hcju/proofwikiqa huggingface dataset:
By Subset of Well-Ordered Set is Well-Ordered, $S_a$ is well-ordered. Suppose that $b \in S_a$. From Ordering on Ordinal is Subset Relation, and the definition of an initial segment, it follows that $b \subset a$. Then: {{begin-eqn}} {{eqn | l = \paren {S_a}_b       | r = \set {x \in S_a: x \subset b}       | c = {{Defof|Initial Segment}} }} {{eqn | r = \set {x \in S: x \subset a \land x \subset b}       | c = {{Defof|Initial Segment}} }} {{eqn | r = \set {x \in S: x \subset b}       | c = as $b \subset a$ }} {{eqn | r = S_b       | c = {{Defof|Initial Segment}} }} {{eqn | r = b       | c = as $S$ is an ordinal }} {{end-eqn}} The result follows from the definition of an ordinal. {{qed}}
Recall that the Ordering on Ordinal is Subset Relation. Let $a$ be the minimal element of $S \setminus T$. By definition of minimal element: :$\forall x \in S: x \subset a$ Hence $x \notin S \setminus T$. By definition of set difference: :$x \in T$ Thus by definition of initial segment: :$S_a \subseteq T$ Now let $b \in T$. Then by definition of ordinal: :$T_b = b = S_b$ Now, if $a \subset b$ then $a \in S_b$. So $a \in T_b$ and hence $a \in T$. But $a \in S \setminus T$, so by definition of set difference, this is not the case. So $a \not \subset b$. We have that an ordinal is well-ordered by $\subseteq$. So by the definition of well-ordering, $\subseteq$ is a total ordering on $S$ So it follows that $b \subseteq a$. But $b \ne a$ since $b \in T$. Hence $b \subset a$, and so $b \in S_a$. This proves that $T \subseteq S_a$. Thus $T = S_a$. {{qed}}
Let $a \in S \cap T$. Then the initial segments $S_a$ and $T_a$ are such that $S_a = a = T_a$. That is: :$\set {x \in S: x \subset a} = a = \set {y \in T: y \subset a}$. So: :$a = \set {z \in S \cap T: z \subset a} = \paren {S \cap T}_a$. {{qed}}
{{improve|Worth removing the technical complexity to make it more accessible<br/>A better approach may be to write a second proof rather than change this one.}} Aiming for a contradiction, suppose that the claim is false. That is, by De Morgan's laws: Conjunction of Negations: :$\left({\neg \left({S \subseteq T}\right)}\right) \land \left({\neg \left({T \subseteq S}\right)}\right)$ where $\neg$ denotes logical not and $\land$ denotes logical and. Now from Intersection is Subset, we have $S \cap T \subseteq S$ and $S \cap T \subseteq T$. Also, by Intersection with Subset is Subset, we have $S \cap T \ne S$ and $S \cap T \ne T$. That is, $S \cap T \subsetneq S$ and $S \cap T \subsetneq T$. Note that by Intersection of Two Ordinals is Ordinal, $S \cap T$ is an ordinal. So by Transitive Set is Proper Subset of Ordinal iff Element of Ordinal, we have: :$S \cap T \in S$ :$S \cap T \in T$ But then $S \cap T \in S \cap T$, which contradicts Ordinal is not Element of Itself. {{qed}} Category:Ordinals l12b3z0c2cmfhnu6sacixy5ivmdb32n
Let $S \cong T$. {{AimForCont}} that $S \ne T$. Then from the corollary to Relation between Two Ordinals, either: :$S$ is an initial segment of $T$ or: :$T$ is an initial segment of $S$. But as $S \cong T$, from No Isomorphism from Woset to Initial Segment, neither $S$ nor $T$ can be an initial segment of the other. From this contradiction it follows that $S = T$. {{qed}}
From No Isomorphism from Woset to Initial Segment, neither $A$ nor $B$ can be an initial segment of the other. By definition, every element of an ordinal is an initial segment of it. Hence, neither $A$ nor $B$ can be an element of the other. By Ordinal Membership is Trichotomy, it follows that $A = B$. {{qed}}
