hcju/proofwikiqa huggingface dataset:
In the following, let $f \subseteq S \times T$ be a relation on $S \times T$. === $(1)$ {{iff}} $(4)$ === $f$ is by definition left-total {{iff}}: :$\forall s \in S: \exists t \in T: \tuple {s, t} \in f$ which means exactly the same thing as: :each element of $S$ is associated with with ''at least'' one element of $T$. $f$ is by definition many-to-one {{iff}}: :$\forall x \in \Dom f: \tuple {x, y_1} \in f \land \tuple {x, y_2} \in f \implies y_1 = y_2$ which means exactly the same thing as: :each element of $S$ is associated with with ''at most'' one element of $T$. Thus: :$f$ is left-total and many-to-one {{iff}}: :each element of $S$ is associated by $f$ with with ''exactly one'' one element of $T$. That is: :$f$ is a mapping by definition 1 {{iff}}: :$f$ is a mapping by definition 4. {{qed|lemma}} === $(2)$ {{iff}} $(4)$ === $f$ is by definition left-total {{iff}}: :$\forall s \in S: \exists t \in T: \tuple {s, t} \in \mathcal R$ and: $f$ is by definition many-to-one {{iff}}: :$\forall x \in \Dom f: \forall y_1, y_2 \in \Cdm f: \tuple {x, y_1} \in f \land \tuple {x, y_2} \in f \implies y_1 = y_2$ which means exactly the same thing as: :$\forall x \in S: \forall y_1, y_2 \in T: \tuple {x, y_1} \in f \land \tuple {x, y_2} \in f \implies y_1 = y_2$ That is: :$f$ is a mapping by definition 2 {{iff}}: :$f$ is a mapping by definition 4. {{qed|lemma}} === $(2)$ {{iff}} $(3)$ === Both definition 2 and definition 3 carry the stipulation that: :$\forall x \in S: \exists y \in T: \tuple {x, y} \in f$ Then we have that definition 2 carries the stipulation that :$\forall x \in S: \forall y_1, y_2 \in T: \tuple {x, y_1} \in f \land \tuple {x, y_2} \in f \implies y_1 = y_2$ By the Rule of Transposition, that means the same as: :$\forall x \in S: \forall y_1, y_2 \in T: \neg \paren {y_1 = y_2} \implies \neg \paren {\tuple {x, y_1} \in f \land \tuple {x, y_2} \in f}$ which in turn is the same as: :$\forall \tuple {x_1, y_1}, \tuple {x_2, y_2} \in f: y_1 \ne y_2 \implies x_1 \ne x_2$ which is the other stipulation of definition 3. That is: :$f$ is a mapping by definition 2 {{iff}}: :$f$ is a mapping by definition 3. {{qed}} Category:Mapping Theory gpz6zqmnt83qtkflh1va6amn7h07yev
{{begin-eqn}} {{eqn	| l = \map \Ci x 	| r = -\gamma - \ln x + \int_0^x \frac {1 - \cos u} u \rd u 	| c = Characterization of Cosine Integral Function }} {{eqn	| r = -\gamma - \ln x + \int_0^x \frac 1 u \paren {1 - \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {u^{2 n} } {\paren {2 n}!} } \rd u 	| c = Power Series Expansion for Cosine Function }} {{eqn	| r = -\gamma - \ln x + \sum_{n \mathop = 1}^\infty \paren {-1}^{n + 1} \int_0^x \frac {u^{2 n - 1} } {\paren {2 n}!} \rd u 	| c = Power Series is Termwise Integrable within Radius of Convergence }} {{eqn	| r = -\gamma - \ln x + \sum_{n \mathop = 1}^\infty \paren {-1}^{n + 1} \frac {u^{2 n} } {\paren {2 n} \times \paren {2 n}!