hcju/proofwikiqa huggingface dataset:
Suppose that $z \in \C$ with $\cmod {z - \xi} < R'$. Then $S' \paren z$ converges absolutely by Existence of Radius of Convergence of Complex Power Series, so: {{begin-eqn}} {{eqn | l = 1       | o = \ge       | r = \limsup_{n \mathop \to \infty} \cmod {n a_n \paren {z - \xi}^{n - 1} }^{1 / n}       | c = by $n$th Root Test }} {{eqn | o = >       | r = \limsup_{n \mathop \to \infty} \cmod {a_n \paren {z - \xi}^n }^{1/n}       | c = as $\cmod {n  \paren {z - \xi}^{n - 1} } > \cmod { \paren {z - \xi}^n }$ for all $n > \cmod {z - \xi}$ }} {{end-eqn}} From the $n$th Root Test, it follows that $S \paren z$ converges absolutely. Hence, $R \ge R'$. {{qed|lemma}} Suppose that $z \in \C$ with $\cmod {z - \xi} < R$. Find $z_o \in \C$ such that $\cmod {z - \xi} < \cmod {z_0 - \xi} < R$, so $S \paren {z_0}$ converges absolutely. From the $n$th Root Test, it follows that $\cmod {a_n \paren {z_0 - \xi}^n }^{1/n} < 1$ for all $n \ge N$ for some $N \in \N$. Then: {{begin-eqn}} {{eqn | l = 1       | r = \limsup_{n \mathop \to \infty} \cmod n^{1/n}       | c = by Limit of Integer to Reciprocal Power }} {{eqn | o = >       | r = \limsup_{n \mathop \to \infty} \cmod n^{1/n} \cmod {\frac {z - \xi} {z_0 - \xi} } }} {{eqn | r = \limsup_{n \mathop \to \infty} \cmod {n \paren {\frac {z - \xi} {z_0 - \xi} }^{n - 1} }^{1/n}        | c = as $\displaystyle \lim_{n \mathop \to \infty} \paren {\paren {\frac {z - \xi} {z_0 - \xi} }^{n - 1} }^{1/n} = 1$  }} {{eqn | o = \ge       | r = \limsup_{n \mathop \to \infty} \cmod {n \paren {\frac {z - \xi} {z_0 - \xi} }^{n - 1} a_n \paren {z_0 - \xi}^{n - 1} }^{1/n}       | c = from the $n$th Root Test, as remarked above }} {{eqn | r = \limsup_{n \to \mathop \infty} \cmod {n a_n \paren {z - \xi}^{n - 1} }^{1/n} }} {{end-eqn}} From the $n$th Root Test, it follows that $S' \paren z$ converges absolutely. Hence, $R' \ge R$, so $R' = R$. {{qed}}
We apply the Method of Truth Tables. $\begin{array}{|c|c||ccc|} \hline p & q & p & \lor & q\\ \hline F & F & F & F & F \\ F & T & F & T & T \\ T & F & T & T & F \\ T & T & T & T & T \\ \hline \end{array}$ As can be seen, whenever either $p$ or $q$ (or both) are true, then so is $p \lor q$. {{qed}}
=== Form 1 === {{:Rule of Addition/Sequent Form/Proof 1/Form 1}} === Form 2 === {{:Rule of Addition/Sequent Form/Proof 1/Form 2}}
We apply the Method of Truth Tables. $\begin{array}{|ccc|ccc|ccc||c|} \hline p & \lor & q & p & \implies & r & q & \implies & r & r \\ \hline F & F & F & F & T & F & F & T & F & F \\ F & F & F & F & T & T & F & T & T & T \\ F & T & T & F & T & F & T & F & F & F \\ F & T & T & F & T & T & T & T & T & T \\ T & T & F & T & F & F & F & T & F & F \\ T & T & F & T & T & T & F & T & T & T \\ T & T & T & T & F & F & T & F & F & F \\ T & T & T & T & T & T & T & T & T & T \\ \hline \end{array}$ As can be seen, when $p \lor q$, $p \implies r$ and $q \implies r$ are all true, then so is $r$. {{qed}}
