id stringlengths 11 15 | title stringlengths 7 171 | problem stringlengths 9 4.33k | solution stringlengths 6 19k | problem_wikitext stringlengths 9 4.42k | solution_wikitext stringlengths 7 19.1k | proof_title stringlengths 9 171 | theorem_url stringlengths 34 198 | proof_url stringlengths 36 198 | categories listlengths 0 9 | theorem_references listlengths 0 36 | proof_references listlengths 0 253 |
|---|---|---|---|---|---|---|---|---|---|---|---|
proofwiki-5800 | Subset of Empty Set | Let $A$ be a class.
Then:
:$A$ is a subset of the empty set $\O$
{{iff}}:
:$A$ is equal to the empty set:
:$A \subseteq \O \iff A = \O$ | {{begin-eqn}}
{{eqn | l = A = \O
| o = \leadsto
| r = A \subseteq \O
| c = {{Defof|Set Equality|index = 2}}
}}
{{end-eqn}}
Conversely:
{{begin-eqn}}
{{eqn | l = A \subseteq \O
| o = \leadsto
| r = A \subseteq \O \land \O \subseteq A
| c = Empty Set is Subset of All Sets
}}
{{eqn | o ... | Let $A$ be a [[Definition:Class (Class Theory)|class]].
Then:
:$A$ is a subset of the [[Definition:Empty Set|empty set]] $\O$
{{iff}}:
:$A$ is equal to the [[Definition:Empty Set|empty set]]:
:$A \subseteq \O \iff A = \O$ | {{begin-eqn}}
{{eqn | l = A = \O
| o = \leadsto
| r = A \subseteq \O
| c = {{Defof|Set Equality|index = 2}}
}}
{{end-eqn}}
[[Definition:Converse Statement|Conversely]]:
{{begin-eqn}}
{{eqn | l = A \subseteq \O
| o = \leadsto
| r = A \subseteq \O \land \O \subseteq A
| c = [[Empty ... | Subset of Empty Set | https://proofwiki.org/wiki/Subset_of_Empty_Set | https://proofwiki.org/wiki/Subset_of_Empty_Set | [
"Empty Set"
] | [
"Definition:Class (Class Theory)",
"Definition:Empty Set",
"Definition:Empty Set"
] | [
"Definition:Converse Statement",
"Empty Set is Subset of All Sets",
"Category:Empty Set"
] |
proofwiki-5801 | Ordinal is Subset of Successor | Let $x$ and $y$ be ordinals.
Let $y^+$ denote the successor of $y$.
Then:
: $x \subseteq y^+ \iff \left({x \subseteq y \lor x = y^+}\right)$ | Let $A \subset B$ denote that $A$ is a proper subset of $B$.
Let $A \in B$ denote that $A$ is an element of $B$.
From Transitive Set is Proper Subset of Ordinal iff Element of Ordinal, $\subset$ and $\in$ can be used interchangeably.
Thus:
{{begin-eqn}}
{{eqn | l = x \subseteq y
| o = \leadsto
| r = x \subs... | Let $x$ and $y$ be [[Definition:Ordinal|ordinals]].
Let $y^+$ denote the [[Definition:Successor Set|successor]] of $y$.
Then:
: $x \subseteq y^+ \iff \left({x \subseteq y \lor x = y^+}\right)$ | Let $A \subset B$ denote that $A$ is a [[Definition:Proper Subset|proper subset]] of $B$.
Let $A \in B$ denote that $A$ is an [[Definition:Element|element]] of $B$.
From [[Transitive Set is Proper Subset of Ordinal iff Element of Ordinal]], $\subset$ and $\in$ can be used interchangeably.
Thus:
{{begin-eqn}}
{{eqn ... | Ordinal is Subset of Successor | https://proofwiki.org/wiki/Ordinal_is_Subset_of_Successor | https://proofwiki.org/wiki/Ordinal_is_Subset_of_Successor | [
"Ordinals"
] | [
"Definition:Ordinal",
"Definition:Successor Mapping/Successor Set"
] | [
"Definition:Proper Subset",
"Definition:Element",
"Transitive Set is Proper Subset of Ordinal iff Element of Ordinal",
"Ordinal is Less than Successor",
"Definition:Converse Statement",
"Transitive Set is Proper Subset of Ordinal iff Element of Ordinal",
"Transitive Set is Proper Subset of Ordinal iff E... |
proofwiki-5802 | Indexed Union Equality | Let $A$, $B_x$, and $C_x$ be classes.
Then:
:$\ds \forall x \in A: B_x = C_x \implies \bigcup_{x \mathop \in A} B_x = \bigcup_{x \mathop \in A} C_x$
{{MissingLinks|$\bigcup_{x \mathop \in A} B_x$ and $\bigcup_{x \mathop \in A} C_x$}} | Proof follows from Indexed Union Subset and definition of set equality.
{{handwaving}}
Category:Set Union
sd06p7qs0nodpnwpepko0zs1snkr7jq | Let $A$, $B_x$, and $C_x$ be [[Definition:Class (Class Theory)|classes]].
Then:
:$\ds \forall x \in A: B_x = C_x \implies \bigcup_{x \mathop \in A} B_x = \bigcup_{x \mathop \in A} C_x$
{{MissingLinks|$\bigcup_{x \mathop \in A} B_x$ and $\bigcup_{x \mathop \in A} C_x$}} | Proof follows from [[Indexed Union Subset]] and definition of [[Definition:Set Equality/Definition 2|set equality]].
{{handwaving}}
[[Category:Set Union]]
sd06p7qs0nodpnwpepko0zs1snkr7jq | Indexed Union Equality | https://proofwiki.org/wiki/Indexed_Union_Equality | https://proofwiki.org/wiki/Indexed_Union_Equality | [
"Set Union"
] | [
"Definition:Class (Class Theory)"
] | [
"Indexed Union Subset",
"Definition:Set Equality/Definition 2",
"Category:Set Union"
] |
proofwiki-5803 | Indexed Union Subset | Let $A$, $B_x$ and $C_x$ be classes.
{{explain|It can be inferred from the context, but the meaning of the subscript on $B_x$ and $C_x$ needs to be explained.}}
Then:
:$\ds \forall x \in A: B_x \subseteq C_x \implies \bigcup_{x \mathop \in A} B_x \subseteq \bigcup_{x \mathop \in A} C_x$
{{MissingLinks|$\bigcup_{x \math... | {{proof wanted}}
Category:Set Union
nnaxj5lt0hhbj4gyixie2ef4rrqsl5e | Let $A$, $B_x$ and $C_x$ be [[Definition:Class (Class Theory)|classes]].
{{explain|It can be inferred from the context, but the meaning of the subscript on $B_x$ and $C_x$ needs to be explained.}}
Then:
:$\ds \forall x \in A: B_x \subseteq C_x \implies \bigcup_{x \mathop \in A} B_x \subseteq \bigcup_{x \mathop \in A}... | {{proof wanted}}
[[Category:Set Union]]
nnaxj5lt0hhbj4gyixie2ef4rrqsl5e | Indexed Union Subset | https://proofwiki.org/wiki/Indexed_Union_Subset | https://proofwiki.org/wiki/Indexed_Union_Subset | [
"Set Union"
] | [
"Definition:Class (Class Theory)"
] | [
"Category:Set Union"
] |
proofwiki-5804 | Ordinal is Less than Sum | Let $x$ and $y$ be ordinals.
Then:
:$x \le \paren {x + y}$
:$x \le \paren {y + x}$ | By Proof by Cases, one of the following holds by Empty Set is Subset of All Sets:
:$\O < y$
:$y = \O$
By Ordinal Addition by Zero:
:$x = \paren {x + \O} = \paren {\O + x}$
By Membership is Left Compatible with Ordinal Addition:
:$\O < y \implies x < \paren {x + y}$
But if $y = \O$, then it is clear the inequality $x \l... | Let $x$ and $y$ be [[Definition:Ordinal|ordinals]].
Then:
:$x \le \paren {x + y}$
:$x \le \paren {y + x}$ | By [[Proof by Cases]], one of the following holds by [[Empty Set is Subset of All Sets]]:
:$\O < y$
:$y = \O$
By [[Ordinal Addition by Zero]]:
:$x = \paren {x + \O} = \paren {\O + x}$
By [[Membership is Left Compatible with Ordinal Addition]]:
:$\O < y \implies x < \paren {x + y}$
But if $y = \O$, then it is cle... | Ordinal is Less than Sum | https://proofwiki.org/wiki/Ordinal_is_Less_than_Sum | https://proofwiki.org/wiki/Ordinal_is_Less_than_Sum | [
"Ordinal Arithmetic"
] | [
"Definition:Ordinal"
] | [
"Proof by Cases",
"Empty Set is Subset of All Sets",
"Ordinal Addition by Zero",
"Membership is Left Compatible with Ordinal Addition",
"Subset is Right Compatible with Ordinal Addition",
"Empty Set is Subset of All Sets",
"Category:Ordinal Arithmetic"
] |
proofwiki-5805 | Limit Ordinals Preserved Under Ordinal Addition | Let $x$ and $y$ be ordinals such that $x$ is a limit ordinal.
Then $y + x$ is a limit ordinal.
That is, letting $K_{II}$ denote the class of all limit ordinals:
:$\forall x \in K_{II}: y + x \in K_{II}$ | {{begin-eqn}}
{{eqn | l = x
| o = \in
| r = K_{II}
}}
{{eqn | ll= \leadsto
| l = x
| o = \ne
| r = \O
| c = {{Defof|Limit Ordinal}}
}}
{{eqn | ll= \leadsto
| l = y + x
| o = \ne
| r = \O
| c = Ordinal is Less than Sum
}}
{{eqn | ll= \leadsto
| l = y + x
... | Let $x$ and $y$ be [[Definition:Ordinal|ordinals]] such that $x$ is a [[Definition:Limit Ordinal|limit ordinal]].
Then $y + x$ is a [[Definition:Limit Ordinal|limit ordinal]].
That is, letting $K_{II}$ denote the [[Definition:Class (Class Theory)|class]] of all [[Definition:Limit Ordinal|limit ordinals]]:
:$\forall... | {{begin-eqn}}
{{eqn | l = x
| o = \in
| r = K_{II}
}}
{{eqn | ll= \leadsto
| l = x
| o = \ne
| r = \O
| c = {{Defof|Limit Ordinal}}
}}
{{eqn | ll= \leadsto
| l = y + x
| o = \ne
| r = \O
| c = [[Ordinal is Less than Sum]]
}}
{{eqn | ll= \leadsto
| l = y ... | Limit Ordinals Preserved Under Ordinal Addition | https://proofwiki.org/wiki/Limit_Ordinals_Preserved_Under_Ordinal_Addition | https://proofwiki.org/wiki/Limit_Ordinals_Preserved_Under_Ordinal_Addition | [
"Ordinal Arithmetic",
"Limit Ordinals"
] | [
"Definition:Ordinal",
"Definition:Limit Ordinal",
"Definition:Limit Ordinal",
"Definition:Class (Class Theory)",
"Definition:Limit Ordinal"
] | [
"Ordinal is Less than Sum",
"Proof by Contradiction",
"Ordinal is Less than Successor",
"Successor of Ordinal Smaller than Limit Ordinal is also Smaller",
"Successor is Less than Successor",
"No Membership Loops",
"Modus Tollendo Ponens"
] |
proofwiki-5806 | Successor is Less than Successor | Let $x$ and $y$ be ordinals and let $x^+$ denote the successor set of $x$.
Then, $x \in y \iff x^+ \in y^+$. | {{begin-eqn}}
{{eqn | l = x \in y
| o = \implies
| r = x^+ \in y^+
| c = Subset is Compatible with Ordinal Successor
}}
{{eqn | l = x \in y
| o = \impliedby
| r = x^+ \in y^+
| c = Sufficient Condition
}}
{{eqn | ll= \leadsto
| l = x \in y
| o = \iff
| r = x^+ \in y... | Let $x$ and $y$ be [[Definition:Ordinal|ordinals]] and let $x^+$ denote the [[Definition:Successor Set|successor set]] of $x$.
Then, $x \in y \iff x^+ \in y^+$. | {{begin-eqn}}
{{eqn | l = x \in y
| o = \implies
| r = x^+ \in y^+
| c = [[Subset is Compatible with Ordinal Successor]]
}}
{{eqn | l = x \in y
| o = \impliedby
| r = x^+ \in y^+
| c = [[Successor is Less than Successor/Sufficient Condition|Sufficient Condition]]
}}
{{eqn | ll= \lead... | Successor is Less than Successor | https://proofwiki.org/wiki/Successor_is_Less_than_Successor | https://proofwiki.org/wiki/Successor_is_Less_than_Successor | [
"Ordinals"
] | [
"Definition:Ordinal",
"Definition:Successor Mapping/Successor Set"
] | [
"Subset is Compatible with Ordinal Successor",
"Successor is Less than Successor/Sufficient Condition",
"Category:Ordinals"
] |
proofwiki-5807 | Successor is Less than Successor | Let $x$ and $y$ be ordinals and let $x^+$ denote the successor set of $x$.
Then, $x \in y \iff x^+ \in y^+$. | Suppose $y^+ \in x^+$.
By the definition of successor, $y^+ \in x \lor y^+ = x$.
Suppose $y^+ = x$.
By Ordinal is Less than Successor, $y \in x$.
Suppose $y^+ \in x$.
By Ordinal is Less than Successor, $y \in y^+$.
By Ordinal is Transitive, $y \in x$. | Let $x$ and $y$ be [[Definition:Ordinal|ordinals]] and let $x^+$ denote the [[Definition:Successor Set|successor set]] of $x$.
Then, $x \in y \iff x^+ \in y^+$. | Suppose $y^+ \in x^+$.
By the definition of [[Definition:Successor Set|successor]], $y^+ \in x \lor y^+ = x$.
Suppose $y^+ = x$.
By [[Ordinal is Less than Successor]], $y \in x$.
Suppose $y^+ \in x$.
By [[Ordinal is Less than Successor]], $y \in y^+$.
By [[Ordinal is Transitive]], $y \in x$. | Successor is Less than Successor/Sufficient Condition/Proof 1 | https://proofwiki.org/wiki/Successor_is_Less_than_Successor | https://proofwiki.org/wiki/Successor_is_Less_than_Successor/Sufficient_Condition/Proof_1 | [
"Ordinals"
] | [
"Definition:Ordinal",
"Definition:Successor Mapping/Successor Set"
] | [
"Definition:Successor Mapping/Successor Set",
"Ordinal is Less than Successor",
"Ordinal is Less than Successor",
"Ordinal is Transitive"
] |
proofwiki-5808 | Successor is Less than Successor | Let $x$ and $y$ be ordinals and let $x^+$ denote the successor set of $x$.
Then, $x \in y \iff x^+ \in y^+$. | First note that by Successor Set of Ordinal is Ordinal, $x^+$ and $y^+$ are ordinals.
Let $x^+ \in y^+$.
Then since $y^+$ is transitive, $x^+ \subseteq y^+$.
Thus $x \in y$ or $x = y$.
If $x = y$ then $x^+ \in x^+$, contradicting Ordinal is not Element of Itself.
Thus $x \in y$. | Let $x$ and $y$ be [[Definition:Ordinal|ordinals]] and let $x^+$ denote the [[Definition:Successor Set|successor set]] of $x$.
Then, $x \in y \iff x^+ \in y^+$. | First note that by [[Successor Set of Ordinal is Ordinal]], $x^+$ and $y^+$ are [[Definition:Ordinal|ordinals]].
Let $x^+ \in y^+$.
Then since $y^+$ is transitive, $x^+ \subseteq y^+$.
Thus $x \in y$ or $x = y$.
If $x = y$ then $x^+ \in x^+$, contradicting [[Ordinal is not Element of Itself]].
