hcju/proofwikiqa huggingface dataset:
We have that $f$ is continuous on $\left({x_{i − 1} \,.\,.\, x_i}\right)$ for every $i \in \left\{ {1, 2, \ldots, n}\right\}$. Since $f$ is piecewise continuous with one-sided limits, the one-sided limits: : $\displaystyle \lim_{x \mathop \to {x_{i−1} }^+} f \left({x}\right)$ and: : $\displaystyle \lim_{x \mathop \to {x_i}^-} f \left({x}\right)$ exist for every $i \in \left\{{1, 2, \ldots, n}\right\}$. Thus, from Extendability Theorem for Function Continuous on Open Interval, a function $f_i$ exists that satisfies: : $f_i$ is defined on $\left[{x_{i − 1} \,.\,.\, x_i}\right]$ : $f_i$ is continuous on $\left[{x_{i − 1} \,.\,.\, x_i}\right]$ : $f_i$ equals $f$ on $\left({x_{i − 1} \,.\,.\, x_i}\right)$ for an arbitrary $i$ in $\left\{{1, 2, \ldots, n}\right\}$ From Continuous Function on Closed Interval is Uniformly Continuous, $f_i$ is uniformly continuous on $\left[{x_{i − 1} \,.\,.\, x_i}\right]$. We have that $\left({x_{i − 1} \,.\,.\, x_i}\right)$ is a subset of $\left[{x_{i − 1} \,.\,.\, x_i}\right]$. Hence $f_i$ is uniformly continuous on $\left({x_{i − 1} \,.\,.\, x_i}\right)$. As $f_i$ equals $f$ on $\left({x_{i − 1} \,.\,.\, x_i}\right)$, $f$ also is uniformly continuous on $\left({x_{i − 1} \,.\,.\, x_i}\right)$. Since $i$ is arbitrary, this results holds for all $i \in \left\{{1, 2, \ldots, n}\right\}$. {{qed}} Category:Piecewise Continuous Functions 1o4dyhf6ae3ngxqoywpr90p0t6neql1
We apply the Principle of Structural Induction on the following statement $\map P \phi$: :The definition is specified for $\phi$ The given hypotheses verify the conditions for the Principle of Structural Induction. It follows that $\map D \phi$ is specified for each WFF $\phi$. Moreover, by virtue of the unique parsability, it is ensured that each WFF $\phi$ has a unique definition $\map D \phi$. {{qed}} Category:Formal Systems 9trc12ncmpr7m2cvo2w8qth75d0tt7m
Let $\phi$ be a WFF of $\LL$. Then $\phi$ is the result of applying finitely many rules of formation of $\LL$. If we can show that the result of each rule of formation satisfies $P$, we will have finished. Suppose now that for a rule of formation $\mathbf R$, all preceding rules have produced WFFs satisfying $P$. By the bottom-up nature of the formal grammar of $\LL$, $\mathbf R$ either: :Introduces a letter as a WFF; or :Constructs a new WFF from previously constructed ones. The two given hypotheses precisely ensure that the WFF resulting from $\mathbf R$ must also satisfy $P$. For the first rule of formation applied, all preceding rules have vacuously produced WFFs satisfying $P$. But now we see that any subsequent rule of formation will satisfy this premise. In particular, it applies to the final rule of formation, and hence $\map P \phi$ is true. {{qed}}
From Integers under Multiplication form Semigroup, $\struct {\Z, \times}$ is a semigroup. From Integer Multiplication Identity is One, $\struct {\Z, \times}$ has an identity element, which is $1$. All the criteria for being a monoid are thus seen to be fulfilled. {{qed}}