For each $a \in S$, let $g_a: S_a \to \map Z a$ be an order isomorphism from $S_a$ to an ordinal $\map Z a$. By Isomorphic Ordinals are Equal and Order Isomorphism between Wosets is Unique, both $\map Z a$ and $g_a$ are unique. So this defines a mapping $Z$ on $S$. Let the image of $Z$ be $W$: :$W = \set {\map Z a: a \in S}$ Now we define $f: S \to W$ as: :$\map f a = \map Z a$ Now we show that $x, y, \in S, x \prec y \implies \map Z x \subset \map Z y$. So, let $x, y, \in S$ such that $x \prec y$. Then: :$(1) \quad g_x: S_x \cong \map Z x$ Also, because: {{begin-eqn}} {{eqn | l = S_x       | r = \set {z \in S: z \prec x}       | c = Initial Segment of Ordinal is Ordinal }} {{eqn | r = \set {z \in S: z \prec y \land z \prec x}       | c =  }} {{eqn | r = \set {z \in S_y: z \prec x}       | c =  }} {{eqn | r = \paren {S_y}_x       | c =  }} {{end-eqn}} we have: :$(2): \quad \map {g_y} {S_x}: S_x \to \paren {\map Z y}_{\map {g_y} x}$ is an order isomorphism So by Isomorphic Ordinals are Equal, we have: :$(3): \quad \map Z x = \paren {\map Z y}_{\map {g_y} x}$ So, in particular: :$\map Z x \subset \map Z y$ By this result, $f: X \to W$ is a bijection. Also by this result, $f: X \to \struct {W, \subseteq}$ is an order isomorphism. This means that $W$ is well-ordered by $\subseteq$. Now we show that $W$ is an ordinal. Let $y \in S$. Since $\map Z y$ is an ordinal, we have: :$x \prec y \implies \paren {\map Z y}_{\map {g_y} x} = \map {g_y} x$ So by $(3)$ above, we have: :$(4) \quad x \prec y \implies \map Z x = \map {g_y} x$ Hence: {{begin-eqn}} {{eqn | l = W_{\map Z y}       | r = \set {\map Z x: \map Z x \subset \map Z y}       | c =  }} {{eqn | r = \set {\map Z x: x \prec y}       | c =  }} {{eqn | r = \set {\map {g_y} x: x \prec y}       | c =  }} {{eqn | r = \map {g_y} {S_y}       | c =  }} {{eqn | r = \map Z y       | c =  }} {{end-eqn}} As $y$ is an arbitrary element of $S$, it follows that $\map Z y$ is an arbitrary element of $W$. So $W$ is an ordinal, as we wanted to prove. {{qed}}
=== Existence === Let $\struct {S, \preceq}$ be a woset. From Condition for Woset to be Isomorphic to Ordinal, it is enough to show that for every $a \in S$, the initial segment $S_a$ of $S$ determined by $a$ is order isomorphic to some ordinal. Let: :$E = \set {a \in S: S_a \text{ is not isomorphic to an ordinal} }$ We will show that $E = \O$. {{AimForCont}} that $E \ne \O$. Let $a$ be the minimal element of $E$. This is bound to exist by definition of woset. So, if $x \prec a$, it follows that $S_x$ is isomorphic to an ordinal. But for $x \prec a$, we have $S_x = \paren {S_a}_x$ from definition of an ordinal. So every segment of $S_a$ is isomorphic to an ordinal. Hence from Condition for Woset to be Isomorphic to Ordinal, $S_a$ itself is isomorphic to an ordinal. This contradicts the supposition that $a \in E$. Hence $E = \O$ and existence has been proved. {{qed|lemma}} === Uniqueness === Uniqueness follows from Isomorphic Ordinals are Equal. Hence the result. {{qed}}
