match for all boolean interpretations. $\begin{array}{|ccc||ccccccccc|} \hline p & \iff & q & (p & \land & q) & \lor & (\neg & p & \land & \neg & q) \\ \hline F & T & F & F & F & F & T & T & F & T & T & F \\ F & F & T & F & F & T & F & T & F & F & F & T \\ T & F & F & T & F & F & F & F & T & F & T & F \\ T & T & T & T & T & T & T & F & T & F & F & T \\ \hline \end{array}$ {{qed}}
{{BeginTableau|p \iff q \vdash \neg p \iff \neg q}} {{Premise|1|p \iff q}} {{BiconditionalElimination|2|1|p \implies q|1|1}} {{SequentIntro|3|1|\neg q \implies \neg p|2|Rule of Transposition}} {{BiconditionalElimination|4|1|q \implies p|1|2}} {{SequentIntro|5|1|\neg p \implies \neg q|4|Rule of Transposition}} {{BiconditionalIntro|6|1|\neg p \iff \neg q|5|3}} {{EndTableau}} {{qed}}
{{BeginTableau|\neg p \iff \neg q \vdash p \iff q}} {{Premise|1|\neg p \iff \neg q}} {{BiconditionalElimination|2|1|\neg p \implies \neg q|1|1}} {{SequentIntro|3|1|\neg \neg q \implies \neg \neg p|2|Rule of Transposition}} {{DoubleNegElimination|4|1|q \implies p|3|(twice)}} {{BiconditionalElimination|5|1|\neg q \implies \neg p|1|2}} {{SequentIntro|6|1|\neg \neg p \implies \neg \neg q|5|Rule of Transposition}} {{DoubleNegElimination|7|1|p \implies q|3|(twice)}} {{BiconditionalIntro|8|1|p \iff q|7|4}} {{EndTableau|qed}} {{LEM|Double Negation Elimination}}
We apply the Method of Truth Tables. As can be seen by inspection, in all cases the truth values under the main connectives match for all boolean interpretations. $\begin{array}{|ccc||ccccc|} \hline p & \iff & q & \neg & p & \iff & \neg & q \\ \hline F & T & F & T & F & T & T & F \\ F & F & T & T & F & F & F & T \\ T & F & F & F & T & F & T & F \\ T & T & T & F & T & T & F & T \\ \hline \end{array}$ {{qed}} Category:Biconditional Equivalent to Biconditional of Negations Category:Truth Table Proofs gi5u0dj3jca25l8ylyo15gtegynwcki
{{BeginTableau|p \iff q \vdash \left({p \lor q}\right) \implies \left({p \land q}\right)}} {{Premise|1|p \iff q}} {{SequentIntro|2|1|\left({p \land q}\right) \lor \left({\neg p \land \neg q}\right)|1|Biconditional as Disjunction of Conjunctions}} {{DeMorgan|3|1|\left({p \land q}\right) \lor \neg \left({p \lor q}\right)|2|Conjunction of Negations}} {{Commutation|4|1|\neg \left({p \lor q}\right) \lor \left({p \land q}\right)|3|Disjunction}} {{SequentIntro|5|1|\left({p \lor q}\right) \implies \left({p \land q}\right)|4|Rule of Material Implication}} {{EndTableau}} {{qed}} Category:Biconditional iff Disjunction implies Conjunction bjqipywjtsnh5w8bf5072djd3n8usvn
{{BeginTableau|\left({p \lor q}\right) \implies \left({p \land q}\right) \vdash p \iff q}} {{Premise|1|\left({p \lor q}\right) \implies \left({p \land q}\right)}} {{SequentIntro|2|1|\neg \left({p \lor q}\right) \lor \left({p \land q}\right)|1|Rule of Material Implication}} {{Commutation|3|1|\left({p \land q}\right) \lor \neg \left({p \lor q}\right)|2|Disjunction}} {{DeMorgan|4|1|\left({p \land q}\right) \lor \left({\neg p \land \neg q}\right)|3|Conjunction of Negations}} {{SequentIntro|5|1|\left ({p \implies q}\right) \land \left ({q \implies p}\right)|4   |Biconditional as Disjunction of Conjunctions}} {{SequentIntro|6|1|p \iff q|5|Rule of Material Equivalence}} {{EndTableau}} {{qed}} {{LEM|Rule of Material Implication}} Category:Biconditional iff Disjunction implies Conjunction fs5pn4li3zbnvspkihv62321uytc5kk
