hcju/proofwikiqa huggingface dataset:
From the definition of the facets, each face is divided into $9$ facets. :600px A facet is either: :on the corner of a face, for example $flu$, $fru$ :on the edge of a face, for example $fu$, $fr$ :in the center of a face, for example $F$. :$(1):\quad$ Each facet can be either in $S_C$ or $S_E$ or $S_Z$ and can not be in more than one. :$(2):\quad$ Each facet can be either in $S_C$ or $S_E$ or $S_Z$ and there are no other possibilities. :$(3):\quad$ None of $S_C$, $S_E$ and $S_Z$ is empty. Thus the criteria for $S = \set {S_C \mid S_E \mid S_Z}$ to be a partition are fulfilled. {{qed}}
Let $n \in \Z$ be an integer. Let $\Bbb O$ be the set of odd integers and $\Bbb E$ be the set of even integers. By the Division Theorem: :$\forall n \in \Z: \exists! q, r \in \Z: n = 2 q + r, 0 \le r < 2$ from which it follows that either: :$n = 2 q \in \Bbb E$ or: :$n = 2 q + 1 \in \Bbb O$ Thus: :$(1): \quad$ each element of $\Z$ is in no more than one of $\Bbb E$ and $\Bbb O$ :$(2): \quad$ each element of $\Z$ is in at least one of $\Bbb E$ and $\Bbb O$ and: :$(3): \quad$ setting $q = 0$ it is seen that $0 \in \Bbb E$ and $1 \in \Bbb O$ and so neither $\Bbb E$ or $\Bbb O$ is empty. Thus $\set {\Bbb E \mid \Bbb O}$ is a partition of $\Z$ by definition. {{qed}}
Let $P$ and $Q$ be propositional functions. Let $S$ and $T$ be subsets of a universe $\Bbb U$ such that: :$S = \set {x \in \Bbb U: \map P x}$ :$T = \set {x \in \Bbb U: \map Q x}$ By the following definitions: {{begin-axiom}} {{axiom | n = 1         | lc= Intersection:         | ml= S \cap T         | mo= :=         | mr= \set {x \in \Bbb U: \map P x \land \map Q x} }} {{axiom | n = 2         | lc= Union:         | ml= S \cup T         | mo= :=         | mr= \set {x \in \Bbb U: \map P x \lor \map Q x} }} {{axiom | n = 3         | lc= Subset:         | ml= S \subseteq T         | mo= :=         | mr= \forall x \in \Bbb U: \map P x \implies \map Q x }} {{axiom | n = 4         | lc= Symmetric Difference:         | ml= S * T         | mo= =         | mr= \set {x \in \Bbb U: \map P x \oplus \map Q x} }} {{axiom | n = 5         | lc= Complement:         | ml= \relcomp {} S         | mo= :=         | mr= \set {x \in \Bbb U: \lnot \map P x} }} {{axiom | n = 6         | lc= Set Equality:         | ml= S = T         | mo= :=         | mr= \forall x \in \Bbb U: \map P x \iff \map Q x }} {{end-axiom}} {{qed}}
The only (strictly) positive integer less than or equal to $1$ is $1$ itself. By Integer is Coprime to 1, $1$ is coprime to itself. Hence, by definition, there is exactly $1$ integer less than or equal to $1$ which is coprime with $1$. Hence the result. {{qed}}
Definition of cardinal: :$(1): \quad \left\vert A \right\vert \sim A$. :$\left\vert{A}\right\vert$ is finite {{iff}} :$\exists n \in \N: \left\vert A \right\vert \sim \N_n$ by definition of finite set {{iff}} : $\exists n \in \N: A \sim \N_n$ by $(1)$ and Set Equivalence is Equivalence Relation {{iff}} : $A$ is finite by definition of finite set. {{qed}}
