hcju/proofwikiqa huggingface dataset:
To save space, we will refer to: : The Well-Ordering Principle as '''WOP''' : The Principle of Finite Induction as '''PFI''' : The Principle of Complete Finite Induction as '''PCI'''. === PFI implies PCI === {{:Equivalence of Well-Ordering Principle and Induction/Proof/PFI implies PCI}}{{qed|lemma}} === PCI implies WOP === {{:Equivalence of Well-Ordering Principle and Induction/Proof/PCI implies WOP}}{{qed|lemma}} === WOP implies PFI === {{:Equivalence of Well-Ordering Principle and Induction/Proof/WOP implies PFI}}{{qed|lemma}} === Final assembly === So, we have that: * PFI implies PCI: The Principle of Mathematical Induction implies the Principle of Complete Induction * PCI implies WOP: The Principle of Complete Induction implies the Well-Ordering Principle * WOP implies PFI: The Well-Ordering Principle implies the Principle of Mathematical Induction. This completes the result. {{Qed}}
By definition of lower bound: :$\forall x \in S: L \preceq x$ But as $\forall y \in T: y \in S$ by definition of subset, it follows that: :$\forall y \in T: L \preceq y$. Hence the result, again by definition of lower bound. {{qed}}
By definition of upper bound: :$\forall x \in S: x \preceq U$ But as $\forall y \in T: y \in S$ by definition of subset, it follows that: :$\forall y \in T: y \preceq U$. Hence the result, again by definition of upper bound. {{qed}}
Let $d = \gcd \set {a, b}$. Then $d \divides a$ and $d \divides b$ by definition. Then from Bézout's Identity, $\exists u, v \in \Z: d = u a + v b$. From Common Divisor Divides Integer Combination, $c \divides a \land c \divides b \implies c \divides u a + v b$ for all $u, v \in \Z$. Thus $c \divides d$. {{qed}}
Let $x$ be an integer. Using Congruence of Powers throughout, we make use of $x \equiv y \pmod 5 \implies x^2 \equiv y^2 \pmod 5$. There are five cases to consider: : $x \equiv 0 \pmod 5$: we have $x^2 \equiv 0^2 \pmod 5 \equiv 0 \pmod 5$. : $x \equiv 1 \pmod 5$: we have $x^2 \equiv 1^2 \pmod 5 \equiv 1 \pmod 5$. : $x \equiv 2 \pmod 5$: we have $x^2 \equiv 2^2 \pmod 5 \equiv 4 \pmod 5$. : $x \equiv 3 \pmod 5$: we have $x^2 \equiv 3^2 \pmod 5 \equiv 4 \pmod 5$. : $x \equiv 4 \pmod 5$: we have $x^2 \equiv 4^2 \pmod 5 \equiv 1 \pmod 5$. {{qed}}
Let $x$ be an integer. Using Congruence of Powers throughout, we make use of: : $x \equiv y \pmod 3 \implies x^2 \equiv y^2 \pmod 3$ There are three cases to consider: : $(1): \quad x \equiv 0 \pmod 3$: we have $x^2 \equiv 0^2 \pmod 3 \equiv 0 \pmod 3$ : $(2): \quad x \equiv 1 \pmod 3$: we have $x^2 \equiv 1^2 \pmod 3 \equiv 1 \pmod 3$ : $(3): \quad x \equiv 2 \pmod 3$: we have $x^2 \equiv 2^2 \pmod 3 \equiv 1 \pmod 3$ {{qed}}
From Integers under Addition form Abelian Group, $\struct {\Z, +}$ is an abelian group. From Integers are Countably Infinite, the set of integers can be placed in one-to-one correspondence with the set of natural numbers. Hence by definition, the underlying set  of $\struct {\Z, +}$ is countably infinite. {{qed}}