We apply the Method of Truth Tables. As can be seen by inspection, in all cases the truth values under the main connectives match for all boolean interpretations. $\begin{array}{|ccc||ccccccc|} \hline p & \iff & q & (p & \lor & q) & \implies & (p & \land & q) \\ \hline F & T & F & F & F & F & T & F & F & F \\ F & F & T & F & T & T & F & F & F & T \\ T & F & F & T & T & F & F & T & F & F \\ T & T & T & T & T & T & T & T & T & T \\ \hline \end{array}$ {{qed}}
{{BeginTableau|\vdash \left({p \iff q}\right) \iff \left({\left({p \lor q}\right) \implies \left({p \land q}\right)}\right)}} {{Assumption|1|p \iff q}} {{SequentIntro|2|1|\left({p \lor q}\right) \implies \left({p \land q}\right)|1|Biconditional iff Disjunction implies Conjunction: Formulation 1}} {{Implication|3||\left({p \iff q}\right) \implies \left({\left({p \lor q}\right) \implies \left({p \land q}\right)}\right)|1|2}} {{Assumption|4|\left({p \lor q}\right) \implies \left({p \land q}\right)}} {{SequentIntro|5|4|p \iff q|4|Biconditional iff Disjunction implies Conjunction: Formulation 1}} {{Implication|6||\left({\left({p \lor q}\right) \implies \left({p \land q}\right)}\right) \implies \left({p \iff q}\right)|4|5}} {{BiconditionalIntro|7||\left({p \iff q}\right) \iff \left({\left({p \lor q}\right) \implies \left({p \land q}\right)}\right)|3|6}} {{EndTableau}} {{qed}} Category:Biconditional iff Disjunction implies Conjunction nd5xg35tyu3oguip3lvunbq029yqv73
We apply the Method of Truth Tables. 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 $T$: $\begin{array}{|ccccccc||ccc|} \hline (p & \iff & q) & \land & (q & \iff & r) & p & \iff & r \\ \hline F & T & F & T & F & T & F & F & T & F \\ F & T & F & F & F & F & T & F & F & T \\ F & F & T & F & T & F & F & F & T & F \\ F & F & T & F & T & T & T & F & F & T \\ T & F & F & F & F & T & F & T & F & F \\ T & F & F & F & F & F & T & T & T & T \\ T & T & T & F & T & F & F & T & F & F \\ T & T & T & T & T & T & T & T & T & T \\ \hline \end{array}$ Hence the result. {{qed}}
{{BeginTableau|p \iff q, q \iff r \vdash p \iff r}} {{Premise|1|p \iff q}} {{Premise|2|q \iff r}} {{BiconditionalElimination|3|1|p \implies q|1|1}} {{BiconditionalElimination|4|2|q \implies r|2|1}} {{SequentIntro|5|1, 2|p \implies r|1, 2|Hypothetical Syllogism: Formulation 1}} {{BiconditionalElimination|6|1|q \implies p|1|2}} {{BiconditionalElimination|7|2|r \implies q|2|2}} {{SequentIntro|8|1, 2|r \implies p|7, 6|Hypothetical Syllogism: Formulation 1}} {{BiconditionalIntro|9|1, 2|p \iff r|5|8}} {{EndTableau}}
From Order of Squares in Ordered Ring, we have: :$x \le y \implies x \circ x \le y \circ y$ To prove the opposite implication, we use a Proof by Contradiction. {{AimForCont}} $x \circ x \le y \circ y$ but $x \not\le y$. Since $\le$ is a total ordering, this means $y < x$. Since $0_R \le y$, Extended Transitivity shows that $0_R < x$, so in particular $x ≠ 0_R$. As $\le$ is compatible with the ring structure $\struct {R, +, \circ}$ and $x$ and $y$ are both positive, we have: :$y \circ y \le y \circ x \le x \circ x$ Since we assume $x \circ x \le y \circ y$, Ordering Cycle implies Equality shows that: :$y \circ y = x \circ x = y \circ x$ Therefore: {{begin-eqn}} {{eqn | l = 0_R       | r = x \circ x + \paren {-\paren {y \circ x} }       | c = {{Defof|Ring Negative}} }} {{eqn | r = x \circ x + \paren {-y} \circ x       | c = Product with Ring Negative }} {{eqn | r = \paren {x + \paren {-y} } \circ x       | c = {{Ring-axiom|D}} }} {{end-eqn}} Since $x \ne y$, we have that $x + \paren {-y} \ne 0_R$; otherwise, by {{Ring-axiom|A1}}, we would have $x = \paren {x + \paren {-y} } + y = y$. We have already shown that $x \ne 0_R$. But then $x$ and $x + \paren {-y}$ are proper zero divisors of $R$, contradicting the hypothesis that $R$ has no proper zero divisors. {{qed}}