} 	| c = Primitive of Power }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn	| l = \frac \d {\d x} \paren {\map \Ci x} 	| r = \frac \d {\d x} \paren {-\gamma - \ln x + \int_0^x \frac {1 - \cos t} t \rd t} 	| c = Characterization of Cosine Integral Function }} {{eqn	| r = -\frac 1 x + \frac 1 x - \frac {\cos x} x 	| c = Derivative of Constant, Derivative of Natural Logarithm, Fundamental Theorem of Calculus: First Part (Corollary) }} {{eqn	| r = -\frac {\cos x} x }} {{end-eqn}} {{qed}} Category:Cosine Integral Function Category:Derivatives 3qppg0agee1s5za559l9vv1uzbekmgv
{{begin-eqn}} {{eqn	| l = \map {\frac \d {\d x} } {\map \Ei x} 	| r = \map {\frac \d {\d x} } {-\gamma - \ln x + \int_0^x \frac {1 - e^{-t} } t \rd t} 	| c = Characterization of Exponential Integral Function }} {{eqn	| r = -\frac 1 x + \frac 1 x - \frac {e^{-x} } x 	| c = Derivative of Constant, Derivative of Natural Logarithm, Fundamental Theorem of Calculus: First Part (Corollary) }} {{eqn	| r = -\frac {e^{-x} } x }} {{end-eqn}} {{qed}} Category:Derivatives Category:Exponential Integral Function r2b3x4o3h45japoer99twgowm3463ne
We have, by the definition of the sine integral function:  :$\displaystyle \map \Si x = \int_0^x \frac {\sin t} t \rd t$ By Fundamental Theorem of Calculus (First Part): Corollary, we  have:  :$\displaystyle \frac \d {\d x} \paren {\map \Si x} = \frac {\sin x} x$ {{qed}} Category:Sine Integral Function Category:Derivatives 1xzciiivsueju0uh896wwyflcyqdzm0
By Derivative of Cosine Integral Function, we have:  :$\displaystyle \frac \d {\d x} \paren {\map \Ci x} = -\frac {\cos x} x$ So: {{begin-eqn}} {{eqn	| l = \int \map \Ci x \rd x 	| r = \int 1 \times \map \Ci \rd x }} {{eqn	| r = x \map \Ci x - \int \paren {-x \frac {\cos x} x} \rd x 	| c = Integration by Parts }} {{eqn	| r = x \map \Ci x + \int \cos x \rd x }} {{eqn	| r = x \map \Ci x + \sin x + C 	| c = Primitive of Cosine Function }} {{end-eqn}} {{qed}} Category:Primitives Category:Cosine Integral Function fs136ioxxwed61mz9ncxmv6tt8nq5y8
By Derivative of Exponential Integral Function, we have:  :$\displaystyle \frac \d {\d x} \paren {\map \Ei x} = -\frac {e^{-x} } x$ So:  {{begin-eqn}} {{eqn	| l = \int \map \Ei x \rd x 	| r = \int 1 \times \map \Ei x \rd x }} {{eqn	| r = x \map \Ei x - \int \paren {-x \frac {e^{-x} } x} \rd x 	| c = Integration by Parts }} {{eqn	| r = x \map \Ei x + \int e^{-x} \rd x }} {{eqn	| r = x \map \Ei x - e^{-x} + C 	| c = Primitive of $e^{a x}$ }} {{end-eqn}} {{qed}} Category:Primitives Category:Exponential Integral Function lx8rwohdp446zk0sgepdwzxze98t7eo
Consider the mapping defined as: :$\forall Y \in \powerset T: \map {f^\gets} Y = \set {s \in S: \exists t \in Y: \map f s = t}$ Let $\Img f \cap Y = \O$. Then: :$\forall t \in T: \neg \exists t \in \Img f \cap Y$ and so: :$\set {s \in S: \exists t \in Y: \map f s = t} = \O$ and so: :$\forall Y \in \powerset T: \map {f^\gets} Y = \begin {cases} \set {s \in S: \exists t \in Y: \map f s = t} & : \Img f \cap Y \ne \O \\ \O & : \Img f \cap Y = \O \end {cases}$ {{qed}} Category:Inverse Image Mappings q33cn7watrniao0nkabo9ep0kadxoev