All derivations can be achieved using applications of the group axioms. === Proof of $(1)$ === {{begin-eqn}} {{eqn | l = a x       | r = b       | c =  }} {{eqn | ll= \leadsto       | l = a^{-1} a x       | r = a^{-1} b       | c =  }} {{eqn | ll= \leadsto       | l = e x       | r = a^{-1} b       | c =  }} {{eqn | ll= \leadsto       | l = x       | r = a^{-1} b       | c =  }} {{end-eqn}} and the converse: {{begin-eqn}} {{eqn | l = x       | r = a^{-1} b       | c =  }} {{eqn | ll= \leadsto       | l = a x       | r = a a^{-1} b       | c =  }} {{eqn | ll= \leadsto       | l = a x       | r = e b       | c =  }} {{eqn | ll= \leadsto       | l = a x       | r = b       | c =  }} {{end-eqn}} {{Qed}} === Proof of $(2)$ === {{begin-eqn}} {{eqn | l = x a       | r = b       | c =  }} {{eqn | ll= \leadsto       | l = x a a^{-1}       | r = b a^{-1}       | c =  }} {{eqn | ll= \leadsto       | l = x e       | r = b a^{-1} b       | c =  }} {{eqn | ll= \leadsto       | l = x       | r = b a^{-1}       | c =  }} {{end-eqn}} and the converse: {{begin-eqn}} {{eqn | l = x       | r = b a^{-1}       | c =  }} {{eqn | ll= \leadsto       | l = x a       | r = b a^{-1} a       | c =  }} {{eqn | ll= \leadsto       | l = x a       | r = b e       | c =  }} {{eqn | ll= \leadsto       | l = x a       | r = b       | c =  }} {{end-eqn}} {{Qed}}
Denote $Z = \map Z G$ and $C = \displaystyle \bigcap_{g \mathop \in G} \map {C_G} g$ for simplicity. By definition of set equality, it suffices to prove $Z \subseteq C$ and $C \subseteq Z$. === $Z$ is contained in $C$ === Suppose that $x \in Z$. Then from the definition of center: :$\forall g \in G: x g = g x$ By definition of centralizer, this corresponds to: :$\forall g \in G: x \in \map {C_G} g$ Therefore we have, by definition of set intersection: :$\displaystyle x \in \bigcap_{g \mathop \in G} \map {C_G} g = C$ Hence $x \in C$. It follows by definition of subset that $Z \subseteq C$. {{qed|lemma}} === $C$ is contained in $Z$ === Suppose now that $x \in C$. Then, by definition of intersection: :$\forall g \in G: x \in \map {C_G} g$ That is, using the definition of centralizer: :$\forall g \in G: x g = g x$ By definition of the center, this means: :$x \in \map Z G = Z$ Hence $x \in Z$. It follows that $C \subseteq Z$. {{qed|lemma}} Therefore, we have established that: :$x \in Z \iff x \in C$ By definition of set equality: : $Z = C$ {{qed}}
{{begin-eqn}} {{eqn | l = \map \phi {e_G}       | r = \map \phi {e_G \circ e_G}       | c = {{Defof|Identity Element}} }} {{eqn | r = \map \phi {e_G} * \map \phi {e_G}       | c = {{Defof|Morphism Property}} }} {{end-eqn}} That is: {{begin-eqn}} {{eqn | l = \map \phi {e_G} * e_H       | r = \map \phi {e_G} * \map \phi {e_G}       | c = {{Defof|Identity Element}} }} {{eqn | ll= \leadsto       | l = e_H       | r = \map \phi {e_G}       | c = Cancellation Laws }} {{end-eqn}} {{qed}}
From Group Homomorphism of Product with Inverse, we have: :$\forall x, y \in G: \map \phi {x \circ y^{-1} } = \map \phi x * \paren {\map \phi y}^{-1}$ Putting $x = y$ we have: {{begin-eqn}} {{eqn | l = \map \phi {e_G}       | r = \map \phi {x \circ x^{-1} }       | c =  }} {{eqn | r = \map \phi x * \paren {\map \phi x}^{-1}       | c =  }} {{eqn | r = e_H       | c =  }} {{end-eqn}} {{qed}}