In order to be classified as a group, the algebraic structure $\struct {\Z, \times}$ needs to fulfil the group axioms. From Integers under Multiplication form Monoid, $\struct {\Z, \times}$ forms a monoid. Therefore group axioms $\text G 0$, $\text G 1$ and $\text G 2$ are satisfied. However, from Invertible Integers under Multiplication, the only integers with inverses under multiplication are $1$ and $-1$. As not all integers have inverses, it follows that $\struct {\Z, \times}$ is not a group. {{qed}}
=== Closure === Integer Multiplication is Closed. {{qed|lemma}} === Associativity === Integer Multiplication is Associative. {{qed|lemma}} The criteria for $\struct {\Z, \times}$ to be a semigroup are seen to be satisfied.  {{Qed}}
Let $\phi$ be a $\mathcal B$-WFF. If $\phi$ is a letter, then it is clearly a $\mathcal T$-WFF. For, it may be formed by replacing the starting metasymbol of $\mathcal T$ by $\phi$. Suppose that $\phi$ is formed from WFFs $\phi_1, \ldots, \phi_n$ by the rule of formation $\mathbf R_{\mathcal B}$ of $\mathcal B$. Suppose also that each of $\phi_1, \ldots, \phi_n$ is also a $\mathcal T$-WFF. Then by applying the rule of formation $\mathbf R$ of $\mathcal T$, we obtain a collation with metasymbols $\psi_1, \ldots, \psi_n$. By assumption, we can apply rules of formation of $\mathcal T$ to each $\psi_i$ to yield the corresponding $\mathcal T$-WFF $\phi_i$. It follows that $\phi$ is also a $\mathcal T$-WFF. By the Principle of Structural Induction, each $\mathcal B$-WFF is also a $\mathcal T$-WFF. Conversely, suppose that $\phi$ is a $\mathcal T$-WFF. Let it be formed by applying the rules of formation $\mathbf R_1, \ldots, \mathbf R_n$, in succession. We will prove that each metasymbol in the inputs of these rules of formation will be replaced by a $\mathcal B$-WFF. In particular, then, $\phi$ will be a $\mathcal B$-WFF. Since $\mathbf R_n$ is the last rule of formation applied, it must replace a metasymbol by a letter. Since letters are $\mathcal B$-WFFs, $\mathbf R_n$ satisfies the assertion. Suppose that we have established that all metasymbols in the result of $\mathbf R_i$ will be replaced by $\mathcal B$-WFFs. Then also the metasymbol $\psi$ that $\mathbf R_i$ replaces by a new collation will be replaced by a WFF. This is because the collation resulting from replacing $\psi$ according to $\mathbf R_i$ contains metasymbols $\psi_i$, which by assumption will be replaced by corresponding $\mathcal B$-WFFs $\phi_i$. Now the rule of formation $\left({\mathbf R_i}\right)_{\mathcal B}$ ensures that $\psi$ will be replaced by a $\mathcal B$-WFF. So each metasymbol $\psi$ in the input of $\mathbf R_i$ will be replaced by a $\mathcal B$-WFF. Hence $\phi$ is a $\mathcal B$-WFF. {{qed}}
=== Definition 1 is equivalent to Definition 3 === By definition of order, the order of $1_R$ is the smallest $p \in \Z_{> 0}$ such that $p \cdot 1_R = 0_R$, the identity of the additive group $\struct {R, +}$ of $\struct {R, +, \circ}$. Hence if the order of $1_R$ is finite, the definitions coincide. If the order of $1_R$ is infinite, there is no such $p$. Then both definitions give $\Char R = 0$, so they coincide as well. {{qed|lemma}} === Definition 1 is equivalent to Definition 2 === Let $g: \Z \to R$ be the initial homomorphism, with $\map g n = n \cdot 1_R$. By Kernel of Ring Homomorphism is Ideal, $\ker g$ is an ideal of $R$. By Ring of Integers is Principal Ideal Domain, there exists a unique $p \in \Z_{\ge 0}$ such that $\ker g$ is the principal