Let row $i$ of $A$ be multiplied by column $i$ of $A^\intercal$. This is the same as multiplying row $i$ of $A$ by row $i$ of $A$. Each row of $A$ has $r$ entries (since any point must be in $r$ blocks). Then: : $ \left({a_{ii}}\right) = r = \sum $ of the all the $1$'s in row $i$ This completes the main diagonal. Let row $i$ of $A$ be multiplied by column $j$ of $A^\intercal$. This is the same as multiplying row $i$ of $A$ by the row $j$ row of $A$. This will give the number of times point $i$ point is the same block as point $j$. Therefore: : $i \ne j \implies \left({a_{ij}}\right) = \lambda$ So: :$A^\intercal \cdot A = \left({a_{ij}}\right) = \left({r - \lambda}\right) I_v + \lambda{J_v}$ {{qed}} {{explain|The order of this matrix is not apparent - it needs to be made more explicit.}} {{explain|Better idea - write a separate page to express this result for a general logical matrix. I think this is a valid construction. Establish the result as the fact that it is a combinatorial matrix.}} Category:Design Theory 8y5b1s3g8bnl9z93k5vclpp7tpd9vo1
Let $A$ be the incidence matrix. By Product of the Incidence Matrix of a BIBD with its Transpose, we have that: :$A^\intercal \cdot A = \begin{bmatrix}  r & \lambda & \cdots & \lambda \\ \lambda & r & \cdots & \lambda \\ \vdots & \vdots & \ddots & \vdots \\ \lambda & \lambda & \cdots & r \\ \end{bmatrix}$ From Necessary Condition for Existence of BIBD: : $r > \lambda$ So we can write $r = \lambda + \mu$ for some $\mu > 0$ and so: :$A^\intercal \cdot A = \begin{bmatrix}  \lambda + \mu & \lambda & \cdots & \lambda \\ \lambda & \lambda + \mu & \cdots & \lambda \\ \vdots & \vdots & \ddots & \vdots \\ \lambda & \lambda & \cdots & \lambda + \mu \\ \end{bmatrix}$ This is a combinatorial matrix of order $v$. So: {{begin-eqn}} {{eqn | l = \map \det {A^\intercal \cdot A}       | r = \mu^{v - 1} \paren {\mu + v \lambda}       | c = Determinant of Combinatorial Matrix }} {{eqn | r = \paren {r + \paren {v - 1} \lambda} \paren {r - \lambda}^{v - 1}       | c =  }} {{eqn | r = r k \paren {r - \lambda}^{v - 1}       | c = Necessary Condition for Existence of BIBD }} {{end-eqn}} {{handwaving|obvious}} Now, since $k < v$ (this is obvious) and using the properties of a BIBD, we get that $r > \lambda$.  So $\map \det {A^\intercal A} \ne 0$. Since $A^\intercal A$ is a $v \times v$ matrix, then the rank, $\rho$, of $A^\intercal A = v$. Using the facts that $\map \rho {A^\intercal A} \le \map \rho A$, and $\map \rho A \le b$ (since $A$ only has $b$ cols), we have that: :$v \le \map \rho A \le b$ {{qed}} {{explain|Some "basic properties" here need to be linked, in order that the general thrust of this can be understood.}} {{explain|We can probably lose most of the fiddly stuff at the end by noting that as $\lambda$ and $\mu$ are both $>0$ then so is the determinant, and go straight to the last-line result about rank.}} {{proofread}} {{Namedfor|Ronald Aylmer Fisher|cat = Fisher}} Category:Design Theory tler7gnjs1h7z4j34l5naobamnjdhs9
{{BeginTableau|\vdash \left({p \implies q}\right) \lor \left({q \implies p}\right)}} {{ExcludedMiddle|1|p \lor \neg p}} {{Assumption|2|p}} {{SequentIntro|3|2|q \implies p|2|True Statement is implied by Every Statement}} {{Addition|4|2|\left({p \implies q}\right) \lor \left({q \implies p}\right)|2|2}} {{Assumption|5|\neg p}} {{SequentIntro|6|5|p \implies q|5|False Statement implies Every Statement}} {{Addition|7|5|\left({p \implies q}\right) \lor \left({q \implies p}\right)|7|1}} {{ProofByCases|8||\left({p \implies q}\right) \lor \left({q \implies p}\right)|1|2|4|5|7}} {{EndTableau}} {{qed}}
We apply the Method of Truth Tables to the proposition. As can be seen by inspection, the truth values under the main connective is true for all boolean interpretations, proving a tautology. $\begin{array}{|ccccccc|} \hline (p & \implies & q) & \lor & (q & \implies & p) \\ \hline F & T & F & T & F & T & F \\ F & T & T & T & T & F & F \\ T & F & F & T & F & T & T \\ T & T & T & T & T & T & T \\ \hline \end{array}$ {{qed}}