Let $x \in G$. Then: {{begin-eqn}} {{eqn | l = \map \phi x * \map \phi {x^{-1} }       | r = \map \phi {x \circ x^{-1} }       | c = {{Defof|Group Homomorphism}} }} {{eqn | r = \map \phi {e_G}       | c = {{Defof|Inverse Element}} }} {{eqn | r = e_H       | c = Group Homomorphism Preserves Identity }} {{end-eqn}} So, by definition, $\map \phi {x^{-1} }$ is the right inverse of $\map \phi x$. {{qed|lemma}} Similarly: {{begin-eqn}} {{eqn | l = \map \phi {x^{-1} } * \map \phi x       | r = \map \phi {x^{-1} \circ x}       | c = {{Defof|Group Homomorphism}} }} {{eqn | r = \map \phi {e_G}       | c = {{Defof|Inverse Element}} }} {{eqn | r = e_H       | c = Group Homomorphism Preserves Identity }} {{end-eqn}} So, again by definition, $\map \phi {x^{-1} }$ is the left inverse of $\map \phi x$. {{qed|lemma}} Finally, as $\map \phi {x^{-1} }$ is both: :a left inverse of $\map \phi x$ and: :a right inverse of $\map \phi x$ it is by definition an inverse. From Inverse in Group is Unique, $\map \phi {x^{-1} }$ is the only such element. Hence the result. {{qed}}
A direct application of Homomorphism to Group Preserves Inverses. {{qed}}
From Group Homomorphism of Product with Inverse, we have: :$\forall x, y \in G: \map \phi {x \circ y^{-1} } = \map \phi x * \paren {\map \phi y}^{-1}$ Putting $x = e_G$ and $y = x$ we have: {{begin-eqn}} {{eqn | l = \map \phi {x^{-1} }       | r = \map \phi {e_G \circ x^{-1} }       | c =  }} {{eqn | r = \map \phi {e_G} * \paren {\map \phi x}^{-1}       | c =  }} {{eqn | r = e_H * \paren {\map \phi x}^{-1}       | c = Group Homomorphism Preserves Identity }} {{eqn | r = \paren {\map \phi x}^{-1}       | c =  }} {{end-eqn}} {{qed}}
It is clear that $e \in N H$, so $N H \ne \O$. Suppose $n_1, n_2 \in N$ and $h_1, h_2 \in H$. Then: {{begin-eqn}} {{eqn | l = \paren {n_1 h_1} \paren {n_2 h_2}       | r = n_1 \paren {h_1 n_2 h_1^{-1} h_1} h_2 }} {{eqn | r = n_1 \paren {h_1 n_2 h_1^{-1} } \paren {h_1 h_2} }} {{end-eqn}} Since $N$ is normal in $G$: :$\exists n \in N: n = h_1 n_2 h_1^{-1}$ Thus: {{begin-eqn}} {{eqn | l = \paren {n_1 h_1} \paren {n_2 h_2}       | r = \paren {n_1 n} \paren {h_1 h_2} }} {{eqn | o = \in       | r = N H }} {{end-eqn}} Also: {{begin-eqn}} {{eqn | l = \paren {n_1 h_1}^{-1}       | r = h_1^{-1} n_1^{-1} }} {{eqn | r = \paren {h_1^{-1} n_1^{-1} h_1} h_1^{-1} }} {{eqn | o = \in       | r = N H }} {{end-eqn}} so from the Two-Step Subgroup Test, $N H$ is a subgroup of $G$. The fact that $N H = H N$ follows from Subset Product of Subgroups. {{qed}}
Follows from Regular Representations in Group are Permutations. Let $h \in H$. Then: : $\lambda_x \left({h}\right) = x h \in x H$ Thus: :$\forall h \in H: \lambda_x h \in x H$ demonstrating that $\lambda_x: H \to x H$ is a mapping. A permutation is a bijection by definition, As Regular Representations in Group are Permutations, it follows that $\lambda_x$ is a bijection. {{qed}} Exactly the same argument applies to $\rho_x$. {{qed}}
By definition, if $\struct {R_1, +_1, \circ_1}$ and $\struct {R_2, +_2, \circ_2}$ are rings then $\struct {R_1, +_1}$ and $\struct {R_2, +_2}$ are groups. Again by definition: :the zero of $\struct {R_1, +_1, \circ_1}$ is the identity of $\struct {R_1, +_1}$ :the zero of $\struct {R_2, +_2, \circ_2}$ is the identity of $\struct {R_2, +_2}$. The result follows from Group Homomorphism Preserves Identity. {{Qed}}
Let $\epsilon > 0$. Then $\dfrac \epsilon K > 0$. If the condition holds as stated, then: :$\exists N \in \N: \forall n \ge N: \size {s_n - l} < K \paren {\dfrac \epsilon K}$ Hence the result by definition of a convergent sequence. {{qed}}
Let $\sequence {a_n}$ be a Cauchy sequence in $\R$. We have the result Real Number Line is Metric Space. Hence by Convergent Subsequence of Cauchy Sequence in Metric Space, it is sufficient to show that $\sequence {a_n}$ has a convergent subsequence. Since $\sequence {a_n}$ is Cauchy, by Real Cauchy Sequence is Bounded, it is also bounded. By the Bolzano-Weierstrass Theorem, $\sequence {a_n}$ has a convergent subsequence. Hence the result. {{qed}}
