avewright/proofwiki-math huggingface dataset:
Theorem: Piecewise Continuous Function with One-Sided Limits is Uniformly Continuous on Each Piece

Let $f$ be a real function defined on a closed interval $\closedint a b$.
Let $f$ be piecewise continuous with one-sided limits:
{{:Definition:Piecewise Continuous Function with One-Sided Limits}}
Then:
:for all $i \in \set {1, 2, \ldots, n}$, $f$ is uniformly continuous on $\openint {x_{i − 1} } {x_i}$.

Proof:
We have that $f$ is continuous on $\openint {x_{i − 1} } {x_i}$ for every $i \in \set {1, 2, \ldots, n}$.
Since $f$ is piecewise continuous with one-sided limits, the one-sided limits:
:$\ds \lim_{x \mathop \to {x_{i−1} }^+} \map f x$
and:
:$\ds \lim_{x \mathop \to {x_i}^-} \map f x$
exist for every $i \in \set {1, 2, \ldots, n}$.
Thus, from Extendability Theorem for Function Continuous on Open Interval, a function $f_i$ exists that satisfies:
:$f_i$ is defined on $\closedint {x_{i − 1} } {x_i}$
:$f_i$ is continuous on $\closedint {x_{i − 1} } {x_i}$
:$f_i$ equals $f$ on $\openint {x_{i − 1} } {x_i}$
for an arbitrary $i$ in $\set {1, 2, \ldots, n}$.
From Continuous Function on Closed Real Interval is Uniformly Continuous, $f_i$ is uniformly continuous on $\closedint {x_{i − 1} } {x_i}$.
We have that $\openint {x_{i − 1} } {x_i}$ is a subset of $\closedint {x_{i − 1} } {x_i}$.
Hence $f_i$ is uniformly continuous on $\openint {x_{i − 1} } {x_i}$.
As $f_i$ equals $f$ on $\openint {x_{i − 1} } {x_i}$, $f$ also is uniformly continuous on $\openint {x_{i − 1} } {x_i}$.
Since $i$ is arbitrary, this results holds for all $i \in \set {1, 2, \ldots, n}$.
{{qed}}
Category:Piecewise Continuous Functions
j23ur0nwew4p7t1y0d18cm8e87qzdhz
Theorem: Principle of Definition by Structural Induction

Let $\LL$ be a formal language.
Let the formal grammar of $\LL$ be a bottom-up grammar with unique parsability.
A definition $\map D \phi$ (in the metalanguage of $\LL$) for all well-formed formulas $\phi$ of $\LL$ is uniquely specified by:
:$(1): \quad$ A definition $\map D a$ for each letter $a$ of $\LL$
:$(2): \quad$ For each rule of formation for $\LL$, a definition $\map D \phi$ of the resultant WFF $\phi$ in terms of the constituent WFFs' definitions $\map D {\phi_1}, \ldots, \map D {\phi_n}$.

Proof:
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 Languages
3qt1s0jkhptpnnuhdnp2hk8nm0tv24b
Theorem: Principle of Structural Induction

Let $\LL$ be a formal language.
Let the formal grammar of $\LL$ be a bottom-up grammar.
Let $\map P \phi$ be a statement (in the metalanguage of $\LL$) about well-formed formulas $\phi$ of $\LL$.
Then $P$ is true for all WFFs of $\LL$ {{iff}} both:
:$\map P a$ is true for all letters $a$ of $\LL$,
and, for each rule of formation of $\LL$, if $\phi$ is a WFF resulting from WFFs $\phi_1, \ldots, \phi_n$ by applying that rule, then:
:$\map P \phi$ is true only if $\map P {\phi_1}, \ldots, \map P {\phi_n}$ are all true.

Proof:
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}}
Theorem: Integers under Multiplication form Monoid

The set of integers under multiplication $\struct {\Z, \times}$ is a monoid.