Let $\map f x = x^2$. Then: :$\map f {x_a} = x_a^2$ and: :$\map f {x_b} = x_b^2$ Then the slope $m$ of the line segment joining $A B$ will be: {{begin-eqn}} {{eqn | l = m        | r = \frac {x_a^2 - x_b^2} {x_b - x_a}       | c = Equation of Straight Line in Plane: Point-Slope Form }} {{eqn | r = \frac {\paren {x_a - x_b} \paren {x_b + x_a} } {x_b - x_a}       | c = Difference of Two Squares }} {{eqn | r = x_b + x_a       | c = cancelling, $x_a \neq x_b$ }} {{end-eqn}} From Equation of Straight Line in Plane: Slope-Intercept Form: :$y = \paren {x_b + x_a} x + c$ where $c$ denotes the $y$-intercept. Substituting the coordinates of point $A = \tuple {x_a, x_a^2}$ for $\tuple {x, y}$: {{begin-eqn}} {{eqn | l = x_a^2        | r = \paren {x_b + x_a} x_a + c }} {{eqn | ll= \leadsto       | l = c       | r = x_a^2 - \paren {x_a + x_b} x_a }} {{eqn | r = x_a^2 - x_a^2 - x_b x_a }} {{eqn | r = -x_b x_a }} {{end-eqn}} {{qed}}
From Integers are Countably Infinite there is a bijection between $\Z$, the set of integers, and $\N$, the set of natural numbers. Let $h: \N \to \Z$ be such a bijection. Let $f: S \to \N$ be a bijection. From Composite of Bijections is Bijection: :$h \circ f: S \to \Z$ is a bijection. Similarly, let $g: S \to \Z$ be a bijection. By Inverse of Bijection is Bijection, $h^{-1}: \Z \to \N$ is a bijection. Again from Composite of Bijections is Bijection: :$h^{-1} \circ g: S \to \N$ is a bijection. Hence the result. {{qed}} Category:Countable Sets bya5obx5auhsemkb9y5lnv8ebjer6c5
Let $f: \Bbb O \to \Z$ be the mapping defined as: :$\forall x \in \Bbb O: \map f x = \dfrac {x + 1} 2$ $f$ is well-defined as $x + 1$ is even and so $\dfrac {x + 1} 2 \in \Z$. Let $x, y \in \Bbb O$ such that $\map f x = \map f y$. Then: {{begin-eqn}} {{eqn | l = \map f x       | r = \map f y       | c =  }} {{eqn | ll= \leadsto       | l = \dfrac {x + 1} 2       | r = \dfrac {y + 1} 2       | c = Definition of $f$ }} {{eqn | ll= \leadsto       | l = x + 1       | r = y + 1       | c =  }} {{eqn | ll= \leadsto       | l = x       | r = y       | c =  }} {{end-eqn}} Thus $f$ is injective by definition. Consider the inverse $f^{-1}$. By inspection: :$\forall x \in \Z: \map {f^{-1} } x = 2 x - 1$ $f^{-1}$ is well-defined, and $2 x - 1$ is odd. Thus $f^{-1}$ is a mapping from $\Z$ to $\Bbb O$. Then: {{begin-eqn}} {{eqn | l = \map {f^{-1} } x       | r = \map {f^{-1} } y       | c =  }} {{eqn | ll= \leadsto       | l = 2 x - 1       | r = 2 y - 1       | c = Definition of $f^{-1}$ }} {{eqn | ll= \leadsto       | l = 2 x       | r = 2 y       | c =  }} {{eqn | ll= \leadsto       | l = x       | r = y       | c =  }} {{end-eqn}} Thus $f^{-1}$ is injective by definition. It follows by the Cantor-Bernstein-Schröder Theorem that there exists a bijection between $\Z$ and $\Bbb O$. {{qed}}
Let $\phi$ be a WFF of Polish notation for $\AA$. Apply the Principle of Mathematical Induction on the length of $\phi$ to prove: :$(1): \quad$ No initial part of $\phi$ is a WFF, except $\phi$ itself; :$(2): \quad$ If the first symbol of $\phi$ has arity $n$, then there exist unique WFFs $\phi_1, \ldots, \phi_n$ such that $\phi = s \phi_1 \cdots \phi_n$. Let $\phi'$ be a WFF that is an initial part of $\phi$. Because all WFFs have at least length $1$, it follows that: :$\phi' = s \phi'_1 \cdots \phi'_n$ Now it must be that either $\phi_1$ is an initial part of $\phi'_1$ or vice