hcju/proofwikips huggingface dataset:
Strictly Increasing Sequence on Ordered Set
Strictly Increasing Sequence on Ordered Set 0 1374 366577 366574 2018-09-20T12:36:57Z KarlFrei 3474 added category wikitext text/x-wiki == Lemma == Let $\left({S, \preceq}\right)$ be a totally ordered set. Let $\left \langle {r_k} \right \rangle_{p \mathop \le k \mathop \le q}$ be a finite sequence of elements of $\left({S, \preceq}\right)$. Then $\left \langle {r_k} \right \rangle_{p \mathop \le k \mathop \le q}$ is strictly increasing : :$\forall k \in \left[{p + 1 \,.\,.\, q}\right]: r_{k - 1} \prec r_k$ == Proof == Let $\left \langle {r_k} \right \rangle_{p \mathop \le k \mathop \le q}$ be strictly increasing. Because $\forall k \in \N_{>0}: k - 1 < k$, it follows directly that: : $\forall k \in \left[{p + 1 \,.\,.\, q}\right]: r_{k - 1} \prec r_k$ For the other direction, we use a Proof by Contraposition. To that end, suppose $\left \langle {r_k} \right \rangle_{p \mathop \le k \mathop \le q}$ is ''not'' strictly increasing. Let $K$ be the set of all $k \in \left[{p \,.\,.\, q}\right]$ such that: : $\exists j \in \left[{p \,.\,.\, q}\right]$ such that $j < k$ and $r_k \preceq r_j$ The set $K$ is not empty because $\left \langle {r_k} \right \rangle_{p \mathop \le k \mathop \le q}$ is not strictly increasing. As $K \subset \N$ and the latter is well-ordered, then so is $K$. Thus $K$ has a minimal element $m$. Thus there exists $j \in \left[{p \,.\,.\, q}\right]$ such that: : $j < m$ and: : $r_m \preceq r_j$ Because $m - 1 < m$: :$j \le m - 1$ and so: :$m - 1 \notin K$ So: : $r_m \preceq r_j\prec r_{m-1}.$ Since orderings are transitive, it follows : $r_m \preceq r_{m-1}.$ From Rule of Transposition it follows that :$\forall k \in \left[{p + 1 \,.\,.\, q}\right]: r_{k - 1} \prec r_k\implies \left \langle {r_k} \right \rangle_{p \mathop \le k \mathop \le q}$ is strictly increasing. The result follows.  == Sources == * {{BookReference|Modern Algebra|1965|Seth Warner|prev=Strictly Increasing Sequence induces Partition|next=Fundamental Principle of Counting}}: $\S 18$: Lemma for Theorem $18.4$
Real Numbers are Close Packed
== Corollary to Between two Real Numbers exists Rational Number ==  :$\forall a, b \in \R: a < b \implies \paren {\exists c \in \R: a < c < b}$  That is, the set of real numbers is close packed. == Proof 1 == {{:Real Numbers are Close Packed/Proof 1}} == Proof 2 == {{:Real Numbers are Close Packed/Proof 2}} == Sources == * {{BookReference|Algebra Volume 1|1982|P.M. Cohn|edpage = Second Edition|ed = 2nd|prev = Quantifier/Examples/Existence of Multiplicative Identity|next = Equivocation of Nothing}}: Chapter $1$: Sets and mappings: $\S 1.1$: The need for logic: Exercise $(8)$
Group Action on Prime Power Order Subset
== Lemma == Let $G$ be a finite group. Let $\mathbb S = \set {S \subseteq G: \card S = p^n}$ where $p$ is prime. That is, the set of all subsets of $G$ whose cardinality is the power of a prime number. Let $G$ act on $\mathbb S$ by the group action defined in Group Action on Sets with k Elements: :$\forall S \in \mathbb S: g * S = g S = \set {x \in G: x = g s: s \in S}$. Then: === Stabilizer is $p$-Subgroup === {{:Group Action on Prime Power Order Subset/Stabilizer is p-Subgroup}} === Stabilizer of Maximal Power Order Subset === {{:Group Action on Prime Power Order Subset/Stabilizer of Maximal Power Order Subset}}
Orbits of Group Action on Sets with Power of Prime Size