Consider the mapping defined as: :$\forall Y \in \powerset T: \map {\mathcal R^\gets} Y = \set {s \in S: \exists t \in Y: \tuple {t, s} \in \mathcal R^{-1} }$ Let $\Img {\mathcal R} \cap Y = \O$. Then: :$\forall t \in T: \neg \exists t \in \Img {\mathcal R} \cap Y$ and so: :$\set {s \in S: \exists t \in Y: \tuple {s, t} \in \mathcal R} = \O$ and so: :$\forall Y \in \powerset T: \map {\mathcal R^\gets} Y = \begin {cases} \set {s \in S: \exists t \in Y: \tuple {t, s} \in \mathcal R^{-1} } & : \Img f \cap Y \ne \O \\ \O & : \Img f \cap Y = \O \end {cases}$ {{qed}} Category:Inverse Image Mappings em2ohif4fartzmykocs7mc5bbelmshh
$f$, being a mapping, is also a relation. Hence Direct Image Mapping of Relation is Mapping can be applied directly. {{qed}}
$f^{-1}$ is a relation. So Inverse Image Mapping of Relation is Mapping applies directly. {{qed}}
By definition, a mapping is a left-total relation. The result then follows from Direct Image Mapping of Left-Total Relation is Empty iff Argument is Empty. {{qed}}
We have, by the definition of the Fresnel sine integral function: :$\displaystyle \map {\operatorname S} x = \sqrt {\frac 2 \pi} \int_0^x \sin u^2 \rd u$ By Fundamental Theorem of Calculus (First Part): Corollary, we therefore have: :$\displaystyle \frac {\d \operatorname S} {\d x} = \sqrt {\frac 2 \pi} \sin x^2$ {{qed}} Category:Fresnel Sine Integral Function Category:Derivatives s59h6l4p0163vzg6ibltkqasnzztki7
We have, by the definition of the Fresnel cosine integral function: :$\displaystyle \map {\operatorname C} x = \sqrt {\frac 2 \pi} \int_0^x \cos u^2 \rd u$ By Fundamental Theorem of Calculus (First Part): Corollary, we therefore have: :$\displaystyle \frac {\d \operatorname C} {\d x} = \sqrt {\frac 2 \pi} \cos x^2$ {{qed}} Category:Fresnel Cosine Integral Function Category:Derivatives gc0m9zlbo4etm7o4yh0l7yfx1dc5nf1
{{begin-eqn}} {{eqn	| l = \map {\operatorname S} x 	| r = \sqrt {\frac 2 \pi} \int_0^x \sin u^2 \rd u 	| c = {{Defof|Fresnel Sine Integral Function}} }} {{eqn	| r = \sqrt {\frac 2 \pi} \int_0^x \paren {\sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {\paren {u^2}^{2 n + 1} } {\paren {2 n + 1}!} } \rd u 	| c = Power Series Expansion for Sine Function }} {{eqn	| r = \sqrt {\frac 2 \pi} \sum_{n \mathop = 0}^\infty \frac {\paren {-1}^n} {\paren {2 n + 1}!} \int_0^x u^{4 n + 2} \rd u 	| c = Power Series is Termwise Integrable within Radius of Convergence }} {{eqn	| r = \sqrt {\frac 2 \pi} \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{4 n + 3} } {\paren {4 n + 3} \paren {2 n + 1}!} 	| c = Primitive of Power }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn	| l = \map {\operatorname C} x  	| r = \sqrt {\frac 2 \pi} \int_0^x \cos u^2 \rd u 	| c = {{Defof|Fresnel Cosine Integral Function}} }} {{eqn	| r = \sqrt {\frac 2 \pi} \int_0^x \paren {\sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {\paren {u^2}^{2 n} } {\paren {2 n}!} } \rd u 	| c = Power Series Expansion for Cosine Function }} {{eqn	| r = \sqrt {\frac 2 \pi} \sum_{n \mathop = 0}^\infty \frac {\paren {-1}^n} {\paren {2 n}!