From Order of Squares in Ordered Ring, we have: :$x \le y \implies x \circ x \le y \circ y$ To prove the reverse implication, suppose that $x \circ x \le y \circ y$. Thus: {{begin-eqn}} {{eqn | l = x \circ x       | o = \le       | r = y \circ y       | c =  }} {{eqn | ll=\leadsto       | l = x \circ x + \paren {-\paren {x \circ x} }       | o = \le       | r = y \circ y + \paren {-\paren {x \circ x} }       | c =  }} {{eqn | ll=\leadsto       | l = 0_R       | o = \le       | r = y \circ y + \paren {-\paren {x \circ x} }       | c =  }} {{eqn | ll=\leadsto       | l = 0_R       | o = \le       | r = \paren {y + \paren {-x} } \circ \paren {y + x}       | c = Difference of Two Squares, which applies because a field is a commutative ring. }} {{end-eqn}} As $0_R \le x, y$ we have $0_R \le x + y$. Hence by the premise we have $0_R \le \paren {x + y}^{-1}$. So as $0_R \le \paren {y + \paren {-x} } \circ \paren {y + x}$ we can multiply both sides by $\paren {x + y}^{-1}$ and get $0_R \le \paren {y + \paren {-x} }$. Adding $-x$ to both sides gives us $x \le y$. {{qed}}
{{BeginTableau|p \vdash p}} {{Premise|1|p}} {{EndTableau}} {{qed}} This is the shortest tableau proof possible.
We apply the Method of Truth Tables (trivially) to the proposition. $\begin{array}{|c|c|} \hline p & p \\ \hline F & F \\ T & T \\ \hline \end{array}$ {{qed}}
We apply the Method of Truth Tables to the proposition. As can be seen by inspection, the truth value under the main connective is $T$ throughout. $\begin{array}{|ccc|} \hline p & \implies & p \\ \hline F & T & F \\ T & T & T \\ \hline \end{array}$ {{qed}}
Let $V \in \tau_1$. For all $z \in V$, we have that $V$ is an open neighborhood of $z$ with respect to $\tau_1$. Then for all $z \in V$, we can find $U_z \in \tau_2$ such that $z \in U_z \subseteq V$. Then: :$\displaystyle V = \bigcup_{z \mathop \in V } \left\{ {z}\right\} \subseteq \bigcup_{z \mathop \in V } U_z \subseteq V$ By definition of set equality, it follows that: : $\displaystyle V = \bigcup_{z \mathop \in V} U_z \in \tau_2$ Hence: : $\tau_1 \subseteq \tau_2$ which by definition means that $\tau_1$ is coarser than $\tau_2$. {{qed}} Category:Topology btap2mwfh280od8bb1mk6thrc4uzu1m
We apply the Method of Truth Tables. As can be seen by inspection, the appropriate truth values match for all boolean interpretations. $\begin{array}{|ccccc||c|} \hline p & \land & (p & \lor & q) & p \\ \hline F & F & F & F & F & F \\ F & F & F & T & T & F \\ T & T & T & T & F & T \\ T & T & T & T & T & T \\ \hline \end{array}$ {{qed}}
By calculation: {{begin-eqn}} {{eqn | l=p \land \left({p \lor q}\right)       | r=\left({p \lor \bot}\right) \land \left({p \lor q}\right)       | c=Disjunction with Contradiction }} {{eqn | r=p \lor \left({\bot \land q}\right)       | c=Disjunction is Left Distributive over Conjunction }} {{eqn | r=p \lor \bot       | c=Conjunction with Contradiction }} {{eqn | r=p       | c=Disjunction with Contradiction }} {{end-eqn}} {{qed}}