== Lemma == Let $G$ be a finite group such that $\order G = k p^n$ where $p \nmid k$. Let $\mathbb S = \set {S \subseteq G: \order S = p^n}$ Let $G$ act on $\mathbb S$ by the group action defined in Group Action on Sets with k Elements: :$\forall S \in \mathbb S: g * S = g S = \set {x \in G: x = g s: s \in S}$ Then: === Orbit Length === {{:Orbits of Group Action on Sets with Power of Prime Size/Orbit Length}} === Orbit whose Length is not Divisible by $p$ === {{:Orbits of Group Action on Sets with Power of Prime Size/Orbit whose Length is not Divisible by p}} === Orbit whose Length is Divisible by $p$ === {{:Orbits of Group Action on Sets with Power of Prime Size/Orbit whose Length is Divisible by p}}
Axiom:Euclid's First Postulate
== Postulate ==  A straight line segment can be drawn joining any two points. {{EuclidPostulateStatement|First|To draw a straight line from any point to any point.}}  == Sources == * {{BookReference|Euclid: The Thirteen Books of The Elements: Volume 1|1926|ed = 2nd|edpage = Second Edition|Sir Thomas L. Heath|prev = Definition:Euclid's Definitions - Book I/23 - Parallel Lines|next = Axiom:Euclid's Second Postulate}}: Book $\text{I}$. Postulates * {{BookReference|Probability Theory|1965|A.M. Arthurs|prev = Definition:Straight Line|next = Definition:Axiom}}: Chapter $2$: Probability and Discrete Sample Spaces: $2.1$ Introduction * {{BookReference|Dictionary of Mathematics|1989|Ephraim J. Borowski|author2 = Jonathan M. Borwein|prev = Axiom:Euclid's Postulates|next = Axiom:Euclid's Second Postulate|entry = Euclid's axioms}} Euclid Postulate 1 n70ocyqo261qg5q7e6kw8y1t8m51qtw
Axiom:Euclid's Second Postulate
== Postulate ==  A straight line segment can be extended indefinitely to form a straight line. {{EuclidPostulateStatement|Second|To produce a finite straight line continuously in a straight line.}}  === Production === {{:Axiom:Euclid's Second Postulate/Production}} == Sources == * {{BookReference|Euclid: The Thirteen Books of The Elements: Volume 1|1926|ed = 2nd|edpage = Second Edition|Sir Thomas L. Heath|prev = Axiom:Euclid's First Postulate|next = Axiom:Euclid's Third Postulate}}: Book $\text{I}$. Postulates * {{BookReference|Dictionary of Mathematics|1989|Ephraim J. Borowski|author2 = Jonathan M. Borwein|prev = Axiom:Euclid's First Postulate|next = Axiom:Euclid's Third Postulate|entry = Euclid's axioms}} Euclid Postulate 2 e9p10asae0xjolxihapf4tieho2npm3
Axiom:Euclid's Third Postulate
== Postulate ==  Given any line segment, a circle can be drawn using the segment as the radius with one endpoint as the center. {{EuclidPostulateStatement|Third|To describe a circle with any centre  and distance.}}  == Sources == * {{BookReference|Euclid: The Thirteen Books of The Elements: Volume 1|1926|ed = 2nd|edpage = Second Edition|Sir Thomas L. Heath|prev = Axiom:Euclid's Second Postulate|next = Axiom:Euclid's Fourth Postulate}}: Book $\text{I}$. Postulates * {{BookReference|Dictionary of Mathematics|1989|Ephraim J. Borowski|author2 = Jonathan M. Borwein|prev = Axiom:Euclid's Second Postulate|next = Axiom:Euclid's Fourth Postulate|entry = Euclid's axioms}} Euclid Postulate 3 6s9apddjmowqegm9i6175yvq7plt7fd
Axiom:Euclid's Fourth Postulate
== Postulate ==  All right angles are congruent. {{EuclidPostulateStatement|Fourth|That all right angles are equal to one another.}}  == Sources == * {{BookReference|Euclid: The Thirteen Books of The Elements: Volume 1|1926|ed = 2nd|edpage = Second Edition|Sir Thomas L. Heath|prev = Axiom:Euclid's Third Postulate|next = Axiom:Euclid's Fifth Postulate}}: Book $\text{I}$. Postulates * {{BookReference|Dictionary of Mathematics|1989|Ephraim J. Borowski|author2 = Jonathan M. Borwein|prev = Axiom:Euclid's Third Postulate|next = Axiom:Euclid's Fifth Postulate|entry = Euclid's axioms}} * {{BookReference|Taming the Infinite|2008|Ian Stewart|prev = Axiom:Euclid's Common Notion 1|next = Definition:Axiom}}: Chapter $2$: The Logic of Shape: Euclid Euclid Postulate 4 gu9wlfhcdl178dovsnycj6mdlqgh6sq
Axiom:Euclid's Common Notions