Thus $x \in y$. | Successor is Less than Successor/Sufficient Condition/Proof 2 | https://proofwiki.org/wiki/Successor_is_Less_than_Successor | https://proofwiki.org/wiki/Successor_is_Less_than_Successor/Sufficient_Condition/Proof_2 | [
"Ordinals"
] | [
"Definition:Ordinal",
"Definition:Successor Mapping/Successor Set"
] | [
"Successor Set of Ordinal is Ordinal",
"Definition:Ordinal",
"Ordinal is not Element of Itself"
] |
proofwiki-5809 | Ordinal Addition is Associative | Ordinal addition is associative, i.e.:
:$\paren {x + y} + z = x + \paren {y + z}$
holds for all ordinals $x$, $y$ and $z$. | By Transfinite Induction on $z$. | [[Definition:Ordinal Addition|Ordinal addition]] is [[Definition:Associative Operation|associative]], i.e.:
:$\paren {x + y} + z = x + \paren {y + z}$
holds for all [[Definition:Ordinal|ordinals]] $x$, $y$ and $z$. | By [[Transfinite Induction/Schema 2|Transfinite Induction]] on $z$. | Ordinal Addition is Associative | https://proofwiki.org/wiki/Ordinal_Addition_is_Associative | https://proofwiki.org/wiki/Ordinal_Addition_is_Associative | [
"Ordinal Arithmetic"
] | [
"Definition:Ordinal Addition",
"Definition:Associative Operation",
"Definition:Ordinal"
] | [
"Transfinite Induction/Schema 2",
"Transfinite Induction/Schema 2",
"Transfinite Induction/Schema 2",
"Transfinite Induction/Schema 2",
"Transfinite Induction/Schema 2"
] |
proofwiki-5810 | Derivative of Composite Function/Corollary | :$\dfrac {\d y} {\d x} = \dfrac {\paren {\dfrac {\d y} {\d u} } } {\paren {\dfrac {\d x} {\d u} } }$
for $\dfrac {\d x} {\d u} \ne 0$. | {{begin-eqn}}
{{eqn | l = \frac {\d y} {\d x} \frac {\d x} {\d u}
| r = \frac {\d y} {\d u}
| c = Derivative of Composite Function
}}
{{eqn | ll= \leadsto
| l = \frac {\d y} {\d x}
| r = \frac {\paren {\dfrac {\d y} {\d u} } } {\paren {\dfrac {\d x}{\d u} } }
| c = dividing both sides by $... | :$\dfrac {\d y} {\d x} = \dfrac {\paren {\dfrac {\d y} {\d u} } } {\paren {\dfrac {\d x} {\d u} } }$
for $\dfrac {\d x} {\d u} \ne 0$. | {{begin-eqn}}
{{eqn | l = \frac {\d y} {\d x} \frac {\d x} {\d u}
| r = \frac {\d y} {\d u}
| c = [[Derivative of Composite Function]]
}}
{{eqn | ll= \leadsto
| l = \frac {\d y} {\d x}
| r = \frac {\paren {\dfrac {\d y} {\d u} } } {\paren {\dfrac {\d x}{\d u} } }
| c = dividing both sides ... | Derivative of Composite Function/Corollary | https://proofwiki.org/wiki/Derivative_of_Composite_Function/Corollary | https://proofwiki.org/wiki/Derivative_of_Composite_Function/Corollary | [
"Derivative of Composite Function"
] | [] | [
"Derivative of Composite Function"
] |
proofwiki-5811 | Product of Subset with Intersection/Corollary | Let $\struct {G, \circ}$ be a group.
Let $X, Y, Z \subseteq G$ such that $X$ is a singleton.
Then:
:$X \circ \paren {Y \cap Z} = \paren {X \circ Y} \cap \paren {X \circ Z}$
:$\paren {Y \cap Z} \circ X = \paren {Y \circ X} \cap \paren {Z \circ X}$
where $X \circ Y$ denotes the subset product of $X$ and $Y$. | Let $X = \set x$.
We have from Product of Subset with Intersection that:
: $X \circ \paren {Y \cap Z} \subseteq \paren {X \circ Y} \cap \paren {X \circ Z}$
Let $g \in \paren {X \circ Y} \cap \paren {X \circ Z}$.
Then by definition of subset product:
: $\exists y \in Y, z \in Z: g = x \circ y = x \circ z$
By the Cancell... | Let $\struct {G, \circ}$ be a [[Definition:Group|group]].
Let $X, Y, Z \subseteq G$ such that $X$ is a [[Definition:Singleton|singleton]].
Then:
:$X \circ \paren {Y \cap Z} = \paren {X \circ Y} \cap \paren {X \circ Z}$
:$\paren {Y \cap Z} \circ X = \paren {Y \circ X} \cap \paren {Z \circ X}$
where $X \circ Y$ denote... | Let $X = \set x$.
We have from [[Product of Subset with Intersection]] that:
: $X \circ \paren {Y \cap Z} \subseteq \paren {X \circ Y} \cap \paren {X \circ Z}$
Let $g \in \paren {X \circ Y} \cap \paren {X \circ Z}$.
Then by definition of [[Definition:Subset Product|subset product]]:
: $\exists y \in Y, z \in Z: g =... | Product of Subset with Intersection/Corollary | https://proofwiki.org/wiki/Product_of_Subset_with_Intersection/Corollary | https://proofwiki.org/wiki/Product_of_Subset_with_Intersection/Corollary | [
"Product of Subset with Intersection"
] | [
"Definition:Group",
"Definition:Singleton",
"Definition:Subset Product"
] | [
"Product of Subset with Intersection",
"Definition:Subset Product",
"Cancellation Laws",
"Definition:Set Equality/Definition 2"
] |
proofwiki-5812 | Single Instruction URM Programs/Zero Function | The zero function $\Zero: \N \to \N$, defined as:
:$\forall n \in \N: \map \Zero n = 0$ | The zero function is computed by the following URM program:
{|
|-
! align="right" | Line !!
! align="left" | Command
|-
| align="right" | $1$ ||
| align="left" | $\map Z 1$
|}
This sets the value $0$ into $R_1$ and then stops.
The output $0$ is in $R_1$ when the program terminates.
{{qed}}
Category:URM Programs
Categor... | The [[Definition:Zero Function|zero function]] $\Zero: \N \to \N$, defined as:
:$\forall n \in \N: \map \Zero n = 0$ | The [[Definition:Zero Function|zero function]] is computed by the following [[Definition:URM Program|URM program]]:
{|
|-
! align="right" | Line !!
! align="left" | Command
|-
| align="right" | $1$ ||
| align="left" | $\map Z 1$
|}
This sets the value $0$ into $R_1$ and then stops.
The [[Definition:Unlimited Registe... | Single Instruction URM Programs/Zero Function | https://proofwiki.org/wiki/Single_Instruction_URM_Programs/Zero_Function | https://proofwiki.org/wiki/Single_Instruction_URM_Programs/Zero_Function | [
"URM Programs",
"Primitive Recursive Functions"
] | [
"Definition:Basic Primitive Recursive Function/Zero Function"
] | [
"Definition:Basic Primitive Recursive Function/Zero Function",
"Definition:Unlimited Register Machine/Program",
"Definition:Unlimited Register Machine",
"Category:URM Programs",
"Category:Primitive Recursive Functions"
] |
proofwiki-5813 | Single Instruction URM Programs/Successor Function | The successor function $\Succ: \N \to \N$, defined as:
:$\forall n \in \N: \map \Succ n = n + 1$ | The successor function is computed by the following URM program:
{|
|-
! align="right" | Line !!
! align="left" | Command
|-
| align="right" | $1$ ||
| align="left" | $\map S 1$
|}
The input $n$ is in $R_1$ when the program starts.
The program adds $1$ to $r_1$ and then stops.
The output $n + 1$ is in $R_1$ when the pr... | The [[Definition:Successor Function|successor function]] $\Succ: \N \to \N$, defined as:
:$\forall n \in \N: \map \Succ n = n + 1$ | The [[Definition:Successor Function|successor function]] is computed by the following [[Definition:URM Program|URM program]]:
{|
|-
! align="right" | Line !!
! align="left" | Command
|-
| align="right" | $1$ ||
| align="left" | $\map S 1$
|}
The [[Definition:Unlimited Register Machine#Input|input]] $n$ is in $R_1$ wh... | Single Instruction URM Programs/Successor Function | https://proofwiki.org/wiki/Single_Instruction_URM_Programs/Successor_Function | https://proofwiki.org/wiki/Single_Instruction_URM_Programs/Successor_Function | [
"URM Programs",
"Primitive Recursive Functions"
] | [
"Definition:Basic Primitive Recursive Function/Successor Function"
] | [
"Definition:Basic Primitive Recursive Function/Successor Function",
"Definition:Unlimited Register Machine/Program",
"Definition:Unlimited Register Machine",
"Definition:Unlimited Register Machine",
"Category:URM Programs",
"Category:Primitive Recursive Functions"
] |
proofwiki-5814 | Single Instruction URM Programs/Projection Function | The projection functions $\pr_j^k: \N^k \to \N$, defined as:
:$\forall j \in \closedint 1 k: \forall \tuple {n_1, n_2, \ldots, n_k} \in \N^k: \map {\pr_j^k} {n_1, n_2, \ldots, n_k} = n_j$ | The projection functions are computed by the following URM program:
{|
|-
! align="right" | Line !!
! align="left" | Command
|-
| align="right" | $1$ ||
| align="left" | $\map C {j, 1}$
|}
The input $\tuple {n_1, n_2, \ldots, n_j, \ldots, n_k}$ is in $R_1, R_2, \ldots, R_j, \ldots, R_k$ when the program starts.
The pro... | The [[Definition:Projection Function|projection functions]] $\pr_j^k: \N^k \to \N$, defined as:
:$\forall j \in \closedint 1 k: \forall \tuple {n_1, n_2, \ldots, n_k} \in \N^k: \map {\pr_j^k} {n_1, n_2, \ldots, n_k} = n_j$ | The [[Definition:Projection Function|projection functions]] are computed by the following [[Definition:URM Program|URM program]]:
{|
|-
! align="right" | Line !!
! align="left" | Command
|-
| align="right" | $1$ ||
| align="left" | $\map C {j, 1}$
|}
The [[Definition:Unlimited Register Machine#Input|input]] $\tuple {... | Single Instruction URM Programs/Projection Function | https://proofwiki.org/wiki/Single_Instruction_URM_Programs/Projection_Function | https://proofwiki.org/wiki/Single_Instruction_URM_Programs/Projection_Function | [
"URM Programs",
"Primitive Recursive Functions"
] | [
"Definition:Basic Primitive Recursive Function/Projection Function"
] | [
"Definition:Basic Primitive Recursive Function/Projection Function",
"Definition:Unlimited Register Machine/Program",
"Definition:Unlimited Register Machine",
"Definition:Unlimited Register Machine",
"Category:URM Programs",
"Category:Primitive Recursive Functions"
] |
proofwiki-5815 | Single Instruction URM Programs/Identity Function | The identity function $I_\N: \N \to \N$ defined as:
:$\forall n \in \N: \map {I_\N} n = n$ | Any of the following URM programs compute the identity function:
{|
|-
! align="right" | Line !!
! align="left" | Command
|-
| align="right" | $1$ ||
| align="left" | $\map Z m$
|}
... where $m \ne 1$.
This sets the value $0$ into $R_m$ and then stops.
{|
|-
! align="right" | Line !!
! align="left" | Command
|-
| align... | The [[Definition:Basic Primitive Recursive Function/Identity Function|identity function]] $I_\N: \N \to \N$ defined as:
:$\forall n \in \N: \map {I_\N} n = n$ | Any of the following [[Definition:URM Program|URM programs]] compute the [[Definition:Basic Primitive Recursive Function/Identity Function|identity function]]:
{|
|-
! align="right" | Line !!
! align="left" | Command
|-
| align="right" | $1$ ||
| align="left" | $\map Z m$
|}
... where $m \ne 1$.
This sets the value $... | Single Instruction URM Programs/Identity Function | https://proofwiki.org/wiki/Single_Instruction_URM_Programs/Identity_Function | https://proofwiki.org/wiki/Single_Instruction_URM_Programs/Identity_Function | [
"URM Programs",
"Primitive Recursive Functions"
] | [
"Definition:Basic Primitive Recursive Function/Identity Function"
] | [
"Definition:Unlimited Register Machine/Program",
"Definition:Basic Primitive Recursive Function/Identity Function",
"Definition:Unlimited Register Machine",
"Definition:Unlimited Register Machine",
"Definition:Unlimited Register Machine",
"Definition:Unlimited Register Machine",
"Definition:Unlimited Re... |
proofwiki-5816 | Antiassociative Operation has no Idempotent Elements | Let $\struct {S, \circ}$ be an algebraic structure.
Let $\circ$ be antiassociative on $S$.
Then no element of $S$ is idempotent under $ \circ$.
That is:
:$\forall x \in S: x \circ x \ne x$ | {{AimForCont}} $a \in S$ such that $a$ is idempotent under $\circ$.
That is:
:$a \circ a = a$
Then:
:$\paren {a \circ a} \circ a = a \circ a$
and
:$a \circ \paren {a \circ a} = a \circ a$
This contradicts our assumption that $\circ$ is antiassociative on $S$.
{{qed}}
Category:Antiassociative Structures
octnc43gusmgbg... | Let $\struct {S, \circ}$ be an [[Definition: Algebraic Structure with One Operation|algebraic structure]].
Let $\circ$ be [[Definition:Antiassociative Operation|antiassociative]] on $S$.
Then no [[Definition: Element|element]] of $S$ is [[Definition:Idempotent Element|idempotent]] under $ \circ$.
That is:
:$\foral... | {{AimForCont}} $a \in S$ such that $a$ is [[Definition:Idempotent Element|idempotent]] under $\circ$.
That is:
:$a \circ a = a$
Then:
:$\paren {a \circ a} \circ a = a \circ a$
and
:$a \circ \paren {a \circ a} = a \circ a$
This [[Proof by Contradiction|contradicts]] our [[Definition:Assumption|assumption]] tha... | Antiassociative Operation has no Idempotent Elements | https://proofwiki.org/wiki/Antiassociative_Operation_has_no_Idempotent_Elements | https://proofwiki.org/wiki/Antiassociative_Operation_has_no_Idempotent_Elements | [
"Antiassociative Structures"
] | [
"Definition: Algebraic Structure with One Operation",
"Definition:Antiassociative Operation",
"Definition: Element",
"Definition:Idempotence/Element"
] | [
"Definition:Idempotence/Element",
"Proof by Contradiction",
"Definition:Assumption",
"Definition:Antiassociative Operation",
"Category:Antiassociative Structures"
] |
proofwiki-5817 | Antiassociative Operation is not Commutative | Let $\struct {S, \circ}$ be an algebraic structure.
Let $\circ$ be antiassociative on $S$.
Then $\circ$ is not commutative on $S$. | We will show there are two elements in $S$ that do not commute.
Let $a \in S$.
From Antiassociative Operation has no Idempotent Elements:
:$a \circ a \ne a$
So for some $b \in S$:
:$a \circ a = b$
Then:
:$\paren {a \circ a} \circ a = b \circ a$
and:
:$a \circ \paren {a \circ a} = a \circ b$
From our assumption, $\circ... | Let $\struct {S, \circ}$ be an [[Definition:Algebraic Structure with One Operation|algebraic structure]].
Let $\circ$ be [[Definition:Antiassociative Operation|antiassociative]] on $S$.
Then $\circ$ is not [[Definition:Commutative Operation|commutative]] on $S$. | We will show there are two [[Definition:Element|elements]] in $S$ that do not [[Definition:Commuting Elements|commute]].
Let $a \in S$.
From [[Antiassociative Operation has no Idempotent Elements]]:
:$a \circ a \ne a$
So for some $b \in S$:
:$a \circ a = b$
Then:
:$\paren {a \circ a} \circ a = b \circ a$
and:
:$a... | Antiassociative Operation is not Commutative | https://proofwiki.org/wiki/Antiassociative_Operation_is_not_Commutative | https://proofwiki.org/wiki/Antiassociative_Operation_is_not_Commutative | [
"Antiassociative Structures"
] | [
"Definition:Algebraic Structure/One Operation",
"Definition:Antiassociative Operation",
"Definition:Commutative/Operation"
] | [
"Definition:Element",
"Definition:Commutative/Elements",
"Antiassociative Operation has no Idempotent Elements",
"Definition:Assumption",
"Definition:Antiassociative Operation",
"Category:Antiassociative Structures"
] |
proofwiki-5818 | Antiassociative Structure of Finite Order | Let $n \in \N$ such that $n > 2$.
Then there exists an algebraic structure $\struct {S, \circ}$ of order $n$ such that $\circ$ is antiassociative on $S$. | Let $S = \set {a_0, \ldots, a_{n - 1} }$.
Let $\circ$ be a binary operation defined on $S$ such that:
:$\forall x \in S: x \circ a_i = a_{\paren {i + 1} \pmod n}$
Then $\forall j, k, m \in \closedint 0 {n - 1}$:
{{begin-eqn}}
{{eqn | l = \paren {a_j \circ a_k} \circ a_m
| r = a_{\paren {m + 1} \pmod n}
}}
{{end-e... | Let $n \in \N$ such that $n > 2$.
Then there exists an [[Definition:Algebraic Structure with One Operation|algebraic structure]] $\struct {S, \circ}$ of [[Definition:Order of Structure|order]] $n$ such that $\circ$ is [[Definition:Antiassociative Operation|antiassociative]] on $S$. | Let $S = \set {a_0, \ldots, a_{n - 1} }$.
Let $\circ$ be a [[Definition:Binary Operation|binary operation]] defined on $S$ such that:
:$\forall x \in S: x \circ a_i = a_{\paren {i + 1} \pmod n}$
Then $\forall j, k, m \in \closedint 0 {n - 1}$:
{{begin-eqn}}
{{eqn | l = \paren {a_j \circ a_k} \circ a_m
| r = ... | Antiassociative Structure of Finite Order | https://proofwiki.org/wiki/Antiassociative_Structure_of_Finite_Order | https://proofwiki.org/wiki/Antiassociative_Structure_of_Finite_Order | [
"Antiassociative Structures"
] | [
"Definition:Algebraic Structure/One Operation",
"Definition:Order of Structure",
"Definition:Antiassociative Operation"
] | [
"Definition:Operation/Binary Operation",
"Category:Antiassociative Structures"
] |
proofwiki-5819 | Example:Antiassociative Structure | Let $\struct {\R_{>0}, \circ}$ be an algebraic structure where $\R_{>0}$ denotes the strictly positive real numbers.