} \int_0^x u^{4 n} \rd u 	| c = Power Series is Termwise Integrable within Radius of Convergence }} {{eqn	| r = \sqrt {\frac 2 \pi} \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{4 n + 1} } {\paren {4 n + 1} \paren {2 n}!} 	| c = Primitive of Power }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn	| l = \map {\operatorname S} {-x} 	| r = \sqrt {\frac 2 \pi} \int_0^{-x} \sin u^2 \rd u 	| c = {{Defof|Fresnel Sine Integral Function}} }} {{eqn	| r = -\sqrt {\frac 2 \pi} \int_0^{-\paren {-x} } \map \sin {\paren {-u}^2} \rd u 	| c = substituting $u \mapsto -u$ }} {{eqn	| r = -\sqrt {\frac 2 \pi} \int_0^x \sin u^2 \rd u }} {{eqn	| r = -\map {\operatorname S} x 	| c = {{Defof|Fresnel Sine Integral Function}} }} {{end-eqn}} {{qed}}
By Fresnel Sine Integral Function is Odd, $\operatorname S$ is an odd function. Therefore, by Odd Function of Zero is Zero: :$\map {\operatorname S} 0 = 0$ {{qed}}
{{begin-eqn}} {{eqn	| l = \lim_{x \mathop \to \infty} \map {\operatorname S} x 	| r = \sqrt {\frac 2 \pi} \lim_{x \mathop \to \infty} \int_0^x \sin u^2 \rd u 	| c = Combination Theorem for Limits of Functions: Multiple Rule, {{Defof|Fresnel Sine Integral Function}} }} {{eqn	| r = \sqrt {\frac 2 \pi} \int_0^\infty \sin u^2 \rd u }} {{eqn	| r = \sqrt {\frac 2 \pi} \times \frac 1 2 \sqrt {\frac \pi 2} 	| c = Definite Integral to Infinity of $\map \sin {a x^2}$ }} {{eqn	| r = \frac 1 2 }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn	| l = \map {\operatorname C} {-x} 	| r = \sqrt {\frac 2 \pi} \int_0^{-x} \cos u^2 \rd u 	| c = {{Defof|Fresnel Cosine Integral Function}} }} {{eqn	| r = -\sqrt {\frac 2 \pi} \int_0^{-\paren {-x} } \map \cos {\paren {-u}^2} \rd u 	| c = substituting $u \mapsto -u$ }} {{eqn	| r = -\sqrt {\frac 2 \pi} \int_0^x \cos u^2 \rd u }} {{eqn	| r = -\map {\operatorname C} x 	| c = {{Defof|Fresnel Cosine Integral Function}} }} {{end-eqn}} {{qed}}
By Fresnel Cosine Integral Function is Odd, $\operatorname C$ is an odd function. Therefore, by Odd Function of Zero is Zero: :$\map {\operatorname C} 0 = 0$ {{qed}}
=== Sufficient Condition === Let $\map {\mathcal R^\to} X = \O$. By definition of direct image mapping: :$\set {t \in T: \exists s \in X: \tuple {s, t} \in \mathcal R} = \O$ That is: :$\neg \exists s \in X: \tuple {s, t} \in \mathcal R$ But as $\mathcal R$ is a left-total relation: :$\forall s \in X: \exists t \in T: \tuple {s, t} \in \mathcal R$ Thus: :$\neg \exists s \in X$ and so: :$X = \O$ {{qed|lemma}} === Sufficient Condition === Let $X = \O$. Then by definition of direct image mapping: :$\map {\mathcal R^\to} X = \O$ {{qed}} Category:Direct Image Mappings 8gxx5mpn9a2u1k9ul4dtd6so413faqd
{{begin-eqn}} {{eqn	| l = \lim_{x \mathop \to \infty} \map {\operatorname C} x 	| r = \sqrt {\frac 2 \pi} \lim_{x \mathop \to \infty} \int_0^x \cos u^2 \rd u 	| c = Combination Theorem for Limits of Functions: Multiple Rule, {{Defof|Fresnel Cosine Integral Function}} }} {{eqn	| r = \sqrt {\frac 2 \pi} \int_0^\infty \cos u^2 \rd u }} {{eqn	| r = \sqrt {\frac 2 \pi} \times \frac 1 2 \sqrt {\frac \pi 2} 	| c = Definite Integral to Infinity of $\map \cos {a x^2}$ }} {{eqn	| r = \frac 1 2 }} {{end-eqn}} {{qed}}
