hcju/proofwikiqa huggingface dataset:
{{begin-eqn}} {{eqn | l = a       | o = <       | r = b       | c =  }} {{eqn | ll= \leadsto       | l = \exists c \in \N_{>0}: a       | r = b + c       | c = {{Defof|Ordering on 1-Based Natural Numbers|$<$ on $\N_{>0}$}} }} {{eqn | ll= \leadsto       | l = a \times n       | r = \paren {b + c} \times n       | c = {{Defof|Binary Operation}} }} {{eqn | r = \paren {b \times n} + \paren {c \times n}       | c = Natural Number Multiplication Distributes over Addition }} {{eqn | ll= \leadsto       | l = a \times n       | o = <        | r = b \times n       | c = {{Defof|Ordering on 1-Based Natural Numbers|$<$ on $\N_{>0}$}}: $c \times n \in \N_{> 0}$ }} {{end-eqn}} {{qed}}
Let $z = a + b - x$. Then: :$\dfrac {\d z} {\d x} = -1$ and: :$x = a \implies z = a + b - a = b$ :$x = b \implies z = a + b - b = a$ So: {{begin-eqn}} {{eqn | l = \int_a^b \map f {a + b - x} \rd x       | r = \int_b^a \map f z \paren {-1} \rd z       | c = Integration by Substitution }} {{eqn | r = \int_a^b \map f z \rd z       | c = Reversal of Limits of Definite Integral }} {{eqn | r = \int_a^b \map f x \rd x       | c = renaming variables }} {{end-eqn}} {{qed}} Category:Definite Integrals 657vwqwz0lc70jyt38jen81j7fstvhr
From the definition of $\N_0$: :$\N_0 = \set {n \in \N: n < 0}$  From the definition of zero, $0$ is the minimal element of $\N$. So there is no element $n$ of $\N$ such that $n < 0$. Thus $\N_0 = \O$. {{qed}} Category:Natural Numbers qbigqp7k4plig58dwsn1gh24fsrjyqb
From the definition of $\N^*_0$: :$\N^*_0 = \set {n \in \N_{>0}: n \le 0}$  From the definition of one, the minimal element of $\N_{>0}$ is $1$. From Zero Strictly Precedes One we know that $0 < 1$. So there is no element $n$ of $\N_{>0}$ such that $n \le 0$. Thus $\N^*_0 = \O$. {{qed}}
By Ordering on $1$-Based Natural Numbers is Trichotomy, one and only one of the following holds: :$a = b$ :$a < b$ :$b < a$ Suppose $a < b$. Then by Ordering on $1$-Based Natural Numbers is Compatible with Addition: :$a + c < b + c$ By Ordering on $1$-Based Natural Numbers is Trichotomy, this contradicts the fact that $a + c = b + c$. Similarly, suppose $b > a$. Then by Ordering on $1$-Based Natural Numbers is Compatible with Addition: :$b + c < a + c$ By Ordering on $1$-Based Natural Numbers is Trichotomy, this also contradicts the fact that $a + c = b + c$. The only other possibility is that $a = b$. So  :$\forall a, b, c \in \N_{>0}: a + c = b + c \implies a = b$ and so $+$ is right cancellable on $\N_{>0}$. From Natural Number Addition is Commutative and Right Cancellable Commutative Operation is Left Cancellable: :$\forall a, b, c \in \N_{>0}: a + b = a + c \implies b = c$ So $+$ is both right cancellable and left cancellable on $\N_{>0}$. Hence the result. {{qed}}
By Ordering on $1$-Based Natural Numbers is Trichotomy, one and only one of the following holds: :$a = b$ :$a < b$ :$b < a$ Let $a + c < b + c$. Suppose $a = b$. Then by Ordering on $1$-Based Natural Numbers is Compatible with Addition: :$a + c = b + c$ By Ordering on $1$-Based Natural Numbers is Trichotomy, this contradicts the fact that $a + c < b + c$. Similarly, suppose $b < a$. Then by Ordering on $1$-Based Natural Numbers is Compatible with Addition: :$b + c < a + c$ By Ordering on $1$-Based Natural Numbers is Trichotomy, this also contradicts the fact that $a + c < b + c$. The only other possibility is that $a < b$. So  :$\forall a, b, c \in \N_{>0}: a + c = b + c \implies a < b$ and so $+$ is right cancellable on $\N_{>0}$ for $<$. Let $a + b < a + c$. Suppose $b = c$. Then by Ordering on $1$-Based Natural Numbers is Compatible with Addition: :$a + b = a + c$ By Ordering on $1$-Based Natural Numbers is Trichotomy, this contradicts the fact that $a + b < a + c$. Similarly, suppose $c < b$. Then by Ordering on $1$-Based Natural Numbers is Compatible with Addition: :$a + c < a + b$ By Ordering on $1$-Based Natural Numbers is Trichotomy, this also contradicts the fact that $a + b < a + c$. The only other possibility is that $b < c$. So  :$\forall a, b, c \in \N_{>0}: a + b < a + c \implies b < c$ and so $+$ is left cancellable on $\N_{>0}$ for $<$. From Natural Number Addition is Commutative and Right Cancellable Commutative Operation is Left Cancellable: :$\forall a, b, c \in \N_{>0}: a + b = a + c \implies b = c$ So $+$ is both right cancellable and left cancellable on $\N_{>0}$. Hence the result. {{qed}}
By Ordering on $1$-Based Natural Numbers is Trichotomy, one and only one of the following holds: :$a = b$ :$a < b$ :$b < a$ Suppose $a < b$. Then by Ordering on $1$-Based Natural Numbers is Compatible with Multiplication: :$a \times c < b \times c$ By Ordering on $1$-Based Natural Numbers is Trichotomy, this contradicts the fact that $a \times c = b \times c$. Similarly, suppose $b > a$. Then by Ordering on $1$-Based Natural Numbers is Compatible with Multiplication: :$b \times c < a \times c$ By Ordering on $1$-Based Natural Numbers is Trichotomy, this also contradicts the fact that $a \times c = b \times c$. The only other possibility is that $a = b$. So  :$\forall a, b, c \in \N_{>0}: a \times c = b \times c \implies a = b$ and so $\times$ is right cancellable on $\N_{>0}$. From Natural Number Multiplication is Commutative and Right Cancellable Commutative Operation is Left Cancellable: :$\forall , b, c \in \N_{>0}: a \times b = a \times c \implies b = c$ So $\times$ is both right cancellable and left cancellable on $\N_{>0}$. Hence the result. {{qed}}
By Ordering on $1$-Based Natural Numbers is Trichotomy, one and only one of the following holds: :$a = b$ :$a < b$ :$b < a$ Let $a \times c < b \times c$. Suppose $a = b$. Then by Ordering on $1$-Based Natural Numbers is Compatible with Multiplication: :$a \times c = b \times c$ By Ordering on $1$-Based Natural Numbers is Trichotomy, this contradicts the fact that $a \times c < b \times c$. Similarly, suppose $b < a$. Then by Ordering on $1$-Based Natural Numbers is Compatible with Multiplication: :$b \times c < a \times c$ By Ordering on $1$-Based Natural Numbers is Trichotomy, this also contradicts the fact that $a \times c < b \times c$. The only other possibility is that $a < b$. So  :$\forall a, b, c \in \N_{>0}: a \times c = b \times c \implies a < b$ and so $\times$ is right cancellable on $\N_{>0}$ for $<$. Let $a \times b < a \times c$. Suppose $b = c$. Then by Ordering on $1$-Based Natural Numbers is Compatible with Multiplication: :$a \times b = a \times c$ By Ordering on $1$-Based Natural Numbers is Trichotomy, this contradicts the fact that $a \times b < a \times c$. Similarly, suppose $c < b$. Then by Ordering on $1$-Based Natural Numbers is Compatible with Multiplication: :$a \times c < a \times b$ By Ordering on $1$-Based Natural Numbers is Trichotomy, this also contradicts the fact that $a \times b < a \times c$. The only other possibility is that $b < c$. So  :$\forall a, b, c \in \N_{>0}: a \times b < a \times c \implies b < c$ and so $\times$ is left cancellable on $\N_{>0}$ for $<$. From Natural Number Multiplication is Commutative and Right Cancellable Commutative Operation is Left Cancellable: :$\forall a, b, c \in \N_{>0}: a \times b = a \times c \implies b = c$ So $+$ is both right cancellable and left cancellable on $\N_{>0}$. Hence the result. {{qed}}