Define:
:$\forall x, y \in \R_{>0}: x \circ y = x y + y$
Then $\circ$ is antiassociative on $\R_{>0}$:
:$\forall x, y, z \in \R_{>0}: \paren {x \circ y} \circ z \ne x \circ \paren {y \circ z}$
and so $\s... | Let $a, b, c \in \R_{>0}$:
Then:
{{begin-eqn}}
{{eqn | l = a \circ \paren {b \circ c}
| r = a \circ \paren {b c + c}
}}
{{eqn | r = a \paren {b c + c} + b c + c
}}
{{eqn | r = a b c + a c + b c + c
}}
{{end-eqn}}
and:
{{begin-eqn}}
{{eqn | l = \paren {a \circ b} \circ c
| r = \paren {a b + b} \circ c
}}
{{e... | Let $\struct {\R_{>0}, \circ}$ be an [[Definition:Algebraic Structure|algebraic structure]] where $\R_{>0}$ denotes the [[Definition:Strictly Positive Real Number|strictly positive real numbers]].
Define:
:$\forall x, y \in \R_{>0}: x \circ y = x y + y$
Then $\circ$ is [[Definition:Antiassociative Operation|antiass... | Let $a, b, c \in \R_{>0}$:
Then:
{{begin-eqn}}
{{eqn | l = a \circ \paren {b \circ c}
| r = a \circ \paren {b c + c}
}}
{{eqn | r = a \paren {b c + c} + b c + c
}}
{{eqn | r = a b c + a c + b c + c
}}
{{end-eqn}}
and:
{{begin-eqn}}
{{eqn | l = \paren {a \circ b} \circ c
| r = \paren {a b + b} \circ c
}... | Example:Antiassociative Structure | https://proofwiki.org/wiki/Example:Antiassociative_Structure | https://proofwiki.org/wiki/Example:Antiassociative_Structure | [
"Examples of Abstract Algebra"
] | [
"Definition:Algebraic Structure",
"Definition:Strictly Positive/Real Number",
"Definition:Antiassociative Operation",
"Definition:Antiassociative Structure"
] | [
"Category:Examples of Abstract Algebra"
] |
proofwiki-5820 | Reciprocal Sequence is Strictly Decreasing | The reciprocal sequence:
:$\sequence {\operatorname {recip} }: \N_{>0} \to \R$: $n \mapsto \dfrac 1 n$
is strictly decreasing. | Follows from Reciprocal Function is Strictly Decreasing and from Restriction of Monotone Function is Monotone.
{{qed}} | The [[Definition:Reciprocal|reciprocal]] [[Definition:Real Sequence|sequence]]:
:$\sequence {\operatorname {recip} }: \N_{>0} \to \R$: $n \mapsto \dfrac 1 n$
is [[Definition:Strictly Decreasing Real Sequence|strictly decreasing]]. | Follows from [[Reciprocal Function is Strictly Decreasing]] and from [[Restriction of Monotone Function is Monotone]].
{{qed}} | Reciprocal Sequence is Strictly Decreasing/Proof 1 | https://proofwiki.org/wiki/Reciprocal_Sequence_is_Strictly_Decreasing | https://proofwiki.org/wiki/Reciprocal_Sequence_is_Strictly_Decreasing/Proof_1 | [
"Reciprocal Sequence is Strictly Decreasing",
"Sequences",
"Real Analysis",
"Reciprocals"
] | [
"Definition:Reciprocal",
"Definition:Real Sequence",
"Definition:Strictly Decreasing/Sequence/Real Sequence"
] | [
"Reciprocal Function is Strictly Decreasing",
"Restriction of Monotone Function is Monotone"
] |
proofwiki-5821 | Reciprocal Sequence is Strictly Decreasing | The reciprocal sequence:
:$\sequence {\operatorname {recip} }: \N_{>0} \to \R$: $n \mapsto \dfrac 1 n$
is strictly decreasing. | Let $n \in \N_{>0}$.
{{begin-eqn}}
{{eqn | l = \frac 1 n - \frac 1 {n + 1}
| r = \frac {\paren {n + 1} - n} {n \paren {n + 1} }
}}
{{eqn | r = \frac 1 {n^2 + n}
}}
{{eqn | o = >
| r = 0
}}
{{eqn | ll= \leadsto
| l = \frac 1 n
| o = >
| r = \frac 1 {n + 1}
}}
{{end-eqn}}
Hence the result, a... | The [[Definition:Reciprocal|reciprocal]] [[Definition:Real Sequence|sequence]]:
:$\sequence {\operatorname {recip} }: \N_{>0} \to \R$: $n \mapsto \dfrac 1 n$
is [[Definition:Strictly Decreasing Real Sequence|strictly decreasing]]. | Let $n \in \N_{>0}$.
{{begin-eqn}}
{{eqn | l = \frac 1 n - \frac 1 {n + 1}
| r = \frac {\paren {n + 1} - n} {n \paren {n + 1} }
}}
{{eqn | r = \frac 1 {n^2 + n}
}}
{{eqn | o = >
| r = 0
}}
{{eqn | ll= \leadsto
| l = \frac 1 n
| o = >
| r = \frac 1 {n + 1}
}}
{{end-eqn}}
Hence the result,... | Reciprocal Sequence is Strictly Decreasing/Proof 2 | https://proofwiki.org/wiki/Reciprocal_Sequence_is_Strictly_Decreasing | https://proofwiki.org/wiki/Reciprocal_Sequence_is_Strictly_Decreasing/Proof_2 | [
"Reciprocal Sequence is Strictly Decreasing",
"Sequences",
"Real Analysis",
"Reciprocals"
] | [
"Definition:Reciprocal",
"Definition:Real Sequence",
"Definition:Strictly Decreasing/Sequence/Real Sequence"
] | [] |
proofwiki-5822 | Two-Step Vector Subspace Test | Let $V$ be a vector space over a division ring $K$.
Let $U \subseteq V$ be a non-empty subset of $V$ such that:
{{begin-eqn}}
{{eqn | n = 1
| q = \forall u \in U, \lambda \in K
| l = \lambda u
| o = \in
| r = U
}}
{{eqn | n = 2
| q = \forall u, v \in U
| l = u + v
| o = \in
... | Suppose that $(1)$ and $(2)$ hold.
From $(1)$, we obtain for every $\lambda \in K$ and $u \in U$ that $\lambda u \in U$.
An application of $(2)$ yields the condition of the One-Step Vector Subspace Test.
Hence $U$ is a vector subspace of $V$.
{{qed}} | Let $V$ be a [[Definition:Vector Space|vector space]] over a [[Definition:Division Ring|division ring]] $K$.
Let $U \subseteq V$ be a non-[[Definition:Empty Set|empty]] [[Definition:Subset|subset]] of $V$ such that:
{{begin-eqn}}
{{eqn | n = 1
| q = \forall u \in U, \lambda \in K
| l = \lambda u
| o... | Suppose that $(1)$ and $(2)$ hold.
From $(1)$, we obtain for every $\lambda \in K$ and $u \in U$ that $\lambda u \in U$.
An application of $(2)$ yields the condition of the [[One-Step Vector Subspace Test]].
Hence $U$ is a [[Definition:Vector Subspace|vector subspace]] of $V$.
{{qed}} | Two-Step Vector Subspace Test | https://proofwiki.org/wiki/Two-Step_Vector_Subspace_Test | https://proofwiki.org/wiki/Two-Step_Vector_Subspace_Test | [
"Linear Algebra",
"Vector Subspaces"
] | [
"Definition:Vector Space",
"Definition:Division Ring",
"Definition:Empty Set",
"Definition:Subset",
"Definition:Vector Subspace"
] | [
"One-Step Vector Subspace Test",
"Definition:Vector Subspace"
] |
proofwiki-5823 | Reciprocal Function is Strictly Decreasing | The reciprocal function:
:$\operatorname{recip}: \R \setminus \set 0 \to \R$, $x \mapsto \dfrac 1 x$
is strictly decreasing:
:on the open interval $\openint 0 \to$
:on the open interval $\openint \gets 0$ | Let $x \in \openint 0 {+\infty}$.
By the definition of negative powers:
:$\dfrac 1 x = x^{-1}$
From Power Rule for Derivatives:
{{begin-eqn}}
{{eqn | l = \frac \d {\d x} x^{-1}
| r = -x^{-2}
}}
{{end-eqn}}
From Square of Real Number is Non-Negative:
: $-x^{-2} < 0$
for all $x$ within the domain.
Thus from Derivat... | The [[Definition:Reciprocal|reciprocal]] [[Definition:Real Function|function]]:
:$\operatorname{recip}: \R \setminus \set 0 \to \R$, $x \mapsto \dfrac 1 x$
is [[Definition:Strictly Decreasing Real Function|strictly decreasing]]:
:on the [[Definition:Open Real Interval|open interval]] $\openint 0 \to$
:on the [[Defi... | Let $x \in \openint 0 {+\infty}$.
By the definition of [[Definition:Power (Algebra)|negative powers]]:
:$\dfrac 1 x = x^{-1}$
From [[Power Rule for Derivatives]]:
{{begin-eqn}}
{{eqn | l = \frac \d {\d x} x^{-1}
| r = -x^{-2}
}}
{{end-eqn}}
From [[Square of Real Number is Non-Negative]]:
: $-x^{-2} < 0$
for ... | Reciprocal Function is Strictly Decreasing/Proof 1 | https://proofwiki.org/wiki/Reciprocal_Function_is_Strictly_Decreasing | https://proofwiki.org/wiki/Reciprocal_Function_is_Strictly_Decreasing/Proof_1 | [
"Real Analysis",
"Reciprocals",
"Reciprocal Function is Strictly Decreasing"
] | [
"Definition:Reciprocal",
"Definition:Real Function",
"Definition:Strictly Decreasing/Real Function",
"Definition:Real Interval/Open",
"Definition:Real Interval/Open"
] | [
"Definition:Power (Algebra)",
"Power Rule for Derivatives",
"Square of Real Number is Non-Negative",
"Definition:Domain (Set Theory)/Mapping",
"Derivative of Monotone Function",
"Definition:Strictly Decreasing/Real Function"
] |
proofwiki-5824 | Reciprocal Function is Strictly Decreasing | The reciprocal function:
:$\operatorname{recip}: \R \setminus \set 0 \to \R$, $x \mapsto \dfrac 1 x$
is strictly decreasing:
:on the open interval $\openint 0 \to$
:on the open interval $\openint \gets 0$ | === Strictly Increasing on $\openint 0 \to$ ===
Let $a, b \in \Dom {\operatorname {recip} }$ such that $a$ and $b$ are both (strictly) positive.
Let $a < b$.
Then $0 < a < b$.
By Properties of Totally Ordered Field:
:$0 < b^{-1} < a^{-1}$
That is:
:$\dfrac 1 b < \dfrac 1 a$
{{qed|lemma}}
=== Strictly Increasing on $\op... | The [[Definition:Reciprocal|reciprocal]] [[Definition:Real Function|function]]:
:$\operatorname{recip}: \R \setminus \set 0 \to \R$, $x \mapsto \dfrac 1 x$
is [[Definition:Strictly Decreasing Real Function|strictly decreasing]]:
:on the [[Definition:Open Real Interval|open interval]] $\openint 0 \to$
:on the [[Defi... | === Strictly Increasing on $\openint 0 \to$ ===
Let $a, b \in \Dom {\operatorname {recip} }$ such that $a$ and $b$ are both [[Definition:Strictly Positive Real Number|(strictly) positive]].
Let $a < b$.
Then $0 < a < b$.
By [[Properties of Totally Ordered Field]]:
:$0 < b^{-1} < a^{-1}$
That is:
:$\dfrac 1 b < \df... | Reciprocal Function is Strictly Decreasing/Proof 2 | https://proofwiki.org/wiki/Reciprocal_Function_is_Strictly_Decreasing | https://proofwiki.org/wiki/Reciprocal_Function_is_Strictly_Decreasing/Proof_2 | [
"Real Analysis",
"Reciprocals",
"Reciprocal Function is Strictly Decreasing"
] | [
"Definition:Reciprocal",
"Definition:Real Function",
"Definition:Strictly Decreasing/Real Function",
"Definition:Real Interval/Open",
"Definition:Real Interval/Open"
] | [
"Definition:Strictly Positive/Real Number",
"Properties of Ordered Field",
"Definition:Strictly Negative/Real Number",
"Inversion Mapping Reverses Ordering in Ordered Group",
"Properties of Ordered Field",
"Negative of Product Inverse",
"Inversion Mapping Reverses Ordering in Ordered Group",
"Inverse ... |
proofwiki-5825 | Coefficients of Polynomial Product | Let $J$ be a set.
Let $p_1, \ldots p_n$ be polynomial forms in the indeterminates $\set {X_j : j \in J}$ over a commutative ring $R$.
Suppose that for each $i$ with $1 \le i \le n$, we have, for appropriate $a_{i, k} \in R$:
:$p_i = \ds \sum_{k \mathop \in Z} a_{i, k} X^k$
where $Z$ comprises the multiindices of natura... | We proceed by induction over $n \ge 1$. | Let $J$ be a [[Definition:Set|set]].
Let $p_1, \ldots p_n$ be [[Definition:Polynomial Form|polynomial forms]] in the [[Definition:Indeterminate (Polynomial Theory)|indeterminates]] $\set {X_j : j \in J}$ over a [[Definition:Commutative Ring|commutative ring]] $R$.
Suppose that for each $i$ with $1 \le i \le n$, we ha... | We proceed by [[Principle of Mathematical Induction|induction]] over $n \ge 1$. | Coefficients of Polynomial Product | https://proofwiki.org/wiki/Coefficients_of_Polynomial_Product | https://proofwiki.org/wiki/Coefficients_of_Polynomial_Product | [
"Polynomial Theory",
"Proofs by Induction"
] | [
"Definition:Set",
"Definition:Polynomial over Ring as Function on Free Monoid on Set",
"Definition:Polynomial Ring/Indeterminate",
"Definition:Commutative Ring",
"Definition:Multiindex",
"Definition:Natural Numbers"
] | [
"Principle of Mathematical Induction",
"Principle of Mathematical Induction"
] |
proofwiki-5826 | Subset Product of Subgroups/Necessary Condition | Let $\left({G, \circ}\right)$ be a group.
Let $H, K$ be subgroups of $G$.
Let $H \circ K$ be a subgroup of $G$.
Then $H$ and $K$ are permutable.
That is:
:$H \circ K = K \circ H$
where $H \circ K$ denotes subset product. | Suppose $H \circ K$ is a subgroup of $G$.
Let $h \circ k \in H \circ K$.
Then $h \circ k$ is the inverse of some element $g$ of $H \circ K$.
Thus we can write $g = h' \circ k'$ for some $h' \in H$ and $k' \in K$.
So:
{{begin-eqn}}
{{eqn | l = h \circ k
| r = g^{-1}
| c = Inverse of Group Inverse: $g$ is the... | Let $\left({G, \circ}\right)$ be a [[Definition:Group|group]].
Let $H, K$ be [[Definition:Subgroup|subgroups]] of $G$.
Let $H \circ K$ be a [[Definition:Subgroup|subgroup]] of $G$.
Then $H$ and $K$ are [[Definition:Permutable Subgroups|permutable]].
That is:
:$H \circ K = K \circ H$
where $H \circ K$ denotes [[De... | Suppose $H \circ K$ is a [[Definition:Subgroup|subgroup]] of $G$.
Let $h \circ k \in H \circ K$.
Then $h \circ k$ is the [[Definition:Inverse Element|inverse]] of some [[Definition:Element|element]] $g$ of $H \circ K$.
Thus we can write $g = h' \circ k'$ for some $h' \in H$ and $k' \in K$.
So:
{{begin-eqn}}
{{eqn |... | Subset Product of Subgroups/Necessary Condition/Proof 1 | https://proofwiki.org/wiki/Subset_Product_of_Subgroups/Necessary_Condition | https://proofwiki.org/wiki/Subset_Product_of_Subgroups/Necessary_Condition/Proof_1 | [
"Subset Product of Subgroups"
] | [
"Definition:Group",
"Definition:Subgroup",
"Definition:Subgroup",
"Definition:Permutable Subgroups",
"Definition:Subset Product"
] | [
"Definition:Subgroup",
"Definition:Inverse (Abstract Algebra)/Inverse",
"Definition:Element",
"Inverse of Group Inverse",
"Definition:Inverse (Abstract Algebra)/Inverse",
"Inverse of Group Product",
"Definition:Group",
"Definition:Subset",
"Inverse of Group Inverse",
"Inverse of Group Product",
... |
proofwiki-5827 | Subset Product of Subgroups/Necessary Condition | Let $\left({G, \circ}\right)$ be a group.
Let $H, K$ be subgroups of $G$.
Let $H \circ K$ be a subgroup of $G$.
Then $H$ and $K$ are permutable.