First note, we have: : $(1): \quad A J = J A = k J$, so $A^\intercal J = \left({J A}\right)^\intercal = \left({k J}\right)^\intercal = k J$, and likewise $J A^\intercal = k J$ : $(2): \quad J^2 = v J$ : $(3): \quad$ If a design is symmetric, then $A A^\intercal = \left({r - \lambda}\right) I + \lambda J = \left({k - \lambda}\right) I + \lambda J$ {{explain|We need to establish these three results. We need to confirm that $J$ is the ones matrix (I think it is, from the context) and then make sure of those above results.}} From $(3)$, we get: {{begin-eqn}} {{eqn | l=\left({A^\intercal - \sqrt{\left({\frac \lambda v}\right) J} }\right) \left({A + \sqrt{\left({\frac{\lambda}v}\right)J} }\right)       | r=A^\intercal A + \sqrt{\frac \lambda v} \left({A^\intercal J - J A}\right) - \frac \lambda v J^2 }} {{eqn | r=A^\intercal A - \lambda J = \left({k - \lambda}\right) I }} {{end-eqn}} We now have that: : $\displaystyle \frac 1 {k - \lambda} \left({A + \sqrt{\left({\frac \lambda v}\right) J}}\right)$ is the inverse of: : $A^\intercal - \sqrt{\left({\dfrac \lambda v}\right) J}$ which implies that they commute with each other. {{explain|... why?}} Thus: {{begin-eqn}} {{eqn |l=\left({k - \lambda}\right) I       |r=\left({A + \sqrt {\frac \lambda v} J}\right) \left({A^\intercal - \sqrt{\frac \lambda v} J}\right) }} {{eqn |r=A A^\intercal + \sqrt{\frac \lambda v} \left({J A^\intercal - A J}\right) - \frac \lambda v J^2 }} {{eqn |r=A A^\intercal - \lambda J }} {{end-eqn}} {{explain|The above steps need justification}} whence: : $A A^\intercal = \left({k - \lambda}\right) + \lambda J = A^\intercal A$ {{qed}} Category:Design Theory iwjr9nexx8v1i2po72j27yez52upfb0
{{AimForCont}} that $G$ contains a subgraph $C$ such that: :$C$ is a cycle graph :$C \ne G$ is non-empty. Then there exists some vertex $v$ that is not in $C$. Let $u$ be any vertex of $C$. Since $G$ is a cycle graph, it is connected. Therefore there is a walk from $u$ to $v$ in $G$. There must be some vertex $x$ that is the last vertex in $C$ along that walk. Therefore, $x$ is adjacent to a vertex not in $C$. Thus it has a degree of at least $3$. But $G$ is a cycle graph and every vertex in a cycle graph has degree $2$. The result follows by Proof by Contradiction. {{Qed}} Category:Graph Theory glgb44drqfuvakgpis4ekwp8aonxm0r
Let $P = p_1 p_2 \ldots p_k$ be the longest path in $G$. If $p_1$ is adjacent to some vertex $v$ not in $P$, then the path $v p_1 p_2 \ldots p_k$ would be longer than $P$, contradicting the choice of $P$. The same argument can be made for $p_k$. So both $p_1$ and $p_k$ are adjacent only to vertices in $P$. Since $\deg(p_1) \ge \dfrac n 2$ and $p_1$ cannot be adjacent to itself, $k \ge \dfrac n 2 +1$. '''Claim:''' There is some value of $j$ ($1 \le j \le k$) such that: * $p_j$ is adjacent to $p_k$, and * $p_{j+1}$ is adjacent to $p_1$. Suppose that the claim is not true. Then since all vertices adjacent to $p_1$ or $p_k$ lie on $P$, there must be at least $\deg(p_1)$ vertices on $P$ not adjacent to $p_k$. Since all the vertices adjacent to $p_k$ and $p_k$ itself also lie on $P$, the path must have at least $\deg(p_1) + \deg(p_k) + 1 \ge n+1$ vertices. But $G$ has only $n$ vertices: a contradiction. This gives a cycle $C = p_{j+1} p_{j+2} \ldots p_{k} p_j p_{j-1} \ldots p_2 p_1 p_{j+1}$. Suppose $G - C$ is nonempty. Then since $G$ is connected, there must be a vertex $v \in G-C$ adjacent to some $p_i$. So the path from $v$ to $p_i$ and then around $C$ to the vertex adjacent to $p_i$ is longer than $P$, contradicting the definition of $P$. Therefore all vertices in $G$ are contained in $C$, making $C$ a Hamilton cycle. {{Qed}}