We apply the Method of Truth Tables. As can be seen by inspection, the appropriate truth values match for all boolean interpretations. $\begin{array}{|ccccc||c|} \hline p & \lor & (p & \land & q) & p \\ \hline F & F & F & F & F & F \\ F & F & F & F & T & F \\ T & T & T & F & F & T \\ T & T & T & T & T & T \\ \hline \end{array}$ {{qed}}
{{begin-eqn}} {{eqn | l=p \lor \left({p \land q}\right)       | r=\left({p \land \top}\right) \lor \left({p \land q}\right)       | c=Conjunction with Tautology }} {{eqn | r=p \land \left({\top \lor q}\right)       | c=Conjunction is Left Distributive over Disjunction }} {{eqn | r=p \land \top       | c=Disjunction with Tautology }} {{eqn | r=p       | c=Conjunction with Tautology }} {{end-eqn}} {{qed}}
$(1)$ follows from the definition of a transitive relation. $(4)$ follows from Reflexive Closure of Transitive Relation is Transitive. Suppose that: :$\paren {a \mathrel \RR b} \land \paren {b \mathrel {\RR^=} c}$ By the definition of reflexive closure: :$b \mathrel \RR c$ or $b = c$ If $b = c$, then since $a \mathrel \RR b$: :$a \mathrel \RR c$ If $b \mathrel \RR c$ then by transitivity of $\RR$: :$a \mathrel \RR c$ Thus $(2)$ holds. A similar argument proves that $(3)$ holds as well: Suppose that: :$\paren {a \mathrel {\RR^=} b} \land \paren {b \mathrel \RR c}$ By the definition of reflexive closure: :$a \mathrel \RR b$ or $a = b$ If $a = b$, then since $b \mathrel \RR c$: :$a \mathrel \RR c$ If $a \mathrel \RR b$ then by transitivity of $\RR$: :$a \mathrel \RR c$ Thus $(3)$ holds. {{qed}} Category:Transitive Relations Category:Reflexive Closures grgr98m0zamvw910geicyuj15dm47uy
We apply the Method of Truth Tables. As can be seen by inspection, the truth values under the main connectives match for each boolean interpretations. $\begin{array}{|c||ccc|} \hline p & p & \lor & p \\ \hline T & T & T & T \\ F & F & F & F \\ \hline \end{array}$ {{qed}}
We apply the Method of Truth Tables. As can be seen by inspection, the truth values under the main connectives match for each boolean interpretations. $\begin{array}{|c||ccc|} \hline p & p & \land & p \\ \hline T & T & T & T \\ F & F & F & F \\ \hline \end{array}$ {{qed}}
Since $\preceq$ is an ordering, it is transitive and antisymmetric. By transitivity, $x_1 \preceq x_3$. Since $x_1 \preceq x_3$ and $x_3 \preceq x_1$, antisymmetry allows us to conclude that $x_1 = x_3$. Since $x_1 = x_3$ and $x_1 \preceq x_2$, we must have $x_3 \preceq x_2$. Since $x_3 \preceq x_2$ and $x_2 \preceq x_3$, antisymmetry allows us to conclude that $x_2 = x_3$. Thus $x_1 = x_2 = x_3$. {{qed}} Category:Order Theory izdvezcs9rfg6c687d0ieugpd04qf8h
Denote the Euclidean topology on $\R^n$ as $\tau$, and denote the Tychonoff topology on $\R^n$ as $\tau'$. Let $U \in \tau$, and let $x = \tuple{x_1, \ldots, x_n} \in U$. Then there exists $\epsilon \in \R_{>0}$ such that the open ball $\map {B_\epsilon} x \subseteq U$. We show that: :$\displaystyle B' = \prod_{i \mathop = 1}^n \left({x_i - \dfrac \epsilon n \,.\,.