ideal $\ideal p$. Suppose there is a smallest $m$ such that $m \cdot 1_R = 0_R$. Then for any $k \in \Z$: :$\map g {k m} = k m \cdot 1_R = k \cdot 0_R = 0_R$. This gives $\ideal m \subseteq \ker g = \ideal p$. Let $a \in \ker g$. Then by the Division Theorem: :$\exists q, r \in \Z, 0 \le r < m: a = m q + r$ As $a, m \in \ker g = \ideal p$ so does $r = a - m q$. Since $m$ is the smallest (strictly) positive integer such that $m \cdot 1_R = 0_R$: :$\forall n \in \Z: 0 < n < m \implies \map g n = n \cdot 1_R \ne 0_R$ This forces $r = 0$. Thus $a = m q \in \ideal m$. Hence we have $\ideal m = \ideal p$, giving $p = m$. Now consider the case where $m$ does not exist. Then for any $k \ne 0$: :$\map g k = k \cdot 1_R \ne 0_R$ And we have: :$\map g 0 = 0 \cdot 1_R = 0_R$ giving $\ker g = \set 0 = \ideal 0$. Both definitions give $\Char R = 0$. Therefore the definitions are equivalent in each case. {{qed}} Category:Rings with Unity 7zcpiooxjpqxzrlej6cbfkafzp4apny
Let $r \in \Q$ be a rational number. Let $\dfrac p q$ and $\dfrac {p'} {q'}$ be two canonical forms of $r$. {{WLOG}}, assume $q \le q'$. By Equality of Rational Numbers: :$p q' = p' q$ Therefore, $q'$ divides $p' q$. As $\dfrac {p'} {q'}$ is a canonical form of $r$, by definition $q'$ is coprime to $p'$. By Euclid's Lemma $q'$ divides $q$. $q$ and $q'$ are both positive. So from Absolute Value of Integer is not less than Divisors: Corollary: :$q' \le q$ As  $q' \le q$ and $q \le q'$ it follows that $q = q'$. It then follows that $p = p'$. Hence $\dfrac p q$ and $\dfrac {p'} {q'}$ are the same unique canonical form of $r$. {{qed}}
We have that the set of rational numbers is the field of quotients of the set of integers. From Divided by Positive Element of Field of Quotients: :$\exists s \in \Z, t \in \Z_{>0}: r = \dfrac s t$ Now if $s \perp t$, our task is complete. Otherwise, let: :$\gcd \set {s, t} = d$ where $\gcd \set {s, t}$ denotes the greatest common divisor of $s$ and $t$. Let $s = p d, t = q d$. We have that $t, d \in \Z_{>0}$ Therefore $q \in \Z_{>0}$ also. From Integers Divided by GCD are Coprime: :$p \perp q$ Also: {{begin-eqn}} {{eqn | l = \frac s t       | r = \frac {p d} {q d}       | c =  }} {{eqn | r = \frac p q \frac d d       | c =  }} {{eqn | r = \frac p q 1       | c =  }} {{eqn | r = \frac p q       | c =  }} {{end-eqn}} Thus: :$r = \dfrac p q$ where $p \perp q$ and $q \in \Z_{>0}$. {{qed}}
By definition of field, both the algebraic structures $\struct {\GF, +}$ and $\struct {\GF^*, \times}$ are (abelian) groups, where $\GF^* := \GF \setminus \set 0$. By definition: :$\struct {\GF, +}$ is of order $4$ :$\struct {\GF^*, \times}$ is of order $3$. From Classification of Groups of Order up to 15, there are only two possibilities: :$(1): \quad \struct {\GF, +} \cong \Z_4$ and $\struct {\GF^*, \times} \cong \Z_3$ :$(2): \quad \struct {\GF, +} \cong \Z_2 \times \Z_2$ and $\struct {\GF^*, \times} \cong \Z_3$. In case $(2)$: :$\forall a \in \GF: a + a = 0_\GF$ but in case $(1)$: :$\exists a, b \in \GF: a + a = b \ne 0_\GF$ where $b + b = 0_\GF$ as there exists an element of $\struct {\GF, +}$ of order $4$. {{AimForCont}} $(1)$ describes a field. Then: {{begin-eqn}} {{eqn | l = \paren {b + b} \times b       | r = 0_\GF \times b       | c = as $b + b = 0_\GF$ from above }} {{eqn | r = 0_\GF       | c = as $0_\GF$ is the zero of $\struct {\GF, +, \times}$ }} {{end-eqn}} But then: {{begin-eqn}} {{eqn | l = \paren {b + b} \times b       | r = \paren {b \times b} + \paren {b \times b}       | c = as $\times$ is distributive over $+$ }} {{end-eqn}} As $\struct {\GF, +} \cong \Z_4$, both $a + a = b$ and $\paren {-a} + \paren {-a} = b$. The only way for $\paren {b \times b} + \paren {b \times b} = 0_\GF$ is for $b \times b = b$ or $b \times b = 0_\GF$. The second case is eliminated as $0_\GF \notin \struct {\GF^*, \times}$. So it must be the case that $b \times b = b$. So as $\struct {\GF^*, \times} \cong \Z_3$ it must follow that $a \times b = a, a \times \paren {-a} = b$ and then $a \times a = -a$. It follows that the Cayley tables of $\struct {\GF, +}$ and $\struct {\GF^*, \times}$ must therefore be as follows: :$\begin{array}{c|cccc} + & 0_\GF & a & b & -a \\ \hline 0_\GF & 0_\GF & a & b & -a \\ a & a & b & -a & 0_\GF \\ b & b & -a & 0_\GF & a \\ -a & -a & 0_\GF & a & b \\ \end{array} \qquad \begin{array}{c|cccc} \times & 0_\GF & b & a & -a \\ \hline 0_\GF & 0_\GF & 0_\GF & 0_\GF & 0_\GF \\ b & 0_\GF & b & a & -a \\ a & 0_\GF & a & -a & b \\ -a & 0_\GF & -a & b & a \\ \end{array}$ {{begin-eqn}} {{eqn | l = \paren {a + b} \times a       | r = \paren {-a} \times a       | c = as $a + b = -a$ from above }} {{eqn | r = b       | c = as $\paren {-a} \times a = b$ from above }} {{end-eqn}} But: {{begin-eqn}} {{eqn | l = \paren {a \times a} + \paren {b \times a}       | r = \paren {-a} + a       | c = from above }} {{eqn | r = 0_\GF       | c = from above }} {{end-eqn}} So: :$\paren {a + b} \times a \ne \paren {a \times a} + \paren {b \times a}$ demonstrating that $\times$ is not distributive over $+$. Thus $\GF$ as has been defined: :$\struct {\GF, +} \cong \Z_4$ and $\struct {\GF^*, \times} \cong \Z_3$ is not a field. It follows that our supposition that $\struct {\GF, +} \cong \Z_4$ was false. Thus, if $\GF$ is a field, then $\struct {\GF, +} \cong \Z_2 \times \Z_2$. That is: :$\forall a \in \GF: a + a = 0_\GF$ as needed to be proved. {{qed}}
Let $\struct {\GF, +, \times}$ be a field of order $4$ whose zero is $0_\GF$. By definition, $\GF$ is a Galois field. The additive group $\struct {\GF, +}$ of $\GF$ can be one of two: :$(1): \quad$ the cyclic group of order $4$, generated by the identity of $\struct {\GF, +}$ which is $0_\GF$ or: :$(2): \quad$ the Klein $4$-group, whose elements are all of the form $a + a = 0_\GF$. {{AimForCont}} $(1)$ is the additive group of a field. Then the characteristic of $\GF$ is $4$. But from Characteristic of Galois Field is Prime it follows that $\GF$ is not a field. From that contradiction it follows that the additive group $\struct {\GF, +}$ of $F$ cannot be the cyclic group of order $4$. Hence for $\GF$ to be a field at all, it is necessary for $\struct {\GF, +}$ to be the Klein $4$-group. {{qed}}
Let $f: \mathbb B^k \to \mathbb B$ be a truth function. The domain of $f$ has $2^k$ elements, from Cardinality of Cartesian Product. The result follows from Cardinality of Set of All Mappings. {{qed}}
