hcju/proofwikiqa huggingface dataset:
First, we prove the result for addition only without multiplying by $\lambda, \mu$: {{begin-eqn}} {{eqn | l = \int_a^b \map f t + \map g t \rd t       | r = \int_a^b \map \Re {\map f t + \map g t} \rd t + i \int_a^b \map \Im {\map f t + \map g t} \rd t       | c = {{Defof|Complex Riemann Integral}} }} {{eqn | r = \int_a^b \map \Re {\map f t} + \map \Re {\map g t} \rd t }} {{eqn | o =       | ro= +       | r = i \int_a^b \map \Im {\map f t} + \map \Im {\map g t} \rd t       | c = Addition of Real and Imaginary Parts }} {{eqn | r = \int_a^b \map \Re {\map f t} \rd t + \int_a^b \map \Re {\map g t} \rd t }}  {{eqn | o =       | ro= +       | r = i \paren {\int_a^b \map \Im {\map f t} \rd t + \int_a^b \map \Im {\map g t} \rd t}       | c = Linear Combination of Integrals }} {{eqn | r = \int_a^b \map f t \rd t + \int_a^b \map g t \rd t }} {{end-eqn}} {{qed|lemma}} Next, we show the result for only one complex integral multiplied by a constant $\lambda$. By definition of complex number, we can find $\lambda_x, \lambda_y \in \R$ so $\lambda = \lambda_x + i \lambda_y$. Then: {{begin-eqn}} {{eqn | l = \int_a^b \lambda \map f t \rd t       | r = \int_a^b \map \Re {\lambda \map f t} \rd t + i \int_a^b \map \Im {\lambda \map f t} \rd t       | c = {{Defof|Complex Riemann Integral}} }} {{eqn | r = \int_a^b \map \Re {\paren {\lambda_x + i \lambda_y} \map f t} \rd t }} {{eqn | o =       | ro= +       | r = i \int_a^b \map \Im {\paren {\lambda_x + i \lambda_y} \map f t} \rd t }} {{eqn | r = \int_a^b \map \Re {\lambda_x \map f t} + \map \Re {i \lambda_y \map f t} \rd t }} {{eqn | o =       | ro= +       | r = i \int_a^b \map \Im {\lambda_x \map f t} + \map \Im {i \lambda_y \map f t} \rd t       | c = Addition of Real and Imaginary Parts }} {{eqn | r = \int_a^b \lambda_x \map \Re {\map f t} - \lambda_y \map \Im {\map f t} \rd t }} {{eqn | o =       | ro= +       | r = i \int_a^b \lambda_x \map \Im {\map f t} + \lambda_y \map \Re {\map f t} \rd t       | c = Multiplication of Real and Imaginary Parts }} {{eqn | r = \lambda_x \int_a^b \map \Re {\map f t} \rd t - \lambda_y \int_a^b \map \Im {\map f t} \rd t }} {{eqn | o =       | ro= +       | r = i \lambda_x \int_a^b \map \Im {\map f t} \rd t + i \lambda_y \int_a^b \map \Re {\map f t} \rd t       | c = Linear Combination of Integrals }} {{eqn | r = \lambda_x \paren {\int_a^b \map \Re {\map f t} \rd t + i \int_a^b \map \Im {\map f t} \rd t} }} {{eqn | o =       | ro= +       | r = i \lambda_y \paren {i \int_a^b \map \Im {\map f t} \rd t + \int_a^b \map \Re {\map f t} \rd t} }} {{eqn | r = \lambda_x \int_a^b \map f t \rd t + i \lambda_y \int_a^b \map f t \rd t       | c = {{Defof|Complex Riemann Integral}} }} {{eqn | r = \lambda \int_a^b \map f t \rd t }} {{end-eqn}} {{qed|lemma}} It follows from the results above that: {{begin-eqn}} {{eqn | l = \lambda \int_a^b \map f t \rd t + \mu \int_a^b \map g t \rd t       | r = \int_a^b \lambda \map f t \rd t + \int_a^b \mu \map g t \rd t }} {{eqn | r = \int_a^b \paren {\lambda \map f t + \mu \map g t} \rd t }} {{end-eqn}} {{qed}}