We apply the Method of Truth Tables. $\begin{array}{|c|ccc||c|} \hline p & p & \implies & q & q\\ \hline F & F & T & F & F \\ F & F & T & T & T \\ T & T & F & F & F \\ T & T & T & T & T \\ \hline \end{array}$ As can be seen, when $p$ is true, and so is $p \implies q$, then $q$ is also true. {{qed}}
Consider the function $\map f x = \map \ln {x^r} - r \ln x$. Then from: :The definition of the natural logarithm :The Fundamental Theorem of Calculus :The Power Rule for Derivatives :The Chain Rule for Derivatives: :$\forall x > 0: \map {f'} x = \dfrac 1 {x^r} r x^{r-1} - \dfrac r x = 0$ Thus from Zero Derivative implies Constant Function, $f$ is constant: :$\forall x > 0: \map \ln {x^r} - r \ln x = c$ To determine the value of $c$, put $x = 1$. From Logarithm of 1 is 0: :$\ln 1 = 0$ Thus: :$c = \ln 1 - r \ln 1 = 0$ {{qed}}
{{begin-eqn}} {{eqn | l = \ln a       | r = b }} {{eqn | ll= \leadstoandfrom       | l = e^b       | r = a }} {{eqn | ll= \leadsto       | l = \paren {e^b}^c       | r = a^c }} {{eqn | ll= \leadsto       | l = e^{c b}       | r = a^c       | c = Exponential of Product }} {{eqn | ll= \leadsto       | l = \ln e^{c b}       | r = \ln a^c }}    {{eqn | ll= \leadsto       | l = c b       | r = \ln a^c       | c = Exponential of Natural Logarithm }} {{end-eqn}} By hypothesis, $\ln a = b$. Multiplying both sides by $c$: :$c \ln a = c b$ But we proved above that: :$c b = \ln a^c$ {{qed}}
Let $y = r \log_a x$. Then: {{begin-eqn}} {{eqn | l = a^y       | r = a^{r \log_a x}       | c =  }} {{eqn | r = \paren {a^{\log_a x} }^r       | c = Exponent Combination Laws }} {{eqn | r = x^r       | c = {{Defof|General Logarithm|Logarithm base $a$}} }} {{end-eqn}} The result follows by taking logs base $a$ of both sides. {{qed}}
Taking the criteria for definition 1 of a Boolean algebra in turn: === $(\text {BA} 0):$ Closure === $\powerset S$ is closed under both $\cup$ and $\cap$: :Power Set is Closed under Intersection :Power Set is Closed under Union :Power Set is Closed under Complement {{qed|lemma}} === $(\text {BA} 1):$ Commutativity === Both $\cup$ and $\cap$ are commutative from Intersection is Commutative and Union is Commutative. {{qed|lemma}} === $(\text {BA} 2):$ Distributivity === Both $\cup$ and $\cap$ distribute over the other, from Union Distributes over Intersection and Intersection Distributes over Union. {{qed|lemma}} === $(\text {BA} 3):$ Identity Elements === Both $\cup$ and $\cap$ have identities: From Power Set with Intersection is Monoid, $S$ is the identity for $\cap$.  From Power Set with Union is Monoid, $\O$ is the identity for $\cup$. {{qed|lemma}} === $(\text {BA} 4):$ Complements === From Union with Complement: :$\forall A \in S: A \cup \map \complement A = S$ which is the identity for $\cap$. From Intersection with Complement: :$\forall A \in S: A \cap \map \complement A = \O$ which is the identity for $\cup$. {{qed|lemma}} All the criteria for a Boolean algebra are therefore fulfilled. {{qed}}