That is:
:$H \circ K = K \circ H$
where $H \circ K$ denotes subset product. | Suppose $H \circ K$ is a subgroup of $G$.
Then:
{{begin-eqn}}
{{eqn | l = H \circ K
| r = \paren {H \circ K}^{-1}
| c = Inverse of Subgroup
}}
{{eqn | r = K^{-1} \circ H^{-1}
| c = Inverse of Product of Subsets of Group
}}
{{eqn | r = K \circ H
| c = Inverse of Subgroup
}}
{{end-eqn}}
{{qed}} | Let $\left({G, \circ}\right)$ be a [[Definition:Group|group]].
Let $H, K$ be [[Definition:Subgroup|subgroups]] of $G$.
Let $H \circ K$ be a [[Definition:Subgroup|subgroup]] of $G$.
Then $H$ and $K$ are [[Definition:Permutable Subgroups|permutable]].
That is:
:$H \circ K = K \circ H$
where $H \circ K$ denotes [[De... | Suppose $H \circ K$ is a [[Definition:Subgroup|subgroup]] of $G$.
Then:
{{begin-eqn}}
{{eqn | l = H \circ K
| r = \paren {H \circ K}^{-1}
| c = [[Inverse of Subgroup]]
}}
{{eqn | r = K^{-1} \circ H^{-1}
| c = [[Inverse of Product of Subsets of Group]]
}}
{{eqn | r = K \circ H
| c = [[Inverse of... | Subset Product of Subgroups/Necessary Condition/Proof 2 | https://proofwiki.org/wiki/Subset_Product_of_Subgroups/Necessary_Condition | https://proofwiki.org/wiki/Subset_Product_of_Subgroups/Necessary_Condition/Proof_2 | [
"Subset Product of Subgroups"
] | [
"Definition:Group",
"Definition:Subgroup",
"Definition:Subgroup",
"Definition:Permutable Subgroups",
"Definition:Subset Product"
] | [
"Definition:Subgroup",
"Inverse of Subgroup",
"Inverse of Product of Subsets of Group",
"Inverse of Subgroup"
] |
proofwiki-5828 | Subset Product of Subgroups/Sufficient Condition | Let $\struct {G, \circ}$ be a group.
Let $H, K$ be subgroups of $G$.
Let $H$ and $K$ be permutable subgroups of $G$.
That is, suppose:
:$H \circ K = K \circ H$
where $H \circ K$ denotes subset product.
Then $H \circ K$ is a subgroup of $G$. | Suppose $H \circ K = K \circ H$.
First note that $H \circ K \ne \O$, as $e_G = e_G \circ e_G \in H \circ K$, from Identity of Subgroup.
Suppose $a_1, a_2 \in H, b_1, b_2 \in K$.
Then:
:$\paren {a_1 \circ b_1} \circ \paren {a_2 \circ b_2} = a_1 \circ \paren {b_1 \circ a_2} \circ b_2$.
Since $H \circ K = K \circ H$, we s... | Let $\struct {G, \circ}$ be a [[Definition:Group|group]].
Let $H, K$ be [[Definition:Subgroup|subgroups]] of $G$.
Let $H$ and $K$ be [[Definition:Permutable Subgroups|permutable subgroups]] of $G$.
That is, suppose:
:$H \circ K = K \circ H$
where $H \circ K$ denotes [[Definition:Subset Product|subset product]].
Th... | Suppose $H \circ K = K \circ H$.
First note that $H \circ K \ne \O$, as $e_G = e_G \circ e_G \in H \circ K$, from [[Identity of Subgroup]].
Suppose $a_1, a_2 \in H, b_1, b_2 \in K$.
Then:
:$\paren {a_1 \circ b_1} \circ \paren {a_2 \circ b_2} = a_1 \circ \paren {b_1 \circ a_2} \circ b_2$.
Since $H \circ K = K \cir... | Subset Product of Subgroups/Sufficient Condition/Proof 1 | https://proofwiki.org/wiki/Subset_Product_of_Subgroups/Sufficient_Condition | https://proofwiki.org/wiki/Subset_Product_of_Subgroups/Sufficient_Condition/Proof_1 | [
"Subset Product of Subgroups"
] | [
"Definition:Group",
"Definition:Subgroup",
"Definition:Permutable Subgroups",
"Definition:Subset Product",
"Definition:Subgroup"
] | [
"Identity of Subgroup",
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Inverse of Group Product",
"Two-Step Subgroup Test",
"Definition:Subgroup"
] |
proofwiki-5829 | Subset Product of Subgroups/Sufficient Condition | Let $\struct {G, \circ}$ be a group.
Let $H, K$ be subgroups of $G$.
Let $H$ and $K$ be permutable subgroups of $G$.
That is, suppose:
:$H \circ K = K \circ H$
where $H \circ K$ denotes subset product.
Then $H \circ K$ is a subgroup of $G$. | Suppose $H \circ K = K \circ H$.
Then:
{{begin-eqn}}
{{eqn | l = \paren {H \circ K} \circ \paren {H \circ K}^{-1}
| r = H \circ K \circ K^{-1} \circ H^{-1}
| c = Inverse of Product of Subsets of Group
}}
{{eqn | r = H \circ K \circ K \circ H
| c = Inverse of Subgroup
}}
{{eqn | r = H \circ K \circ H
... | Let $\struct {G, \circ}$ be a [[Definition:Group|group]].
Let $H, K$ be [[Definition:Subgroup|subgroups]] of $G$.
Let $H$ and $K$ be [[Definition:Permutable Subgroups|permutable subgroups]] of $G$.
That is, suppose:
:$H \circ K = K \circ H$
where $H \circ K$ denotes [[Definition:Subset Product|subset product]].
Th... | Suppose $H \circ K = K \circ H$.
Then:
{{begin-eqn}}
{{eqn | l = \paren {H \circ K} \circ \paren {H \circ K}^{-1}
| r = H \circ K \circ K^{-1} \circ H^{-1}
| c = [[Inverse of Product of Subsets of Group]]
}}
{{eqn | r = H \circ K \circ K \circ H
| c = [[Inverse of Subgroup]]
}}
{{eqn | r = H \circ K... | Subset Product of Subgroups/Sufficient Condition/Proof 2 | https://proofwiki.org/wiki/Subset_Product_of_Subgroups/Sufficient_Condition | https://proofwiki.org/wiki/Subset_Product_of_Subgroups/Sufficient_Condition/Proof_2 | [
"Subset Product of Subgroups"
] | [
"Definition:Group",
"Definition:Subgroup",
"Definition:Permutable Subgroups",
"Definition:Subset Product",
"Definition:Subgroup"
] | [
"Inverse of Product of Subsets of Group",
"Inverse of Subgroup",
"Product of Subgroup with Itself",
"Definition:By Hypothesis",
"Product of Subgroup with Itself",
"Definition:Set Equality/Definition 2",
"One-Step Subgroup Test using Subset Product"
] |
proofwiki-5830 | Characteristics of Eulerian Graph/Necessary Condition | Let $G$ be a finite (undirected) graph.
Let $G$ be Eulerian.
Then $G$ is connected and each vertex of $G$ is even.
Note that the definition of graph here includes:
* Simple graph
* Loop-graph
* Multigraph
* Loop-multigraph
but does not include directed graph. | Suppose that $G$ is Eulerian.
Then $G$ contains an Eulerian circuit, that is, a circuit that uses each vertex and passes through each edge exactly once.
Since a circuit must be connected, $G$ is connected.
Beginning at a vertex $v$, follow the Eulerian circuit through $G$.
As the circuit passes through each vertex, it ... | Let $G$ be a [[Definition:Finite Graph|finite]] [[Definition:Undirected Graph|(undirected) graph]].
Let $G$ be [[Definition:Eulerian Graph|Eulerian]].
Then $G$ is [[Definition:Connected Graph|connected]] and each [[Definition:Vertex of Graph|vertex]] of $G$ is [[Definition:Even Vertex of Graph|even]].
Note that th... | Suppose that $G$ is [[Definition:Eulerian Graph|Eulerian]].
Then $G$ contains an [[Definition:Eulerian Circuit|Eulerian circuit]], that is, a [[Definition:Circuit (Graph Theory)|circuit]] that uses each [[Definition:Vertex of Graph|vertex]] and passes through each [[Definition:Edge of Graph|edge]] exactly once.
Since... | Characteristics of Eulerian Graph/Necessary Condition | https://proofwiki.org/wiki/Characteristics_of_Eulerian_Graph/Necessary_Condition | https://proofwiki.org/wiki/Characteristics_of_Eulerian_Graph/Necessary_Condition | [
"Characteristics of Eulerian Graph"
] | [
"Definition:Finite Graph",
"Definition:Undirected Graph",
"Definition:Eulerian Graph",
"Definition:Connected (Graph Theory)/Graph",
"Definition:Graph (Graph Theory)/Vertex",
"Definition:Even Vertex of Graph",
"Definition:Graph (Graph Theory)",
"Definition:Simple Graph",
"Definition:Loop-Graph",
"D... | [
"Definition:Eulerian Graph",
"Definition:Eulerian Circuit",
"Definition:Circuit (Graph Theory)",
"Definition:Graph (Graph Theory)/Vertex",
"Definition:Graph (Graph Theory)/Edge",
"Definition:Circuit (Graph Theory)",
"Definition:Connected (Graph Theory)/Graph",
"Definition:Connected (Graph Theory)/Grap... |
proofwiki-5831 | Characteristics of Eulerian Graph/Sufficient Condition | Let $G$ be a finite (undirected) graph which is connected
Let each vertex of $G$ be even.
Then $G$ is an Eulerian graph. | === Proof 1 ===
{{:Characteristics of Eulerian Graph/Sufficient Condition/Proof 1}}
=== Proof 2 ===
{{:Characteristics of Eulerian Graph/Sufficient Condition/Proof 2}} | Let $G$ be a [[Definition:Finite Graph|finite]] [[Definition:Undirected Graph|(undirected) graph]] which is [[Definition:Connected Graph|connected]]
Let each [[Definition:Vertex of Graph|vertex]] of $G$ be [[Definition:Even Vertex of Graph|even]].
Then $G$ is an [[Definition:Eulerian Graph|Eulerian graph]]. | === [[Characteristics of Eulerian Graph/Sufficient Condition/Proof 1|Proof 1]] ===
{{:Characteristics of Eulerian Graph/Sufficient Condition/Proof 1}}
=== [[Characteristics of Eulerian Graph/Sufficient Condition/Proof 2|Proof 2]] ===
{{:Characteristics of Eulerian Graph/Sufficient Condition/Proof 2}} | Characteristics of Eulerian Graph/Sufficient Condition | https://proofwiki.org/wiki/Characteristics_of_Eulerian_Graph/Sufficient_Condition | https://proofwiki.org/wiki/Characteristics_of_Eulerian_Graph/Sufficient_Condition | [
"Eulerian Graphs"
] | [
"Definition:Finite Graph",
"Definition:Undirected Graph",
"Definition:Connected (Graph Theory)/Graph",
"Definition:Graph (Graph Theory)/Vertex",
"Definition:Even Vertex of Graph",
"Definition:Eulerian Graph"
] | [
"Characteristics of Eulerian Graph/Sufficient Condition/Proof 1",
"Characteristics of Eulerian Graph/Sufficient Condition/Proof 2"
] |
proofwiki-5832 | Characteristics of Eulerian Graph/Sufficient Condition | Let $G$ be a finite (undirected) graph which is connected
Let each vertex of $G$ be even.
Then $G$ is an Eulerian graph. | Suppose that an (undirected) graph $G$ is connected and its vertices all have even degree.
If there is more than one vertex in $G$, then each vertex must have degree greater than $0$.
Begin at a vertex $v$.
From Graph with Even Vertices Partitions into Cycles, we know that $v$ will be on at least one cycle.
Since $G$ i... | Let $G$ be a [[Definition:Finite Graph|finite]] [[Definition:Undirected Graph|(undirected) graph]] which is [[Definition:Connected Graph|connected]]
Let each [[Definition:Vertex of Graph|vertex]] of $G$ be [[Definition:Even Vertex of Graph|even]].
Then $G$ is an [[Definition:Eulerian Graph|Eulerian graph]]. | Suppose that an [[Definition:Undirected Graph|(undirected) graph]] $G$ is [[Definition:Connected Graph|connected]] and its [[Definition:Vertex of Graph|vertices]] all have [[Definition:Even Vertex of Graph|even degree]].
If there is more than one [[Definition:Vertex of Graph|vertex]] in $G$, then each [[Definition:Ver... | Characteristics of Eulerian Graph/Sufficient Condition/Proof 1 | https://proofwiki.org/wiki/Characteristics_of_Eulerian_Graph/Sufficient_Condition | https://proofwiki.org/wiki/Characteristics_of_Eulerian_Graph/Sufficient_Condition/Proof_1 | [
"Eulerian Graphs"
] | [
"Definition:Finite Graph",
"Definition:Undirected Graph",
"Definition:Connected (Graph Theory)/Graph",
"Definition:Graph (Graph Theory)/Vertex",
"Definition:Even Vertex of Graph",
"Definition:Eulerian Graph"
] | [
"Definition:Undirected Graph",
"Definition:Connected (Graph Theory)/Graph",
"Definition:Graph (Graph Theory)/Vertex",
"Definition:Even Vertex of Graph",
"Definition:Graph (Graph Theory)/Vertex",
"Definition:Graph (Graph Theory)/Vertex",
"Definition:Degree of Vertex",
"Definition:Graph (Graph Theory)/V... |
proofwiki-5833 | Characteristics of Eulerian Graph/Sufficient Condition | Let $G$ be a finite (undirected) graph which is connected
Let each vertex of $G$ be even.
Then $G$ is an Eulerian graph. | Suppose that an (undirected) graph $G$ is connected and its vertices all have even degree.
From Graph with Even Vertices Partitions into Cycles, we can split $G$ into a number of cycles $\mathbb S = C_1, C_2, \ldots, C_k$.
Start at any vertex $v$ on cycle $C_1$ and traverse its edges until we encounter a vertex of anot... | Let $G$ be a [[Definition:Finite Graph|finite]] [[Definition:Undirected Graph|(undirected) graph]] which is [[Definition:Connected Graph|connected]]
Let each [[Definition:Vertex of Graph|vertex]] of $G$ be [[Definition:Even Vertex of Graph|even]].
Then $G$ is an [[Definition:Eulerian Graph|Eulerian graph]]. | Suppose that an [[Definition:Undirected Graph|(undirected) graph]] $G$ is [[Definition:Connected Graph|connected]] and its [[Definition:Vertex of Graph|vertices]] all have [[Definition:Even Vertex of Graph|even degree]].
From [[Graph with Even Vertices Partitions into Cycles]], we can split $G$ into a number of [[Defi... | Characteristics of Eulerian Graph/Sufficient Condition/Proof 2 | https://proofwiki.org/wiki/Characteristics_of_Eulerian_Graph/Sufficient_Condition | https://proofwiki.org/wiki/Characteristics_of_Eulerian_Graph/Sufficient_Condition/Proof_2 | [
"Eulerian Graphs"
] | [
"Definition:Finite Graph",
"Definition:Undirected Graph",
"Definition:Connected (Graph Theory)/Graph",
"Definition:Graph (Graph Theory)/Vertex",
"Definition:Even Vertex of Graph",
"Definition:Eulerian Graph"
] | [
"Definition:Undirected Graph",
"Definition:Connected (Graph Theory)/Graph",
"Definition:Graph (Graph Theory)/Vertex",
"Definition:Even Vertex of Graph",
"Graph with Even Vertices Partitions into Cycles",
"Definition:Cycle (Graph Theory)",
"Definition:Graph (Graph Theory)/Vertex",
"Definition:Cycle (Gr... |
proofwiki-5834 | Characterization of Pre-Measures | Let $X$ be a set, and let $\SS \subseteq \powerset X$ be a collection of subsets of $X$.
Let $\O \in \SS$.
Denote $\overline \R_{\ge 0}$ for the set of positive extended real numbers.
A mapping $\mu: \SS \to \overline \R_{\ge 0}$ is a pre-measure {{iff}}:
{{begin-itemize}}
{{item|(1):|$\map \mu \O {{=}} 0$}}
{{item|(2)... | {{ProofWanted|will be very similar to Characterization of Measures}}
Category:Pre-Measures
64jdmjdy4nopm4tffbkzqu4uysyt9vd | Let $X$ be a [[Definition:Set|set]], and let $\SS \subseteq \powerset X$ be a collection of [[Definition:Subset|subsets]] of $X$.
Let $\O \in \SS$.
Denote $\overline \R_{\ge 0}$ for the set of [[Definition:Positive Real Number|positive]] [[Definition:Extended Real Number Line|extended real numbers]].