By Scalar Multiple of Simple Function is Simple Function, $- g = -1 \cdot g$ is a simple function. By Pointwise Sum of Simple Functions is Simple Function, so is $f - g$. {{qed}} Category:Simple Functions mpoikn4qgke6a5sf5msokunhl9flqe3
=== Necessary condition === Let $r = \cmod z$. If $z = 0$, we have: :$z = 0e^{i \theta} = re^{i \theta}$ Suppose $z \ne 0$ and $\theta = \map \arg z$. By definition of argument, the following two equations hold: :$(1): \quad \dfrac {\map \Re z} r = \cos \theta$ :$(2): \quad \dfrac {\map \Im z} r = \sin \theta$ where: :$\map \Re z$ denotes the real part of $z$ :$\map \Im z$ denotes the imaginary part of $z$. Then: {{begin-eqn}} {{eqn | l = z       | r = \map \Re z + i \map \Im z       | c = {{Defof|Complex Number}} }} {{eqn | r = r \cos \theta + i r \sin \theta       | c = from $(1)$ and $(2)$ }} {{eqn | r = r \paren {\cos \theta + i \sin \theta} }} {{eqn | r = re^{i \theta}       | c = Euler's Formula }} {{end-eqn}} {{qed|lemma}} === Sufficient condition === Let $z = re^{i \theta}$. From the equations above, we find: :$\map \Re {re^{i \theta} } = r \cos \theta$ :$\map \Im {re^{i \theta} } = r \sin \theta$ Then: {{begin-eqn}} {{eqn | l = \cmod {r e^{i \theta} }       | r = \sqrt {\paren {r \cos \theta}^2 + \paren {r \sin \theta}^2}       | c = {{Defof|Modulus of Complex Number}} }} {{eqn | r = r \sqrt {\cos^2 \theta + \sin^2 \theta}       | c = as $r \ge 0$ }} {{eqn | r = r         | c = Sum of Squares of Sine and Cosine }} {{end-eqn}} If $r \ne 0$, we find $\map \arg {r e^{i \theta} }$ by solving the two equations by definition of argument: :$(1): \quad \dfrac {r \cos \theta} r = \map \cos {\map \arg {r e^{i \theta} } }$ :$(2): \quad \dfrac {r \sin \theta} r = \map \sin {\map \arg {r e^{i \theta} } }$ We find: :$\map \arg {r e^{i \theta} } = \theta$ {{qed}}
We will first prove that there is an injection from $X$ to $Y$. Let $x \in X$. By Expression of Vector as Linear Combination from Basis is Unique: General Result, there is a unique finite subset $C_x$ of $R \times Y$ such that: :$\displaystyle x = \sum_{\tuple {r, v} \mathop \in C_x} r \cdot v$ and :$\forall \tuple {r, v} \in C_x: r \ne 0_R$ Define $\Phi: X \to \powerset Y$ by: :$\map \Phi x := \Img {C_x}$ We next show that $\sequence {\map \Phi x}_{x \mathop \in X}$ satisfies the '''marriage condition'''.  That is, for every finite subset $F$ of $X$: :$\displaystyle \card F \le \card {\bigcup \map \Phi F}$ Since $X$ is a basis, it is linearly independent. By Subset of Linearly Independent Set is Linearly Independent, $F$ is also linearly independent. By the definition of $\Phi$: :$\displaystyle F \subseteq \operatorname {span} \bigcup \map \Phi F$. By Finite Union of Finite Sets is Finite, $\displaystyle \bigcup \map \Phi F$ is finite. Thus by Size of Linearly Independent Subset is at Most Size of Finite Generator: :$\displaystyle \size F \le \size {\bigcup \map \Phi F}$ By Hall's Marriage Theorem, there is an injection from $X$ into $Y$. Precisely the same argument with $X$ and $Y$ interchanged shows that there is an injection from $Y$ into $X$ as well. Thus by the Cantor-Bernstein-Schröder Theorem, $X$ and $Y$ are equinumerous. {{qed}} {{BPI|Hall's Marriage Theorem/General Set|3}}