By definition of cardinality and the trivial subgroup: :$\order {\set e} = 1$ From Lagrange's Theorem: :$\index G {\set e} = \dfrac {\order G} {\order {\set e} } = \dfrac {\order G} 1 = \order G$ {{qed}}
For finite groups, we can apply Lagrange's Theorem: :$\index G H = \dfrac {\order G} {\order H}$ But then: : $\dfrac {\order G} {\order H} = 1 \iff \order G = \order H$ Hence the result. For the general case (including infinite groups) we need to consider the (left) coset space $G / H$. Note that we must have $e H = H \in G / H$. Hence: :$\index G H = 1 \iff G / H = \set H \iff G = H$ {{qed}}
{{begin-eqn}} {{eqn | l = \dfrac {1 + \cos 2 x} 2       | r = \dfrac 1 2 \left({1 + \dfrac {e^{2 i x} + e^{-2 i x} } 2}\right)       | c = Cosine Exponential Formulation }} {{eqn | r = \dfrac 1 4 \left({e^{2 i x} + 2 + e^{-2 i x} }\right)       | c = simplifying }} {{eqn | r = \dfrac 1 4 \left({e^{2 i x} + 2 \left({e^{i x} }\right) \left({e^{-i x} }\right) + e^{-2 i x} }\right)       | c =  }} {{eqn | r = \left({\dfrac {e^{i x} + e^{-i x} } 2}\right)^2       | c = Square of Sum }} {{eqn | r = \cos^2 x       | c = Cosine Exponential Formulation }} {{end-eqn}} {{qed}}
Checking in turn each of the criteria for equivalence: === Reflexivity === Let $A \in \powerset S$. From Set Equals Itself: :$A = A$ So set equality has been shown to be reflexive on $\powerset S$. {{qed|lemma}} === Symmetry === Let $A, B \in \powerset S$. Let $A = B$. Then by definition of set equality: :$A \subseteq B$ :$B \subseteq A$ from which it follows by definition of set equality that $B = A$. So set equality has been shown to be symmetric on $\powerset S$. {{qed|lemma}} === Transitivity === Let $A, B, C \in \powerset S$. Let $A = B$ and $B = C$. Then by definition of set equality: :$(1): \quad A \subseteq B$ :$(2): \quad B \subseteq C$ :$(3): \quad C \subseteq B$ :$(4): \quad B \subseteq A$ From $(1)$ and $(2)$ and Subset Relation is Transitive: :$A \subseteq C$ From $(3)$ and $(4)$ and Subset Relation is Transitive: :$C \subseteq A$ from which it follows by definition of set equality that $A = C$. So set equality has been shown to be transitive on $\powerset S$. {{qed|lemma}} Set equality has been shown to be reflexive, symmetric and transitive on $\powerset S$. Hence by definition it is an equivalence relation on $\powerset S$. {{qed}}
Let $f: S \to T$ be an injection by definition 1. Thus: :$\forall x_1, x_2 \in S: \map f {x_1} = \map f {x_2} \implies x_1 = x_2$ Let $x_1 \ne x_2$. {{AimForCont}} that: :$\map f {x_1} = \map f {x_2}$ Then by definition 1: :$x_1 = x_2$ This contradicts the assumption that $x_1 \ne x_2$. Thus by Proof by Contradiction: :$\forall x_1, x_2 \in S: x_1 \ne x_2 \implies \map f {x_1} \ne \map f {x_2}$ Thus $f$ is an injection by definition 1 a. {{qed|lemma}} Let $f: S \to T$ be an injection by definition 1 a. Thus: :$\forall x_1, x_2 \in S: x_1 \ne x_2 \implies \map f {x_1} \ne \map f {x_2}$ Let $\map f {x_1} = \map f {x_2}$. {{AimForCont}} that: :$x_1 \ne x_2$ Then by definition 1 a: :$\map f {x_1} \ne \map f {x_2}$ This contradicts the assumption that $\map f {x_1} = \map f {x_2}$. Thus by Proof by Contradiction: :$\forall x_1, x_2 \in S: \map f {x_1} = \map f {x_2} \implies x_1 = x_2$ Thus $f$ is an injection by definition 1. {{qed}}