Proof:
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}}
Theorem: Integers under Multiplication do not form Group

The set of integers under multiplication $\struct {\Z, \times}$ does not form a group.

Proof:
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-axiom|0}}, {{Group-axiom|1}} and {{Group-axiom|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}}
Theorem: Integers under Multiplication form Semigroup

The set of integers under multiplication $\struct {\Z, \times}$ is a semigroup.

Proof:
=== {{Semigroup-axiom|0|nolink}} ===
Integer Multiplication is Closed, fulfilling {{Semigroup-axiom|0}}.
{{qed|lemma}}
Theorem: Bottom-Up Form of Top-Down Grammar defines same Formal Language

Let $\LL$ be a formal language.
Let $\TT$ be a top-down grammar for $\LL$.
Let $\BB$ be the bottom-up form of $\TT$.
Then $\BB$ is also a formal grammar for $\LL$.

Proof:
Let $\phi$ be a $\BB$-WFF.
If $\phi$ is a letter, then it is {{afortiori}} a $\TT$-WFF.
For, it may be formed by replacing the starting metasymbol of $\TT$ by $\phi$.
Suppose that $\phi$ is formed from WFFs $\phi_1, \ldots, \phi_n$ by the rule of formation $\mathbf R_\BB$ of $\BB$.
Suppose also that each of $\phi_1, \ldots, \phi_n$ is also a $\TT$-WFF.
Then by applying the rule of formation $\mathbf R$ of $\TT$, we obtain a collation with metasymbols $\psi_1, \ldots, \psi_n$.
By assumption, we can apply rules of formation of $\TT$ to each $\psi_i$ to yield the corresponding $\TT$-WFF $\phi_i$.
It follows that $\phi$ is also a $\TT$-WFF.
By the Principle of Structural Induction, each $\BB$-WFF is also a $\TT$-WFF.
Conversely, suppose that $\phi$ is a $\TT$-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 $\BB$-WFF.
In particular, then, $\phi$ will be a $\BB$-WFF.
Since $\mathbf R_n$ is the last rule of formation applied, it must replace a metasymbol by a letter.
Since letters are $\BB$-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 $\BB$-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 $\BB$-WFFs $\phi_i$.
Now the rule of formation $\paren {\mathbf R_i}_\BB$ ensures that $\psi$ will be replaced by a $\BB$-WFF.
So each metasymbol $\psi$ in the input of $\mathbf R_i$ will be replaced by a $\BB$-WFF.
Hence $\phi$ is a $\BB$-WFF.
{{qed}}
Theorem: Equivalence of Definitions of Characteristic of Ring

Let $\struct {R, +, \circ}$ be a ring with unity whose zero is $0_R$ and whose unity is $1_R$.
{{TFAE|def = Characteristic of Ring}}

Proof:
=== 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}}
Theorem: Canonical Form of Rational Number is Unique

The canonical form of a rational number is unique.

Proof:
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}}
Theorem: Existence of Canonical Form of Rational Number

Let $r \in \Q$.
Then:
:$\exists p \in \Z, q \in \Z_{>0}: r = \dfrac p q, p \perp q$
That is, every rational number can be expressed in its canonical form.

Proof:
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}}
Theorem: Existence of Dyadic Rational between two Rationals

Let $a$ and $b$ be rational numbers such that $a < b$.
Then there exist integers $m$ and $r$ such that:
:$a < \dfrac m {2^r} < b$
That is, there exists a dyadic rational between any pair of rational numbers.