From Count of Truth Functions there are $2^{\paren {2^1} } = 4$ distinct truth functions on $1$ variable. These can be depicted in a truth table as follows: :$\begin{array}{|c|cccc|} \hline p & \circ_1 & \circ_2 & \circ_3 & \circ_4 \\ \hline \T & \T & \T & \F & \F \\ \F & \T & \F & \T & \F \\ \hline \end{array}$ $\circ_1$: Whether $p = \T$ or $p = \F$, $\map {\circ_1} p = \T$. Thus $\circ_1$ is the constant function $\map {\circ_1} p = \T$. $\circ_2$: We have: :$(1): \quad p = \T \implies \map {\circ_2} p = \T$ :$(2): \quad p = \F \implies \map {\circ_2} p = \F$ Thus $\circ_2$ is the identity function $\map {\circ_2} p = p$. $\circ_3$: We have: :$(1): \quad p = \T \implies \map {\circ_3} p = \F$ :$(2): \quad p = \F \implies \map {\circ_3} p = \T$ Thus $\circ_3$ is the logical not function $\map {\circ_3} p = \neg p$. $\circ_4$: Whether $p = \T$ or $p = \F$, $\map {\circ_4} p = \F$. Thus $\circ_4$ is the constant function $\map {\circ_4} p = \F$. All four have been examined, and there are no other unary truth functions. {{qed}}
From Count of Truth Functions there are $2^{\paren {2^2} } = 16$ distinct truth functions on $2$ variables. These can be depicted in a truth table as follows: $\begin{array}{|r|cccc|} \hline p                                & \T & \T & \F & \F \\ q                                & \T & \F & \T & \F \\ \hline \map {f_\T} {p, q}               & \T & \T & \T & \T \\ p \lor q                         & \T & \T & \T & \F \\ p \impliedby q                   & \T & \T & \F & \T \\ \map {\pr_1} {p, q}              & \T & \T & \F & \F \\ p \implies q                     & \T & \F & \T & \T \\ \map {\pr_2} {p, q}              & \T & \F & \T & \F \\ p \iff q                         & \T & \F & \F & \T \\ p \land q                        & \T & \F & \F & \F \\ p \uparrow q                     & \F & \T & \T & \T \\ \map \neg {p \iff q}             & \F & \T & \T & \F \\ \map {\overline {\pr_2} } {p, q} & \F & \T & \F & \T \\ \map \neg {p \implies q}         & \F & \T & \F & \F \\ \map {\overline {\pr_1} } {p, q} & \F & \F & \T & \T \\ \map \neg {p \impliedby q}       & \F & \F & \T & \F \\ p \downarrow q                   & \F & \F & \F & \T \\ \map {f_\F} {p, q}               & \F & \F & \F & \F \\ \hline \end{array}$ That accounts for all $16$ of them. {{qed}}
Let $a, b \in A_p$. Then $a = \dfrac {p n_1} {d_1}, b = \dfrac {p n_1} {d_1}$ where: :$n_1, n_2 \in \Z$ :$d_1, d_2 \in \Z_{>0}$ :$p n_1 \perp d_1, p n_2 \perp d_2$ Then: {{begin-eqn}} {{eqn | l = a + b       | r = \frac {p n_1} {d_1} + \frac {p n_2} {d_2}       | c =  }} {{eqn | r = \frac {p n_1 d_2 + p n_2 d_1} {d_1 d_2}       | c = {{Defof|Rational Addition}} }} {{eqn | r = \frac {p \left({n_1 d_2 + n_2 d_1}\right)} {d_1 d_2}       | c =  }} {{end-eqn}} From Euclid's Lemma for Prime Divisors, if $p \divides d_1 d_2$ then either $p \divides d_1$ or $p \divides d_2$. But neither of these is the case, so $p \nmid d_1 d_2$. Hence by Prime not Divisor implies Coprime: :$p \perp d_1 d_2$ where $\perp$ denotes coprimeness. So when $\dfrac {p \left({n_1 d_2 + n_2 d_1}\right)} {d_1 d_2}$ is expressed in canonical form, $p$ will still be a divisor of the numerator. Hence the result. {{qed}}