Let $\sequence {a_n}$ be a Cauchy sequence in $\R$. By Real Cauchy Sequence is Bounded, $\sequence {a_n}$ is bounded. By the Bolzano-Weierstrass Theorem, $\sequence {a_n}$ has a convergent subsequence $\sequence {a_{n_r} }$. Let $a_{n_r} \to l$ as $r \to \infty$. It is to be shown that $a_n \to l$ as $n \to \infty$. Let $\epsilon \in \R_{>0}$ be a (strictly) positive real number. Then $\dfrac \epsilon 2 > 0$. Hence: :$(1): \quad \exists R \in \R: \forall r > R: \size {a_{n_r} - l} < \dfrac \epsilon 2$ We have that $\sequence {a_n}$ is a Cauchy sequence. Hence: :$(2): \quad \exists N \in \R: \forall m > N, n > N: \size {x_m - x_n} \le \dfrac \epsilon 2$ Let $n > N$. Let $r \in \N$ be sufficiently large that: :$n_r > N$ and: :$r > R$ Then $(1)$ is satisfied, and $(2)$ is satisfied with $m = n_r$. So: {{begin-eqn}} {{eqn | lo= \forall n > N:       | l = \size {a_n - l}       | r = \size {a_n - a_{n_r} + a_{n_r} - l}       | c =  }} {{eqn | o = \le       | r = \size {a_n - a_{n_r} } + \size{a_{n_r} - l}       | c = Triangle Inequality for Real Numbers }} {{eqn | o = <       | r = \dfrac \epsilon 2 + \dfrac \epsilon 2       | c =  }} {{eqn | r = \epsilon       | c =  }} {{end-eqn}} So, given $\epsilon > 0$, we have found $n \in \R$ such that: :$\forall n > N: \size {a_n - l} < \epsilon$ Thus: :$x_n \to l$ as $n \to \infty$. {{qed}}
=== Necessary Condition === Suppose $\sequence {x_n}$ is convergent. From Convergent Sequence in Metric Space is Cauchy Sequence, we have that every convergent sequence in a metric space is a Cauchy sequence. We also have that the real number line is a metric space. Hence $\sequence {x_n}$ is a Cauchy sequence. {{qed|lemma}} === Sufficient Condition === Suppose $\sequence {x_n}$ is a Cauchy sequence. We have the result that a Cauchy Sequence Converges on Real Number Line. Hence $\sequence {x_n}$ is convergent. {{qed|lemma}} The conditions have been shown to be equivalent. Hence the result. {{qed}}
Let $f: S \to T$ be an injection by definition 1. Thus: :$\forall x_1, x_2 \in S: \map f {x_1} = \map f {x_2} \implies x_1 = x_2$ {{AimForCont}} $f^{-1} \sqbrk {\set y}$ has more than one element. That is: :$\exists y \in T: x_1, x_2 \in \map {f^{-1} } y, x_1 \ne x_2$ Then we have: :$\map f {x_1} = \map f {x_2}$ but: :$x_1 \ne x_2$ This contradicts our initial hypothesis that $f$ is an injection by definition 1. From this contradiction it follows that $f^{-1} \sqbrk {\set y}$ has no more than one element. That is, $f$ is an injection by definition 4. {{qed|lemma}} Let $f: S \to T$ be an injection by definition 4. That is, let $\map {f^{-1} } y$ be a singleton for all $y \in T$. {{AimForCont}} it is not the case that: :$\forall x_1, x_2 \in S: \map f {x_1} = \map f {x_2} \implies x_1 = x_2$ Then by definition: :$\exists x_1, x_2 \in S, x_1 \ne x_2: \map f {x_1} = \map f {x_2} = y$ By definition of preimage of $y \in T$: :$x_1 \in \map {f^{-1} } y, x_2 \in \map {f^{-1} } y$ and so: $\set {x_1, x_2} \subseteq \map {f^{-1} } y$ Thus $\map {f^{-1} } y$ has more than one element for at least one $y \in T$. This contradicts our initial hypothesis that $f$ is an injection by definition 4. Thus: :$\forall x_1, x_2 \in S: \map f {x_1} = \map f {x_2} \implies x_1 = x_2$ So $f$ is an injection by definition 1. {{Qed}}