Let $f: S \to T$ be an injection by definition 1. Thus: :$(1): \quad \forall x_1, x_2 \in S: f \left({x_1}\right) = f \left({x_2}\right) \implies x_1 = x_2$ Let $y \in T: y = f \left({x_1}\right) = f \left({x_2}\right)$. Thus $(1)$ can be rewritten in the language of relations as: :$\forall x_1, x_2 \in S: \left({x_1, y}\right) \in f \land  f \left({x_2, y}\right) \in f \implies x_1 = x_2$ So, by definition, $f$ is a one-to-many relation. By definition, $f$ is also mapping. So, by definition, $f$ is a relation which is both many-to-one and left-total. A relation which is both many-to-one and one-to-many is by definition a one-to-one relation. Thus $f$ is a relation which is one-to-one and left-total. So $f$ is an injection by definition 2. {{qed|lemma}} Let $f: S \to T$ be an injection by definition 2. That is, $f$ is a relation which is one-to-one and left-total. A one-to-one relation is a relation which is both many-to-one and one-to-many. So $f$ is a relation which is both many-to-one and left-total. Thus $f$ is a mapping which is one-to-many: :$\forall x_1, x_2 \in S: \left({x_1, y}\right) \in f \land  f \left({x_2, y}\right) \in f \implies x_1 = x_2$ Setting $y = f \left({x_1}\right) = f \left({x_2}\right)$: :$\forall x_1, x_2 \in S: f \left({x_1}\right) = f \left({x_2}\right) \implies x_1 = x_2$ So $f$ is an injection by definition 1. {{qed}}
=== Definition $1$ iff Definition $1 \ \text{a}$ === {{:Equivalence of Definitions of Injection/Definition 1 iff Definition 1 a}} === Definition $1$ iff Definition $2$ === {{:Equivalence of Definitions of Injection/Definition 1 iff Definition 2}} === Definition $1$ iff Definition $3$ === {{:Equivalence of Definitions of Injection/Definition 1 iff Definition 3}} === Definition $1$ iff Definition $4$ === {{:Equivalence of Definitions of Injection/Definition 1 iff Definition 4}} === Definition $1$ iff Definition $5$ === {{:Equivalence of Definitions of Injection/Definition 1 iff Definition 5}} === Definition $1$ iff Definition $6$ === {{:Equivalence of Definitions of Injection/Definition 1 iff Definition 6}} Category:Injections Category:Equivalence of Definitions of Injection 2jvkpdszokgwj9y31nq4mhdxrlpt4yc
This is a direct statement of the definition of set equality: :$R = S := \paren {R \subseteq S} \land \paren {S \subseteq R}$ {{qed}}
Suppose such a $\mathcal U$ exists. Using the Axiom of Specification, we can create the set: :$R = \left\{{x \in \mathcal U: x \notin x}\right\}$ But from Russell's Paradox, this set cannot exist. Thus: :$R \notin \mathcal U$ and so $\mathcal U$ cannot contain everything. {{qed}}
Let $x \in \set a \cup \set b$. Then by the Axiom of Unions: :$x = a \lor x = b$ It follows from the Axiom of Pairing that: :$x \in \set {a, b}$ Thus by definition of subset: :$\set a \cup \set b \subseteq \set {a, b}$ {{qed|lemma}} Let $x \in \set {a, b}$. Then by the Axiom of Pairing: :$x = a \lor x = b$ So by the Axiom of Unions: :$x \in \set a \cup \set b$ Thus by definition of subset: :$\set {a, b} \subseteq \set a \cup \set b$ {{qed|lemma}} The result follows by definition of set equality. {{qed}}