versa. By the induction hypothesis on $(1)$, it must be that $\phi_1 = \phi'_1$ because $\phi_1$ and $\phi'_1$ are both strictly shorter than $\phi$. Now if, given $1 < j \le n$: :$\phi_i = \phi'_i$ for all $i < j$ it follows that $\phi_j$ and $\phi'_j$ start at the same position in $\phi$. Hence again, inductively, it follows that $\phi_j = \phi'_j$. Thus $\phi_i = \phi'_i$ for all $1 \le i \le n$; that is: :$\phi = \phi'$ The result follows by the Principle of Mathematical Induction. {{qed}}
An identity matrix, by definition, has instances of $1$ on the main diagonal and $0$ elsewhere. Each diagonal element is by definition on one row and one column of the matrix. Also by definition, each diagonal element is on a different row and  column from each other diagonal element. The result follows by definition of permutation matrix. {{qed}}
Let $\mathbf A$ be a full rook matrix. By definition, $\mathbf A$ is an instance of a permutation matrix. By Determinant of Permutation Matrix, it follows that $\det \mathbf A = \pm 1$. By Matrix is Invertible iff Determinant has Multiplicative Inverse: :$\mathbf A$ is invertible. {{qed}}
An element $a b_{ij}$ of $\mathbf {A B}$ is formed by multiplying each element of row $i$ of $\mathbf A$ by its corresponding element of column $j$ of $\mathbf B$. No more than $1$ element of row $i$ equals $1$, and the rest equal $0$. No more than $1$ column $k$ of $\mathbf B$ contains $1$ in its $i$th element, and the rest contain $0$. So of all the elements of row $i$ of $\mathbf {A B}$, only $a b_{ik}$ is $1$, and the rest are $0$. By the same argument, each row of $\mathbf {A B}$ contains no more than one $1$, and all the rest of the elements are $0$. In $\mathbf B$, each column $j$ has no more than one element equal to $1$, and all are in a different row $k$. Thus each row contains its $1$, if it has one, in a different column from all the other rows. So there is no more than one $1$ in each column of $\mathbf {A B}$. Hence the result by definition of rook matrix. {{qed}}
We have to prove Open Set Axioms $(O1)-(O3)$: :$(O1): \quad$ The union of an arbitrary subset of $\tau$ is an element of $\tau$. Let $\mathcal F \subseteq \tau$. Define by definition of $\tau$ a family $\left({\mathcal G_A}\right)_{A \in \mathcal F}$ such that :$\forall A \in \mathcal F: A = \bigcup \mathcal G_A \land \mathcal G_A \subseteq \mathcal B$. By General Distributivity of Set Union: :$\displaystyle \bigcup \bigcup_{A \in \mathcal F} \mathcal G_A = \bigcup_{A \in \mathcal F} \bigcup \mathcal G_A = \bigcup \mathcal F$ By Union of Subsets is Subset/Family of Sets: :$\displaystyle \bigcup_{A \in \mathcal F} \mathcal G_A \subseteq \mathcal B$ Thus by definition of $\tau$ :$\bigcup \mathcal F \in \tau$ :$(O2): \quad$ The intersection of any two elements of $\tau$ is an element of $\tau$. Let $A$ and $B$ be elements of $\tau$. By definition of $\tau$ there exist subsets $\mathcal G_A$ and $\mathcal G_B$ of $\mathcal B$ such that: :$A = \bigcup \mathcal G_A$ and $B = \bigcup \mathcal G_B$ and $\mathcal G_A, \mathcal G_B \subseteq \mathcal B$ Set $\mathcal G_C = \left\{{C \in \mathcal B: C \subseteq A \cap B}\right\}$ By Union of Subsets is Subset: :$\bigcup \mathcal