=== Necessary Condition === We use a Proof by Contraposition. To that end, suppose: :$\exists t \in T: f^{-1} \paren t = \O$ That is: :$\neg \paren {\forall t \in T: \exists s \in S: f \paren s = t}$ So, by definition, $f: S \to T$ is not a surjection. From Rule of Transposition it follows that if $f$ is a surjection: $\neg \exists t \in T: f^{-1} \paren t = \O$ {{qed|lemma}} === Sufficient Condition === We again use a Proof by Contraposition. To that end, suppose $f$ is not a surjection. Then by definition: :$\exists t \in T: \neg \paren {\exists s \in S: f \paren s = t}$ That is: :$\exists t \in T: f^{-1} \paren t = \O$ From Rule of Transposition it follows that if $\neg \exists t \in T: f^{-1} \paren t = \O$, then $f$ is a surjection. {{Qed}}
Suppose that there is no $t\in T$ such that $f^{-1}\paren t$ is empty. By Denial of Existence, this is equivalent to saying that for all $t\in T$, $f^{-1}\paren t$ is not empty. This is equivalent to the statement that $f^{-1}\paren t$ contains at least one element for each $t\in T$. In other words, for each $t\in T$, there exists an $s\in S$ such that $f \paren s = t$. This is the definition of $f$ being surjective. Thus if there is no $t\in T$ such that $f^{-1}\paren t$ is empty, then $f$ is surjective. Since this proof only uses statements of equivalence, it also shows that $f$ being surjective implies that there is no $t\in T$ such that $f^{-1}\paren t$ is empty. {{qed}}
Let $T = P \setminus s \sqbrk P$. From Axiom $(\text P 4)$ we know that $T \ne \O$. Now suppose that $t_1 \in T$ and $t_2 \in T$. Aiming for a contradiction, suppose that $t_1 \ne t_2$. Define $A = P \setminus \set {t_2}$. Thus $t_1 \in A \ne P$. Moreover, by the nature of $t_2$: :$x \in A \implies \map s x \in A$ Thus, by the induction axiom $(\text P 5)$, $A = P$. From this contradiction it follows that $P \setminus s \sqbrk P$ cannot contain two different elements. {{qed}}
=== Formulation 1 implies Formulation 2 === For $(\text P 3)$, there is nothing to prove. Next, axiom $(\text P 4)$ of Formulation 2. Recall the definition of the image of a mapping as applicable to $s$: :$\Img s = \set {n \in P: \exists m \in P: \map s m = n}$ From this definition, it follows that $0 \notin \Img s$, and hence: :$\Img s \ne P$ Lastly, axiom $(\text P 5)$ of Formulation 2. By Non-Successor Element of Peano Structure is Unique, we know that there is exactly one element of $P$ that is not in the image of $s$. By axiom $(\text P 3)$ of Formulation 1, we know this element is $0$. Therefore, the premises: :$0 \in A$ and: :$\exists x \in A: \neg \exists y \in P: x = \map s y$ are logically equivalent, and hence so are both forms of axiom $(\text P 5)$. {{qed|lemma}} === Formulation 2 implies Formulation 1 === By Non-Successor Element of Peano Structure is Unique, we know that there is exactly one element of $P$ that is not in the image of $s$. We will identify this element with the distinguished element $0$. This implies that axioms $(\text P 3)$ and $(\text P 4)$ of Formulation 1 are satisfied. Lastly, axiom $(\text P 5)$ of Formulation 1. Since $0$ satisfies the premise: :$\neg \exists y \in P: 0 = \map s y$ we conclude that axiom $(\text P 5)$ of Formulation 1 follows a fortiori from axiom $(\text P 5)$ of Formulation 2. {{qed}}