Take any two non-adjacent vertices $u, v \in G$. Then: :$\displaystyle \deg u + \deg v \ge \frac n 2 + \frac n 2 = n$ The result follows by a direct application of Ore's Theorem. {{qed}}
We apply the Method of Truth Tables to the proposition. As can be seen by inspection, in each case, the truth values under the main connectives match for all boolean interpretations. $\begin{array}{|ccccc||ccccc|} \hline (p & \iff & q) & \iff & r & (p & \oplus & q) & \oplus & r \\ \hline F & T & F & F & F & F & F & F & F & F \\ F & T & F & T & T & F & F & F & T & T \\ F & F & T & T & F & F & T & T & T & F \\ F & F & T & F & T & F & T & T & F & T \\ T & F & F & T & F & T & T & F & T & F \\ T & F & F & F & T & T & T & F & F & T \\ T & T & T & F & F & T & F & T & F & F \\ T & T & T & T & T & T & F & T & T & T \\ \hline \end{array}$ {{qed}} From Biconditional is Associative and Exclusive Or is Associative, we have that both $\iff$ and $\oplus$ are associative, which justifies the rendition of this result without parentheses.
In NAND is Functionally Complete it is demonstrated that NAND is functionally complete. In NOR is Functionally Complete it is demonstrated that NOR is functionally complete.
: $1:$ Suppose $\phi$ is an atom $p$. Then we need to show that $p \vdash p$ and $\neg p \vdash \neg p$. These are proved in one line in the proof of the Law of Identity. : $2:$ Suppose $\phi$ is of the form $\neg \phi_1$. There are two cases to consider: * Suppose $\phi$ evaluates to $T$. {{finish|More hard work. I'll come back to this once I've got some more basics done.}} Category:Propositional Logic odpu8tmuf1zc50moxxiynr9im73pl6r
We refer to the rules of formation. From $\mathbf W: TF$, $\top$ and $\bot$ (both of length 1) are WFFs. From $\mathbf W: \mathcal P_0$, all elements of $\mathcal P_0$ (all of length 1) are WFFs. Every other rule of formation of the formal grammar of propositional logic consists of an existing WFF in addition to at least one other primitive symbol. Hence the result. {{qed}}
We will prove by strong induction on $n$ that: :All WFFs of length $n$ are balanced. Let $l \left({\mathbf A}\right)$ denote the number of left brackets in a string $\mathbf A$. Let $r \left({\mathbf A}\right)$ denote the number of right brackets in a string $\mathbf A$. === Induction Basis === The only WFFs of PropLog of Length 1 are letters of the alphabet and the symbols $\bot$ and $\top$. By definition, none of these consist of brackets. So every WFF $\mathbf A$ of length $1$ has $l \left({\mathbf A}\right) = r \left({\mathbf A}\right) = 0$. So every WFF of length $1$ is balanced. This provides the induction basis. === Induction Hypothesis === Now, suppose all WFFs of propositional logic of length at most $n$ are balanced. This provides the induction hypothesis. === Induction Step === Let $\mathbf A$ be a WFF of propositional logic of length $n + 1$. The following cases are to be considered (with $\mathbf B$ and $\mathbf C$ WFFs): :$\mathbf A = \neg \mathbf B$ :$\mathbf A = \left({\mathbf B \circ \mathbf C}\right)$, where $\circ$ is one of the binary connectives Suppose $\mathbf A = \neg \mathbf B$. Then $\mathbf B$ is