A [[Definition... | {{ProofWanted|will be very similar to [[Characterization of Measures]]}}
[[Category:Pre-Measures]]
64jdmjdy4nopm4tffbkzqu4uysyt9vd | Characterization of Pre-Measures | https://proofwiki.org/wiki/Characterization_of_Pre-Measures | https://proofwiki.org/wiki/Characterization_of_Pre-Measures | [
"Pre-Measures"
] | [
"Definition:Set",
"Definition:Subset",
"Definition:Positive/Real Number",
"Definition:Extended Real Number Line",
"Definition:Mapping",
"Definition:Pre-Measure",
"Definition:Additive Function (Measure Theory)",
"Definition:Increasing Sequence of Sets",
"Definition:Limit of Increasing Sequence of Set... | [
"Characterization of Measures",
"Category:Pre-Measures"
] |
proofwiki-5835 | Power Set is Magma of Sets | Let $X$ be a set.
Let $\family {\phi_i}: i \in I$ be an indexed family of mappings.
Then $\powerset X$, the power set of $X$, is a magma of sets for $\family {\phi_i}: i \in I$ on $X$. | For every $i \in I$, let $J_i$ be an index set, and let:
:$\phi_i: \powerset X^{J_i} \to \powerset X$
be a partial mapping.
For each $i \in I$, for each $\family {S_{j_i} }_{j_i \mathop \in J_i} \in \powerset X^{J_i} \cap \DD_i$, that:
:$\map {\phi_i} {\family {S_{j_i} }_{j_i \mathop \in J_i} } \in \powerset X$
{{expla... | Let $X$ be a [[Definition:Set|set]].
Let $\family {\phi_i}: i \in I$ be an [[Definition:Indexed Family of Sets|indexed family]] of [[Definition:Mapping|mappings]].
Then $\powerset X$, the [[Definition:Power Set|power set]] of $X$, is a [[Definition:Magma of Sets|magma of sets]] for $\family {\phi_i}: i \in I$ on $X$... | For every $i \in I$, let $J_i$ be an [[Definition:Index Set|index set]], and let:
:$\phi_i: \powerset X^{J_i} \to \powerset X$
be a [[Definition:Partial Mapping|partial mapping]].
For each $i \in I$, for each $\family {S_{j_i} }_{j_i \mathop \in J_i} \in \powerset X^{J_i} \cap \DD_i$, that:
:$\map {\phi_i} {\family... | Power Set is Magma of Sets | https://proofwiki.org/wiki/Power_Set_is_Magma_of_Sets | https://proofwiki.org/wiki/Power_Set_is_Magma_of_Sets | [
"Magmas of Sets"
] | [
"Definition:Set",
"Definition:Indexing Set/Family of Sets",
"Definition:Mapping",
"Definition:Power Set",
"Definition:Magma of Sets"
] | [
"Definition:Indexing Set",
"Definition:Many-to-One Relation",
"Definition:Codomain (Set Theory)/Mapping",
"Definition:Indexing Set/Family of Sets",
"Definition:Domain (Set Theory)/Relation",
"Definition:Magma of Sets",
"Category:Magmas of Sets"
] |
proofwiki-5836 | Existence and Uniqueness of Magma of Sets Generated by Collection of Subsets | Let $X$ be a set, and let $\Phi := \set {\phi_i: i \in I}$ be a collection of partial mappings with codomain $\powerset X$, the power set of $X$.
Let $\GG \subseteq \powerset X$ be a collection of subsets of $X$.
Then the magma of sets generated by $\GG$ exists and is unique. | === Uniqueness ===
Suppose that both $\SS$ and $\TT$ are magmas of sets for $\Phi$ generated by $\GG$.
Applying condition $(2)$ for these twice, we obtain:
:$\SS \subseteq \TT$
:$\TT \subseteq \SS$
By definition of set equality:
:$\SS = \TT$
{{qed|lemma}} | Let $X$ be a [[Definition:Set|set]], and let $\Phi := \set {\phi_i: i \in I}$ be a collection of [[Definition:Partial Mapping|partial mappings]] with [[Definition:Codomain of Relation|codomain]] $\powerset X$, the [[Definition:Power Set|power set]] of $X$.
Let $\GG \subseteq \powerset X$ be a collection of [[Definitio... | === Uniqueness ===
Suppose that both $\SS$ and $\TT$ are [[Definition:Magma of Sets Generated by Collection of Subsets|magmas of sets for $\Phi$ generated by $\GG$]].
Applying condition $(2)$ for these twice, we obtain:
:$\SS \subseteq \TT$
:$\TT \subseteq \SS$
By definition of [[Definition:Set Equality|set equalit... | Existence and Uniqueness of Magma of Sets Generated by Collection of Subsets | https://proofwiki.org/wiki/Existence_and_Uniqueness_of_Magma_of_Sets_Generated_by_Collection_of_Subsets | https://proofwiki.org/wiki/Existence_and_Uniqueness_of_Magma_of_Sets_Generated_by_Collection_of_Subsets | [
"Magmas of Sets"
] | [
"Definition:Set",
"Definition:Many-to-One Relation",
"Definition:Codomain (Set Theory)/Relation",
"Definition:Power Set",
"Definition:Subset",
"Definition:Magma of Sets Generated by Collection of Subsets"
] | [
"Definition:Magma of Sets Generated by Collection of Subsets",
"Definition:Set Equality",
"Definition:Magma of Sets",
"Definition:Magma of Sets",
"Definition:Magma of Sets",
"Definition:Magma of Sets Generated by Collection of Subsets"
] |
proofwiki-5837 | Intersection of Magmas of Sets is Magma of Sets | Let $X$ be a set.
Let $\Phi := \set {\phi_i: i \in I}$ be a collection of partial mappings with codomain $\powerset X$, the power set of $X$.
Let $\SS_j$ be a magma of sets for $\Phi$, for each $j \in J$, for some index set $J$.
Then:
:$\SS := \ds \bigcap_{j \mathop \in J} \SS_j$
is also a magma of sets for $\Phi$. | {{explain|the notation in the below}}
For each $i \in I$ and $j \in J$, we have:
:$\map {\phi_i} {\sequence {S_{j, j_i} }_{j_i \mathop \in J_i} } \in \SS_j$
Thus, if for each $j \in J$, one has:
:$(1): \quad \sequence {S_{j_i} }_{j_i \mathop \in J_i} \in \SS_j^{J_i}$
it follows from definition of set intersection that:... | Let $X$ be a [[Definition:Set|set]].
Let $\Phi := \set {\phi_i: i \in I}$ be a collection of [[Definition:Partial Mapping|partial mappings]] with [[Definition:Codomain of Relation|codomain]] $\powerset X$, the [[Definition:Power Set|power set]] of $X$.
Let $\SS_j$ be a [[Definition:Magma of Sets|magma of sets]] for $... | {{explain|the notation in the below}}
For each $i \in I$ and $j \in J$, we have:
:$\map {\phi_i} {\sequence {S_{j, j_i} }_{j_i \mathop \in J_i} } \in \SS_j$
Thus, if for each $j \in J$, one has:
:$(1): \quad \sequence {S_{j_i} }_{j_i \mathop \in J_i} \in \SS_j^{J_i}$
it follows from definition of [[Definition:Set ... | Intersection of Magmas of Sets is Magma of Sets | https://proofwiki.org/wiki/Intersection_of_Magmas_of_Sets_is_Magma_of_Sets | https://proofwiki.org/wiki/Intersection_of_Magmas_of_Sets_is_Magma_of_Sets | [
"Magmas of Sets"
] | [
"Definition:Set",
"Definition:Many-to-One Relation",
"Definition:Codomain (Set Theory)/Relation",
"Definition:Power Set",
"Definition:Magma of Sets",
"Definition:Indexing Set",
"Definition:Magma of Sets"
] | [
"Definition:Set Intersection",
"Definition:Magma of Sets",
"Definition:Magma of Sets",
"Category:Magmas of Sets"
] |
proofwiki-5838 | Sigma-Algebra as Magma of Sets | The concept of $\sigma$-algebra is an instance of a magma of sets. | It will suffice to define partial mappings such that the axiom for a magma of sets crystallises into the axioms for a $\sigma$-algebra.
Let $X$ be any set, and let $\powerset X$ be its power set.
Let $I$ be an index set.
For $S \in \powerset X$, define:
:$\phi_1: \powerset X \to \powerset X: \map {\phi_1} S := X$
:$\ph... | The concept of [[Definition:Sigma-Algebra|$\sigma$-algebra]] is an instance of a [[Definition:Magma of Sets|magma of sets]]. | It will suffice to define [[Definition:Partial Mapping|partial mappings]] such that the axiom for a [[Definition:Magma of Sets|magma of sets]] crystallises into the axioms for a [[Definition:Sigma-Algebra|$\sigma$-algebra]].
Let $X$ be any [[Definition:Set|set]], and let $\powerset X$ be its [[Definition:Power Set|pow... | Sigma-Algebra as Magma of Sets | https://proofwiki.org/wiki/Sigma-Algebra_as_Magma_of_Sets | https://proofwiki.org/wiki/Sigma-Algebra_as_Magma_of_Sets | [
"Sigma-Algebras",
"Magmas of Sets"
] | [
"Definition:Sigma-Algebra",
"Definition:Magma of Sets"
] | [
"Definition:Many-to-One Relation",
"Definition:Magma of Sets",
"Definition:Sigma-Algebra",
"Definition:Set",
"Definition:Power Set",
"Definition:Indexing Set",
"Definition:Sigma-Algebra",
"Definition:Real Interval/Closed",
"Definition:Indexing Set",
"Definition:Many-to-One Relation",
"Union of S... |
proofwiki-5839 | Ordinal Multiplication is Closed | Let $x$ and $y$ be ordinals.
Let $\On$ denote the class of all ordinals.
:$x \cdot y \in \On$ | The proof proceeds by transfinite induction on $y$. | Let $x$ and $y$ be [[Definition:Ordinal|ordinals]].
Let $\On$ denote the [[Definition:Class of All Ordinals|class of all ordinals]].
:$x \cdot y \in \On$ | The proof proceeds by [[Transfinite Induction/Schema 2|transfinite induction]] on $y$. | Ordinal Multiplication is Closed | https://proofwiki.org/wiki/Ordinal_Multiplication_is_Closed | https://proofwiki.org/wiki/Ordinal_Multiplication_is_Closed | [
"Ordinal Arithmetic"
] | [
"Definition:Ordinal",
"Definition:Class of All Ordinals"
] | [
"Transfinite Induction/Schema 2",
"Transfinite Induction/Schema 2",
"Transfinite Induction/Schema 2",
"Transfinite Induction/Schema 2"
] |
proofwiki-5840 | Natural Number Multiplication is Closed | Let $m$ and $n$ be natural numbers.
Then:
:$m \times n \in \N$
where $\times$ denotes natural number multiplication. | Let $g: \N \to \N$ be defined by:
:$\map g n = m + n$
Applying the Principle of Recursive Definition to $0$ and $g$, we get the following function $f: \N \to \N$:
:$\map f n = \begin{cases} 0 & : n = 0 \\ m + \map f k & : n = k + 1 \end{cases}$
which is seen to be equivalent to $m \times n$ for all $m,n \in \N$.
Hence ... | Let $m$ and $n$ be [[Definition:Natural Numbers|natural numbers]].
Then:
:$m \times n \in \N$
where $\times$ denotes [[Definition:Natural Number Multiplication|natural number multiplication]]. | Let $g: \N \to \N$ be defined by:
:$\map g n = m + n$
Applying the [[Principle of Recursive Definition]] to $0$ and $g$, we get the following function $f: \N \to \N$:
:$\map f n = \begin{cases} 0 & : n = 0 \\ m + \map f k & : n = k + 1 \end{cases}$
which is seen to be equivalent to $m \times n$ for all $m,n \in \N$... | Natural Number Multiplication is Closed | https://proofwiki.org/wiki/Natural_Number_Multiplication_is_Closed | https://proofwiki.org/wiki/Natural_Number_Multiplication_is_Closed | [
"Natural Number Multiplication"
] | [
"Definition:Natural Numbers",
"Definition:Multiplication/Natural Numbers"
] | [
"Principle of Recursive Definition"
] |
proofwiki-5841 | Ordinal Multiplication by Zero | Let $x$ be an ordinal.
{{begin-eqn}}
{{eqn | l = x \cdot \O
| r = \O
| c =
}}
{{eqn | l = \O \cdot x
| r = \O
| c =
}}
{{end-eqn}} | {{begin-eqn}}
{{eqn | l = x \cdot \O
| r = \O
| c = {{Defof|Ordinal Multiplication}}
}}
{{end-eqn}}
For $\O \cdot x = \O$, the proof shall proceed by Transfinite Induction on $x$. | Let $x$ be an [[Definition:Ordinal|ordinal]].
{{begin-eqn}}
{{eqn | l = x \cdot \O
| r = \O
| c =
}}
{{eqn | l = \O \cdot x
| r = \O
| c =
}}
{{end-eqn}} | {{begin-eqn}}
{{eqn | l = x \cdot \O
| r = \O
| c = {{Defof|Ordinal Multiplication}}
}}
{{end-eqn}}
For $\O \cdot x = \O$, the proof shall proceed by [[Transfinite Induction/Schema 2|Transfinite Induction]] on $x$. | Ordinal Multiplication by Zero | https://proofwiki.org/wiki/Ordinal_Multiplication_by_Zero | https://proofwiki.org/wiki/Ordinal_Multiplication_by_Zero | [
"Ordinal Arithmetic"
] | [
"Definition:Ordinal"
] | [
"Transfinite Induction/Schema 2",
"Transfinite Induction/Schema 2",
"Transfinite Induction/Schema 2",
"Transfinite Induction/Schema 2"
] |
proofwiki-5842 | Ordinal Multiplication by One | Let $x$ be an ordinal.
Let $1$ denote the ordinal one.
{{begin-eqn}}
{{eqn | l = x \cdot 1
| r = x
}}
{{eqn | l = 1 \cdot x
| r = x
}}
{{end-eqn}} | {{begin-eqn}}
{{eqn | l = x \cdot 1
| r = x \cdot \O^+
| c = {{Defof|One (Ordinal)}}
}}
{{eqn | r = \paren {x \cdot \O} + x
| c = {{Defof|Ordinal Multiplication}}
}}
{{eqn | r = \O + x
| c = {{Defof|Ordinal Multiplication}}
}}
{{eqn | r = x
| c = Ordinal Addition by Zero
}}
{{end-eqn}}
{{q... | Let $x$ be an [[Definition:Ordinal|ordinal]].
Let $1$ denote the [[Definition:One (Ordinal)|ordinal one]].
{{begin-eqn}}
{{eqn | l = x \cdot 1
| r = x
}}
{{eqn | l = 1 \cdot x
| r = x
}}
{{end-eqn}} | {{begin-eqn}}
{{eqn | l = x \cdot 1
| r = x \cdot \O^+
| c = {{Defof|One (Ordinal)}}
}}
{{eqn | r = \paren {x \cdot \O} + x
| c = {{Defof|Ordinal Multiplication}}
}}
{{eqn | r = \O + x
| c = {{Defof|Ordinal Multiplication}}
}}
{{eqn | r = x
| c = [[Ordinal Addition by Zero]]
}}
{{end-eqn}}... | Ordinal Multiplication by One | https://proofwiki.org/wiki/Ordinal_Multiplication_by_One | https://proofwiki.org/wiki/Ordinal_Multiplication_by_One | [
"Ordinal Arithmetic"
] | [
"Definition:Ordinal",
"Definition:One (Ordinal)"
] | [
"Ordinal Addition by Zero",
"Transfinite Induction/Schema 2",
"Transfinite Induction/Schema 2",
"Transfinite Induction/Schema 2",
"Transfinite Induction/Schema 2"
] |
proofwiki-5843 | Ordinal Addition/Examples/Ordinal Addition by One | Let $x$ be an ordinal.
Let $x^+$ denote the successor of $x$.
Let $1$ denote (ordinal) one, the successor of the zero ordinal $\O$.
Then:
:$x + 1 = x^+$
where $+$ denotes ordinal addition. | {{begin-eqn}}
{{eqn | l = x + 1
| r = x + \O^+
| c = {{Defof|One (Ordinal)|(ordinal) $1$}}
}}
{{eqn | r = \paren {x + \O}^+
| c = {{Defof|Ordinal Addition}}
}}
{{eqn | r = x^+
| c = Ordinal Addition by Zero
}}
{{end-eqn}}
{{qed}} | Let $x$ be an [[Definition:Ordinal|ordinal]].
Let $x^+$ denote the [[Definition:Successor Set|successor]] of $x$.
Let $1$ denote [[Definition:One (Ordinal)|(ordinal) one]], the [[Definition:Successor Set|successor]] of the [[Definition:Zero Ordinal|zero ordinal]] $\O$.