We apply the Method of Truth Tables. $\begin{array}{|c|c||ccc|} \hline p & q & p & \implies & q\\ \hline F & F & F & T & F \\ F & T & F & T & T \\ T & F & T & F & F \\ T & T & T & T & T \\ \hline \end{array}$ As can be seen by inspection, only when $p$ is true and $q$ is false, then so is $p \implies q$ also false. {{qed}}
Let $\epsilon \in \R_{>0}$. From Tail of Convergent Series tends to Zero, it follows that there exists $M \in \N$ such that: :$\displaystyle \sum_{n \mathop = M + 1}^\infty \cmod {a_n} < \dfrac \epsilon 2$ and: :$\displaystyle \sum_{n \mathop = M + 1}^\infty \cmod {b_n} < \dfrac \epsilon 2$ For all $m \ge M$, it follows that: {{begin-eqn}} {{eqn | l = \cmod {\sum_{n \mathop = 1}^\infty a_n + \sum_{n \mathop = 1}^\infty b_n - \sum_{n \mathop = 1}^m \paren {a_n + b_n} }       | r = \cmod {\sum_{n \mathop = m + 1}^\infty a_n + \sum_{n \mathop = m + 1}^\infty b_n} }} {{eqn | o = \le       | r = \sum_{n \mathop = m + 1}^\infty \cmod {a_n} + \sum_{n \mathop = m + 1}^\infty \cmod {b_n}       | c = by Triangle Inequality }} {{eqn | o = \le       | r = \sum_{n \mathop = M + 1}^\infty \cmod {a_n} + \sum_{n \mathop = M + 1}^\infty \cmod {b_n} }} {{eqn | o = <       | r = \epsilon }} {{end-eqn}} By definition of convergent series, it follows that: {{begin-eqn}} {{eqn | l = \sum_{n \mathop = 1}^\infty a_n + \sum_{n \mathop = 1}^\infty b_n       | r = \lim_{m \mathop \to \infty} \sum_{n \mathop = 1}^m \paren {a_n + b_n} }} {{eqn | r = \sum_{n \mathop = 1}^\infty \paren {a_n + b_n} }} {{end-eqn}} To show that $\displaystyle \sum_{n \mathop = 1}^\infty \paren {a_n + b_n}$ is absolutely convergent, note that: {{begin-eqn}} {{eqn | l = \sum_{n \mathop = 1}^\infty \cmod {a_n} + \sum_{n \mathop = 1}^\infty \cmod {b_n}       | r = \sum_{n \mathop = 1}^\infty \paren {\cmod {a_n} + \cmod {b_n} }       | c = as shown above }} {{eqn | o = \ge       | r = \sum_{n \mathop = 1}^\infty \cmod {a_n + b_n}       | c = by Triangle Inequality }} {{end-eqn}} {{qed}}
{{BeginTableau|\left({p \vdash q}\right) \vdash p \implies q}} {{Premise|1|p}} |- | align="right" | 2 || | align="right" | 1 | $q$ | By hypothesis | 1 | as $p \vdash q$ {{Implication|3|1|p \implies q|1|2}} {{EndTableau}} {{Qed}}
=== Form 1 === {{:Rule of Simplification/Sequent Form/Formulation 2/Proof 1/Form 1}} === Form 2 === {{:Rule of Simplification/Sequent Form/Formulation 2/Proof 1/Form 2}}
By the definition of a complement of $x$: :$\neg x \vee x = \top$ :$\neg x \wedge x = \bot$ Since $\vee$ and $\wedge$ are commutative: :$x \vee \neg x = \top$ :$x \wedge \neg x = \bot$ Thus by the definition of complement, $x$ is a complement of $\neg x$. By the definition of a uniquely complemented lattice, $x = \neg \neg x$. {{qed}} Category:Bounded Lattices 762w9prslhhnoeouofbbnd3v1cv3473