G_C \subseteq A \cap B$ We will prove the inclusion: $A \cap B \subseteq \bigcup \mathcal G_C$ Let $x \in A \cap B$. Then by definition of intersection: : $v \in A$ and $x \in B$ Hence by definition of union: :$\exists D \in \mathcal G_A: x \in D$ Analogically: :$\exists E \in \mathcal G_B: x \in E$ By definition of subset $D, E \in \mathcal B$ and by definition of intersection $x \in D \cap E$. Then by $(B1)$: :$\exists U \in \mathcal B: x \in U \subseteq D \cap E$ By Set is Subset of Union/Set of Sets: :$D \subseteq A$ and $E \subseteq B$ Then by Set Intersection Preserves Subsets: :$D \cap E \subseteq A \cap B$ Hence by Subset Relation is Transitive: :$U \subseteq A \cap B$ Then by definition of $\mathcal G_C$: :$U \in \mathcal G_C$ Thus by definition of union: :$x \in \bigcup \mathcal G_C$ This ends the proof of inclusion. Then by definition of set equality: :$A \cap B = \bigcup \mathcal G_C$ By definition of subset: :$\mathcal G_C \subseteq \mathcal B$ Thus by definition of $\tau$: :$A \cap B \in \tau$ :$(O3): \quad S$ is an element of $\tau$. By $(B2)$ and definition of union: :$\bigcup \mathcal B = X$ Because $\mathcal B \subseteq \mathcal B$ by definition of $\tau$: :$S \in \tau$ It remains to prove that $\mathcal B$ is a basis of $T$. Let $U$ be an open set of $S$. Let $x$ be a point of $S$ such that $x \in U$ By definition of $\tau$ there exists $\mathcal G \subseteq \mathcal B$ such that: :$ U = \bigcup \mathcal G$ By definition of union: :$\exists A \in \mathcal G: x \in A$ By definition of subset: $A \in \mathcal B$ Thus by Set is Subset of Union: :$x \in A \subseteq U$ Thus the result by definition of basis. {{qed}}
Follows directly from Matrix is Invertible iff Determinant has Multiplicative Inverse. {{qed}} Category:Matrix Theory 2movrglnpt2ymh2r3u019rajq7tq63p
Let $x_1, x_2, z \in \Z$. Let $x_1 \equiv x_2 \pmod z$ as defined by the equal remainder after division: :$\RR_z = \set {\tuple {x, y} \in \Z \times \Z: \exists k \in \Z: x = y + k z}$ Let $\tuple {x_1, x_2} \in \RR_z$. Then by definition: :$\exists k \in \Z: x_1 = x_2 + k z$ So, by definition of the modulo operation: {{begin-eqn}} {{eqn | l = x_1 \mod z       | r = \paren {x_2 + k z} - z \floor {\frac {x_2 + k z} z}       | c =  }} {{eqn | r = \paren {x_2 + k z} - z \floor {\frac {x_2} z + k}       | c =  }} {{eqn | r = \paren {x_2 + k z} - z \floor {\frac {x_2} z} + k z       | c =  }} {{eqn | r = x_2 - z \floor {\frac {x_2} z}       | c =  }} {{eqn | r = x_2 \mod z       | c =  }} {{end-eqn}} So: :$x_1 \equiv x_2 \pmod z$ in the sense of definition by modulo operation. {{qed|lemma}} Now let $x_1 \equiv x_2 \pmod z$ in the sense of definition by modulo operation. That is: :$x_1 \equiv x_2 \pmod z \iff x_1 \mod z = x_2 \mod z$ Let $z = 0$. Then by definition, $x_1 \mod 0 = x_1$ and $x_2 \mod 0 = x_2$. So as $x_1 \mod 0 = x_2 \mod 0$ we have that $x_1 = x_2$. So: :$x_1 - x_2 = 0 = 0.z$ and so $x_1 \equiv x_2 \pmod z$ in the sense of definition by integer multiple. Now suppose $z \ne 0$. Then from definition of the modulo operation: :$x_1 \mod z = x_1 - z \floor {\dfrac {x_1} z}$ :$x_2 \mod z = x_2 - z \floor {\dfrac {x_2} z}$ Thus: :$x_1 - z \floor {\dfrac {x_1} z} = x_2 - z \floor {\dfrac {x_2} z}$ and so: :$x_1 - x_2 = z \paren {\floor {\dfrac {x_1} z} - \floor {\dfrac {x_2} z} }$ From the definition of the floor function, we see that both $\floor {\dfrac {x_1} z}$ and $\floor {\dfrac {x_2} z}$ are integers. Therefore, so is $\floor {\dfrac {x_1} z} - \floor {\dfrac {x_2} z}$ an integer. So $\exists k \in \Z: x_1 - x_2 = k z$. Thus $x_1 - x_2 = k z$ and: :$x_1 \equiv x_2 \pmod z$ in the sense of definition by integer multiple. {{qed|lemma}} Now let $x_1 \equiv x_2 \pmod z$ in the sense of definition by integer multiple. That is, $\exists k \in \Z: x_1 - x_2 = k z$. Then $x_1 = x_2 + k z$ and so $\tuple {x_1, x_2} \in \RR_z$ where: :$\RR_z = \set {\tuple {x, y} \in \R \times \R: \exists k \in \Z: x = y + k z}$ and so :$x_1 \equiv x_2 \pmod z$ in the sense of definition by equivalence relation. {{qed|lemma}} So all three definitions are equivalent: $(1) \implies (2) \implies (3) \implies (1)$. {{qed}}
From the formal definition of Polish notation, it follows that $a$ must be introduced by the rule of formation: :$a \mathbf A_1 \cdots \mathbf A_n$ for some well-formed formulas $\mathbf A_1, \ldots, \mathbf A_n$. By Unique Readability for Polish Notation, the $\mathbf A_i$ are uniquely determined. Then $\mathbf A' = a \mathbf A_1 \cdots \mathbf A_n$ is a well-formed part of $\mathbf A$. Moreover, any well-formed part of $\mathbf A$ containing $a$ must contain $\mathbf A'$. Hence $\mathbf A'$ is the scope of $a$. {{qed}}
Let $\struct {P, \subseteq}$ be the ordered set consisting of all proper ideals of $R$, ordered by inclusion. The theorem is proved by applying Zorn's Lemma to $P$. First, we check that the conditions for Zorn's Lemma are met: $P$ must be non-empty, and every non-empty chain in $P$ must have an upper bound. === $P$ is non-empty === Since $R$ is non-null, the zero ideal is a proper ideal of $R$, and thus an element of $P$. === Every non-empty chain in $P$ has an upper bound in $P$ === Let $\sequence {I_\alpha}_{\alpha \mathop \in A}$ be a non-empty chain of ideals in $P$. Let $\displaystyle I = \bigcup_{\alpha \mathop \in A} I_\alpha$. We will show that $I$ is an upper bound in $P$ for the chain $\sequence {I_\alpha}_{\alpha \mathop \in A}$. ==== $I$ is a proper ideal of $R$ ==== By By Union of Chain of Proper Ideals is Proper Ideal, $I$ is a proper ideal of $R$. ==== $I$ is an upper bound for the chain $\sequence {I_\alpha}$ ==== Since $I$ is a proper ideal of $R$, it is an element of our ordered set $P$. $I$ is the union of the $I_\alpha$, so $I_\alpha \subseteq I$ for all $\alpha \in A$. This means that $I$ is an upper bound in $P$ for the chain $\sequence {I_\alpha}_{\alpha \mathop \in A}$. === Applying Zorn's Lemma === We have shown that the conditions for Zorn's Lemma are met: :$(1): \quad P$ is non-empty :$(2): \quad$ every non-empty chain in $P$ has an upper bound. Applying Zorn's Lemma to $\struct {P, \subseteq}$ gives us a maximal element $M$. This $M$ is a proper ideal of $R$ which is not contained in any other proper ideal. So by definition, $M$ is a maximal ideal of $R$. {{qed}} {{AoC|Zorn's Lemma}} {{Namedfor|Wolfgang Krull|cat = Krull}} Category:Ideal Theory Category:Maximal Ideals of Rings qz6mfnvslwx2t1vsb0tyu1x03xkh4md