There exists no $n \in \N$ such that $n + 1 = 0$. Thus $\map f 0$ has no preimage. The result follows by definition of surjection. {{qed}}
Let $T \subseteq C$. We have: {{begin-eqn}} {{eqn | l = \map {\paren {f \circ g}^\gets} T       | r = \begin {cases} \set {x \in A: \map g {\map f x} \in T} & : \Img {g \circ f} \cap T \ne \O \\ \O & : \Img {g \circ f} \cap T = \O \end {cases}       | c =  }} {{eqn-intertext|and}} {{eqn | l = \map {f^\gets \circ g^\gets} T       | r = \begin {cases} \set {x \in A: \map f x \in \map {g^\gets} T} & : \Img f \cap \map {g^\gets} T \ne \O \\ \O & : \Img f \cap \map {g^\gets} T = \O \end {cases}       | c =  }} {{eqn | r = \begin {cases} \set {x \in A: \map g {\map f x} \in T} & : \Img f \cap \map {g^\gets} T \ne \O \\ \O & : \Img f \cap \map {g^\gets} T = \O \end {cases}       | c =  }} {{end-eqn}} It remains to be shown that: :$\Img f \cap \map {g^\gets} T = \O \iff \Img {g \circ f} \cap T = \O$ So: {{begin-eqn}} {{eqn | l = \Img f \cap \map {g^\gets} T       | r = \O       | c =  }} {{eqn | ll= \leadstoandfrom       | l = \Img f       | o = \subseteq       | r = \relcomp B {\map {g^\gets} T}       | c = Empty Intersection iff Subset of Complement }} {{eqn | r = \map {g^\gets} {\relcomp C T}       | c = Complement of Preimage equals Preimage of Complement }} {{eqn | ll= \leadstoandfrom       | l = \Img {g \circ f}       | r = \map {g^\gets} {\Img f}       | c = Intersection with Complement is Empty iff Subset }} {{eqn | o = \subseteq       | r = \relcomp C T       | c = Subset of Preimage under Relation is Preimage of Subset: Corollary }} {{eqn | ll= \leadstoandfrom       | l = \Img {g \circ f} \cap T       | r = \O       | c = Empty Intersection iff Subset of Complement }} {{end-eqn}} {{qed}}
From Structure Induced by Ring with Unity Operations is Ring with Unity, $\struct {R^S, +', \circ'}$ has a unity $f_{1_R}$ defined by: :$\forall x \in S: \map {f_{1_R}} x = 1_R$ === Unit of Ring of Mappings implies Image is Subset of Ring Units === {{:Unit of Ring of Mappings iff Image is Subset of Ring Units/Unit of Ring of Mappings implies Image is Subset of Ring Units}}{{qed|lemma}} === Image is Subset of Ring Units implies Unit of Ring of Mappings === {{:Unit of Ring of Mappings iff Image is Subset of Ring Units/Image is Subset of Ring Units implies Unit of Ring of Mappings}}{{qed}} Category:Rings of Mappings Category:Rings with Unity Category:Unit of Ring of Mappings iff Image is Subset of Ring Units 7btfcu9b1kohilia70pthfc5xcdcxs6