=== $(1)$ implies $(2)$ === Suppose: :$\forall p, q \in S: \neg p \vee \neg q = \neg \left({p \wedge q}\right)$ Then applying this to $\neg p$ and $\neg q$: :$\neg \neg p \vee \neg \neg q = \neg \left({\neg p \wedge \neg q}\right)$ By Complement of Complement in Uniquely Complemented Lattice, $\neg \neg p = p$ and $\neg \neg q = q$. Thus: :$p \vee q = \neg \left({\neg p \wedge \neg q}\right)$. Taking complements of both sides: :$\neg \left({p \vee q}\right) = \neg \neg \left({\neg p \wedge \neg q}\right)$ Again applying Complement of Complement in Uniquely Complemented Lattice: :$\neg \left({p \vee q}\right) = \neg p \wedge \neg q$ {{qed|lemma}} === $(2)$ implies $(1)$ === By Dual Pairs (Order Theory), $\wedge$ and $\vee$ are dual. Thus this implication follows from the above by Duality. {{qed|lemma}} === $(1)$ implies $(3)$ === By the definition of a lattice: :$p \preceq q \iff p \vee q = q$ Applying this to $\neg q$ and $\neg p$: :$\neg q \preceq \neg p \iff \neg q \vee \neg p = \neg p$ By $(1)$: :$\neg q \vee \neg p = \neg \left({q \wedge p}\right)$ So: :$\neg q \preceq \neg p \iff \neg \left({q \wedge p}\right) = \neg p$ Taking the complements of both sides of the equation on the right, and applying Complement of Complement in Uniquely Complemented Lattice: :$\neg q \preceq \neg p \iff q \wedge p = p$ But the right side is equivalent to $p \preceq q$ {{explain|We define the ordering on a lattice (definition 3) based on joins. We need an equivalent one based on meets, or maybe we have it somewhere already.}} Therefore: :$\neg q \preceq \neg p \iff p \preceq q$ {{qed|lemma}} === $(3)$ implies $(1)$ === {{improve|This is ugly}} {{MissingLinks}} Suppose that $p \preceq q \iff \neg q \preceq \neg p$ By the definition of join: :$\neg p, \neg q \preceq \neg p \vee \neg q$ Thus $\neg \left({\neg p \vee \neg q}\right) \preceq p, q$. By the definition of meet: :$\neg \left({\neg p \vee \neg q}\right) \preceq p \wedge q$ Thus: :$\neg\left({p \wedge q}\right) \preceq \neg\neg \left({\neg p \vee \neg q}\right)$ By Complement of Complement in Uniquely Complemented Lattice: $*\quad \neg\left({p \wedge q}\right) \preceq \neg p \vee \neg q$ Dually: :$\neg x \wedge \neg y \preceq \neg \left({x \vee y}\right)$ Letting $x = \neg p$ and $y = \neg q$: :$\neg \neg p \wedge \neg \neg q \preceq \neg \left({\neg p \vee \neg q}\right)$ By Complement of Complement in Uniquely Complemented Lattice: :$p \wedge q \preceq \neg \left({\neg p \vee \neg q}\right)$ By the premise and Complement of Complement in Uniquely Complemented Lattice, then: $**\quad \neg p \vee \neg q \preceq \neg \left({p \wedge q}\right)$ By $*$ and $**$: $\quad \neg\left({p \wedge q}\right) = \neg p \vee \neg q$ {{qed|lemma}} === $(1)$, $(2)$, and $(3)$ together imply $(4)$ === $b, c \preceq b \vee c$, so :$a \wedge b \preceq a \wedge \left({b \vee c}\right)$ :$a \wedge c \preceq a \wedge \left({b \vee c}\right)$ By the definition of join: :$\left({a \wedge b}\right) \vee \left({a \wedge c}\right) \preceq a \wedge \left({b \vee c}\right)$ {{finish}}
Since $\bot$ is the identity for $\vee$, the first condition for $\neg \bot$: :$\bot \vee \neg \bot = \top$ implies that $\neg \bot = \top$ is the only possibility. Since $\top$ is the identity for $\wedge$, it follows that: :$\bot \wedge \top = \bot$ and we conclude that: :$\neg \bot = \top$ as desired. {{qed}}