== Common Notions ==  This is a set of axiomatic statements that appear at the start of Book $\text{I}$ of . === Common Notion 1 === {{:Axiom:Euclid's Common Notion 1}} === Common Notion 2 === {{:Axiom:Euclid's Common Notion 2}} === Common Notion 3 === {{:Axiom:Euclid's Common Notion 3}} === Common Notion 4 === {{:Axiom:Euclid's Common Notion 4}} === Common Notion 5 === {{:Axiom:Euclid's Common Notion 5}} == Historical Note == {{:Axiom:Euclid's Common Notions/Historical Note}} == References == <references/> == Sources == * {{BookReference|Euclid: The Thirteen Books of The Elements: Volume 1|1926|ed = 2nd|edpage = Second Edition|Sir Thomas L. Heath|prev = Axiom:Euclid's Postulates|next = Axiom:Playfair's Axiom}}: Book $\text{I}$. Common Notions * {{BookReference|Calculus Gems|1992|George F. Simmons|prev = Axiom:Euclid's Postulates|next = Definition:Axiom}}: Chapter $\text {A}.4$: Euclid (flourished ca. $300$ B.C.) * {{BookReference|The Penguin Dictionary of Mathematics|1998|David Nelson|ed = 2nd|edpage = Second Edition|prev = Definition:Euclidean Geometry|next = Axiom:Euclid's Postulates|entry = Euclidean geometry}} * {{BookReference|The Penguin Dictionary of Mathematics|2008|David Nelson|ed = 4th|edpage = Fourth Edition|prev = Definition:Euclidean Geometry|next = Axiom:Euclid's Postulates|entry = Euclidean geometry}} * {{BookReference|Taming the Infinite|2008|Ian Stewart|prev = Definition:Euclid's Definitions|next = Axiom:Euclid's Postulates}}: Chapter $2$: The Logic of Shape: Euclid * {{BookReference|The Concise Oxford Dictionary of Mathematics|2014|Christopher Clapham|author2 = James Nicholson|ed = 5th|edpage = Fifth Edition|prev = Axiom:Euclid's Postulates|next = Mathematician:Eudoxus of Cnidus|entry = Euclid's axioms}}
Closure of Subset of Metric Space by Convergent Sequence
== Lemma == Let $M$ be a metric space. Let $C \subseteq M$. Let $x \in M$. Let $\map \cl C$ denote the closure of $C$ in $T$. Then $x \in \map \cl C$  there exists a sequence $\sequence {x_n}$ in $C$ which converges to $x$. == Proof == === Necessary Condition === Suppose there exists a sequence $\sequence {x_n}$ in $C$ which converges to $x$. Let $\epsilon > 0$. Then by definition: :$\exists N \in \N: \forall n > N: x_n \in \map {B_\epsilon} x$ where $\map {B_\epsilon} x$ is the open $\epsilon$-ball of $x$ in $M$. Since $\forall n: x_n \in C$, it follows that: :$\forall \epsilon > 0: \map {B_\epsilon} x \cap C \ne \O$ Hence $x \in \map \cl C$. {{qed|lemma}} === Sufficient Condition === Now suppose $x \in \map \cl C$. By definition of closure: :$\forall n \in \N: \exists x_n \in C \cap \map {B_{1 / n} } x$ Thus clearly $\sequence {x_n}$ converges to $x$. {{handwaving|Lose the weasel word "clearly"}}  == Sources == * {{BookReference|Introduction to Metric and Topological Spaces|1975|W.A. Sutherland|prev = Definition:Sequentially Compact Space|next = Definition:Cauchy Sequence (Metric Space)}}: $7.2$: Sequential compactness: Lemma $7.2.2$
Series Law for Extremal Length/Rho is Well Defined
{{MissingLinks|throughout}} == Lemma for Series Law for Extremal Length == {{explain|The details of what is to be proved needs to be stated here, in order that this page be appropriately self-contained.}} == Proof == To see that $\rho$ is a well-defined metric, we need to check that it transforms correctly when changing local coordinates. Let $z = \map z t$ and $w = \map w t$ be charts on the Riemann surface $X$. Let $\map {\rho_1^z} t$ and $\map {\rho_1^w} t$ be the coefficient functions when $\rho_1$ is expressed in the local coordinates $z$ and $w$, respectively. We use the analogous notation for $\rho_2$ and $\rho$. Since $\rho_j$ is a metric for $j \in \set {1, 2}$, we have: :$\map {\rho_j^w} t = \map {\rho_j^z} t \cdot \size {\dfrac {\d z} {\d w} }$ where $\dfrac {\d z} {\d w}$ denotes, the derivative of the coordinate change $z \circ w^{-1}$. Thus we have: {{begin-eqn}} {{eqn | l = \map {\rho^w} t       | r = \max \set {\map {\rho_1^w} t, \map {\rho_2^w} t}       | c =  }} {{eqn | r = \max \set {\map {\rho_1^z} t, \map {\rho_2^z} t} \cdot \size {\frac {\d z} {\d w} }       | c =  }} {{eqn | r = \map {\rho^z} t \cdot \size {\dfrac {\d z} {\d w} }       | c =  }} {{end-eqn}} This means that $\rho$ transforms correctly and is a metric, as desired.  bp2yqxpiygxf9e7xgb07wg13fk5vska