Then:
:$x + 1 = x^+$
where $+$ denotes [[Def... | {{begin-eqn}}
{{eqn | l = x + 1
| r = x + \O^+
| c = {{Defof|One (Ordinal)|(ordinal) $1$}}
}}
{{eqn | r = \paren {x + \O}^+
| c = {{Defof|Ordinal Addition}}
}}
{{eqn | r = x^+
| c = [[Ordinal Addition by Zero]]
}}
{{end-eqn}}
{{qed}} | Ordinal Addition/Examples/Ordinal Addition by One | https://proofwiki.org/wiki/Ordinal_Addition/Examples/Ordinal_Addition_by_One | https://proofwiki.org/wiki/Ordinal_Addition/Examples/Ordinal_Addition_by_One | [
"Ordinal Addition",
"1"
] | [
"Definition:Ordinal",
"Definition:Successor Mapping/Successor Set",
"Definition:One (Ordinal)",
"Definition:Successor Mapping/Successor Set",
"Definition:Zero (Ordinal)",
"Definition:Ordinal Addition"
] | [
"Ordinal Addition by Zero"
] |
proofwiki-5844 | Closed Set Measurable in Borel Sigma-Algebra | Let $\struct {S, \tau}$ be a topological space.
Let $\map \BB \tau$ be the associated Borel $\sigma$-algebra.
Let $C$ be a closed set in $\struct {S, \tau}$.
Then $C$ is $\map \BB \tau$-measurable. | Since $C$ is closed, $S \setminus C$ is open.
By definition of Borel $\sigma$-algebra, $S \setminus C \in \map \BB \tau$.
By axiom $(2)$ for $\sigma$-algebras:
:$S \setminus \paren {S \setminus C} \in \map \BB \tau$
and this set equals $C$ by Set Difference with Set Difference since $C \subseteq S$.
The result follows ... | Let $\struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
Let $\map \BB \tau$ be the associated [[Definition:Borel Sigma-Algebra|Borel $\sigma$-algebra]].
Let $C$ be a [[Definition:Closed Set (Topology)|closed set]] in $\struct {S, \tau}$.
Then $C$ is [[Definition:Measurable Set|$\map \BB \ta... | Since $C$ is [[Definition:Closed Set (Topology)|closed]], $S \setminus C$ is [[Definition:Open Set (Topology)|open]].
By definition of [[Definition:Borel Sigma-Algebra|Borel $\sigma$-algebra]], $S \setminus C \in \map \BB \tau$.
By axiom $(2)$ for [[Definition:Sigma-Algebra|$\sigma$-algebras]]:
:$S \setminus \paren... | Closed Set Measurable in Borel Sigma-Algebra | https://proofwiki.org/wiki/Closed_Set_Measurable_in_Borel_Sigma-Algebra | https://proofwiki.org/wiki/Closed_Set_Measurable_in_Borel_Sigma-Algebra | [
"Sigma-Algebras",
"Borel Sigma-Algebras",
"Borel Sigma-Algebras",
"Closed Sets"
] | [
"Definition:Topological Space",
"Definition:Borel Sigma-Algebra",
"Definition:Closed Set/Topology",
"Definition:Measurable Set"
] | [
"Definition:Closed Set/Topology",
"Definition:Open Set/Topology",
"Definition:Borel Sigma-Algebra",
"Definition:Sigma-Algebra",
"Set Difference with Set Difference",
"Definition:Measurable Set",
"Category:Borel Sigma-Algebras",
"Category:Closed Sets"
] |
proofwiki-5845 | Euler Phi Function preserves Divisibility | :$d \divides n \implies \map \phi d \divides \map \phi n$
where $d \divides n$ denotes that $d$ is a divisor of $n$. | Let $d \divides n$.
We can write $n$ as $n = d p_1 p_2 p_3 \cdots p_r$, where $p_1, p_2, \ldots, p_r$ are all the primes (not necessarily distinct) which divide $\dfrac n d$.
Thus, repeatedly using Euler Phi Function of Product with Prime:
{{begin-eqn}}
{{eqn | l = \map \phi d
| o = \divides
| r = \map \phi... | :$d \divides n \implies \map \phi d \divides \map \phi n$
where $d \divides n$ denotes that $d$ is a [[Definition:Divisor of Integer|divisor]] of $n$. | Let $d \divides n$.
We can write $n$ as $n = d p_1 p_2 p_3 \cdots p_r$, where $p_1, p_2, \ldots, p_r$ are all the [[Definition:Prime Number|primes]] (not necessarily distinct) which [[Definition:Divisor of Integer|divide]] $\dfrac n d$.
Thus, repeatedly using [[Euler Phi Function of Product with Prime]]:
{{begin-eqn... | Euler Phi Function preserves Divisibility | https://proofwiki.org/wiki/Euler_Phi_Function_preserves_Divisibility | https://proofwiki.org/wiki/Euler_Phi_Function_preserves_Divisibility | [
"Euler Phi Function"
] | [
"Definition:Divisor (Algebra)/Integer"
] | [
"Definition:Prime Number",
"Definition:Divisor (Algebra)/Integer",
"Euler Phi Function of Product with Prime",
"Divisor Relation on Positive Integers is Partial Ordering",
"Category:Euler Phi Function"
] |
proofwiki-5846 | Commutative B-Algebra is Entropic Structure | Let $\struct {G, *}$ be a commutative $B$-algebra.
Then $\struct {G, *}$ is an entropic structure. | From Commutative $B$-Algebra Induces Abelian Group we have that there exists an abelian group $\struct {G, \circ}$ such that:
:$\forall a, b \in G: a \circ b^{-1} = a * b$
where $a * b$ is defined by the binary operation in $\struct {G, *}$.
From Abelian Group Induces Entropic Structure, we have directly that $\struct... | Let $\struct {G, *}$ be a [[Definition:Commutative B-Algebra|commutative $B$-algebra]].
Then $\struct {G, *}$ is an [[Definition:Entropic Structure|entropic structure]]. | From [[Commutative B-Algebra Induces Abelian Group|Commutative $B$-Algebra Induces Abelian Group]] we have that there exists an [[Definition:Abelian Group|abelian group]] $\struct {G, \circ}$ such that:
:$\forall a, b \in G: a \circ b^{-1} = a * b$
where $a * b$ is defined by the [[Definition:Binary Operation|binary ... | Commutative B-Algebra is Entropic Structure | https://proofwiki.org/wiki/Commutative_B-Algebra_is_Entropic_Structure | https://proofwiki.org/wiki/Commutative_B-Algebra_is_Entropic_Structure | [
"B-Algebras",
"Entropic Structures"
] | [
"Definition:Commutative B-Algebra",
"Definition:Entropic Structure"
] | [
"Commutative B-Algebra Induces Abelian Group",
"Definition:Abelian Group",
"Definition:Operation/Binary Operation",
"Abelian Group Induces Entropic Structure",
"Definition:Entropic Structure",
"Category:B-Algebras",
"Category:Entropic Structures"
] |
proofwiki-5847 | Commutative B-Algebra Induces Abelian Group | Let $\left({X, \circ }\right)$ be a commutative $B$-algebra.
Let $*$ be the binary operation on $X$ defined as:
:$\forall a, b \in X: a * b := a \circ \left({0 \circ b}\right)$
Then the algebraic structure $\left({X, *}\right)$ is an abelian group such that:
:$\forall x \in X: 0 \circ x$ is the inverse element of $x$ u... | From B-Algebra Induces Group, the algebraic structure $\left({X, *}\right)$ is a group such that:
:$\forall x \in X: 0 \circ x$ is the inverse element of $x$ under $*$.
It remains to show that $*$ is a commutative operation.
Let $x, y \in X$:
{{begin-eqn}}
{{eqn | l=x * y
| r=x \circ \left({0 \circ y}\right)
... | Let $\left({X, \circ }\right)$ be a [[Definition:Commutative B-Algebra|commutative $B$-algebra]].
Let $*$ be the [[Definition:Binary Operation|binary operation]] on $X$ defined as:
:$\forall a, b \in X: a * b := a \circ \left({0 \circ b}\right)$
Then the [[Definition:Algebraic Structure|algebraic structure]] $\left... | From [[B-Algebra Induces Group]], the [[Definition:Algebraic Structure|algebraic structure]] $\left({X, *}\right)$ is a [[Definition:Group|group]] such that:
:$\forall x \in X: 0 \circ x$ is the [[Definition:Inverse Element|inverse element]] of $x$ under $*$.
It remains to show that $*$ is a [[Definition:Commutative... | Commutative B-Algebra Induces Abelian Group | https://proofwiki.org/wiki/Commutative_B-Algebra_Induces_Abelian_Group | https://proofwiki.org/wiki/Commutative_B-Algebra_Induces_Abelian_Group | [
"B-Algebras"
] | [
"Definition:Commutative B-Algebra",
"Definition:Operation/Binary Operation",
"Definition:Algebraic Structure",
"Definition:Abelian Group",
"Definition:Inverse (Abstract Algebra)/Inverse"
] | [
"B-Algebra Induces Group",
"Definition:Algebraic Structure",
"Definition:Group",
"Definition:Inverse (Abstract Algebra)/Inverse",
"Definition:Commutative/Operation",
"Definition:Commutative B-Algebra"
] |
proofwiki-5848 | Membership is Left Compatible with Ordinal Multiplication | Let $x$, $y$, and $z$ be ordinals.
Then:
:$\paren {x < y \land z > 0} \iff \paren {z \cdot x} < \paren {z \cdot y}$ | === Sufficient Condition ===
The proof of the sufficient condition shall proceed by Transfinite Induction on $y$.
{{qed|lemma}} | Let $x$, $y$, and $z$ be [[Definition:Ordinal|ordinals]].
Then:
:$\paren {x < y \land z > 0} \iff \paren {z \cdot x} < \paren {z \cdot y}$ | === Sufficient Condition ===
The proof of the [[Definition:Sufficient Condition|sufficient condition]] shall proceed by [[Transfinite Induction/Schema 2|Transfinite Induction]] on $y$.
{{qed|lemma}} | Membership is Left Compatible with Ordinal Multiplication | https://proofwiki.org/wiki/Membership_is_Left_Compatible_with_Ordinal_Multiplication | https://proofwiki.org/wiki/Membership_is_Left_Compatible_with_Ordinal_Multiplication | [
"Ordinal Arithmetic"
] | [
"Definition:Ordinal"
] | [
"Definition:Conditional/Sufficient Condition",
"Transfinite Induction/Schema 2",
"Transfinite Induction/Schema 2",
"Transfinite Induction/Schema 2",
"Transfinite Induction/Schema 2"
] |
proofwiki-5849 | Ordinal Multiplication is Left Cancellable | Let $x, y, z$ be ordinals.
Let $z \ne 0$.
Then:
:$\ds \paren {z \cdot x} = \paren {z \cdot y} \implies x = y$
That is, ordinal multiplication is left cancellable. | For the proof, $<$, $\in$, and $\subsetneq$ will be used interchangeably.
This is justified by Transitive Set is Proper Subset of Ordinal iff Element of Ordinal.
Since $z \ne 0$ and $z \not < 0$, $0 < z$ by Ordinal Membership is Trichotomy.
Note that:
{{begin-eqn}}
{{eqn | l = x < y \land z > 0
| o = \implies
... | Let $x, y, z$ be [[Definition:Ordinal|ordinals]].
Let $z \ne 0$.
Then:
:$\ds \paren {z \cdot x} = \paren {z \cdot y} \implies x = y$
That is, [[Definition:Ordinal Multiplication|ordinal multiplication]] is [[Definition:Left Cancellable Operation|left cancellable]]. | For the proof, $<$, $\in$, and $\subsetneq$ will be used interchangeably.
This is justified by [[Transitive Set is Proper Subset of Ordinal iff Element of Ordinal]].
Since $z \ne 0$ and $z \not < 0$, $0 < z$ by [[Ordinal Membership is Trichotomy]].
Note that:
{{begin-eqn}}
{{eqn | l = x < y \land z > 0
| o =... | Ordinal Multiplication is Left Cancellable | https://proofwiki.org/wiki/Ordinal_Multiplication_is_Left_Cancellable | https://proofwiki.org/wiki/Ordinal_Multiplication_is_Left_Cancellable | [
"Ordinal Arithmetic"
] | [
"Definition:Ordinal",
"Definition:Ordinal Multiplication",
"Definition:Left Cancellable Operation"
] | [
"Transitive Set is Proper Subset of Ordinal iff Element of Ordinal",
"Ordinal Membership is Trichotomy",
"Membership is Left Compatible with Ordinal Multiplication",
"Membership is Left Compatible with Ordinal Multiplication",
"No Membership Loops",
"Definition:Conditional/Consequent",
"Rule of Transpos... |
proofwiki-5850 | Subset is Right Compatible with Ordinal Multiplication | Let $x, y, z$ be ordinals.
Then:
:$x \le y \implies \paren {x \cdot z} \le \paren {y \cdot z}$ | The proof shall proceed by Transfinite Induction on $z$. | Let $x, y, z$ be [[Definition:Ordinal|ordinals]].
Then:
:$x \le y \implies \paren {x \cdot z} \le \paren {y \cdot z}$ | The proof shall proceed by [[Transfinite Induction/Schema 2|Transfinite Induction]] on $z$. | Subset is Right Compatible with Ordinal Multiplication | https://proofwiki.org/wiki/Subset_is_Right_Compatible_with_Ordinal_Multiplication | https://proofwiki.org/wiki/Subset_is_Right_Compatible_with_Ordinal_Multiplication | [
"Ordinal Arithmetic"
] | [
"Definition:Ordinal"
] | [
"Transfinite Induction/Schema 2",
"Transfinite Induction/Schema 2",
"Transfinite Induction/Schema 2",
"Transfinite Induction/Schema 2"
] |
proofwiki-5851 | Ordinals have No Zero Divisors | Let $x$ and $y$ be ordinals.
Then:
:$\paren {x \cdot y} = 0 \iff \paren {x = 0 \lor y = 0}$ | === Necessary Condition ===
Suppose that $\paren {x \cdot y} = 0$ and that $x \ne 0$.
By Ordinal Multiplication by Zero:
: $\paren {x \cdot 0} = 0$
Therefore:
: $\paren {x \cdot y} = \paren {x \cdot 0}$
Thus, by Ordinal Multiplication is Left Cancellable, we have that $y = 0$.
{{qed|lemma}} | Let $x$ and $y$ be [[Definition:Ordinal|ordinals]].
Then:
:$\paren {x \cdot y} = 0 \iff \paren {x = 0 \lor y = 0}$ | === Necessary Condition ===
Suppose that $\paren {x \cdot y} = 0$ and that $x \ne 0$.
By [[Ordinal Multiplication by Zero]]:
: $\paren {x \cdot 0} = 0$
Therefore:
: $\paren {x \cdot y} = \paren {x \cdot 0}$
Thus, by [[Ordinal Multiplication is Left Cancellable]], we have that $y = 0$.
{{qed|lemma}} | Ordinals have No Zero Divisors | https://proofwiki.org/wiki/Ordinals_have_No_Zero_Divisors | https://proofwiki.org/wiki/Ordinals_have_No_Zero_Divisors | [
"Ordinal Arithmetic"
] | [
"Definition:Ordinal"
] | [
"Ordinal Multiplication by Zero",
"Ordinal Multiplication is Left Cancellable",
"Ordinal Multiplication by Zero"
] |
proofwiki-5852 | Sigma-Algebra Closed under Union/Corollary | Let $A_1, \ldots, A_n \in \Sigma$.
Then $\ds \bigcup_{k \mathop = 1}^n A_k \in \Sigma$. | Define for $k \in \N, k > n: A_k = \O$.
Then by Sigma-Algebra Contains Empty Set, axiom $(3)$ of a $\sigma$-algebra applies.
Hence:
:$\ds \bigcup_{k \mathop \in \N} A_k = \bigcup_{k \mathop = 1}^n A_k \in \Sigma$
{{qed}}
Category:Sigma-Algebras
lp4zopmk2bg2sfnv9r613ojhqz3lmda | Let $A_1, \ldots, A_n \in \Sigma$.
Then $\ds \bigcup_{k \mathop = 1}^n A_k \in \Sigma$. | Define for $k \in \N, k > n: A_k = \O$.
Then by [[Sigma-Algebra Contains Empty Set]], axiom $(3)$ of a [[Definition:Sigma-Algebra|$\sigma$-algebra]] applies.
Hence:
:$\ds \bigcup_{k \mathop \in \N} A_k = \bigcup_{k \mathop = 1}^n A_k \in \Sigma$
{{qed}}
[[Category:Sigma-Algebras]]
lp4zopmk2bg2sfnv9r613ojhqz3lmda | Sigma-Algebra Closed under Union/Corollary | https://proofwiki.org/wiki/Sigma-Algebra_Closed_under_Union/Corollary | https://proofwiki.org/wiki/Sigma-Algebra_Closed_under_Union/Corollary | [
"Sigma-Algebras"
] | [] | [
"Sigma-Algebra Contains Empty Set",
"Definition:Sigma-Algebra",
"Category:Sigma-Algebras"
] |
proofwiki-5853 | Measure is Subadditive/Corollary | Let $E_1, \ldots, E_n \in \Sigma$.
Then:
:$\ds \map \mu {\bigcup_{k \mathop = 1}^n E_k} \le \sum_{k \mathop = 1}^n \map \mu {E_k}$. | We have Measure is Subadditive.
The result follows by an application of Finite Union of Sets in Subadditive Function.
{{qed}}
Category:Measure is Subadditive
mzzr5mjrak44ofvq192b71j3b7l43k5 | Let $E_1, \ldots, E_n \in \Sigma$.