We need to check that all of Peano's axioms hold for $\left({P', s', s \left({0}\right)}\right)$. Now from Restriction of Injection is Injection, because $s$ is an injection then so is $s'$. So Axiom $(P3)$ holds. Suppose that for some $n \in P'$, we would have: :$s' \left({n}\right) = s \left({0}\right)$ Then by definition of $s'$ as the restriction of $s$: :$s \left({n}\right) = s \left({0}\right)$ However, since $0 \notin P'$, it follows that $n \ne 0$. This contradicts the fact that $s$ is an injection. Hence for all $n \in P'$: :$s' \left({n}\right) \ne s \left({0}\right)$ So Axiom $(P4)$ holds as well. Let $A \subseteq P'$ be such that $s \left({0}\right) \in A$ and: :$\forall z \in A: s' \left({z}\right) \in A$ Let $A_0$ be defined as: :$A_0 = A \cup \left\{{0}\right\}$ Then $0 \in A_0$, and since: :$\forall z \in A: s' \left({z}\right) \in A$ :$s \left({0}\right) \in A$ it follows that: :$\forall z \in A_0: s \left({z}\right) \in A_0$ Therefore, since $\left({P, s, 0}\right)$ is a Peano structure: :$A_0 = P$ Therefore, since $s \left({n}\right) \ne 0$ for all $n \in P$: :$\forall n \in P: s \left({n}\right) \in A$ Hence, by definition of $P'$: :$P' \subseteq A$ By definition of set equality: :$A = P'$ meaning that Axiom $(P5)$ holds as well. Having verified all axioms, it follows that $\left({P', s', s \left({0}\right)}\right)$ is a Peano structure. {{qed}} Category:Peano's Axioms thpgny4pfzgn622rwi4plokz2kh4kd6
First we show that $f \restriction_{X \times Y}$ is a mapping from $X$ to $Y$. By Restriction of Mapping is Mapping, $f \restriction_{X \times T}$ is a mapping from $X$ to $T$. If $x \in X$, then by the definition of image: :$\map f x \in f \sqbrk X$ {{explain|More explanation required.}} Since $f \sqbrk X \subseteq Y$, $f \restriction_{X \times Y}$ is a mapping from $X$ to $Y$. By definition of an injection: :$\forall s_1, s_2 \in S: \map f {s_1} = \map f {s_2} \implies s_1 = s_2$. {{AimForCont}} $f \restriction_{X \times Y}: X \to Y$ were ''not'' an injection. Then: :$\exists x_1, x_2 \in X: x_1 \ne x_2, \map f {x_1} = \map f {x_2}$ But then: :$\exists x_1, x_2 \in S: x_1 \ne x_2, \map f {x_1} = \map f {x_2}$ So $f: S \to T$ would not then be an injection. Hence the result. {{qed}} Category:Injections cjlhkrj0b6mhp1hbfnelm8lste85t5v
Let $S$ be a set and let $\Delta_S$ be the diagonal relation on $S$. Let $\tuple {x, y_1} \in \Delta_S \land \tuple {x, y_2} \in \Delta_S$. From the definition of the diagonal relation: :$\tuple {x, y_1} = \tuple {x, x}$ :$\tuple {x, y_2} = \tuple {x, x}$ and so $y_1 = y_2$. {{qed}}
{{AimForCont}} $f$ is not a many-to-one relation. Then there exists $y_1 \in \R$ and $y_2 \in \R$ where $y_1 \ne y_2$ such that: :$\exists x \in \R: \map f x = y_1, \map f x = y_2$ There are two possibilities: {{begin-eqn}} {{eqn | n = 1       | l = x       | o = \ge       | r = 0       | c =  }} {{eqn | n = 2       | l = x       | o = <       | r = 0       | c =  }} {{end-eqn}} Suppose $x \ge 0$. Then: {{begin-eqn}} {{eqn | l = y_1 = \map f x       | r = x }} {{eqn | l = y_2 = \map f x       | r = x }} {{end-eqn}} That is: :$y_2 = y_1 = x$ Suppose $x < 0$. Then: {{begin-eqn}} {{eqn | l = y_1 = \map f {x_1}       | r = -x }} {{eqn | l = y_2 = \map f {x_2}       | r = -x }} {{end-eqn}} That is: :$y_2 = y_1 = -x$ So by Proof by Cases we have that $y_1 = y_2$ whatever $x$ may be. This contradicts our assertion that $y_1 \ne y_2$. Hence it follows by Proof by Contradiction that $f$ is many-to-one. {{qed}}