\, x_i + \dfrac \epsilon n}\right) \subseteq \map {B_\epsilon} x$ For if $y = \tuple{y_1, \ldots, y_n} \in B'$, then: {{begin-eqn}} {{eqn | l = \map d {x, y}       | r = \paren{\sum_{i \mathop = 1}^n \size{x_i - y_i}^2}^{1/2}       | c = where $d$ denotes the Euclidean metric }} {{eqn | o = \le       | r = \sum_{i \mathop = 1}^n \size{x_i - y_i}       | c = by Minkowski's Inequality for Sums }} {{eqn | o = <       | r = \sum_{i \mathop = 1}^n \dfrac \epsilon n }} {{eqn | r = \epsilon }} {{end-eqn}} From Natural Basis of Tychonoff Topology of Finite Product, $B' \in \tau'$. As $B' \subseteq \map {B_\epsilon} x \subseteq U$, Neighborhood Condition for Coarser Topology shows that $\tau' \subseteq \tau$. {{qed|lemma}} Let $U' \in \tau'$. Let $x = \tuple{x_1, \ldots, x_n} \in U'$. From Natural Basis of Tychonoff Topology of Finite Product, sets of the type $\displaystyle \prod_{i \mathop = 1}^n U'_i$ with $U'_i \in \tau_1$ form an analytic basis for $\tau'$. From Equivalence of Definitions of Analytic Basis, it follows that we can select $U'_1, \ldots, U'_n \in \tau_1$ such that $\displaystyle x \in \prod_{i \mathop = 1}^n U'_i \subseteq U'$. By definition of open set, it follows that for all $i \in \set {1, \ldots, n}$, we can find $\epsilon_i \in \R_{>0}$ such that $\openint {x_i - \epsilon_i} {x_i + \epsilon_i} \subseteq U'_i$. Put $\epsilon = \min \set{\epsilon_i : i = 1, \ldots, n}$. We show that the open ball $\map {B_\epsilon} x \subseteq U'$. For if $y = \tuple{y_1, \ldots, y_n} \in \map {B_\epsilon} x$, then $y_i \in \openint {x_i - \epsilon_i} {x_i + \epsilon_i }$, as: :$\size{x_i - y_i} < \epsilon \le \epsilon_i$ It follows that: :$\displaystyle \map {B_\epsilon} x \subseteq \prod_{i \mathop = 1}^n \openint {x_i - \epsilon_i} {x_i + \epsilon_i } \subseteq \prod_{i \mathop = 1}^n U'_i \subseteq U'$ Then Neighborhood Condition for Coarser Topology shows that $\tau \subseteq \tau'$. {{qed|lemma}} It follows that $\tau = \tau'$. {{qed}}
As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations. $\begin{array}{|cccc||cccc|} \hline \neg & (p & \land & q) & p & \implies & \neg & q \\ \hline T & F & F & F & F & T & T & F \\ T & F & F & T & F & T & F & T \\ T & T & F & F & T & T & T & F \\ F & T & T & T & T & F & F & T \\ \hline \end{array}$ {{qed}}
{{BeginTableau|\vdash \left({\neg \left({p \land q}\right)}\right) \iff \left({p \implies \neg q}\right)}} {{Assumption|1|\neg \left({p \land q}\right)}} {{SequentIntro|2|1|p \implies \neg q|1|Modus Ponendo Tollens: Formulation 1: Forward Implication}} {{Implication|3||\left({\neg \left({p \land q}\right)}\right) \implies \left({p \implies \neg q}\right)|1|2}} {{Assumption|4|p \implies \neg q}} {{SequentIntro|5|4|\neg \left({p \land q}\right)|4|Modus Ponendo Tollens: Formulation 1: Reverse Implication}} {{Implication|6||\left({p \implies \neg q}\right) \implies \left({\neg \left({p \land q}\right)}\right)|4|5}} {{BiconditionalIntro|7||\left({\neg \left({p \land q}\right)}\right) \iff \left({p \implies \neg q}\right)|3|6}} {{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. $\begin{array}{|ccc||ccccc|} \hline p & \implies & q & \neg & (p & \land & \neg & q) \\ \hline F & T & F & T & F & F & T & F \\ F & T & T & T & F & F & F & T \\ T & F & F & F & T & T & T & F \\ T & T & T & T & T & F & F & T \\ \hline \end{array}$ {{qed}}
{{BeginTableau|\neg \left({p \land \neg q}\right) \vdash p \implies q}} {{Premise|1|\neg \left({p \land \neg q}\right)}} {{DeMorgan|2|1|\neg p \lor \neg \neg q|1|Disjunction of Negations}} {{SequentIntro|3|1|p \implies \neg \neg q|2|Rule of Material Implication}} {{Assumption|4|p}} {{ModusPonens|5|1, 4|\neg \neg q|3|4}} {{DoubleNegElimination|6|1, 4|q|5}} {{Implication|7|1|p \implies q|4|6}} {{EndTableau}} {{qed}} {{LEM|Double Negation Elimination}}