=== Sufficient Condition === Let $f$ be a bijection. Thus a fortiori $f$ is: :a surjection :a one-to-many relation From One-to-Many Image of Set Difference: Corollary 2 we have: :$\forall X \in \powerset S: \map {\paren {f^\to \circ \complement_S} } X = \map {\paren {\complement_{\Img f} \circ f^\to} } X$ By definition of surjection: :$\Img f = T$ and so: :$\forall X \in \powerset S: \map {\paren {f^\to \circ \complement_S} } X = \map {\paren {\complement_T \circ f^\to} } X$ That is: :$f^\to \circ \complement_S = \complement_T \circ f^\to$ {{qed|lemma}} === Necessary Condition === Let $f$ be a mapping such that: :$f^\to \circ \complement_S = \complement_T \circ f^\to$ We have that: {{begin-eqn}} {{eqn | l = \map {f^\to} S       | r = \map {f^\to} {\relcomp S \O}       | c = {{Defof|Relative Complement}} }} {{eqn | r = \relcomp T {\map {f^\to} \O}       | c = by hypothesis }} {{eqn | r = \relcomp T \O       | c = {{Defof|Direct Image Mapping of Mapping}}: $\map {f^\to} \O = \O$ }} {{eqn | r = T       | c = {{Defof|Relative Complement}} }} {{end-eqn}} Hence, by definition, $f$ is a surjection. Now consider $A, B \subseteq S$: {{begin-eqn}} {{eqn | l = \map {f^\to \circ \complement_S} {A \cap B}       | r = \map {f^\to} {\relcomp S {A \cap B} }       | c = {{Defof|Composition of Mappings}} }} {{eqn | r = \map {f^\to} {\relcomp S A \cup \relcomp S B}       | c = De Morgan's Laws: Relative Complement of Intersection }} {{eqn | r = \map {f^\to} {\relcomp S A} \cup \map {f^\to} {\relcomp S B}       | c = Image of Union under Mapping }} {{eqn | r = \map {f^\to \circ \complement_S} A \cup \map {f^\to \circ \complement_S} B       | c = {{Defof|Composition of Mappings}} }} {{eqn | r = \map {\complement_T \circ f^\to} A \cup \map {\complement_T \circ f^\to} B       | c = by hypothesis }} {{eqn | r = \relcomp T {\map {f^\to} A} \cup \relcomp T {\map {f^\to} B}       | c = {{Defof|Composition of Mappings}} }} {{eqn | r = \relcomp T {\map {f^\to} A \cap \map {f^\to} B}       | c = De Morgan's Laws: Relative Complement of Intersection }} {{eqn | ll= \leadsto       | l = \complement_T \circ \map {f^\to} {A \cap B}       | r = \relcomp T {\map {f^\to} A \cap \map {f^\to} B}       | c = by hypothesis }} {{eqn | ll= \leadsto       | l = \map {f^\to} {A \cap B}       | r = \map {f^\to} A \cap \map {f^\to} B       | c = Relative Complement Mapping on Powerset is Bijection, and so Bijection iff Left and Right Cancellable }} {{end-eqn}} It follows from Image of Intersection under Injection that $f^\to$ is an injection. Hence from Mapping is Injection if its Direct Image Mapping is Injection: :$f$ is an injection. We have therefore that $f$ is both an injection and a surjection. Hence, by definition, $f$ is a bijection. {{qed}}
Let $f: \powerset S \to \powerset S$ be a mapping defined as: :$\forall T \in \powerset S: \map f T = \relcomp S T$ It is to be demonstrated that $f$ is a bijection. By definition of relative complement: :$\relcomp S T = S \setminus T = \set {x \in S: x \notin T}$ and so it can be seen that $f$ is well-defined. Let $T_1, T_2 \in \powerset S: \map f {T_1} = \map f {T_2}$. Then: {{begin-eqn}} {{eqn | l = \map f {T_1}       | r = \map f {T_2}       | c =  }} {{eqn | ll= \leadsto       | l = \relcomp S {T_1}       | r = \relcomp S {T_2}       | c = by definition of $f$ }} {{eqn | ll= \leadsto       | l = \relcomp S {\relcomp S {T_1} }       | r = \relcomp S {\relcomp S {T_2} }       | c =  }} {{eqn | ll= \leadsto       | l = T_1       | r = T_2       | c = Relative Complement of Relative Complement }} {{end-eqn}} That is, $f$ is an injection. Also: {{begin-eqn}} {{eqn | l = T       | o = \in       | r = \powerset S       | c =  }} {{eqn | ll= \leadsto       | lo= \exists X \in \powerset S:       | l = X       | r = \relcomp S T }} {{eqn | ll= \leadsto       | l = \relcomp S X       | r = \relcomp S {\relcomp S T} }} {{eqn | ll= \leadsto       | l = \relcomp S X       | r = T       | c = Relative Complement of Relative Complement }} {{end-eqn}} That is, for all $T \in \powerset S$ there exists an $X$ such that $T = \map f X$. This demonstrates that $f$ is a surjection. The result follows by definition of bijection. {{qed}}
{{begin-eqn}} {{eqn	| l = {}_2 \map {F_1} {-p, 1; 1; -x} 	| r = \sum_{n \mathop = 0}^\infty \frac {\paren {-p}^{\overline n} 1^{\overline n} } {1^{\overline n} } \frac {\paren {-x}^n} {n!} 	| c = {{Defof|Gaussian Hypergeometric Function}} }} {{eqn	| r = \sum_{n \mathop = 0}^\infty \paren {-1}^n \paren {\prod_{j \mathop = 0}^{n - 1} \paren {-p + j} } \frac {x^n} {n!} 	| c = {{Defof|Rising Factorial}} }} {{eqn	| r = \sum_{n \mathop = 0}^\infty \paren {\prod_{j \mathop = 0}^{n - 1} \paren {-1} } \paren {\prod_{j \mathop = 0}^{n - 1} \paren {-p + j} } \frac {x^n} {n!} }} {{eqn	| r = \sum_{n \mathop = 0}^\infty \paren {\prod_{j \mathop = 0}^{n - 1} \paren {p - j} } \frac {x^n} {n!} 	| c = Product of Products }} {{eqn	| r = \paren {1 + x}^p 	| c = General Binomial Theorem }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn	| l = x \, {}_2 \map {F_1} {1, 1; 2; -x} 	| r = x \sum_{n \mathop = 0}^\infty \frac {\paren {1^{\overline n} }^2} {2^{\overline n} } \frac {\paren {-x}^n} {n!} 	| c = {{Defof|Gaussian Hypergeometric Function}} }} {{eqn	| r = \sum_{n \mathop = 0}^\infty \paren {-1}^n \paren {n!}^2 \paren {\frac {\paren {2 - 1}!} {\paren {2 + n - 1}!} } \frac {x^{n + 1} } {n!} 	| c = Rising Factorial as Quotient of Factorials, One to Integer Rising is Integer Factorial }} {{eqn	| r = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {\paren {n!}^2} {\paren {n + 1}! n!} x^{n + 1} }} {{eqn	| r = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {\paren {n!}^2} {\paren {n + 1} \paren {n!}^2} x^{n + 1} 	| c = {{Defof|Factorial/Definition 1|Factorial}} }} {{eqn	| r = \sum_{n \mathop = 1}^\infty \paren {-1}^{n - 1} \frac {x^n} n 	| c = shifting the index }} {{eqn	| r = \map \ln {1 + x} 	| c = Power Series Expansion for $\map \ln {1 + x}$ }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn	| l = \map {\laptrans {e^{b t} \sin a t} } s 	| r = \map {\laptrans {\sin a t} } {s - b} 	| c = Laplace Transform of Exponential times Function }} {{eqn	| r = \frac a {\paren {s - b}^2 + a^2} 	| c = Laplace Transform of Sine }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn	| l = \map {\laptrans {e^{b t} \cos a t} } s 	| r = \map {\laptrans {\cos a t} } {s - b} 	| c = Laplace Transform of Exponential times Function }} {{eqn	| r = \frac {s - b} {\paren {s - b}^2 + a^2} 	| c = Laplace Transform of Cosine }} {{end-eqn}} {{qed}}