We know that $\bot$ is the identity for $\vee$. Therefore, from the condition that: :$\bot \vee a = \top$ for a complement $a$ of $\bot$, it follows that $a = \top$ is the only possibility. Since also: :$\bot \wedge \top = \bot$ as $\top$ is the identity for $\wedge$, the result follows. {{qed}}
By Dual Pairs (Order Theory), $\top$ is dual to $\bot$. The result follows from the Duality Principle and Complement of Bottom. {{qed}}
Since $\top$ is the identity for $\wedge$, the second condition for $\neg \top$: :$\top \wedge \neg \top = \bot$ implies that $\neg \top = \bot$ is the only possibility. Since $\bot$ is the identity for $\vee$, it follows that: :$\top \vee \bot = \top$ and we conclude that: :$\neg \top = \bot$ as desired. {{qed}}
Let us take the axioms of a Boolean algebra $\struct {S, \wedge, \vee}$: {{begin-axiom}} {{axiom | n = \text {BA} 0         | lc=         | t = $S$ is closed under both $\vee$ and $\wedge$ }} {{axiom | n = \text {BA} 1         | lc=         | t = Both $\vee$ and $\wedge$ are commutative }} {{axiom | n = \text {BA} 2         | lc=         | t = Both $\vee$ and $\wedge$ distribute over the other }} {{axiom | n = \text {BA} 3         | lc=         | t = Both $\vee$ and $\wedge$ have identities $\bot$ and $\top$ respectively }} {{axiom | n = \text {BA} 4         | lc=         | t = $\forall a \in S: \exists \neg a \in S: a \vee \neg a = \top, a \wedge \neg a = \bot$ }} {{end-axiom}} It can be seen by inspection, that exchanging $\wedge$ and $\vee$, and $\bot$ and $\top$ throughout does not change the axioms. Thus, what you get is a Boolean algebra again. Hence the result. {{qed}}
Let $x \in S$. Then: {{begin-eqn}} {{eqn | l = x \vee \top       | r = \paren {x \vee \top} \wedge \top       | c = Boolean Algebra: Axiom $(\text {BA}_1 3)$: $\top$ is the identity of $\wedge$ }} {{eqn | r = \paren {x \vee \top} \wedge \paren {x \vee \neg x}       | c = Boolean Algebra: Axiom $(\text {BA}_1 4)$: $x \vee x' = \top$ }} {{eqn | r = x \vee \paren {\top \wedge x'}       | c = Boolean Algebra: Axiom $(\text {BA}_1 2)$: both $\vee$ and $\wedge$ distribute over the other }} {{eqn | r = x \vee x'       | c = Boolean Algebra: Axiom $(\text {BA}_1 3)$: $\top$ is the identity of $\wedge$ }} {{eqn | r = \top       | c = Boolean Algebra: Axiom $(\text {BA}_1 4)$ $x \vee x' = \top$ }} {{end-eqn}} So $x \vee \top = \top$. {{qed|lemma}} The result $x \wedge \bot = \bot$ follows from the Duality Principle. {{qed}}
By virtue of Complement in Boolean Algebra is Unique, it will suffice to verify: :$\paren {a \vee b} \wedge \paren {\neg a \wedge \neg b} = \bot$ :$\paren {a \vee b} \vee \paren {\neg a \wedge \neg b} = \top$ For the first of these, compute: {{begin-eqn}} {{eqn | l = \paren {a \vee b} \wedge \paren {\neg a \wedge \neg b}       | r = \paren {\paren {a \vee b} \wedge \neg a} \wedge \neg b       | c = $\wedge$ is associative }} {{eqn | r = \paren {\paren {a \wedge \neg a} \vee \paren {b \wedge \neg a} } \wedge \neg b       | c = $\wedge$ distributes