Let $u, v: \set {\tuple {x, y} \in \R^2: x + i y = z \in D} \to \R$ be the two real-valued functions defined in the Cauchy-Riemann Equations: :$\map u {x, y} = \map \Re {\map f z}$ :$\map v {x, y} = \map \Im {\map f z}$ By the Cauchy-Riemann Equations: :$f' = \dfrac {\partial f} {\partial x} = \dfrac {\partial u} {\partial x} + i \dfrac {\partial v} {\partial x}$ :$f' = -i \dfrac {\partial f} {\partial y} = \dfrac {\partial v} {\partial y} - i \dfrac {\partial u} {\partial y}$ As $f' = 0$ by assumption, this implies: :$0 = \dfrac {\partial u} {\partial x} = \dfrac {\partial u} {\partial y} = \dfrac {\partial v} {\partial x} = \dfrac {\partial v} {\partial y}$ Then Zero Derivative implies Constant Function shows that $\map u {x + t, y} = \map u {x, y}$ for all $t \in \R$. Similar results apply for the other three partial derivatives. Let $z, w \in D$. From Connected Domain is Connected by Staircase Contours, it follows that there exists a staircase contour $C$ in $D$ with endpoints $z$ and $w$. The contour $C$ is a concatenation of directed smooth curves that can be parameterized as line segments on one of these two forms: :$(1): \quad \map \gamma t = z_0 + t r$ :$(2): \quad \map \gamma t = z_0 + i t r$ for some $z_0 \in D$ and $r \in \R$ for all $t \in \closedint 0 1$. If $z_1 \in D$ lies on the same line segment as $z_0$, it follows that for parameterizations of type $(1)$: {{begin-eqn}} {{eqn | l = \map f {z_1}       | r = \map u {z_0 + t r} + \map v {z_0 + t r}       | c = for some $t \in \closedint 0 1$ }} {{eqn | r = \map u {z_0} + \map v {z_0}       | c = Zero Derivative implies Constant Function }} {{eqn | r = \map f {z_0} }} {{end-eqn}} Similarly, for parameterizations of type $(2)$: {{begin-eqn}} {{eqn | l = \map f {z_1}       | r = \map u {z_0 + i t r} + \map v {z_0 + i t r}       | c = for some $t \in \closedint 0 1$ }} {{eqn | r = \map f {z_0}       | c = Zero Derivative implies Constant Function }} {{end-eqn}} As the image of $C$ is connected by these line segments, it follows that for all $z_0$ and $z_1$ in the image of $C$: :$\map f {z_0} = \map f {z_1}$ In particular: :$\map f z = \map f w$ so $f$ is constant on $D$. {{qed}}
By definition of the Chebyshev distance on $\R^n$, we have: :$\displaystyle \map {d_\infty} {x, y} = \max_{i \mathop = 1}^n \size {x_i - y_i}$ where $x = \tuple {x_1, x_2, \ldots, x_n}$ and $y = \tuple {y_1, y_2, \ldots, y_n}$. Let $j$ be chosen so that: :$\displaystyle \size {x_j - y_j} = \max_{i \mathop = 1}^n \size {x_i - y_i}$ Then: {{begin-eqn}} {{eqn | l = \map {d_\infty} {x, y}       | r = \paren {\size {x_j - y_j}^p}^{1/p} }} {{eqn | o = \le       | r = \paren {\sum_{i \mathop = 1}^n \size {x_i - y_i}^p}^{1/p} }} {{eqn | r = \map {d_p} {x, y} }} {{eqn | o = \le       | r = \paren {n \size {x_j - y_j}^p}^{1/p} }} {{eqn | r = n^{1/p} \size {d_\infty} {x, y} }} {{end-eqn}}
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. $\begin{array}{|ccccc||cccc|} \hline \neg & p & \lor & \neg & q & \neg & (p & \land & q) \\ \hline T & F & T & T & F & T & F & F & F \\ T & F & T & F & T & T & F & F & T \\ F & T & T & T & F & T & T & F & F \\ F & T & F & F & T & F & T & T & T \\ \hline \end{array}$ {{qed}}