{{begin-eqn}} {{eqn | l = S * T       | r = \paren {S \setminus T} \cup \paren {T \setminus S}       | c = {{Defof|Symmetric Difference|index = 1}} }} {{eqn | r = \paren {\paren {S \cup T} \setminus T} \cup \paren {\paren {S \cup T} \setminus S}       | c = Set Difference with Union is Set Difference }} {{eqn | r = \paren {S \cup T} \setminus \paren {T \cap S}       | c = De Morgan's Laws: Difference with Intersection }} {{eqn | r = \paren {S \cup T} \setminus \paren {S \cap T}       | c = Intersection is Commutative }} {{end-eqn}}
{{begin-eqn}} {{eqn | l = S * T       | r = \paren {S \setminus T} \cup \paren {T \setminus S}       | c = {{Defof|Symmetric Difference|index = 1}} }} {{eqn | r = \paren {S \cap \overline T} \cup \paren {\overline S \cap T}       | c = Set Difference as Intersection with Complement }} {{end-eqn}}
{{begin-eqn}} {{eqn | l = S * T       | r = \paren {S \cup T} \setminus \paren {S \cap T}       | c = {{Defof|Symmetric Difference|index = 2}} }} {{eqn | r = \paren {S \cup T} \cap \paren {\overline {S \cap T} }       | c = Set Difference as Intersection with Complement }} {{eqn | r = \paren {S \cup T} \cap \paren {\overline S \cup \overline T}       | c = De Morgan's Laws: Complement of Intersection }} {{end-eqn}}
Let $S$ be a countable set. Let $T \subseteq S$. By definition, there exists an injection $f: S \to \N$. Let $i: T \to S$ be the inclusion mapping. We have that $i$ is an injection. Because the composite of injections is an injection, it follows that $f \circ i: T \to \N$ is an injection. Hence, $T$ is countable. {{qed}}
Recall the definition of the sine function: {{begin-eqn}} {{eqn | l = \sin z       | r = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {z^{2 n + 1} } {\paren {2 n + 1}!}       | c =  }} {{eqn | r = z - \frac {z^3} {3!} + \frac {z^5} {5!} - \frac {z^7} {7!} + \cdots + \paren {-1}^n \frac {z^{2 n + 1} } {\paren {2 n + 1}!} + \cdots       | c =  }} {{end-eqn}} Recall the definition of the exponential as a power series: {{begin-eqn}} {{eqn | l = \exp z       | r = \sum_{n \mathop = 0}^\infty \frac {z^n} {n!}       | c =  }} {{eqn | r = 1 + \frac z {1!} + \frac {z^2} {2!} + \frac {z^3} {3!} + \cdots + \frac {z^n} {n!} + \cdots       | c =  }} {{end-eqn}} Then, starting from the {{RHS}}: {{begin-eqn}} {{eqn | l = \frac {\exp \paren {i z} - \exp \paren {-i x} } {2 i}       | r = \frac 1 {2 i} \paren {\sum_{n \mathop = 0}^\infty \frac {\paren {i z}^n} {n!} - \sum_{n \mathop = 0}^\infty \frac {\paren {-i z}^n} {n!} }       | c =  }} {{eqn | r = \frac 1 {2 i} \sum_{n \mathop = 0}^\infty \paren {\frac {\paren {i z}^n - \paren {-i z}^n} {n!} }       | c =  }} {{eqn | r = \frac 1 {2 i} \sum_{n \mathop = 0}^\infty \paren {\frac {\paren {i z}^{2 n} - \paren {-i z}^{2 n} } {\paren {2 n}!} + \frac {\paren {i z}^{2 n + 1} - \paren {-i z}^{2 n + 1} } {\paren {2 n + 1}!