Goldbach Conjecture
== Conjecture ==  Every even integer greater than $2$ is the sum of two primes.  === Marginal Conjecture === {{:Goldbach Conjecture/Marginal}} {{Hilbert23|8b}} == Landau's Problems == This is the $1$st of Landau's problems. == Also see == * Goldbach's Weak Conjecture * Goldbach Conjecture implies Goldbach's Marginal Conjecture {{Namedfor|Christian Goldbach|cat = Goldbach}} == Historical Note == {{:Goldbach Conjecture/Historical Note}} == Sources == * {{BookReference|Algebra Volume 1|1982|P.M. Cohn|edpage = Second Edition|ed = 2nd|prev = Proof by Counterexample|next = Definition:Sufficient Condition}}: Chapter $1$: Sets and mappings: $\S 1.1$: The need for logic * {{BookReference|Curious and Interesting Numbers|1986|David Wells|prev = Tamref's Last Theorem/Mistake|next = Definition:Binary Notation}}: $2$ * {{BookReference|Curious and Interesting Numbers|1997|David Wells|ed = 2nd|edpage = Second Edition|prev = 5^x + 12^y equals 13^z has Unique Solution|next = Definition:Binary Notation}}: $2$ * {{BookReference|The Penguin Dictionary of Mathematics|1998|David Nelson|ed = 2nd|edpage = Second Edition|prev = Gödel's Incompleteness Theorems/Historical Note|next = Goldbach Conjecture/Historical Note|entry = Goldbach's conjecture}} * {{BookReference|The Penguin Dictionary of Mathematics|2008|David Nelson|ed = 4th|edpage = Fourth Edition|prev = Gödel's Incompleteness Theorems/Historical Note|next = Goldbach Conjecture/Historical Note|entry = Goldbach's conjecture}} * {{BookReference|Taming the Infinite|2008|Ian Stewart|prev = Goldbach Conjecture/Marginal/Historical Note|next = Goldbach Conjecture/Historical Note}}: Chapter $7$: Patterns in Numbers: Fermat
Euclid's Lemma for Prime Divisors
{{Previous POTW|2 April 2009|10 April 2009}} == Lemma ==  Let $p$ be a prime number. Let $a$ and $b$ be integers such that: : $p \divides a b$ where $\divides$ means '''is a divisor of'''. Then $p \divides a$ or $p \divides b$.  === General Result === {{:Euclid's Lemma for Prime Divisors/General Result}} === Corollary === {{:Euclid's Lemma for Prime Divisors/Corollary}} == Proof 1 == {{:Euclid's Lemma for Prime Divisors/Proof 1}} == Proof 2 == {{:Euclid's Lemma for Prime Divisors/Proof 2}} == Proof 3 == {{:Euclid's Lemma for Prime Divisors/Proof 3}} == Also presented as == Some sources present this as: Let $p$ be a prime number. Let $a$ and $b$ be integers such that: :$a b \equiv 0 \pmod p$ Then either $a \equiv 0 \pmod p$ or $b \equiv 0 \pmod p$. == Also known as == Some sources give the name of this as '''Euclid's first theorem'''. {{Namedfor|Euclid}} == Also see == Some sources use this property to '''define''' a prime number. == Sources == * {{BookReference|Sets and Groups|1965|J.A. Green|prev = Modulo Arithmetic/Examples/Residue of 2^512 Modulo 5|next = Converse of Euclid's Lemma does not Hold}}: Chapter $2$. Equivalence Relations: Exercise $7$ * {{BookReference |An Introduction to the Theory of Numbers|1979|G.H. Hardy|author2 = E.M. Wright|ed = 5th|edpage = Fifth Edition|prev = Fundamental Theorem of Arithmetic|next = Euclid's Lemma for Prime Divisors/General Result}}: $\text I$: The Series of Primes: $1.3$ Statement of the fundamental theorem of arithmetic: Theorem $3$ * {{BookReference|Calculus Gems|1992|George F. Simmons|prev = Expression for Integer as Product of Primes is Unique|next = Definition:Irrational Number/Historical Note}}: Chapter $\text {B}.16$: The Sequence of Primes: Problem $1 \ \text{(b)}$