As $\mathcal R$ is a relation, then so is its inverse $\mathcal R^{-1}$. Let $\mathcal R \sqbrk X \subseteq Y$. Thus: {{begin-eqn}} {{eqn | l = \mathcal R \sqbrk X       | o = \subseteq       | r = Y       | c =  }} {{eqn | ll= \leadsto       | l = \mathcal R^{-1} \sqbrk {\mathcal R \sqbrk X}       | o = \subseteq       | r = \mathcal R^{-1} \sqbrk Y       | c = Image of Subset under Relation is Subset of Image: Corollary 1 }} {{eqn | ll= \leadsto       | l = X       | o = \subseteq       | r = \mathcal R^{-1} \sqbrk Y       | c = Image of Preimage under Relation is Subset, Subset Relation is Transitive }} {{end-eqn}} So: :$\mathcal R \sqbrk X \subseteq Y \implies X \subseteq \mathcal R^{-1} \sqbrk Y$ {{qed|lemma}} Now let $X \subseteq \mathcal R^{-1} \sqbrk Y$. The same argument applies: {{begin-eqn}} {{eqn | l = X       | o = \subseteq       | r = \mathcal R^{-1} \sqbrk Y       | c =  }} {{eqn | ll= \leadsto       | l = \mathcal R \sqbrk X       | o = \subseteq       | r = \mathcal R \sqbrk {\mathcal R^{-1} \sqbrk Y}       | c = Image of Subset under Relation is Subset of Image }} {{eqn | ll= \leadsto       | l = \mathcal R \sqbrk X       | o = \subseteq       | r = Y       | c = Image of Preimage under Relation is Subset, Subset Relation is Transitive }} {{end-eqn}} So: :$X \subseteq \mathcal R^{-1} \sqbrk Y \implies \mathcal R \sqbrk X \subseteq Y$ {{qed|lemma}} Thus we have: :$X \subseteq \mathcal R^{-1} \sqbrk Y \implies \mathcal R \sqbrk X \subseteq Y$ :$\mathcal R \sqbrk X \subseteq Y \implies X \subseteq \mathcal R^{-1} \sqbrk Y$ Hence the result. {{qed}} Category:Subsets Category:Preimages under Relations onk7azzmohjl3o7y1t8ttw51dsigz2t
=== Sufficient Condition === Let $g$ be such that: :$\forall A \subseteq S: A = \paren {f^{-1} \circ f} \sqbrk A$ Then by Subset equals Preimage of Image implies Injection, $f$ is an injection. {{qed|lemma}} === Necessary Condition === Let $f$ be an injection. Then by Preimage of Image of Subset under Injection equals Subset: :$\forall A \subseteq S: A = \paren {f^{-1} \circ f} \sqbrk A$ {{qed}}
=== Sufficient Condition === Let $f$ be such that: :$\forall B \in \powerset T: B = \map {\paren {f^\to \circ f^\gets} } B$ From Subset equals Image of Preimage implies Surjection, $f$ is a surjection. {{qed|lemma}} === Necessary Condition === Let $f$ be a surjection. Then by Image of Preimage of Subset under Surjection equals Subset: :$B = \map {f^\to} {\map {f^\gets} B}$ {{qed}}
Let $f: S \to T$ be an injection. We have by definition of inverse relation that: :$f^{-1} = \set {\tuple {t, s}: t = \map f s}$ Let $f: S \to T$ be an injection. Let $\tuple {t, s_1} \in f^{-1}$ and $\tuple {t, s_2} \in f^{-1}$. By definition, we have that $\map f {s_1} = t = \map f {s_2}$. But as $f$ is an injection: :$\map f {s_1} = \map f {s_2} \implies s_1 = s_2$ So we have that: :$\tuple {t, s_1} \in f^{-1} \land \tuple {t, s_2} \in f^{-1} \implies s_1 = s_2$ and so by definition, $f^{-1}$ is a many-to-one relation. {{qed}} Category:Injections Category:Inverse Relations ixecpmse2fy6f1ur9mnk60mdnw8boe6
Follows directly from: :Injection iff Left Cancellable: $f$ is an injection {{iff}} $f$ is left cancellable :Surjection iff Right Cancellable: $f$ is a surjection {{iff}} $f$ is right cancellable. {{qed}}