} }       | c = split into even and odd $n$ }} {{eqn | r = \frac 1 {2 i} \sum_{n \mathop = 0}^\infty \frac {\paren {i z}^{2 n + 1} - \paren {-i z}^{2 n + 1} } {\paren {2 n + 1}!}       | c = as $\paren {-i z}^{2 n} = \paren {i z}^{2 n}$ }} {{eqn | r = \frac 1 {2 i} \sum_{n \mathop = 0}^\infty \frac {2 \paren {i z}^{2 n + 1} } {\paren {2 n + 1}!}       | c = as $\paren {-1}^{2 n + 1} = -1$ }} {{eqn | r = \frac 1 i \sum_{n \mathop = 0}^\infty \frac {\paren {i z}^{2 n + 1} } {\paren {2 n + 1}!}       | c = cancel $2$ }} {{eqn | r = \frac 1 i \sum_{n \mathop = 0}^\infty \frac {i \paren {-1}^n z^{2 n + 1} } {\paren {2 n + 1}!}       | c = as $i^{2 n + 1} = i \paren {-1})^n $ }} {{eqn | r = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {z^{2 n + 1} } {\paren {2 n + 1!} }       | c = cancel $i$ }} {{eqn | r = \sin z }} {{end-eqn}} {{qed}}
Recall Euler's Formula: :$\exp \paren {i z} = \cos z + i \sin z$ Then, starting from the {{RHS}}: {{begin-eqn}} {{eqn | l = \frac {\exp \paren {i z} - \exp \paren {-i z} }{2 i}       | r = \frac {\paren {\cos z + i \sin z} - \paren {\cos \paren {-z} + i \sin \paren {-z} } } {2 i} }} {{eqn | r = \frac {\paren {\cos z + i \sin z - \cos z - i \sin \paren {-z} } } {2 i}       | c = Cosine Function is Even }} {{eqn | r = \frac {i \sin z - i \sin \paren {-z} } {2 i}       | c =  }} {{eqn | r = \frac {i \sin z - i \paren {-\sin \paren {-z} } } {2 i}       | c = Sine Function is Odd }} {{eqn | r = \frac {2 i \sin z} {2 i}       | c =  }} {{eqn | r = \sin z       | c =  }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | n = 1       | l = \exp \paren {i z}       | r = \cos z + i \sin z       | c = Euler's Formula }} {{eqn | n = 2       | l = \exp \paren {-i z}       | r = \cos z - i \sin z       | c = Euler's Formula: Corollary }} {{eqn | ll= \leadsto       | l = \exp \paren {i z} - \exp \paren {-i z}       | r = \paren {\cos z + i \sin z} - \paren {\cos z - i \sin z}       | c = $(1) - (2)$ }} {{eqn | r = 2 i \sin z       | c = simplifying }} {{eqn | ll= \leadsto       | l = \frac {\exp \paren {i z} - \exp \paren {-i z} } {2 i}       | r = \sin z       | c =  }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \map \cos {a + b} + i \, \map \sin {a + b}       | r = e^{i \paren {a + b} }       | c = Euler's Formula }} {{eqn | r = e^{i a} e^{i b}       | c = Exponential of Sum }} {{eqn | r = \paren {\cos a + i \sin a} \paren {\cos b + i \sin b}       | c = Euler's Formula }} {{eqn | r = \paren {\cos a \cos b - \sin a \sin b} + i \paren {\sin a \cos b + \cos a \sin b}       | c = Complex Numbers form Field }} {{end-eqn}} By equating the imaginary parts, the result follows. {{qed}}