over $\vee$ }} {{eqn | r = \paren {\bot \vee \paren {b \wedge \neg a} } \wedge \neg b       | c = Axiom $(\text {BA} \ 4)$ for Boolean algebras }} {{eqn | r = \paren {b \wedge \neg a} \wedge \neg b       | c = $\bot$ is the identity for $\vee$ }} {{eqn | r = \neg a \wedge \paren {b \wedge \neg b}       | c = $\wedge$ is commutative and associative }} {{eqn | r = \neg a \wedge \bot       | c = Axiom $(\text {BA} \ 4)$ }} {{eqn | r = \bot       | c = Identities of Boolean Algebra also Zeroes }} {{end-eqn}} By the Duality Principle, we also conclude: :$\paren {a \wedge b} \vee \paren {\neg a \vee \neg b} = \top$ Substituting $\neg a$ and $\neg b$ for $a$ and $b$, respectively, this becomes: :$\paren {\neg a \wedge \neg b} \vee \paren {a \vee b} = \top$ where Complement of Complement (Boolean Algebras) was applied. Since $\vee$ is commutative, we have also derived the second of the two equations above. Hence: :$\neg \paren {a \vee b} = \neg a \wedge \neg b$ {{qed|lemma}} Now by again the Duality Principle: :$\neg \paren {a \wedge b} = \neg a \vee \neg b$ follows immediately. {{qed}}
Suppose $b, c \in S$ both satisfy the identities. Then: {{begin-eqn}} {{eqn | l = b       | r = b \wedge \top       | c = $\top$ is the identity for $\wedge$ }} {{eqn | r = b \wedge \left({a \vee c}\right)       | c = by hypothesis }} {{eqn | r = \left({b \wedge a}\right) \vee \left({b \wedge c}\right)       | c = $\wedge$ distributes over $\vee$ }} {{eqn | r = \bot \vee \left({b \wedge c}\right)       | c = by hypothesis }} {{eqn | r = \left({a \wedge c}\right) \vee \left({b \wedge c}\right)       | c = Axiom $(BA \ 4)$ }} {{eqn | r = \left({a \vee b}\right) \wedge c       | c = $\wedge$ distributes over $\vee$ }} {{eqn | r = \top \wedge c       | c = by hypothesis }} {{eqn | r = c       | c = $\top$ is the identity for $\wedge$ }} {{end-eqn}} That is, $b = c$. The result follows. {{qed}} Category:Boolean Algebras 4vs574sm211kodsk7csg2129n0tmgmh
{{begin-eqn}} {{eqn | l = a       | r = a \vee \bot       | c = $\bot$ is the identity for $\vee$ }} {{eqn | r = a \vee \left({c \wedge \neg c}\right)       | c = Boolean Algebra: Axiom $(BA_1 \ 4)$ }} {{eqn | r = \left({a \vee c}\right) \wedge \left({a \vee \neg c}\right)       | c = $\vee$ distributes over $\wedge$ }} {{eqn | r = \left({b \vee c}\right) \wedge \left({b \vee \neg c}\right)       | c = by hypothesis }} {{eqn | r = b \vee \left({c \wedge \neg c}\right)       | c = $\vee$ distributes over $\wedge$ }} {{eqn | r = b \vee \bot       | c = Boolean Algebra: Axiom $(BA_1 \ 4)$ }} {{eqn | r = b       | c = $\bot$ is the identity for $\vee$ }} {{end-eqn}} Hence the result. {{qed}}
Follows from Cancellation of Join in Boolean Algebra through the Duality Principle {{qed}}