Proof:
As $a < b$ it follows that $a \ne b$ and so $b - a \ne 0$.
Thus:
:$\dfrac 1 {b - a} \in \R$
By the Axiom of Archimedes:
:$\exists r \in \N: r > \dfrac 1 {b - a}$
Notice that $2^r > r$.
Thus we also have:
:$2^r > \dfrac 1 {b - a}$
Let $M := \set {x \in \Z: x > a 2^r}$.
By Set of Integers Bounded Below has Smallest Element, there exists $m \in \Z$ such that $m$ is the smallest element of $M$.
That is:
:$m > a 2^r$
and, by definition of smallest element:
:$m - 1 \le a 2^r$
As $2^r > \dfrac 1 {b - a}$, it follows from Ordering of Reciprocals that:
:$\dfrac 1 {2^r} < b - a$
Thus:
{{begin-eqn}}
{{eqn | l = m - 1
      | o = \le
      | r = a 2^r
      | c = 
}}
{{eqn | ll= \leadsto
      | l = m
      | o = \le
      | r = a 2^r + 1
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac m {2^r}
      | o = \le
      | r = a + \frac 1 {2^r}
      | c = 
}}
{{eqn | o = <
      | r = a + \paren {b - a}
      | c = 
}}
{{eqn | r = b
      | c = 
}}
{{end-eqn}}
Thus we have shown that $a < \dfrac m {2^r} < b$.
{{qed}}
Theorem: Field with 4 Elements has only Order 2 Elements

Let $\struct {\GF, +, \times}$ be a field which has exactly $4$ elements.
Then:
:$\forall a \in \GF: a + a = 0_\GF$
where $0_\GF$ is the zero of $\GF$.

Proof:
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}}
Theorem: Count of Truth Functions

There are $2^{\paren {2^k} }$ distinct truth functions on $k$ variables.

Proof:
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 of Finite Sets.
The result follows from Cardinality of Set of All Mappings.
{{qed}}
Theorem: Unary Truth Functions

There are $4$ distinct unary truth functions:
:$(1): \quad$ The constant function $\map f p = \F$
:$(2): \quad$ The constant function $\map f p = \T$
:$(3): \quad$ The identity function $\map f p = p$
:$(4): \quad$ The logical not function $\map f p = \neg p$

Proof:
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:
:<nowiki>$\begin{array}{|c|cccc|} \hline
p & \circ_1 & \circ_2 & \circ_3 & \circ_4 \\
\hline
\T & \T & \T & \F & \F \\
\F & \T & \F & \T & \F \\
\hline
\end{array}$</nowiki>
$\circ_1$: 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:
{{begin-eqn}}
{{eqn | n = 1
      | l = p = \T
      | o = \implies
      | r = \map {\circ_2} p = \T
}}
{{eqn | n = 2
      | l = p = \F
      | o = \implies
      | r = \map {\circ_2} p = \F
}}
{{end-eqn}}
Thus $\circ_2$ is the identity function $\map {\circ_2} p = p$.
$\circ_3$: We have:
{{begin-eqn}}
{{eqn | n = 1
      | l = p = \T
      | o = \implies
      | r = \map {\circ_3} p = \F
}}
{{eqn | n = 2
      | l = p = \F
      | o = \implies
      | r = \map {\circ_3} p = \T
}}
{{end-eqn}}
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}}
Theorem: Binary Truth Functions

There are $16$ distinct binary truth functions:
* Two constant operations:
** $\map {f_\F} {p, q} = \F$
** $\map {f_\T} {p, q} = \T$
* Two projections:
** $\map {\pr_1} {p, q} = p$
** $\map {\pr_2} {p, q} = q$
* Two negated projections:
** $\map {\overline {\pr_1} } {p, q} = \neg p$
** $\map {\overline {\pr_2} } {p, q} = \neg q$
* The conjunction: $p \land q$
* The disjunction: $p \lor q$
* Two conditionals:
** $p \implies q$
** $q \implies p$
* The biconditional: $p \iff q$
* The exclusive or: $\map \neg {p \iff q}$
* Two negated conditionals:
** $\map \neg {p \implies q}$
** $\map \neg {q \implies p}$
* The NAND: $p \uparrow q$
* The NOR: $p \downarrow q$

Proof:
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}}
Theorem: Set of Rational Numbers whose Numerator Divisible by p is Closed under Addition

Let $p$ be a prime number.
Let $A_p$ be the set of all rational numbers which, when expressed in canonical form has a numerator which is divisible by $p$.
Then $A_p$ is closed under rational addition.