{{begin-eqn}}  {{eqn | l = \sin a \cos b + \cos a \sin b       | r = \paren {\frac {e^{i a} - e^{-i a} }{2 i} } \cos b + \cos a \paren {\frac {e^{i b} - e^{-i b} }{2 i} }       | c = Sine Exponential Formulation }} {{eqn | r = \paren {\frac {e^{i a} - e^{-i a} } {2 i} } \paren {\frac {e^{i b} + e^{-i b} } 2}           + \paren {\frac {e^{i a} + e^{-i a} } 2} \paren {\frac {e^{i b} - e^{-i b} } {2 i} }       | c = Cosine Exponential Formulation }} {{eqn | r = \frac {e^{i a} e^{i b} + e^{i a} e^{-i b} - e^{-i a} e^{i b} - e^{-i a} e^{-i b} } {4 i}       | c = expanding }} {{eqn | o =        | ro= +       | r = \frac {e^{i a} e^{i b} - e^{i a} e^{-i b} + e^{-i a} e^{i b} - e^{-i a} e^{-i b} } {4 i} }} {{eqn | r = \frac {e^{i a} e^{i b} - e^{-i a} e^{-i b} } {2 i}       | c = sinplifying }} {{eqn | r = \frac {e^{i \paren {a + b} } - e^{-i \paren {a + b} } } {2 i}       | c = Exponential of Sum }} {{eqn | r = \map \sin {a + b}       | c = Sine Exponential Formulation }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \sin \left({a + b}\right)       | r = \cos \left({\frac \pi 2 - \left({a + b}\right)}\right)       | c = Cosine of Complement equals Sine }} {{eqn | r = \cos \left({\left({\frac \pi 2 - a}\right) - b}\right) }} {{eqn | r = \cos \left({\frac \pi 2 - a}\right) \cos b + \sin \left({\frac \pi 2 - a}\right) \sin b       | c = Cosine of Difference }} {{eqn | r = \sin a \cos b + \sin \left({\frac \pi 2 - a}\right) \sin b       | c = Cosine of Complement equals Sine }} {{eqn | r = \sin a \cos b + \cos a \sin b       | c = Sine of Complement equals Cosine }} {{end-eqn}} {{qed}}
Recall the analytic definitions of sine and cosine: :$\displaystyle \sin x = \sum_{n \mathop = 0}^\infty \left({-1}\right)^n \frac {x^{2 n + 1}} {\left({2 n + 1}\right)!}$ :$\displaystyle \cos x = \sum_{n \mathop = 0}^\infty \left({-1}\right)^n \frac {x^{2 n}} {\left({2 n}\right)!}$ Let: {{begin-eqn}} {{eqn | l = g \left({a}\right)       | r = \sin \left({a + b}\right) - \sin a \cos b - \cos a \sin b }} {{eqn | l = h \left({a}\right)       | r = \cos \left({a + b}\right) - \cos a \cos b + \sin a \sin b }} {{end-eqn}} Let us differentiate these with respect to $a$, keeping $b$ constant. Then from Derivative of Sine Function and Derivative of Cosine Function, we have: {{begin-eqn}} {{eqn | l = g' \left({a}\right)       | r = \cos \left({a + b}\right) - \cos a \cos b + \sin a \sin b = h \left({a}\right) }} {{eqn | l = h' \left({a}\right)       | r = - \sin \left({a + b}\right) + \sin a \cos b + \cos a \sin b = - g \left({a}\right) }} {{end-eqn}} Hence: {{begin-eqn}} {{eqn | l = D_a \left({\left({g \left({a}\right)}\right)^2 + \left({h \left({a}\right)}\right)^2}\right)       | r = 2 g \left({a}\right) g' \left({a}\right) + 2 h \left({a}\right) h' \left({a}\right) }} {{eqn | r = 0 }} {{end-eqn}} Thus from Derivative of Constant: :$\forall a \in \R: g \left({a}\right)^2 + h \left({a}\right)^2 = c$ In particular, it is true for $a = 0$, and so: :$g \left({0}\right)^2 + h \left({0}\right)^2 = 0$ So: :$g \left({a}\right)^2 + h \left({a}\right)^2 = 0$ But from Square of Real Number is Non-Negative: :$g \left({a}\right)^2 \ge 0$ and $h \left({a}\right)^2 \ge 0$ So it follows that: :$g \left({a}\right) = 0$ and: :$h \left({a}\right) = 0$ Hence the result. {{qed}}