Continued Fraction Algorithm
== Algorithm ==  The '''Continued Fraction Algorithm''' is a method for finding the continued fraction expansion for any irrational number to as many partial quotients as desired. Let $x_0$ be the irrational number in question. The steps are: {{begin-eqn}} {{eqn | n = 1       | l = k       | r = 0       | c = initialise }} {{eqn | n = 2       | l = a_k       | r = \floor {x_k}       | c = the $k$th partial quotient (that is, $a_k$) is the integer part of $x_k$ }} {{eqn | n = 3       | l = x_{k + 1}       | r = \frac 1 {x_k - a_k}       | c = the subsequent term is the reciprocal of the fractional part of $x_k$ }} {{eqn | n = 4       | l = k       | r = k + 1       | c = increase $k$ by $1$ then go to step $(2)$ }} {{end-eqn}} Then $x_0 = \sqbrk {a_0, a_1, a_2, \ldots}$ is the required continued fraction expansion.  == Examples == {{:Continued Fraction Algorithm/Examples}} == Proof 1 == {{:Continued Fraction Algorithm/Proof 1}}
Clear Registers Program
== URM Program == Let $a, b \in \N$ be natural numbers such that $0 < a$. Then we define the URM program $Z \left({a, b}\right)$ to be: {| |- ! align="right" | Line !! ! align="left" | Command |- | align="right" | $1$ || | align="left" | $Z \left({a}\right)$ |- | align="right" | $2$ || | align="left" | $Z \left({a + 1}\right)$ |- | align="right" | $3$ || | align="left" | $Z \left({a + 2}\right)$ |- | align="right" | $\vdots$ || | align="left" | $\vdots$ |- | align="right" | $b - a + 1$ || | align="left" | $Z \left({b}\right)$ |} This program clears (that is, sets to $0$) all the registers from $R_a$ through to $R_b$. If $a > b$ then $Z \left({a, b}\right)$ is the null URM program. The length of $Z \left({a, b}\right)$ is: :$\lambda \left({Z \left({a, b}\right)}\right) = \begin{cases} 0 & : a > b \\ b - a + 1 & : a \le b \end{cases}$ == Proof == Self-evident.
Biconditional Properties
== Theorems == === Biconditional is Commutative === {{:Biconditional is Commutative}} === Biconditional is Associative === {{:Biconditional is Associative}} === Biconditional is Reflexive === {{:Biconditional is Reflexive}}
Finished Set Lemma
== Lemma == Let $\Delta$ be a finished set of WFFs of propositional logic. Then any model of the set of basic WFFs in $\Delta$ is a model of $\Delta$. === Corollary === {{:Finished Set Lemma/Corollary}} == Proof == Let $\mathcal M$ be a model of the set of basic WFFs in $\Delta$. We need to show that $\mathcal M \models \Delta$. That is, that $\mathcal M \models \mathbf C$ for each $\mathbf C \in \Delta$. Now, let $R \left({n}\right)$ be a propositional function on the set of natural numbers $\N$ such that: :$R \left({n}\right) = T$ iff: for every WFF $\mathbf C$, if $\mathbf C \in \Delta$ and $\mathbf C$ has length at most $n$, then $\mathcal M \models \mathbf C$. It is clear that $R \left({0}\right), R \left({1}\right), R \left({2}\right)$ are true because every WFF length 2 or less is basic, and $\mathcal M$ models every basic WFF in $\Delta$. So, assume $R \left({k}\right)$ is true for some $k \in \N$. Suppose $\mathbf C$ has length at most $k+1$ and belongs to $\Delta$. By examining each of the cases in the definition of finished set, we see that since $\mathcal M$ models every WFF in $\Delta$ of length at most $k$, then $\mathcal M$ models $\mathbf C$. Thus $R \left({k+1}\right)$ is true. Thus by strong induction, $R \left({n}\right)$ is true for all $n \in \N$. Hence the result.  == Sources == * {{BookReference|Mathematical Logic and Computability|1996|H. Jerome Keisler|author2 = Joel Robbin|prev = Finished Set Lemma/Corollary|next = Completeness Theorem for Propositional Tableaus and Boolean Interpretations}}: $\S 1.9$: Finished Sets: Lemma $1.9.1$