Let $a, b, c \in S$. Let: :$x = a \wedge \paren {b \wedge c}$ :$y = \paren {a \wedge b} \wedge c$ Then: {{begin-eqn}} {{eqn | l = a \vee x       | r = a \vee \paren {a \wedge \paren {b \wedge c} }       | c =  }} {{eqn |r = \paren {a \vee a} \wedge \paren {a \vee \paren {b \wedge c} }       | c = Boolean Algebra: Axiom $(\text {BA}_1 \ 2)$: both $\vee$ and $\wedge$ distribute over the other }} {{eqn | r = a \wedge \paren {a \vee \paren {b \wedge c} }       | c = Operations of Boolean Algebra are Idempotent }} {{eqn | r = a       | c = Absorption Laws (Boolean Algebras) }} {{end-eqn}} Similarly: {{begin-eqn}} {{eqn | l = a \vee y       | r = a \vee \paren {\paren {a \wedge b} \wedge c}       | c =  }} {{eqn | r = \paren {a \vee \paren {a \wedge b} } \wedge \paren {a \vee c}       | c = Boolean Algebra: Axiom $(\text {BA}_1 \ 2)$: both $\vee$ and $\wedge$ distribute over the other }} {{eqn | r = a \wedge \paren {a \vee c}       | c = Absorption Laws (Boolean Algebras) }} {{eqn | r = a       | c = Absorption Laws (Boolean Algebras) }} {{end-eqn}} Thus we see we have $a \vee x = a \vee y$. Next: {{begin-eqn}} {{eqn | l = \neg a \vee x       | r = \neg a \vee \paren {a \wedge \paren {b \wedge c} }       | c =  }} {{eqn | r = \paren {\neg a \vee a} \wedge \paren {\neg a \vee \paren {b \wedge c} }       | c = Boolean Algebra: Axiom $(\text {BA}_1 \ 2)$: both $\vee$ and $\wedge$ distribute over the other }} {{eqn | r = \top \wedge \paren {\neg a \vee \paren {b \wedge c} }       | c = Boolean Algebra: Axiom $(\text {BA}_1 \ 4)$: $\neg a \vee a = \top$ }} {{eqn | r = \neg a \vee \paren {b \wedge c}       | c = Boolean Algebra: Axiom $(\text {BA}_1 \ 3)$: $\top$ is the identity of $\wedge$ }} {{end-eqn}} Similarly: {{begin-eqn}} {{eqn | l = \neg a \vee y       | r = \neg a \vee \paren {\paren {a \wedge b} \wedge c}       | c =  }} {{eqn | r = \paren {\neg a \vee \paren {a \wedge b} } \wedge \paren {\neg a \vee c}       | c = Boolean Algebra: Axiom $(\text {BA}_1 \ 2)$: both $\vee$ and $\wedge$ distribute over the other }} {{eqn | r = \paren {\paren {\neg a \vee a} \wedge \paren {\neg a \vee b} } \wedge \paren {\neg a \vee c}       | c = Boolean Algebra: Axiom $(\text {BA}_1 \ 2)$: both $\vee$ and $\wedge$ distribute over the other }} {{eqn | r = \paren {\top \wedge \paren {\neg a \vee b} } \wedge \paren {\neg a \vee c}       | c = Boolean Algebra: Axiom $(\text {BA}_1 \ 4)$: $\neg a \vee a = \top$ }} {{eqn | r = \paren {\neg a \vee b} \wedge \paren {\neg a \vee c}       | c = Boolean Algebra: Axiom $(\text {BA}_1 \ 3)$: $\top$ is the identity of $\wedge$ }} {{eqn | r = \neg a \vee \paren {b \wedge c}       | c = Boolean Algebra: Axiom $(\text {BA}_1 \ 2)$: both $\vee$ and $\wedge$ distribute over the other }} {{end-eqn}} Thus we see we have $\neg a \vee x = \neg a \vee y$. In conclusion, we have: :$a \vee x = a \vee y$ :$\neg a \vee x = \neg a \vee y$ Hence $x = y$ by Cancellation of Join in Boolean Algebra, that is: :$\paren {a \wedge b} \wedge c = a \wedge \paren {b \wedge c}$ {{qed|lemma}} The result: :$\paren {a \vee b} \vee c = a \vee \paren {b \vee c}$ follows from the Duality Principle. {{qed}}