{{begin-eqn}} {{eqn | l = p \oplus q       | o = \dashv \vdash       | r = \paren {p \lor q} \land \neg \paren {p \land q}       | c = {{Defof|Exclusive Or}} }} {{eqn | o = \dashv \vdash       | r = \paren {p \lor q} \land \paren {\neg p \lor \neg q}       | c = De Morgan's Laws: Disjunction of Negations }} {{end-eqn}} {{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. $\begin{array}{|ccc||ccccccccc|} \hline p & \oplus & q & (p & \lor & q) & \land & (\neg & p & \lor & \neg & q) \\ \hline F & F & F & F & F & F & F & T & F & T & T & F \\ F & T & T & F & T & T & T & T & F & T & F & T  \\ T & T & F & T & T & F & T & F & T & T & T & F  \\ T & F & T & T & T & T & F & F & T & F & F & T  \\ \hline \end{array}$ {{qed}}
{{begin-eqn}} {{eqn | l = p \uparrow q       | o = \dashv \vdash       | r = \neg \left({p \land q}\right)       | c = Definition of Logical NAND }} {{eqn | o = \dashv \vdash       | r = \neg p \lor \neg q       | c = De Morgan's Laws: Disjunction of Negations }} {{end-eqn}} {{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. $\begin{array}{|ccc||ccccc|} \hline p & \uparrow & q & \neg & p & \lor & \neg & q \\ \hline F & T & F & T & F & T & T & F \\ F & T & T & T & F & T & F & T  \\ T & T & F & F & T & T & T & F  \\ T & F & T & F & T & F & F & T  \\ \hline \end{array}$ {{qed}}
We apply the Method of Truth Tables. $\begin{array}{|c||ccc|} \hline p & p & \lor & q \\ \hline F & F & F & F \\ F & F & T & T \\ T & T & T & F \\ T & T & T & T \\ \hline \end{array}$ As can be seen, when $p$ is true so is $p \lor q$. {{qed}}
We apply the Method of Truth Tables. $\begin{array}{|c||ccc|} \hline q & p & \lor & q \\ \hline F & F & F & F \\ T & F & T & T \\ F & T & T & F \\ T & T & T & T \\ \hline \end{array}$ As can be seen, when $q$ is true so is $p \lor q$. {{qed}}
=== Form 1 === {{:Rule of Addition/Sequent Form/Formulation 2/Proof 1/Form 1}} === Form 2 === {{:Rule of Addition/Sequent Form/Formulation 2/Proof 1/Form 2}}
{{BeginTableau|\vdash \left({p \implies \neg p}\right) \implies \neg p|Instance 2 of the Hilbert-style systems}} {{TableauLine  |n = 1  |f = \left({p \lor p}\right) \implies p  |rlnk = Definition:Hilbert Proof System/Instance 2  |rtxt = Axiom $A1$ }} {{TableauLine  |n = 2  |f = \left({\neg p \lor \neg p}\right) \implies \neg p  |rlnk = Definition:Hilbert Proof System/Instance 2  |rtxt = Rule $RST \, 1$  |dep = 1  |c = $\neg p \, / \, p$ }} {{TableauLine  |n = 3  |f = \left({p \implies \neg p}\right) \implies \neg p  |rlnk = Definition:Hilbert Proof System/Instance 2  |rtxt = Rule $RST \, 2 \, (2)$  |dep = 2 }} {{EndTableau}} {{Qed}}
Let us use the following abbreviations {{begin-eqn}} {{eqn | l = \phi       | o = \text{ for }       | r = p \implies q       | c =  }} {{eqn | l = \psi       | o = \text{ for }       | r = q \implies r       | c =  }} {{eqn | l = \chi       | o = \text{ for }       | r = p \implies r       | c =  }} {{end-eqn}} From Hypothetical Syllogism: Formulation 3 we have: : $(1): \quad \vdash \paren {\paren {p \implies q} \land \paren {q \implies r} } \implies \paren {p \implies r}$ {{BeginTableau|\paren {q \implies r} \implies \paren {\paren {p \implies q} \implies \paren {p \implies r} } }} {{Assumption|1|\psi \land \phi}} {{Commutation|2|1|\phi \land \psi|1|Conjunction}} {{SequentIntro|3|1|\chi|2|Hypothetical Syllogism: Formulation 3 (see $(1)$ above)}} {{Implication|4||\paren {\psi \land \phi} \implies \chi|1|3}} {{SequentIntro|5||\psi \implies \paren {\phi \implies \chi}|4|Rule of Exportation}} {{EndTableau}} Expanding the abbreviations leads us back to: :$\vdash \paren {q \implies r} \implies \paren {\paren {p \implies q} \implies \paren {p \implies r} }$ {{qed}}