Proof:
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 \paren {n_1 d_2 + n_2 d_1} } {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 $\dfrac {p \paren {n_1 d_2 + n_2 d_1} } {d_1 d_2}$ is expressed in canonical form, $p$ will still be a divisor of the numerator.
Hence the result.
{{qed}}
Theorem: Inverse Hyperbolic Sine Logarithmic Formulation

{{finish|The true story is more complicated than this. The inverse hyperbolic sine is a multifunction. So is $\ln$ in the complex plane. And $\sqrt {z^2 + 1}$ has two values, pos and neg, so you need to justify which one is taken. The best context to put this page is probably into Definition:Inverse Hyperbolic Sine/Complex/Principal Branch, where an explanation needs to be made as to why the principal value is taken thus.}}
For any complex number $z \in \C$:
:$\arsinh z = \map \ln {z + \sqrt {z^2 + 1} }$
where $\arsinh z$ is the inverse hyperbolic sine.

Proof:
{{begin-eqn}}
{{eqn | l = z
      | r = \sinh \arsinh z
}}
{{eqn | ll= \leadstoandfrom
      | l = z
      | r = \frac {e^{\arsinh z} - e^{-\arsinh z} } 2
      | c = {{Defof|Inverse Hyperbolic Sine}}
}}
{{eqn | ll= \leadstoandfrom
      | l = 2 z e^{\arsinh z}
      | r = e^{2 \arsinh z} - 1
      | c = Multiplication by $2 e^{\arsinh z}$
}}
{{eqn | ll= \leadstoandfrom
      | l = 0
      | r = e^{2 \arsinh z} - 2 z e^{\arsinh z} - 1
}}
{{eqn | ll= \leadstoandfrom
      | l = e^{\arsinh z}
      | r = z + \sqrt {z^2 + 1}
      | c = Quadratic Formula, $e^z > 0, \sqrt {z^2 + 1} > z$
}}
{{eqn | ll= \leadstoandfrom
      | l = \arsinh z
      | r = \map \ln {z + \sqrt {z^2 + 1 } }
}}
{{end-eqn}}
{{qed}}
Category:Inverse Hyperbolic Sine
9j2q4i3i3rbdfssfiktcb77eayvwgs7
Theorem: Integers under Addition form Abelian Group

The set of integers under addition $\struct {\Z, +}$ forms an abelian group.

Proof:
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.
Theorem: Piecewise Continuous Function with One-Sided Limits is Darboux Integrable

Let $f$ be a real function defined on a closed interval $\closedint a b$.
Let $f$ be piecewise continuous with one-sided limits on $\closedint a b$.
Then $f$ is Darboux integrable on $\closedint a b$.

Proof:
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}}
Theorem: Piecewise Continuous Function with One-Sided Limits is Darboux Integrable

Let $f$ be a real function defined on a closed interval $\closedint a b$.
Let $f$ be piecewise continuous with one-sided limits on $\closedint a b$.
Then $f$ is Darboux integrable on $\closedint a b$.

Proof:
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 $\openint {x_{i - 1} } {x_i}$ 
:$\ds \lim_{x \mathop \to {x_{i - 1} }^+} \map f x$ and $\ds \lim_{x \mathop \to {x_i}^-} \map f x$ 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 $\openint {x_{i - 1} } {x_i}$
:$\ds \lim_{x \mathop \to {x_{i - 1} }^+} \map f x$ and $\ds \lim_{x \mathop \to {x_i}^-} \map f x$ 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 $\ds \lim_{x \mathop \to x_{i - 1}^+} \map f x$ and $\ds \lim_{x \mathop \to x_i^-} \map f x$ 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}}