Then:
:$\ds \map \mu {\bigcup_{k \mathop = 1}^n E_k} \le \sum_{k \mathop = 1}^n \map \mu {E_k}$. | We have [[Measure is Subadditive]].
The result follows by an application of [[Finite Union of Sets in Subadditive Function]].
{{qed}}
[[Category:Measure is Subadditive]]
mzzr5mjrak44ofvq192b71j3b7l43k5 | Measure is Subadditive/Corollary | https://proofwiki.org/wiki/Measure_is_Subadditive/Corollary | https://proofwiki.org/wiki/Measure_is_Subadditive/Corollary | [
"Measure Theory",
"Measure is Subadditive",
"Measure is Subadditive"
] | [] | [
"Measure is Subadditive",
"Finite Union of Sets in Subadditive Function",
"Category:Measure is Subadditive"
] |
proofwiki-5854 | Convergence in Norm Implies Convergence in Measure | Let $\struct {X, \Sigma, \mu}$ be a measure space, and let $p \in \R, p \ge 1$.
Let $\sequence {f_n}_{n \mathop \in \N}, f_n : X \to \R$ be a sequence of $p$-integrable functions.
Also, let $f: X \to \R$ be a $p$-integrable function.
Suppose that $f_n$ converges in norm to $f$ (in the $L^p$-norm).
Then $f_n$ converges ... | Let $\epsilon > 0$.
Then:
{{begin-eqn}}
{{eqn | l = \map \mu {\set {x \in X : \size {\map {f_n} x - \map f x} \ge \epsilon } }
| r = \map \mu {\set {x \in X : \size {\map {f_n} x - \map f x}^p \ge \epsilon^p } }
}}
{{eqn | o = \le
| r = \frac{1}{\epsilon^p} \int \size {f_n - f}^p \rd \mu
| c = Markov'... | Let $\struct {X, \Sigma, \mu}$ be a [[Definition:Measure Space|measure space]], and let $p \in \R, p \ge 1$.
Let $\sequence {f_n}_{n \mathop \in \N}, f_n : X \to \R$ be a [[Definition:Sequence|sequence]] of [[Definition:P-Integrable Function|$p$-integrable functions]].
Also, let $f: X \to \R$ be a [[Definition:P-Inte... | Let $\epsilon > 0$.
Then:
{{begin-eqn}}
{{eqn | l = \map \mu {\set {x \in X : \size {\map {f_n} x - \map f x} \ge \epsilon } }
| r = \map \mu {\set {x \in X : \size {\map {f_n} x - \map f x}^p \ge \epsilon^p } }
}}
{{eqn | o = \le
| r = \frac{1}{\epsilon^p} \int \size {f_n - f}^p \rd \mu
| c = [[Mark... | Convergence in Norm Implies Convergence in Measure | https://proofwiki.org/wiki/Convergence_in_Norm_Implies_Convergence_in_Measure | https://proofwiki.org/wiki/Convergence_in_Norm_Implies_Convergence_in_Measure | [
"Measure Theory"
] | [
"Definition:Measure Space",
"Definition:Sequence",
"Definition:Integrable Function/p-Integrable",
"Definition:Integrable Function/p-Integrable",
"Definition:Convergence in Norm",
"Definition:Lp Norm",
"Definition:Convergence in Measure"
] | [
"Markov's Inequality"
] |
proofwiki-5855 | Pointwise Convergence Implies Convergence in Measure on Finite Measure Space | Let $\struct {X, \Sigma, \mu}$ be a finite measure space.
Let $\sequence {f_n}_{n \mathop \in \N}, f_n : X \to \R$ be a sequence of measurable functions.
Also, let $f: X \to \R$ be a measurable function.
Suppose that $f$ is the pointwise limit of the $f_n$ $\mu$-almost everywhere.
Then $f_n$ converges in measure to $f$... | Since $\sequence {f_n}_{n \mathop \in \N}$ converges almost everywhere, we have:
:$\set {x \in X : \sequence {\map {f_n} x}_{n \mathop \in \N} \text { does not converge to } \map f x}$ is a $\mu$-null set.
We aim to deduce that:
:$\ds \lim_{n \mathop \to \infty} \map \mu {\set {x \in X : \size {\map {f_n} x - \map f ... | Let $\struct {X, \Sigma, \mu}$ be a [[Definition:Finite Measure Space|finite measure space]].
Let $\sequence {f_n}_{n \mathop \in \N}, f_n : X \to \R$ be a [[Definition:Sequence|sequence]] of [[Definition:Measurable Function|measurable functions]].
Also, let $f: X \to \R$ be a [[Definition:Measurable Function|measura... | Since $\sequence {f_n}_{n \mathop \in \N}$ [[Definition:Convergence Almost Everywhere|converges almost everywhere]], we have:
:$\set {x \in X : \sequence {\map {f_n} x}_{n \mathop \in \N} \text { does not converge to } \map f x}$ is a [[Definition:Null Set|$\mu$-null set]].
We aim to deduce that:
:$\ds \lim_{n \ma... | Pointwise Convergence Implies Convergence in Measure on Finite Measure Space | https://proofwiki.org/wiki/Pointwise_Convergence_Implies_Convergence_in_Measure_on_Finite_Measure_Space | https://proofwiki.org/wiki/Pointwise_Convergence_Implies_Convergence_in_Measure_on_Finite_Measure_Space | [
"Convergence in Measure"
] | [
"Definition:Finite Measure Space",
"Definition:Sequence",
"Definition:Measurable Function",
"Definition:Measurable Function",
"Definition:Pointwise Limit",
"Definition:Almost Everywhere",
"Definition:Convergence in Measure"
] | [
"Definition:Convergence Almost Everywhere",
"Definition:Null Set",
"Definition:Decreasing Sequence of Sets",
"Definition:Finite Measure",
"Measure of Limit of Decreasing Sequence of Measurable Sets",
"Definition:Pointwise Convergence",
"Null Sets Closed under Subset",
"Set is Subset of Union",
"Meas... |
proofwiki-5856 | Convergence in Sigma-Finite Measure | Let $\struct {X, \Sigma, \mu}$ be a $\sigma$-finite measure space.
Let $\sequence {f_n}_{n \mathop \in \N}: f_n: X \to \R$ be a sequence of measurable functions.
Also, let $f, g: X \to \R$ be measurable functions.
Suppose that $f_n$ converges in measure to both $f$ and $g$ (in $\mu$).
Then $f$ and $g$ are equal $\mu$-a... | As $\mu$ is $\sigma$-finite, there is an exhausting sequence $\sequence {E_m}_m$ in $\Sigma$ such that:
:$\forall m : \map \mu {E_m} < \infty$
By triangle inequality:
:$\forall x \in X :\quad \cmod {\map f x - \map g x} \le \cmod {\map f x - \map {f_n} x} + \cmod {\map {f_n} x - \map g x}$
For all $m, k \in \N_{>0}$... | Let $\struct {X, \Sigma, \mu}$ be a [[Definition:Sigma-Finite Measure Space|$\sigma$-finite measure space]].
Let $\sequence {f_n}_{n \mathop \in \N}: f_n: X \to \R$ be a [[Definition:Sequence|sequence]] of [[Definition:Measurable Function|measurable functions]].
Also, let $f, g: X \to \R$ be [[Definition:Measurable F... | As $\mu$ is [[Definition:Sigma-Finite Measure|$\sigma$-finite]], there is an [[Definition:Exhausting Sequence of Sets|exhausting sequence]] $\sequence {E_m}_m$ in $\Sigma$ such that:
:$\forall m : \map \mu {E_m} < \infty$
By [[Triangle Inequality for Complex Numbers|triangle inequality]]:
:$\forall x \in X :\quad \cmo... | Convergence in Sigma-Finite Measure | https://proofwiki.org/wiki/Convergence_in_Sigma-Finite_Measure | https://proofwiki.org/wiki/Convergence_in_Sigma-Finite_Measure | [
"Measure Theory"
] | [
"Definition:Sigma-Finite Measure Space",
"Definition:Sequence",
"Definition:Measurable Function",
"Definition:Measurable Function",
"Definition:Convergence in Measure",
"Definition:Almost Everywhere"
] | [
"Definition:Sigma-Finite Measure",
"Definition:Exhausting Sequence of Sets",
"Triangle Inequality/Complex Numbers",
"Intersection Distributes over Union",
"Measure is Subadditive",
"Monotone Convergence Theorem",
"Measure is Subadditive"
] |
proofwiki-5857 | Vitali's Theorem | Let $\struct {X, \Sigma, \mu}$ be a $\sigma$-finite measure space, and let $p \in \R, p \ge 1$.
Let $\sequence {f_n}_{n \mathop \in \N}: f_n: X \to \R$ be a sequence of $p$-integrable functions.
Also, let $f: X \to \R$ be a measurable function.
Suppose that $\ds \operatorname*{\mu-\!\lim\,} \limits_{n \mathop \to \inft... | {{ProofWanted}}
{{Namedfor|Giuseppe Vitali|cat = Vitali}} | Let $\struct {X, \Sigma, \mu}$ be a [[Definition:Sigma-Finite Measure Space|$\sigma$-finite measure space]], and let $p \in \R, p \ge 1$.
Let $\sequence {f_n}_{n \mathop \in \N}: f_n: X \to \R$ be a [[Definition:Sequence|sequence]] of [[Definition:P-Integrable Function|$p$-integrable functions]].
Also, let $f: X \to ... | {{ProofWanted}}
{{Namedfor|Giuseppe Vitali|cat = Vitali}} | Vitali's Theorem | https://proofwiki.org/wiki/Vitali's_Theorem | https://proofwiki.org/wiki/Vitali's_Theorem | [
"Measure Theory"
] | [
"Definition:Sigma-Finite Measure Space",
"Definition:Sequence",
"Definition:Integrable Function/p-Integrable",
"Definition:Measurable Function",
"Definition:Convergence in Measure",
"Definition:P-Seminorm",
"Definition:Uniformly Integrable",
"Definition:Sigma-Finite Measure Space",
"Definition:Conve... | [] |
proofwiki-5858 | Element Commutes with Product of Commuting Elements/General Theorem | Let $\struct {S, \circ}$ be a semigroup.
Let $\sequence {a_k}_{1 \mathop \le k \mathop \le n}$ be a sequence of terms of $S$.
Let $b \in S$ such that $b$ commutes with $a_k$ for each $k \in \closedint 1 n$.
Then $b$ commutes with $a_1 \circ \cdots \circ a_n$. | The proof proceeds by induction on $n$, the length of $\sequence {a_k}_{1 \mathop \le k \mathop \le n}$. | Let $\struct {S, \circ}$ be a [[Definition:Semigroup|semigroup]].
Let $\sequence {a_k}_{1 \mathop \le k \mathop \le n}$ be a [[Definition:Sequence|sequence of terms]] of $S$.
Let $b \in S$ such that $b$ [[Definition:Commuting Elements|commutes]] with $a_k$ for each $k \in \closedint 1 n$.
Then $b$ [[Definition:Comm... | The proof proceeds by [[Principle of Mathematical Induction|induction]] on $n$, the [[Definition:Length of Sequence|length]] of $\sequence {a_k}_{1 \mathop \le k \mathop \le n}$. | Element Commutes with Product of Commuting Elements/General Theorem | https://proofwiki.org/wiki/Element_Commutes_with_Product_of_Commuting_Elements/General_Theorem | https://proofwiki.org/wiki/Element_Commutes_with_Product_of_Commuting_Elements/General_Theorem | [
"Element Commutes with Product of Commuting Elements"
] | [
"Definition:Semigroup",
"Definition:Sequence",
"Definition:Commutative/Elements",
"Definition:Commutative/Elements"
] | [
"Principle of Mathematical Induction",
"Definition:Length of Sequence",
"Definition:Length of Sequence",
"Definition:Length of Sequence",
"Definition:Length of Sequence",
"Principle of Mathematical Induction"
] |
proofwiki-5859 | Trivial Subgroup is Subgroup | Let $\struct {G, \circ}$ be a group whose identity is $e$.
Then the trivial subgroup $\struct {\set e, \circ}$ is indeed a subgroup of $\struct {G, \circ}$. | Using the One-Step Subgroup Test:
: $(1): \quad e \in \set e \leadsto \set e \ne \O$
: $(2): \quad e \in \set e \leadsto e \circ e^{-1} = e \in \set e$
{{qed}} | Let $\struct {G, \circ}$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
Then the [[Definition:Trivial Subgroup|trivial subgroup]] $\struct {\set e, \circ}$ is indeed a [[Definition:Subgroup|subgroup]] of $\struct {G, \circ}$. | Using the [[One-Step Subgroup Test]]:
: $(1): \quad e \in \set e \leadsto \set e \ne \O$
: $(2): \quad e \in \set e \leadsto e \circ e^{-1} = e \in \set e$
{{qed}} | Trivial Subgroup is Subgroup | https://proofwiki.org/wiki/Trivial_Subgroup_is_Subgroup | https://proofwiki.org/wiki/Trivial_Subgroup_is_Subgroup | [
"Subgroups"
] | [
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Trivial Subgroup",
"Definition:Subgroup"
] | [
"One-Step Subgroup Test"
] |
proofwiki-5860 | Trivial Subgroup is Normal | Let $\struct {G, \circ}$ be a group whose identity is $e$.
Then the trivial subgroup $\struct {\set e, \circ}$ of $G$ is a normal subgroup in $G$. | First, by Trivial Subgroup is Subgroup, $\struct {\set e, \circ}$ is a subgroup of $G$.
To show $\struct {\set e, \circ}$ is normal in $G$:
:$\forall a \in G: a \circ e \circ a^{-1} = a \circ a^{-1} = e$
Hence the result.
{{qed}} | Let $\struct {G, \circ}$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
Then the [[Definition:Trivial Subgroup|trivial subgroup]] $\struct {\set e, \circ}$ of $G$ is a [[Definition:Normal Subgroup|normal subgroup]] in $G$. | First, by [[Trivial Subgroup is Subgroup]], $\struct {\set e, \circ}$ is a [[Definition:Subgroup|subgroup]] of $G$.
To show $\struct {\set e, \circ}$ is [[Definition:Normal Subgroup|normal]] in $G$:
:$\forall a \in G: a \circ e \circ a^{-1} = a \circ a^{-1} = e$
Hence the result.
{{qed}} | Trivial Subgroup is Normal | https://proofwiki.org/wiki/Trivial_Subgroup_is_Normal | https://proofwiki.org/wiki/Trivial_Subgroup_is_Normal | [
"Normal Subgroups"
] | [
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Trivial Subgroup",
"Definition:Normal Subgroup"
] | [
"Trivial Subgroup is Subgroup",
"Definition:Subgroup",
"Definition:Normal Subgroup"
] |
proofwiki-5861 | Group is Normal in Itself | Let $\struct {G, \circ}$ be a group.
Then $\struct {G, \circ}$ is a normal subgroup of itself. | First, by Group is Subgroup of Itself, $\struct {G, \circ}$ is a subgroup of itself.
To show $\struct {G, \circ}$ is normal in $G$:
:$\forall a, g \in G: a \circ g \circ a^{-1} \in G$
as $G$ is closed by definition.
Hence the result.
{{qed}} | Let $\struct {G, \circ}$ be a [[Definition:Group|group]].
Then $\struct {G, \circ}$ is a [[Definition:Normal Subgroup|normal subgroup]] of itself. | First, by [[Group is Subgroup of Itself]], $\struct {G, \circ}$ is a [[Definition:Subgroup|subgroup]] of itself.
To show $\struct {G, \circ}$ is [[Definition:Normal Subgroup|normal]] in $G$:
:$\forall a, g \in G: a \circ g \circ a^{-1} \in G$
as $G$ is [[Definition:Closed Algebraic Structure|closed]] by definition.
... | Group is Normal in Itself | https://proofwiki.org/wiki/Group_is_Normal_in_Itself | https://proofwiki.org/wiki/Group_is_Normal_in_Itself | [
"Normal Subgroups"
] | [
"Definition:Group",
"Definition:Normal Subgroup"
] | [
"Group is Subgroup of Itself",
"Definition:Subgroup",
"Definition:Normal Subgroup",
"Definition:Closure (Abstract Algebra)/Algebraic Structure"
] |
proofwiki-5862 | Relativisation is Standard Model | Let $P$ be a well-formed formula.
Let $A$ be a finite set such that $x \in A$ {{iff}} $x$ is a free variable in $P$.
Then:
:$A \subseteq B \implies \paren {B \models P \iff P^B}$
{{explain|Definition of $P^B$}} | The proof proceeds by Induction on Well-Formed Formulas of $P$.
$P$ must be of the form:
:$x \in y$
:$\paren {Q \land R}$
:$\neg Q$
or:
:$\forall x: Q$
for some propositions $Q$ and $R$. | Let $P$ be a [[Definition:Well-Formed Formula|well-formed formula]].
Let $A$ be a finite set such that $x \in A$ {{iff}} $x$ is a [[Definition:Free Variable|free variable]] in $P$.