{{begin-eqn}} {{eqn | n = 1       | l = 2 \sin a \cos b       | r = \sin \paren {a + b} + \sin \paren {a - b}       | c = Simpson's Formula for Sine by Cosine: Proof 2 }} {{eqn | n = 2       | l = 2 \cos a \sin b       | r = \sin \paren {a + b} - \sin \paren {a - b}       | c = Simpson's Formula for Cosine by Sine: Proof 2 }} {{eqn | ll= \leadsto       | l = 2 \sin \paren {a + b}       | r = 2 \sin a \cos b + 2 \cos a \sin b       | c = $(1) + (2)$ }} {{eqn | ll= \leadsto       | l = \sin \paren {a + b}       | r = \sin a \cos b + \cos a \sin b       | c =  }} {{end-eqn}} {{qed}}
This proof is based on the definition of the complex exponential as the unique solution of the differential equation: :$\dfrac \d {\d z} \exp = \exp$ which satisfies the initial condition $\map \exp 0 = 1$. Define the complex function  $f: \C \to \C$ by: :$\map f z = \map \exp z \, \map \exp {z_1 + z_2 - z}$ Then find its derivative: {{begin-eqn}} {{eqn | l = D_z \, \map f z       | r = \paren {D_z \, \map \exp z} \map \exp {z_1 + z_2 - z} + \map \exp z \paren {D_z \map \exp {z_1 + z_2 - z} }       | c = Derivative of Complex Composite Function }} {{eqn | r = \map \exp z \, \map \exp {z_1 + z_2 - z} + \map \exp z \, \map \exp {z_1 + z_2 - z} \map {D_z} {z_1 + z_2 - z}       | c = as $\exp$ is its own derivative }} {{eqn | r = \map \exp z \, \map \exp {z_1 + z_2 - z} - \map \exp z \, \map \exp {z_1 + z_2 - z}       | c = Derivative of Complex Power Series }} {{eqn | r = 0 }} {{end-eqn}} From Zero Derivative implies Constant Complex Function, it follows that $f$ is constant. Then: {{begin-eqn}} {{eqn | l = \map \exp {z_1} \, \map \exp {z_2}       | r = \map f {z_1} }} {{eqn | r = \map f 0       | c = as $f$ is constant }} {{eqn | r = \map \exp 0 \, \map \exp {z_1 + z_2} }} {{eqn | r = \map \exp {z_1 + z_2}       | c = as $\map \exp 0 = 1$  }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \map \tanh {-x}       | r = \frac {\map \sinh {-x} } {\map \cosh {-x} }        | c = {{Defof|Hyperbolic Tangent|index = 2}} }} {{eqn | r = \frac {-\sinh x} {\map \cosh {-x} }        | c = Hyperbolic Sine Function is Odd }} {{eqn | r = \frac {-\sinh x} {\cosh x}        | c = Hyperbolic Cosine Function is Even }} {{eqn | r = -\tanh x       | c = {{Defof|Hyperbolic Tangent|index = 2}} }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \map \sin {i z}       | r = \frac {e^{i \paren {i z} } - e^{i \paren {-i z} } } {2 i}       | c = Sine Exponential Formulation }} {{eqn | r = \paren {-i} \frac {e^{-z} - e^z} 2       | c = $i^2 = -1$ }} {{eqn | r = i \frac {e^z - e^{-z} } 2       | c = $i^2 = -1$ }} {{eqn | r = i \sinh z       | c = {{Defof|Hyperbolic Sine}} }} {{end-eqn}} {{qed}}
{{begin-eqn}}  {{eqn | l = \map \cos {i z}       | r = \frac {e^{i \paren {i z} } + e^{-i \paren {i z} } } 2       | c = Cosine Exponential Formulation }} {{eqn | r = \frac {e^{-z} + e^z} 2       | c = $i^2 = -1$ }} {{eqn | r = \cosh z       | c = {{Defof|Hyperbolic Cosine}} }} {{end-eqn}} {{qed}}
{{begin-eqn}}  {{eqn | l = \map \tan {i z}       | r = \frac {\map \sin {i z} } {\map \cos {i z} }       | c = {{Defof|Complex Tangent Function}} }} {{eqn | r = \frac {i \sinh z} {\map \cos {i z} }       | c = Hyperbolic Sine in terms of Sine }} {{eqn | r = \frac {i \sinh z} {\cosh z}       | c = Hyperbolic Cosine in terms of Cosine }} {{eqn | r = i \tanh z       | c = {{Defof|Hyperbolic Tangent}} }} {{end-eqn}} {{qed}}