{{begin-eqn}} {{eqn | l = z       | r = \sinh \operatorname {arsinh} z }} {{eqn | ll= \leadstoandfrom       | l = z       | r = \frac {e^{\operatorname{arsinh} z} - e^{-\operatorname{arsinh} z} } 2       | c = {{Defof|Inverse Hyperbolic Sine}} }} {{eqn | ll= \leadstoandfrom       | l = 2 z e^{\operatorname{arsinh} z}       | r = e^{2 \operatorname{arsinh} z} - 1       | c = Multiplication by $2 e^{\operatorname{arsinh} z}$ }} {{eqn | ll= \leadstoandfrom       | l = 0       | r = e^{2 \operatorname{arsinh} z} - 2 z e^{\operatorname{arsinh} z} - 1 }} {{eqn | ll= \leadstoandfrom       | l = e^{\operatorname{arsinh} z}       | r = z + \sqrt {z^2 + 1}       | c = Quadratic Formula, $e^z > 0, \sqrt {z^2 + 1} > z$ }} {{eqn | ll= \leadstoandfrom       | l = \operatorname{arsinh} z       | r = \map \ln {z + \sqrt {z^2 + 1 } } }} {{end-eqn}} {{qed}} Category:Inverse Hyperbolic Sine mkdlysytrenv0xs7a0zsgl31zghtzvk
From the definition of the integers, the algebraic structure $\struct {\Z, +}$ is an isomorphic copy of the inverse completion of $\struct {\N, +}$. From Natural Numbers under Addition form Commutative Semigroup, $\struct {\N, +}$ is a commutative semigroup. From Natural Number Addition is Cancellable all elements of $\struct {\N, +}$ are cancellable. The result follows from Inverse Completion of Commutative Semigroup is Abelian Group. Thus addition on $\Z$ is well-defined, closed, associative and commutative on $\Z$. {{qed|lemma}} Let us define $\eqclass {\tuple {a, b} } \boxminus$ as in the formal definition of integers. That is, $\eqclass {\tuple {a, b} } \boxminus$ is an equivalence class of ordered pairs of natural numbers under the congruence relation $\boxminus$. $\boxminus$ is the congruence relation defined on $\N \times \N$ by: :$\tuple {x_1, y_1} \boxminus \tuple {x_2, y_2} \iff x_1 + y_2 = x_2 + y_1$ In order to streamline the notation, we will use $\eqclass {a, b} {}$ to mean $\eqclass {\tuple {a, b} } \boxminus$, as suggested. === Identity is Zero === From Construction of Inverse Completion: Identity of Quotient Structure, the identity of $\struct {\Z, +}$ is $\eqclass {c, c} {}$ for any $c \in \N$: {{begin-eqn}} {{eqn | ll= \forall a, b, c \in \N:       | o =        | r = \eqclass {a, b} {} + \eqclass {c, c} {}       | c =  }} {{eqn | r = \eqclass {a, b} {}       | c =  }} {{eqn | r = \eqclass {c, c} {} + \eqclass {a, b} {}       | c =  }} {{end-eqn|lemma}} $\eqclass {c, c} {}$ is the equivalence class of pairs of elements $\N \times \N$ whose difference is zero. Thus the identity of $\struct {\Z, +}$ is seen to be $0$. Note that a perfectly good representative of $\eqclass {c, c} {}$ is $\eqclass {0, 0} {}$. This usually keeps to a minimum the complexity of any arithmetic that is needed. {{qed|lemma}} === Construction of Inverses === From Construction of Inverse Completion: Invertible Elements in Quotient Structure, we see that every element of $\struct {\Z, +}$ has an inverse. We can see that: {{begin-eqn}} {{eqn | ll= \forall a, b \in \N:       | l = \eqclass {a, b} {} + \eqclass {b, a} {}       | r = \eqclass {a + b, b + a} {}       | c =  }} {{eqn | r = \eqclass {a + b, a + b} {}       | c = Natural Number Addition is Commutative }} {{end-eqn}} The above construction is valid because $a$ and $b$ are both in $\N$ and hence cancellable. From Construction of Inverse Completion: Identity of Quotient Structure, $\eqclass {a + b, a + b} {}$ is a member of the equivalence class which is the identity of $\struct {\Z, +}$. Thus the inverse of $\eqclass {a, b} {}$ is $\eqclass {b, a} {}$. {{qed}}