As $f$ is an injection, it is a fortiori a one-to-many relation. From Image of Set Difference under Relation: Corollary 2 we have: :$\forall A \in \powerset S: \map {\paren {\complement_{\Img {\mathcal R} } \circ \mathcal R^\to} } A = \map {\paren {\mathcal R^\to \circ \complement_S} } A$ where $\mathcal R \subseteq S \times T$ is a one-to-many relation on $S \times T$. Substituting $f$ for $\mathcal R$ gives the result: :$\forall A \in \powerset S: \map {\paren {\complement_{\Img f} \circ f^\to} } A = \map {\paren {f^\to \circ \complement_S} } A$ {{Qed}}
{{begin-eqn}} {{eqn | lo= \forall A \in \powerset S:       | l = A       | o = \subseteq       | r = \map {\paren {f^\gets \circ f^\to} } A       | c = Subset of Domain is Subset of Preimage of Image }} {{eqn | ll= \leadsto       | lo= \forall A \in \powerset S:       | l = \map {f^\to} A       | o = \subseteq       | r = \map {\paren {f^\to \circ f^\gets \circ f^\to} } A       | c = Image of Subset under Mapping is Subset of Image }} {{eqn-intertext|and:}} {{eqn | lo= \forall B \in \powerset T:       | l = \map {\paren {f^\to \circ f^\gets} } B       | o = \subseteq       | r = B       | c = Subset of Codomain is Superset of Image of Preimage }} {{eqn | ll= \leadsto       | lo= \forall A \in \powerset S:       | l = \map {\paren {f^\to \circ f^\gets} } {\map {f^\to} A}       | o = \subseteq       | r = \map {f^\to} A       | c = Image of Subset under Mapping is Subset of Image, setting $B = \map {f^\to} A \in \powerset T$ }} {{end-eqn}} Thus we have: :$\forall A \in \powerset S: \map {f^\to} A \subseteq \map {\paren {f^\to \circ f^\gets \circ f^\to} } A \subseteq \map {f^\to} A$ and the result follows. {{qed}}
{{begin-eqn}} {{eqn | lo= \forall A \in \powerset S:       | l = A       | o = \subseteq       | r = \map {\paren {f^\gets \circ f^\to} } A       | c = Subset of Domain is Subset of Preimage of Image }} {{eqn | ll= \leadsto       | lo= \forall B \in \powerset T:       | l = \map {f^\gets} B       | o = \subseteq       | r = \map {\paren {f^\gets \circ f^\to} } {\map {f^\gets} B}       | c = Image of Subset under Mapping is Subset of Image, setting $A = \map {f^\gets} B \in \powerset S$ }} {{eqn-intertext|and:}} {{eqn | lo= \forall B \in \powerset T:       | l = \map {\paren {f^\to \circ f^\gets} } B       | o = \subseteq       | r = B       | c = Subset of Codomain is Superset of Image of Preimage }} {{eqn | ll= \leadsto       | lo= \forall A \in \powerset S:       | l = \map {\paren {f^\gets \circ f^\to \circ f^\gets} } B       | o = \subseteq       | r = \map {f^\gets} B       | c = Image of Subset under Mapping is Subset of Image }} {{end-eqn}} Thus we have: :$\forall B \in \powerset T: \map {f^\gets} B \subseteq \map {\paren {f^\gets \circ f^\to \circ f^\gets} } B \subseteq \map {f^\gets} B$ and the result follows. {{qed}}