Finite Main Lemma of Propositional Tableaus
{{mergeto|Main Lemma of Propositional Tableaus}} == Lemma == Let $\mathbf H$ be a finite set of WFFs of propositional logic. Either $\mathbf H$ has a tableau confutation or $\mathbf H$ has a model. == Proof == Let $\mathbf H$ be a finite set of WFFs of propositional logic which does not have a tableau confutation. By the Tableau Extension Lemma, the tableau which consists only of a root node with hypothesis set $\mathbf H$ can be extended into a finite finished tableau $T$. The tableau $T$ still has root $\mathbf H$. Since $T$ is not a confutation, it has a finished branch $\Gamma$. By the Corollary to the Finished Branch Lemma, $\Gamma$ has a model, $\mathcal M$, say. In particular, $\mathcal M$ is a model of $\mathbf H$ as required.  == Comment == From Tableau Confutation implies Unsatisfiable, we already know that $\mathbf H$ can not have both a tableau confutation ''and'' a model. This result gives us that $\mathbf H$ has a tableau confutation iff $\mathbf H$ does not have a model. == Sources == * {{BookReference|Mathematical Logic and Computability|1996|H. Jerome Keisler|author2=Joel Robbin|prev=Completeness Theorem for Propositional Tableaus and Boolean Interpretations|next=Definition:Finished Branch of Propositional Tableau}}: $\S 1.10$: Completeness: Lemma $1.10.1$
Tableau Extension Lemma
== Lemma == Let $T$ be a finite propositional tableau. Let its hypothesis set $\mathbf H$ be finite. Then $T$ can be extended to a finished finite propositional tableau $T'$ which also has hypothesis set $\mathbf H$. === General Statement === {{:Tableau Extension Lemma/General Statement}} == Proof == Let $\mathbf A$ be a WFF of propositional logic at a node $t$ such that: * $\mathbf A$ is not a basic WFF; * There is a non-contradictory branch through $t$ on which $\mathbf A$ is not used. Such a WFF $\mathbf A$ will be called '''unused'''. Note that a tableau is finished iff there are no unused WFFs in the tableau. Let $u \left({T}\right)$ be the length of the longest unused WFF in $T$. If $T$ is finished, we set $u \left({T}\right) = 0$. Since there are only finitely many WFFs in $T$, the number $u \left({T}\right)$ must exist. Let $R \left({n}\right)$ be the proposition that every finite propositional tableau with hypothesis set $\mathbf H$ and with $u \left({T}\right) < n$ can be extended to a finite finished tableau. That is, $R \left({n}\right)$ is an assertion that this lemma is true whenever $u \left({T}\right) < n$. The statement $R \left({1}\right)$ is true, because a tableau $T$ such that $u \left({T}\right) < 1$ is already finished. So, let us assume the truth of $R \left({k}\right)$. Suppose that $u \left({T}\right) < k+1$. We extend $T$ to a new tableau $T'$ by using every unused WFF $\mathbf A$ in $T$ once on every non-contradictory branch through $\mathbf A$. Each of the unused WFFs in $T$ is used in the new tableau $T'$. Also, each new WFF which was added when forming $T'$ has a length less than $u \left({T}\right)$, because the added WFFs are always shorter than the used WFFs. So $u \left({T'}\right) < u \left({T}\right) < k + 1$, so $u \left({T'}\right) < k$. So, by the induction hypothesis $R \left({k}\right)$, there is a finite finished extension $T''$ of $T'$. $T''$ is also a finished extension of $T$. Hence $R \left({k+1}\right)$, and the result follows by strong induction.  == Sources == * {{BookReference|Mathematical Logic and Computability|1996|H. Jerome Keisler|author2 = Joel Robbin|prev = Definition:Extension of Propositional Tableau/Definition 1|next = Extended Completeness Theorem for Propositional Tableaus and Boolean Interpretations}}: $\S 1.10$: Completeness: Lemma $1.10.2$