Let $\preceq$ be an ordering on the set $S$. If $\preceq$ is a total ordering, the result is complete. Suppose, then, that $\preceq$ is not a total ordering. Let $T$ be the set of orderings on $S$ that extend $\preceq$, ordered by inclusion. Let $C$ be a chain in $T$. By Union of Nest of Orderings is Ordering, $\bigcup C$ is an ordering. Thus every chain in $T$ has an upper bound in $T$. By Zorn's Lemma, $T$ has a maximal element, $\le$. $\le$ is seen to be a total ordering, as follows: Suppose that $a, b \in S$, $a \not \le b$, and $b \not \le a$. Let $\le'$ be the relation defined as: :$\le' \mathop {:=} \le \cup \left\{{(a, b)}\right\}$ Let $\le'^-$ be the transitive closure of $\le'$. Then by Ordering can be Expanded to compare Additional Pair, $\le'^-$ is an ordering. But $\le'^- \mathop {\supsetneq} \le$, contradicting the maximality of $\le$. Thus, $\le$ is a total ordering. {{qed}}
Let $\prec$ be the reflexive reduction of $\preceq$. By Reflexive Reduction of Ordering is Strict Ordering, $\prec$ is a strict ordering. By the strict form of the Order-Extension Principle, there exists a strict total ordering $<$ on $S$ such that: :$\forall a, b \in S: \left({a \prec b \implies a < b}\right)$ Let $\le$ be the reflexive closure of $<$. Let $a, b \in S$. If $a \preceq b$, then by Law of Excluded Middle either $a \prec b$ or $a = b$. If $a = b$, then by the definition of reflexive closure, $a \le b$. If $a \prec b$, then by the choice of $<$, $a < b$ so $a \le b$. Thus for all $a, b \in S$, $a \preceq b \implies a < b$. By Reflexive Closure of Strict Total Ordering is Total Ordering, $\le$ is a total ordering. {{qed}}
Let $\preceq$ be an arbitrary element of $C$. Let $\sim \, = \, \bigcup C$. Checking in turn each of the criteria for an ordering: Let $a, b \in S$. Let $a \sim b$ and $b \sim a$. Since $a \sim b$ and $b \sim a$, there exist $\preceq_1, \preceq_2 \, \in \, C$ such that $a \preceq_1 b$ and $b \preceq_2 a$. Since $C$ is a chain: :$\preceq_1 \, \subset \, \preceq_2$ or: :$\preceq_2 \, \subset \, \preceq_1$ {{WLOG}}, suppose $\preceq_1 \, \subset \, \preceq_2$. Then it must hold that $a \preceq_2 b$. Since: :$a \preceq_2 b$ :$b \preceq_2 a$ :$\preceq_2 \, \in \, T$ is an ordering it follows that: :$a = b$ and so $\sim$ is antisymmetric. {{qed|lemma}} === Reflexivity === Let $a \in S$. Because $\preceq$ is an ordering: :$a \preceq a$ As $\preceq \, \subseteq \, \sim$: : $a \sim a$ and so $\sim$ is reflexive. {{qed|lemma}} === Transitivity === Let $a, b, c \in S$. Let $a \sim b$ and $b \sim c$. Thus there are elements $\preceq_1$ and $\preceq_2$ of $C$ such that: :$a \preceq_1 b$ and $b \preceq_2 c$ Since $C$ is a chain, it must hold (as above) that: :$a \preceq_2 b$ or $b \preceq_1 c$ {{WLOG}}, suppose $a \preceq_2 b$ Since: :$a \preceq_2 b$ :$b \preceq_2 c$ :$\preceq_2 \, \in \, T$ is an ordering it follows that: :$a \preceq_2 c$ Since $\preceq_2 \, \subseteq \, \sim$: :$a \sim c$ So $\sim$ has been shown to be transitive. {{qed|lemma}} $\sim$ has been shown to be reflexive, transitive and antisymmetric. Hence by definition it is an ordering. {{qed}} Category:Set Theory Category:Order Theory qytyf4e5w8hyzajk2ab51r9gy6muw1n