a WFF of length $n$. By the induction hypothesis, it is hence balanced. Since $\mathbf A$ contains the same brackets as $\mathbf B$, it must also be balanced. Suppose now $\mathbf A = \left({\mathbf B \circ \mathbf C}\right)$, where $\circ$ is one of the binary connectives. Then: :$l \left({\mathbf A}\right) = l \left({\mathbf B}\right) + l \left({\mathbf C}\right) + 1$ :$r \left({\mathbf A}\right) = r \left({\mathbf B}\right) + r \left({\mathbf C}\right) + 1$. Now, $\mathbf B$ and $\mathbf C$ are both of length strictly less than $n + 1$, and thus balanced, by the induction hypothesis. That is: :$l \left({\mathbf B}\right) = r \left({\mathbf B}\right)$ :$l \left({\mathbf C}\right) = r \left({\mathbf C}\right)$. It follows that also: :$l \left({\mathbf A}\right) = r \left({\mathbf A}\right)$ which precisely states that $\mathbf A$ is balanced. We assumed that all WFFs of length $n$ and less are balanced, and demonstrated that as a consequence all WFFs of length $n+1$ are balanced. {{qed|lemma}} The result now follows by strong induction. {{qed}}
The transversal $AD$ cuts the lines $BC$ and $AE$ and makes $\angle DAE = \angle ADC$. From Equal Alternate Angles implies Parallel Lines it follows that $EA \parallel BC$. {{qed}} {{Euclid Note|31|I}}
:300px Let $\triangle ABC$ be a triangle. Let $BC$ be extended to a point $D$. From External Angle of Triangle equals Sum of other Internal Angles: : $\angle ACD = \angle ABC + \angle BAC$ Bby by Euclid's Second Common Notion: : $\angle ACB + \angle ACD = \angle ABC + \angle BAC + \angle ACB$ But from Two Angles on Straight Line make Two Right Angles, $\angle ACB + \angle ACD$ equals two right angles. So by Euclid's First Common Notion, $\angle ABC + \angle BAC + \angle ACB$ equals two right angles. {{qed}}
:500px Let $\Delta ABC$ be a triangle. Let $DAE$ be a line such that $DE \parallel BC$. By Parallelism implies Equal Alternate Angles, $\angle DAB = \angle ABC$ and $\angle EAC = \angle ACB$. Therefore, the sum of the three angles is $\angle ABC + \angle BCA + \angle CAB = \angle DAB + \angle BAC + \angle CAE = 180^\circ$. {{qed}}
This directly follows from: * The Modus Ponendo Ponens: $p \implies q, p \vdash q$ * The Rule of Implication: $\left({p \vdash q}\right) \vdash p \implies q$. {{qed}}
This is a corollary of the Extended Soundness Theorem for Propositional Tableaus and Boolean Interpretations: Let $\mathbf H$ be a countable set of propositional formulas. Let $\mathbf A$ be a propositional formula. If $\mathbf H \vdash \mathbf A$, then $\mathbf H \models \mathbf A$. In this case, we have $\mathbf H = \varnothing$. Hence the result. {{qed}}
Suppose that $v$ models $\mathbf A$: :$v \models \mathbf A$ Then $v \left({\mathbf A}\right) = T$ by definition of models. By definition of boolean interpretation, $v \left({\neg \mathbf A}\right) = F$. In particular, $v (\neg \mathbf A) \ne T$, so that: :$v \not\models \neg \mathbf A$ Hence the result. {{qed}} Category:Propositional Logic r1x63wssaxbv1ldoku4z649dfz2wow9