The only element of $T^\O$ is the null relation: :$\varnothing \times T$ From Null Relation is Mapping iff Domain is Empty Set, $\O \times T$ is a mapping from $\O$ to $T$. The result follows from Empty Mapping is Unique. That is: :$\card {T^\O} = 1$ {{qed}}
From Null Relation is Mapping iff Domain is Empty Set, the null relation: :$\mathcal R = \O \subseteq S \times T$ is not a mapping unless $S = \O$. So if $S \ne \O$: :$\card {\O^S} = 0$ If $S = \O$: :$\card {\O^S} = 1$ Hence the result. {{qed}}
{{begin-eqn}} {{eqn | l = \bigcup_{\alpha \mathop \in I} \paren {A_\alpha \cap B}       | r = \paren {\bigcup_{\alpha \mathop \in I} A_\alpha} \cap B       | c = Intersection Distributes over Union: Family of Sets }} {{eqn | ll= \leadsto       | l = \bigcup_{\alpha \mathop \in I} \paren {A_\alpha \cap \paren {\bigcup_{\beta \mathop \in J} B_\beta} }       | r = \paren {\bigcup_{\alpha \mathop \in I} A_\alpha} \cap \paren {\bigcup_{\beta \mathop \in J} B_\beta}       | c = setting $\displaystyle B = \paren {\bigcup_{\beta \mathop \in J} B_\beta}$ }} {{eqn | ll= \leadsto       | l = \bigcup_{\alpha \mathop \in I} \paren {\bigcup_{\beta \mathop \in J} \paren {A_\alpha \cap B_\beta} }       | r = \paren {\bigcup_{\alpha \mathop \in I} A_\alpha} \cap \paren {\bigcup_{\beta \mathop \in J} B_\beta}       | c = Intersection Distributes over Union: Family of Sets }} {{eqn | ll= \leadsto       | l = \bigcup_{\paren {\alpha, \beta} \mathop \in I \times J} \paren {A_\alpha \cap B_\beta}       | r = \paren {\bigcup_{\alpha \mathop \in I} A_\alpha} \cap \paren {\bigcup_{\beta \mathop \in J} B_\beta}       | c =  }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \bigcap_{\alpha \mathop \in I} \paren {A_\alpha \cup B}       | r = \paren {\bigcap_{\alpha \mathop \in I} A_\alpha} \cup B       | c = Intersection Distributes over Union: Family of Sets }} {{eqn | ll= \leadsto       | l = \bigcap_{\alpha \mathop \in I} \paren {A_\alpha \cup \paren {\bigcap_{\beta \mathop \in J} B_\beta} }       | r = \paren {\bigcap_{\alpha \mathop \in I} A_\alpha} \cup \paren {\bigcap_{\beta \mathop \in J} B_\beta}       | c = setting $\displaystyle B = \paren {\bigcap_{\beta \mathop \in J} B_\beta}$ }} {{eqn | ll= \leadsto       | l = \bigcap_{\alpha \mathop \in I} \paren {\bigcap_{\beta \mathop \in J} \paren {A_\alpha \cup B_\beta} }       | r = \paren {\bigcap_{\alpha \mathop \in I} A_\alpha} \cup \paren {\bigcap_{\beta \mathop \in J} B_\beta}       | c = Intersection Distributes over Union: Family of Sets }} {{eqn | ll= \leadsto       | l = \bigcap_{\paren {\alpha, \beta} \mathop \in I \times J} \paren {A_\alpha \cup B_\beta}       | r = \paren {\bigcap_{\alpha \mathop \in I} A_\alpha} \cup \paren {\bigcap_{\beta \mathop \in J} B_\beta}       | c =  }} {{end-eqn}} {{qed}}
=== Necessary Condition === Let $f$ be a surjection. Let $B \subseteq T$ such that $B \ne \varnothing$. Then: :$\exists t \in T: t \in B$ From Preimages All Exist iff Surjection: :$\map {f^{-1} } t \ne \O$ As $t \in B$ it follows from Preimage of Subset is Subset of Preimage that: :$f^{-1} \sqbrk B \ne \O$ $B$ is arbitrary, so: :$\forall B \subseteq T, B \ne \O: f^{-1} \sqbrk B \ne \O$ {{qed|lemma}} === Sufficient Condition === Suppose that: :$\forall B \subseteq T, B \ne \O: f^{-1} \sqbrk B \ne \O$ {{AimForCont}} $f$ is not a surjection. Then by definition: :$\exists t \in T: \neg \paren {\exists s \in S: \map f s = t}$ That is: :$\exists \set t \subseteq T: f^{-1} \sqbrk {\set t} = \O$ which contradicts the hypothesis. So by Proof by Contradiction, $f$ is a surjection. Hence the result. {{Qed}}