Let $a, b \in S$. Then: {{begin-eqn}} {{eqn | l = a \vee \paren {a \wedge b}       | r = \paren {a \wedge \top} \vee \paren {a \wedge b}       | c = Boolean Algebra: Axiom $(BA_1 \ 3)$: $\top$ is the identity for $\wedge$ }} {{eqn | r = a \wedge \paren {\top \vee b}       | c = Boolean Algebra: Axiom $(BA_1 \ 2)$: $\wedge$ distributes over $\vee$ }} {{eqn | r = a \wedge \top       | c = Identities of Boolean Algebra also Zeroes }} {{eqn | r = a       | c = Boolean Algebra: Axiom $(BA_1 \ 3)$ $\top$ is the identity for $\wedge$ }} {{end-eqn}} as desired. {{qed|lemma}} The result: :$a = a \wedge \paren {a \vee b}$ follows from the Duality Principle. {{qed}}
{{BeginTableau   |\neg p \lor q \vdash p \implies q}} {{Premise        |1|\neg p \lor q}} {{DeMorgan       |2|1|\neg \paren {\neg \neg p \land \neg q}|1|Disjunction}} {{SequentIntro   |3|1   |\neg \neg p \implies q   |2   |Implication Equivalent to Negation of Conjunction with Negative: $\neg \paren {p \land \neg q} \vdash p \implies q$ }} {{Assumption     |4|p}} {{DoubleNegIntro |5|4|\neg \neg p|4}} {{ModusPonens    |6|1, 4|q|3|5}} {{Implication    |7|1|p \implies q|4|6}} {{EndTableau}} {{qed}}
The series $\displaystyle \sum_{n \mathop = 1}^\infty \paren {-b_n}$ is absolutely convergent, as $\cmod {-b_n} = \cmod {b_n}$ for all $n \in \N$. Then: {{begin-eqn}} {{eqn | l = \sum_{n \mathop = 1}^\infty a_n - \sum_{n \mathop = 1}^\infty b_n       | r = \sum_{n \mathop = 1}^\infty a_n + \paren {-1} \sum_{n \mathop = 1}^\infty b_n }} {{eqn | r = \sum_{n \mathop = 1}^\infty a_n + \sum_{n \mathop = 1}^\infty \paren {-1} b_n       | c = Manipulation of Absolutely Convergent Series: Scale Factor }} {{eqn | r = \sum_{n \mathop = 1}^\infty \paren {a_n - b_n}       | c = Sum of Absolutely Convergent Series }} {{end-eqn}} {{qed}}
{{BeginTableau|p \vdash q \implies p}} {{Premise|1|p}} {{Addition|2|1|\neg q \lor p|1|2}} {{SequentIntro|3|1|q \implies p|1|Rule of Material Implication}} {{EndTableau}} {{qed}}
We apply the Method of Truth Tables. As can be seen by inspection, where the truth value in the relevant column on the {{LHS}} is $T$, that under the one on the {{RHS}} is also $T$: $\begin{array}{|c||ccc|} \hline p & q & \implies & p \\ \hline F & F & T & F \\ F & T & F & F \\ T & F & T & T \\ T & T & T & T \\ \hline \end{array}$ {{qed}}
{{BeginTableau|\neg p \vdash p \implies q}} {{Premise|1|\neg p}} {{Addition|2|1|\neg p \lor q|1|1}} {{SequentIntro|3|1|p \implies q|2|Rule of Material Implication}} {{EndTableau}} {{qed}}
We apply the Method of Truth Tables. As can be seen by inspection, where the truth value in the relevant column on the {{LHS}} is $T$, that under the one on the {{RHS}} is also $T$: $\begin{array}{|cc||ccc|} \hline \neg & p & p & \implies & q \\ \hline T & F & F & T & F \\ T & F & F & T & T \\ F & F & T & F & F \\ F & F & T & T & T \\ \hline \end{array}$ {{qed}}
{{BeginTableau|p \lor q, \neg q \vdash p}} {{Premise|1|p \lor q}} {{Premise|2|\neg q}} {{Assumption|3|q}} {{SequentIntro|4|2|q \implies p|2|False Statement implies Every Statement}} {{ModusPonens|5|2, 3|p|4|3}} {{Assumption|6|p}} {{ProofByCases|7|1, 2|p|1|3|5|6|6}} {{EndTableau}} {{qed}}