By definition of tableau proof, $\mathbf H \vdash_{\mathrm{PT}} \mathbf A$ means: :There exists a tableau confutation of $\mathbf H \cup \set {\neg\mathbf A}$. By Tableau Confutation implies Unsatisfiable, it follows that $\mathbf H \cup \set {\neg\mathbf A}$ is unsatisfiable for boolean interpretations. Therefore, if some boolean interpretation $v$ models $\mathbf H$: :$v \models_{\mathrm{BI}} \mathbf H$ then since $\mathbf H \cup \set {\neg\mathbf A}$ is unsatisfiable: :$v \not\models_{\mathrm{BI}} \neg\mathbf H$ Now by definition of the relation $\models_{\mathrm{BI}}$, it must be that: :$\map v {\neg \mathbf A} = F$ By the truth table for $\neg$, this implies: :$\map v {\mathbf A} = T$ which is to say $v \models_{\mathrm{BI}} \mathbf A$. Hence: :$v \models_{\mathrm{BI}} \mathbf H$ implies $v \models_{\mathrm{BI}} \mathbf A$ that is, $\mathbf A$ is a $\mathrm{BI}$-semantic consequence of $\mathbf H$: :$\mathbf H \models_{\mathrm{BI}} \mathbf A$ which was to be shown. {{qed}}
Let $\mathbf A$ be a semantic consequence of $\mathbf H$ for boolean interpretations. That is, if $v \models_{\mathrm{BI}} \mathbf H$, also $v \models_{\mathrm{BI}} \mathbf A$. By the truth function for $\neg$, it follows that for such $v$: :$v \not\models_{\mathrm{BI}} \neg \mathbf A$ Therefore, $\mathbf H' := \mathbf H \cup \left\{{\mathbf A}\right\}$ is unsatisfiable for boolean interpretations. Since $\mathbf H'$ is countable, it follows from the Compactness Theorem for Boolean Interpretations that: :Some finite $\mathbf H'' \subseteq \mathbf H'$ is unsatisfiable. By the Tableau Extension Lemma, there exists a finite finished tableau $T$ for $\mathbf H''$. By definition of finished tableau, every branch of $T$ is finished or contradictory. From the Corollary to the Finished Branch Lemma, $\Phi \left[{\Gamma}\right]$ is satisfiable for any finished branch $\Gamma$. But since $\mathbf H'' \subseteq \Phi \left[{\Gamma}\right]$, this would imply $\mathbf H''$ is also satisfiable, which is a contradiction. It follows that every branch of $T$ is contradictory. Since $\mathbf H'' \subseteq \mathbf H'$, replacing the hypothesis set $\mathbf H'$ of $T$ with $\mathbf H''$ yields another propositional tableau $T'$. Since every branch of $T'$ is contradictory, $T'$ is a tableau confutation of $\mathbf H'$. Recalling that $\mathbf H' = \mathbf H \cup \left\{{\neg\mathbf A}\right\}$, we conclude that $T'$ is a tableau proof of $\mathbf A$ from $\mathbf H$: :$\mathbf H \vdash_{\mathrm{PT}} \mathbf A$ {{qed}}
This is a corollary of the Extended Completeness Theorem for Propositional Tableaus and Boolean Interpretations. Namely, it is the special case $\mathbf H = \varnothing$. Hence the result. {{qed}}
We will show that we can choose an infinite sequence of nodes $t_0, t_1, t_2, \ldots$ of $T$ such that: * $t_0$ is the root node * $t_{n+1}$ is a child of $t_n$ Then the sequence $t_0, t_1, t_2, \ldots$ is such a branch of infinite length. Let $I$ be the set of all nodes in $T$ that have infinitely many descendants. Define a relation $\mathcal R$ on $I$ by letting $x \mathrel{\mathcal R} y$ {{iff}} $y$ is a child of $x$. Since $T$ has infinitely many nodes, its root node is in $I$. Let $t \in I$. By the premise, $t$ has a finite number of children. Suppose for the sake of contradiction that all of these childen had a finite number of descendants. Then by Finite Union of Finite Sets is Finite, that would mean that $t$ had a finite number of descendants, contradicting the fact that $t$ is in $I$. So $t$ has at least one child with infinitely many descendants. That is, $\mathcal R \left({t}\right)$ is non-empty for each $t \in I$. That is, $\mathcal R$ is a left-total relation. Thus by Dependent Choice for Finite Sets, there is an infinite sequence $\left\langle{t_n}\right\rangle$ such that: : $t_0$ is the root of $T$. : For all $n \in \N$, $t_n \mathrel{\mathcal R} t_{n+1}$. That is, $t_{n+1}$ is a child of $t_n$. {{qed}}