Let $B \subseteq T$. Then: {{begin-eqn}} {{eqn | l = y       | o = \in       | r = \paren {\mathcal R \circ \mathcal R^{-1} } \sqbrk B }} {{eqn | ll= \leadsto       | l = y       | o = \in       | r = \mathcal R \sqbrk {\mathcal R^{-1} \sqbrk B}       | c = Definition of Composition of Relations }} {{eqn | ll= \leadsto       | l = \exists x \in \mathcal R^{-1} \sqbrk B: \tuple {x, y}       | o = \in       | r = \mathcal R }} {{eqn | ll= \leadsto       | l = y       | o = \in       | r = B       | c = }} {{end-eqn}} So by definition of subset: :$B \subseteq T \implies \paren {\mathcal R \circ \mathcal R^{-1} }\sqbrk B \subseteq B$ {{qed}}
Proof by Counterexample: Let $S = T = \set {0, 1}$. Let $\mathcal R = \set {\tuple {0, 0}, \tuple {0, 1} }$. We have that: :$\mathcal R^{-1} = \set {\tuple {0, 0}, \tuple {1, 0} }$ :$\powerset S = \powerset T = \set {\O, \set 0, \set 1, \set {0, 1} }$ Thus, by inspection: {{begin-eqn}} {{eqn | l = \map {\mathrel R^\to} \O       | r = \O       | c =  }} {{eqn | l = \map {\mathrel R^\to} {\set 0}       | r = \set {0, 1}       | c =  }} {{eqn | l = \map {\mathrel R^\to} {\set 1}       | r = \O       | c =  }} {{eqn | l = \map {\mathrel R^\to} {\set {0, 1} }       | r = \set {0, 1}       | c =  }} {{end-eqn}} Note that $\paren {\mathrel R^\to}^{-1}$ is the inverse of a mapping which is neither an injection nor a surjection, and so is itself not a mapping from $\powerset T$ to $\powerset S$. {{begin-eqn}} {{eqn | l = \map {\paren {\mathrel R^\to}^{-1} } \O       | r = \set {\O, \set 1}       | c =  }} {{eqn | l = \map {\paren {\mathrel R^\to}^{-1} } {\set 0}       | r = \O       | c =  }} {{eqn | l = \map {\paren {\mathrel R^\to}^{-1} } {\set 1}       | r = \O       | c =  }} {{eqn | l = \map {\paren {\mathrel R^\to}^{-1} } {\set {0, 1} }       | r = \set {\set 0, \set {0, 1} }       | c =  }} {{end-eqn}} This can be seen to be completely different from $\mathrel R^\gets$, which can be determined by inspection to be: {{begin-eqn}} {{eqn | l = \map {\mathrel R^\gets} \O       | r = \O       | c =  }} {{eqn | l = \map {\mathrel R^\gets} {\set 0}       | r = \set 0       | c =  }} {{eqn | l = \map {\mathrel R^\gets} {\set 1}       | r = \set 0       | c =  }} {{eqn | l = \map {\mathrel R^\gets} {\set {0, 1} }       | r = \set 0       | c =  }} {{end-eqn}} {{qed}} Category:Direct Image Mappings Category:Inverse Image Mappings 35arzqf6eutbis2oqntixzfkvhvw995
As $f$, being a mapping, is also a relation, we can apply Preimage of Union under Relation: Family of Sets: :$\displaystyle \RR^{-1} \sqbrk {\bigcup_{i \mathop \in I} T_i} = \bigcup_{i \mathop \in I} \RR^{-1} \sqbrk {T_i}$ where $\RR^{-1} \sqbrk {T_i}$ denotes the preimage of $T_i$ under $\RR^{-1}$. {{qed}}
We have that $f$ is a mapping, and so also a relation. Thus its inverse $f^{-1}$ is also a relation. Hence we can apply Image of Union under Relation: Family of Sets: :$\displaystyle \RR \sqbrk {\bigcup_{i \mathop \in I} T_i} = \bigcup_{i \mathop \in I} \RR \sqbrk {T_i}$ where $\RR \sqbrk {T_i}$ denotes the image of $T_i$ under $\RR$. {{qed}}