{{BeginTableau |\vdash \left({q \implies r}\right) \implies \left({\left({p \implies q}\right) \implies \left({p \implies r}\right)}\right)|Instance 2 of the Hilbert-style systems}} {{TableauLine  |n = 1  |f = \left({q \implies r}\right) \implies \left({\left({p \lor q}\right) \implies \left ({p \lor r}\right)}\right)  |rlnk = Definition:Hilbert Proof System/Instance 2  |rtxt = Axiom $A4$ }} {{TableauLine  |n = 2  |f = \left({q \implies r}\right) \implies \left({\left({\neg p \lor q}\right) \implies \left ({\neg p \lor r}\right)}\right)  |rlnk = Definition:Hilbert Proof System/Instance 2  |rtxt = Rule $RST \, 1$  |dep = 1  |c = $\neg p \, / \, p$ }} {{TableauLine  |n = 3  |f = \left({q \implies r}\right) \implies \left({\left({p \implies q}\right) \implies \left({p \implies r}\right)}\right)  |rlnk = Definition:Hilbert Proof System/Instance 2  |rtxt = Rule $RST \, 2 \, (2)$  |dep = 2 }} {{EndTableau}} {{Qed}}
{{BeginTableau|\vdash p \lor \neg p|Instance 2 of the Hilbert-style systems}} {{TheoremIntro|1|\left({q \implies r}\right) \implies \left({\left({p \implies q}\right) \implies \left({p \implies r}\right)}\right)|Hypothetical Syllogism}} {{TableauLine  |n = 2  |f = \left({\left({p \lor p}\right) \implies p}\right) \implies \left({\left({p \implies \left({p \lor p}\right)}\right) \implies \left({p \implies p}\right)}\right)  |rlnk = Definition:Hilbert Proof System/Instance 2  |rtxt = Rule $RST \, 1$  |dep = 1  |c = $p \lor p \, / \, q$, $p \, / \, r$ }} {{TableauLine  |n = 3  |f = (p \lor p) \implies p  |rlnk = Definition:Hilbert Proof System/Instance 2  |rtxt = Axiom $A1$ }} {{TableauLine  |n = 4  |f = \left({p \implies \left({p \lor p}\right)}\right) \implies \left({p \implies p}\right)  |rlnk = Definition:Hilbert Proof System/Instance 2  |rtxt = Rule $RST \, 3$  |dep = 2, 3 }} {{TableauLine  |n = 5  |f = p \implies (p \lor p)  |rlnk = Definition:Hilbert Proof System/Instance 2  |rtxt = Axiom $A2$, Rule $RST \, 1$  |c = $p \, / \, q$ }} {{TableauLine  |n = 6  |f = p \implies p  |rlnk = Definition:Hilbert Proof System/Instance 2  |rtxt = Rule $RST \, 3$  |dep = 4, 5 }} {{TableauLine  |n = 7  |f = \neg p \lor p  |rlnk = Definition:Hilbert Proof System/Instance 2  |rtxt = Rule $RST \, 2 \, (2)$  |dep = 6 }} {{TableauLine  |n = 8  |f = (p \lor q) \implies (q \lor p)  |rlnk = Definition:Hilbert Proof System/Instance 2  |rtxt = Axiom $A3$ }} {{TableauLine  |n = 9  |f = (\neg p \lor p) \implies (p \lor \neg p)  |rlnk = Definition:Hilbert Proof System/Instance 2  |rtxt = Rule $RST \, 1$  |dep = 8  |c = $p \, / \, q$, $\neg p \, / \, p$ }} {{TableauLine  |n = 10  |f = p \lor \neg p  |rlnk = Definition:Hilbert Proof System/Instance 2  |rtxt = Rule $RST \, 3$  |dep = 7, 9 }} {{EndTableau}} {{Qed}}