Let $\prec$ be the reflexive reduction of $\preceq$. Let $\prec' = {\prec} \cup \set {\tuple {a, b} }$. By Reflexive Reduction of Ordering is Strict Ordering, $\prec$ is a strict ordering. Define a relation $\prec'_2$ by letting $p \prec'_2 q$ {{iff}}: :$p \prec q$ or :$p \preceq a$ and $b \preceq q$ By Strict Ordering can be Expanded to Compare Additional Pair, $\prec'_2$ is a strict ordering and is the transitive closure of $\prec'$. Then the reflexive closure of $\prec'_2$, ${\prec'_2}^=$ is the transitive reflexive closure of $\prec'$. From Equivalence of Definitions of Reflexive Transitive Closure, ${\prec'_2}^=$ is the transitive closure of the reflexive closure of $\prec'$. The reflexive closure of $\prec'$ is $\preceq'$, so ${\prec'_2}^=$ is the transitive closure of $\preceq'$. But ${\prec'_2}^=$ is clearly the relation defined by the two conditions in the theorem statement. {{qed}}
We must show that: * $\mathcal R \subseteq \mathcal R^+$ * $\mathcal R^+$ is transitive * $\mathcal R^+$ is the smallest relation with both of those characteristics. === Proof of Subset === $\mathcal R \subseteq \mathcal R^+$: $\mathcal R^+$ contains all of the $\mathcal R_i$, so in particular $\mathcal R^+$ contains $\mathcal R$. === Proof of Transitivity === Every element of $\mathcal R^+$ is in one of the $\mathcal R_i$. From the method of construction of $\mathcal R^+$, we have that $\forall i, j \in \N: \mathcal R_i \subseteq \mathcal R_{\max \left({i, j}\right)}$. Suppose $\left({s_1, s_2}\right) \in \mathcal R_j$ and $\left({s_2, s_3}\right) \in \mathcal R_k$. Then as: : $\mathcal R_j \subseteq \mathcal R_{\max \left({j, k}\right)}$ and: : $\mathcal R_k \subseteq \mathcal R_{\max \left({j, k}\right)}$ it follows that: : $\left({s_1, s_2}\right) \in \mathcal R_{\max \left({j, k}\right)}$ and $\left({s_2, s_3}\right) \in \mathcal R_{\max \left({j, k}\right)}$ It follows from the method of construction that $\left({s_1, s_3}\right) \in \mathcal R_{\max \left({j, k}\right)}$. Hence as $\mathcal R_{\max \left({j, k}\right)} \subseteq \mathcal R^+$, it follows that $\mathcal R^+$ is transitive. === Proof of being the smallest such relation === Let $\mathcal R'$ be any transitive relation containing $\mathcal R$. We want to show that $\mathcal R^+ \subseteq \mathcal R'$. It is sufficient to show that $\forall i \in \N: \mathcal R_i \subseteq \mathcal R'$. Since $\mathcal R \subseteq \mathcal R'$, we have that $\mathcal R_0 \subseteq \mathcal R'$. Suppose $\mathcal R_i \subseteq \mathcal R'$. From the method of construction, as $\mathcal R'$ is transitive, $\mathcal R_{i+1} \subseteq \mathcal R'$. Therefore, by induction, $\forall i \in \N: \mathcal R_i \subseteq \mathcal R'$. So $\mathcal R^+ \subseteq \mathcal R'$, and hence the result. {{qed}} Category:Transitive Closures 3idvvcqj8zmsoq42p8z4rh1hhhpr1m4