=== Definition 1 implies Definition 2 === Let $x$ be an maximal element by definition 1. That is: :$(1): \quad \forall y \in T: x \preceq y \implies x = y$ Aiming for a contradiction, suppose that: :$\exists y \in T: x \prec y$ Then by definition: :$x \preceq y \land x \ne y$ which contradicts $(1)$. Thus by Proof by Contradiction: :$\nexists y \in T: x \prec y$ That is $x$ is a maximal element by definition 2. {{qed|lemma}} === Definition 2 implies Definition 1 === Let $x$ be a maximal element by definition 2. That is: :$(2): \quad \nexists y \in T: x \prec y$ Aiming for a contradiction, suppose that: :$\exists y \in T: x \preceq y: x \ne y$ That is: :$\exists y \in T: x \prec y$ which contradicts $(2)$. Thus: :$\forall y \in T: x \preceq y \implies x = y$ Thus $x$ is a maximal element by definition 1. {{qed}}
=== Definition 1 implies Definition 2 === Let $x$ be an minimal element by definition 1. That is: :$(1): \quad \forall y \in T: y \preceq x \implies x = y$ {{AimForCont}}: :$\exists y \in T: y \prec x$ Then by definition: :$y \preceq x \land x \ne y$ which contradicts $(1)$. Thus by Proof by Contradiction: :$\nexists y \in T: y \prec x$ That is $x$ is a minimal element by definition 2. {{qed|lemma}} === Definition 2 implies Definition 1 === Let $x$ be a minimal element by definition 2. That is: :$(2): \quad \nexists y \in T: y \prec x$ {{AimForCont}}: :$\exists y \in T: y \preceq x: x \ne y$ That is: :$\exists y \in T: y \prec x$ which contradicts $(2)$. Thus: :$\forall y \in T: y \preceq x \implies x = y$ Thus $x$ is a minimal element by definition 1. {{qed}}
Let $y \in \mathcal C$ such that $y \subseteq \left\{{x}\right\}$. We have that $\varnothing \notin \mathcal C$. Therefore: :$\exists z \in S: z \in y$ But as $y \subseteq \left\{{x}\right\}$ it follows that: :$z \in \left\{{x}\right\}$ and so by definition of singleton: :$z = x$ and so: :$y = \left\{{x}\right\}$ and so: :$y = x$ Thus, by definition, $\left\{{x}\right\}$ is a minimal element of $\left({\mathcal C, \subseteq}\right)$. {{qed}}
=== Necessary Condition === Let $S$ not be a singleton. Then $\exists x, y \in S: x \ne y$. Let $Z \in \mathcal C$ be the smallest element of $\mathcal C$. Then: :$\forall T \in \mathcal C: Z \subseteq T$ But by Singleton of Power Set less Empty Set is Minimal Subset, both $\left\{{x}\right\}$ and $\left\{{y}\right\}$ are minimal elements of $\left({\mathcal C, \subseteq}\right)$. Therefore it cannot be the case that $Z \subseteq \left\{{x}\right\}$ and $Z \subseteq \left\{{y}\right\}$. Therefore $\left({\mathcal C, \subseteq}\right)$ has no smallest element. {{qed|lemma}} === Sufficient Condition === Let the ordered structure $\left({\mathcal C, \subseteq}\right)$ has no smallest element. Aiming for a contradiction, suppose $S$ is a singleton. Let $S = \left\{{x}\right\}$. Then: :$\mathcal C = \left\{{\left\{{x}\right\}}\right\}$ Then: :$\forall y \in \mathcal C: y \subseteq \left\{{x}\right\}$ trivially. Thus $\left({\mathcal C, \subseteq}\right)$ has a smallest element which is $\left\{{x}\right\}$. By Proof by Contradiction it follows that $S$ is not a singleton. {{qed}}
=== Closure === We have that Natural Number Multiplication is Closed. That is, $\struct {\N, \times}$ is closed. {{qed|lemma}} === Associativity === We have that Natural Number Multiplication is Associative. {{qed|lemma}} Thus the criteria are fulfilled for $\struct {\N, \times}$ to form a semigroup. {{qed}}
This is a specific instance of Natural Numbers Bounded Below under Addition form Commutative Semigroup. {{qed}}