We are given that $f$ is piecewise continuous with one-sided limits on $\closedint a b$. Therefore, there exists a finite subdivision $\set {x_0, x_1, \ldots, x_n}$ of $\closedint a b$, where $x_0 = a$ and $x_n = b$, such that for all $i \in \set {1, 2, \ldots, n}$: :$f$ is continuous on $\openint {x_{i - 1} } {x_i}$  :$\displaystyle \lim_{x \mathop \to {x_{i - 1} }^+} \map f x$ and $\displaystyle \lim_{x \mathop \to {x_i}^-} \map f x$ exist. Note that $n$ is the number of intervals $\openint {x_{i - 1} } {x_i}$ defined from the (finite) subdivision $\set {x_0, x_1, \ldots, x_n}$. We shall use proof by induction on the intervals. For all $k \in \set {1, 2, \ldots, n}$, let $\map P k$ be the proposition: :$f$ is Darboux integrable on $\closedint {x_0} {x_k}$. === Basis for the Induction === $\map P 1$ is the case: :$f$ is Darboux integrable on $\closedint {x_{i - 1} } {x_i}$ for an arbitrary $i \in \set {1, 2, \ldots, k}$.  Piecewise continuity with one-sided limits of $f$ for the case $n = 1$ means that: :$f$ is continuous on $\openint {x_{i - 1} } {x_i}$ :$\displaystyle \lim_{x \mathop \to {x_{i - 1} }^+} \map f x$ and $\displaystyle \lim_{x \mathop \to {x_i}^-} \map f x$ exist. By Integrability Theorem for Functions Continuous on Open Intervals, $f$ is Darboux integrable on $\closedint {x_{i - 1} } {x_i}$. Thus $\map P 1$ is seen to hold. This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true. So this is the induction hypothesis: :$f$ is Darboux integrable on $\openint {x_0} {x_k}$. from which it is to be shown that: :$f$ is Darboux integrable on $\openint {x_0} {x_{k + 1} }$. === Induction Step === This is the induction step: By definition of a piecewise continuous function with one-sided limits, for every $i \in \set {1, 2, \ldots, k, k + 1}$: :$f$ is continuous on $\openint {x_{i - 1} } {x_i}$ :the one-sided limits $\displaystyle \lim_{x \mathop \to x_{i - 1}^+} \map f x$ and $\displaystyle \lim_{x \mathop \to x_i^-} \map f x$ exist. By the induction hypothesis, $f$ is Darboux integrable on $\closedint {x_0} {x_k}$. From the basis for the induction, $f$ is Darboux integrable on $\closedint {x_k} {x_{k + 1} }$. We have that $f$ is Darboux integrable on $\closedint {x_0} {x_k}$ and $\closedint {x_k} {x_{k + 1} }$. Therefore, by Existence of Integral on Union of Adjacent Intervals: :$f$ is Darboux integrable on $\closedint {x_0} {x_k} \cup \closedint {x_k} {x_{k + 1} }$. We have that: :$\closedint {x_0} {x_{k + 1} } = \closedint {x_0} {x_k} \cup \closedint {x_k} {x_{k + 1} }$ Accordingly, $f$ is Darboux integrable on $\closedint {x_0} {x_{k + 1} }$. So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$f$ is Darboux integrable on $\closedint a b$. {{qed}}
We are given that $f$ is piecewise continuous with one-sided limits on $\closedint a b$. From Piecewise Continuous Function with One-Sided Limits is Bounded, $f$ is a bounded piecewise continuous function. The result follows from Bounded Piecewise Continuous Function is Darboux Integrable. {{qed}}
Suppose there exists an infinite sequence $\sequence {a_n}$ in $S$ such that: :$\forall n \in \N: a_{n + 1} \prec a_n$ We let $T = \set {a_0, a_1, a_2, \ldots}$. Clearly $T$ has no minimal element. Thus by definition $S$ is not well-founded. {{qed}}