Then:
:$A \subseteq B \implies \paren {B \models P \iff P^B}$
{{explain|Definition of $P^B$}} | The proof proceeds by [[Induction on Well-Formed Formulas]] of $P$.
$P$ must be of the form:
:$x \in y$
:$\paren {Q \land R}$
:$\neg Q$
or:
:$\forall x: Q$
for some [[Definition:Proposition|propositions]] $Q$ and $R$. | Relativisation is Standard Model | https://proofwiki.org/wiki/Relativisation_is_Standard_Model | https://proofwiki.org/wiki/Relativisation_is_Standard_Model | [
"Inner Model Theory"
] | [
"Definition:Well-Formed Formula",
"Definition:Free Variable"
] | [
"Induction on Well-Formed Formulas",
"Definition:Proposition"
] |
proofwiki-5863 | Induction on Well-Formed Formulas | Let $\FF$ be a formal language with a bottom-up grammar.
Let $\Phi$ be a proposition about the well-formed formulas of $\FF$.
Suppose that $\Phi$ is true for all letters of $\FF$.
Suppose further that every rule of formation preserves $\Phi$, i.e. when fed well-formed formulas satisfying $\Phi$, it yields new well-form... | By definition of bottom-up grammar, the well-formed formulas of $\FF$ comprise:
:letters of $\FF$;
:expressions resulting from rules of formation.
Either case is dealt with by the assumptions on $\Phi$.
Hence the result, from Proof by Cases.
{{qed}}
Category:Formal Languages
dtt1zuemgcw8nkikkml6zac4n15jrf7 | Let $\FF$ be a [[Definition:Formal Language|formal language]] with a [[Definition:Bottom-Up Grammar|bottom-up grammar]].
Let $\Phi$ be a [[Definition:Proposition|proposition]] about the [[Definition:Well-Formed Formula|well-formed formulas]] of $\FF$.
Suppose that $\Phi$ is true for all [[Definition:Letter of Formal... | By definition of [[Definition:Bottom-Up Grammar|bottom-up grammar]], the [[Definition:Well-Formed Formula|well-formed formulas]] of $\FF$ comprise:
:[[Definition:Letter of Formal Language|letters]] of $\FF$;
:expressions resulting from [[Definition:Rule of Formation|rules of formation]].
Either case is dealt with by... | Induction on Well-Formed Formulas | https://proofwiki.org/wiki/Induction_on_Well-Formed_Formulas | https://proofwiki.org/wiki/Induction_on_Well-Formed_Formulas | [
"Formal Languages"
] | [
"Definition:Formal Language",
"Definition:Formal Grammar/Bottom-Up",
"Definition:Proposition",
"Definition:Well-Formed Formula",
"Definition:Formal Language/Alphabet/Letter",
"Definition:Rule of Formation",
"Definition:Well-Formed Formula",
"Definition:Well-Formed Formula",
"Definition:Well-Formed F... | [
"Definition:Formal Grammar/Bottom-Up",
"Definition:Well-Formed Formula",
"Definition:Formal Language/Alphabet/Letter",
"Definition:Rule of Formation",
"Proof by Cases",
"Category:Formal Languages"
] |
proofwiki-5864 | Substitution of Elements | Let $a$, $b$, and $x$ be sets.
:$a = b \implies \paren {a \in x \iff b \in x}$ | By the {{Axiom-link|Extension|Sets}}:
:$a = b \implies \paren {a \in x \implies b \in x}$
Equality is Symmetric, so also by the {{Axiom-link|Extension|Sets}}:
:$a = b \implies \paren {b \in x \implies a \in x}$
{{qed}} | Let $a$, $b$, and $x$ be [[Definition:Set|sets]].
:$a = b \implies \paren {a \in x \iff b \in x}$ | By the {{Axiom-link|Extension|Sets}}:
:$a = b \implies \paren {a \in x \implies b \in x}$
[[Equality is Symmetric]], so also by the {{Axiom-link|Extension|Sets}}:
:$a = b \implies \paren {b \in x \implies a \in x}$
{{qed}} | Substitution of Elements | https://proofwiki.org/wiki/Substitution_of_Elements | https://proofwiki.org/wiki/Substitution_of_Elements | [
"Set Theory"
] | [
"Definition:Set"
] | [
"Equality is Symmetric"
] |
proofwiki-5865 | Limit Ordinals Preserved Under Ordinal Multiplication | Let $x$ and $y$ be ordinals.
Let $x$ be non-empty.
Let $y$ be a limit ordinal.
It follows that the ordinal product $\left({x \times y}\right)$ is a limit ordinal. | $y$ is a limit ordinal and thus is nonzero, by definition.
$x$ and $y$ are both nonzero.
So by Ordinals have No Zero Divisors:
:$x \times y \ne 0$
So by definition of limit ordinal:
:$x \times y \in K_{II} \lor \exists z \in \On: x \times y = z^+$
{{AimForCont}} $x \times y = z^+$ for some ordinal $z$.
{{begin-eqn}}
{{... | Let $x$ and $y$ be [[Definition:Ordinal|ordinals]].
Let $x$ be [[Definition:Non-Empty Set|non-empty]].
Let $y$ be a [[Definition:Limit Ordinal|limit ordinal]].
It follows that the [[Definition:Ordinal Multiplication|ordinal product]] $\left({x \times y}\right)$ is a [[Definition:Limit Ordinal|limit ordinal]]. | $y$ is a limit ordinal and thus is nonzero, by definition.
$x$ and $y$ are both nonzero.
So by [[Ordinals have No Zero Divisors]]:
:$x \times y \ne 0$
So by definition of [[Definition:Limit Ordinal|limit ordinal]]:
:$x \times y \in K_{II} \lor \exists z \in \On: x \times y = z^+$
{{AimForCont}} $x \times y = z^+$ ... | Limit Ordinals Preserved Under Ordinal Multiplication | https://proofwiki.org/wiki/Limit_Ordinals_Preserved_Under_Ordinal_Multiplication | https://proofwiki.org/wiki/Limit_Ordinals_Preserved_Under_Ordinal_Multiplication | [
"Ordinal Arithmetic",
"Limit Ordinals"
] | [
"Definition:Ordinal",
"Definition:Non-Empty Set",
"Definition:Limit Ordinal",
"Definition:Ordinal Multiplication",
"Definition:Limit Ordinal"
] | [
"Ordinals have No Zero Divisors",
"Definition:Limit Ordinal",
"Successor is Less than Successor",
"Ordinal Addition/Examples/Ordinal Addition by One",
"Subset is Right Compatible with Ordinal Addition",
"Successor of Ordinal Smaller than Limit Ordinal is also Smaller"
] |
proofwiki-5866 | Ordinal is Less than Ordinal times Limit | Let $y$ be a limit ordinal.
Let $x$ and $z$ be ordinals.
Then:
:$z < x \times y \iff \exists w \in y: z < x \times w$ | {{begin-eqn}}
{{eqn | l = z
| o = <
| r = x \times y
}}
{{eqn | ll= \leadstoandfrom
| l = z
| o = <
| r = \bigcup_{w \mathop \in y} \paren {x \times w}
| c = {{Defof|Ordinal Multiplication}}
}}
{{eqn | ll= \leadstoandfrom
| q = \exists w \in y
| l = z
| o = <
... | Let $y$ be a [[Definition:Limit Ordinal|limit ordinal]].
Let $x$ and $z$ be [[Definition:Ordinal|ordinals]].
Then:
:$z < x \times y \iff \exists w \in y: z < x \times w$ | {{begin-eqn}}
{{eqn | l = z
| o = <
| r = x \times y
}}
{{eqn | ll= \leadstoandfrom
| l = z
| o = <
| r = \bigcup_{w \mathop \in y} \paren {x \times w}
| c = {{Defof|Ordinal Multiplication}}
}}
{{eqn | ll= \leadstoandfrom
| q = \exists w \in y
| l = z
| o = <
... | Ordinal is Less than Ordinal times Limit | https://proofwiki.org/wiki/Ordinal_is_Less_than_Ordinal_times_Limit | https://proofwiki.org/wiki/Ordinal_is_Less_than_Ordinal_times_Limit | [
"Ordinal Arithmetic"
] | [
"Definition:Limit Ordinal",
"Definition:Ordinal"
] | [] |
proofwiki-5867 | Class Equality is Reflexive | Let $A$ be a class.
Then:
:$A = A$
where $=$ denotes class equality. | {{NotZFC}}
:$\forall x: \left({ x \in A \iff x \in A }\right)$
{{explain|More explanation is needed as to why this constitutes a proof.}}
{{qed}} | Let $A$ be a [[Definition:Class (Class Theory)|class]].
Then:
:$A = A$
where $=$ denotes [[Definition:Class Equality|class equality]]. | {{NotZFC}}
:$\forall x: \left({ x \in A \iff x \in A }\right)$
{{explain|More explanation is needed as to why this constitutes a proof.}}
{{qed}} | Class Equality is Reflexive | https://proofwiki.org/wiki/Class_Equality_is_Reflexive | https://proofwiki.org/wiki/Class_Equality_is_Reflexive | [
"Class Theory"
] | [
"Definition:Class (Class Theory)",
"Definition:Class Equality"
] | [] |
proofwiki-5868 | Class Equality is Symmetric | Let $A$ and $B$ be classes.
Let $=$ denote class equality.
Then:
:$A = B \implies B = A$ | {{NotZFC}}
From Biconditional is Commutative:
:$\forall x: \left({ x \in A \iff x \in B }\right) \implies \forall x: \left({ x \in B \iff x \in A }\right)$
Hence the result by definition of class equality. | Let $A$ and $B$ be [[Definition:Class (Class Theory)|classes]].
Let $=$ denote [[Definition:Class Equality|class equality]].
Then:
:$A = B \implies B = A$ | {{NotZFC}}
From [[Biconditional is Commutative]]:
:$\forall x: \left({ x \in A \iff x \in B }\right) \implies \forall x: \left({ x \in B \iff x \in A }\right)$
Hence the result by definition of [[Definition:Class Equality|class equality]]. | Class Equality is Symmetric | https://proofwiki.org/wiki/Class_Equality_is_Symmetric | https://proofwiki.org/wiki/Class_Equality_is_Symmetric | [
"Class Theory"
] | [
"Definition:Class (Class Theory)",
"Definition:Class Equality"
] | [
"Biconditional is Commutative",
"Definition:Class Equality"
] |
proofwiki-5869 | Class Equality is Transitive | Let $A$, $B$, and $C$ be classes.
Let $=$ denote class equality.
Then
:$\left({ A = B \land B = C }\right) \implies B = A$ | {{NotZFC}}
By Universal Generalisation and Biconditional is Transitive:
:$\forall x: \left({ \left({ x \in A \iff x \in B }\right) \land \left({ x \in B \iff x \in C }\right) }\right) \implies \forall x: \left({ x \in A \iff x \in C }\right)$ by Universal Generalisation and Biconditional is Transitive
{{explain|Further... | Let $A$, $B$, and $C$ be [[Definition:Class (Class Theory)|classes]].
Let $=$ denote [[Definition:Class Equality|class equality]].
Then
:$\left({ A = B \land B = C }\right) \implies B = A$ | {{NotZFC}}
By [[Universal Generalisation]] and [[Biconditional is Transitive]]:
:$\forall x: \left({ \left({ x \in A \iff x \in B }\right) \land \left({ x \in B \iff x \in C }\right) }\right) \implies \forall x: \left({ x \in A \iff x \in C }\right)$ by [[Universal Generalisation]] and [[Biconditional is Transitive]]
... | Class Equality is Transitive | https://proofwiki.org/wiki/Class_Equality_is_Transitive | https://proofwiki.org/wiki/Class_Equality_is_Transitive | [
"Class Theory"
] | [
"Definition:Class (Class Theory)",
"Definition:Class Equality"
] | [
"Universal Generalisation",
"Biconditional is Transitive",
"Universal Generalisation",
"Biconditional is Transitive"
] |
proofwiki-5870 | Substitutivity of Class Equality | Let $A$ and $B$ be classes.
Let $\map P A$ be a well-formed formula of the language of set theory.
Let $\map P B$ be the same proposition $\map P A$ with all instances of $A$ replaced with instances of $B$.
Let $=$ denote class equality.
:$A = B \implies \paren {\map P A \iff \map P B}$ | By induction on the well-formed parts of $\map P A$.
The proof shall use $\implies$ and $\neg$ as the primitive connectives. | Let $A$ and $B$ be [[Definition:Class (Class Theory)|classes]].
Let $\map P A$ be a [[Definition:Well-Formed Formula|well-formed formula]] of the [[Definition:Language of Set Theory|language of set theory]].
Let $\map P B$ be the same proposition $\map P A$ with all instances of $A$ replaced with instances of $B$.
L... | By [[Induction on Well-Formed Formulas|induction on the well-formed parts]] of $\map P A$.
The proof shall use $\implies$ and $\neg$ as the [[Definition:Language of Set Theory|primitive connectives]]. | Substitutivity of Class Equality | https://proofwiki.org/wiki/Substitutivity_of_Class_Equality | https://proofwiki.org/wiki/Substitutivity_of_Class_Equality | [
"Class Theory"
] | [
"Definition:Class (Class Theory)",
"Definition:Well-Formed Formula",
"Definition:Language of Set Theory",
"Definition:Class Equality"
] | [
"Induction on Well-Formed Formulas",
"Definition:Language of Set Theory"
] |
proofwiki-5871 | Class is Extensional | Let $A$ be a class.
Then:
:$A = \set {x : x \in A}$
That is, $A$ is extensional. | We have:
:$x \in \set {x : x \in A} \iff x \in A$
by Characterization of Class Membership (applied to $\set {x : x \in A}$).
By Universal Generalisation, it follows that:
:$\forall x: \paren {x \in A \iff x \in \set {x : x \in A} }$
Hence the result, by definition of class equality.
{{qed}} | Let $A$ be a [[Definition:Class (Class Theory)|class]].
Then:
:$A = \set {x : x \in A}$
That is, $A$ is [[Axiom:Axiom of Extension (Classes)|extensional]]. | We have:
:$x \in \set {x : x \in A} \iff x \in A$
by [[Characterization of Class Membership]] (applied to $\set {x : x \in A}$).
By [[Universal Generalisation]], it follows that:
:$\forall x: \paren {x \in A \iff x \in \set {x : x \in A} }$
Hence the result, by definition of [[Definition:Class Equality|class equa... | Class is Extensional | https://proofwiki.org/wiki/Class_is_Extensional | https://proofwiki.org/wiki/Class_is_Extensional | [
"Class Theory"
] | [
"Definition:Class (Class Theory)",
"Axiom:Axiom of Extension/Class Theory"
] | [
"Characterization of Class Membership",
"Universal Generalisation",
"Definition:Class Equality"
] |
proofwiki-5872 | Set is Small Class | Let $x$ be a set.
Then $x$ is a small class. | :$x = x$
Therefore by Existential Generalisation:
:$\exists y: x = y$
{{qed}}
{{explain|Add the appropriate words to link the above statements to the fact being proved, e.g. "Therefore by definition of ..." etc.}} | Let $x$ be a [[Definition:Set|set]].
Then $x$ is a [[Definition:Small Class|small class]]. | :$x = x$
Therefore by [[Existential Generalisation]]:
:$\exists y: x = y$
{{qed}}
{{explain|Add the appropriate words to link the above statements to the fact being proved, e.g. "Therefore by definition of ..." etc.}} | Set is Small Class | https://proofwiki.org/wiki/Set_is_Small_Class | https://proofwiki.org/wiki/Set_is_Small_Class | [
"Set Theory",
"Class Theory"
] | [
"Definition:Set",
"Definition:Small Class"
] | [
"Existential Generalisation"
] |
proofwiki-5873 | Class Member of Class Builder | Let $A$ be a class.
Let $x$ be a set.
Let $\map P x$ be a well-formed formula in the language of set theory.
Let $\map P A$ denote the formula $\map P x$ with all free instances of $x$ replaced with instances of $A$.
Let $\set {x: \map P x}$ be a class specified using class builder notation.
Then:
:$A \in \set {x: \map... | {{begin-eqn}}
{{eqn | l = A
| o = \in
| m = \leftset x
| mo= :
| r = \rightset {\map P x}
}}
{{eqn | ll= \leadsto
| q = \exists x \in \set {x: \map P x}
| l = A
| m = x
| c = {{Defof|Class (Zermelo-Fraenkel)}}
}}
{{eqn | ll= \leadsto
| q = \exists x
| l = x
... | Let $A$ be a [[Definition:Class (Class Theory)|class]].
Let $x$ be a [[Definition:Set|set]].
Let $\map P x$ be a [[Definition:Well-Formed Formula|well-formed formula]] in the [[Definition:Language of Set Theory|language of set theory]].
Let $\map P A$ denote the formula $\map P x$ with all [[Definition:Free Variabl... | {{begin-eqn}}
{{eqn | l = A
| o = \in
| m = \leftset x
| mo= :
| r = \rightset {\map P x}
}}
{{eqn | ll= \leadsto
| q = \exists x \in \set {x: \map P x}
| l = A
| m = x
| c = {{Defof|Class (Zermelo-Fraenkel)}}
}}
{{eqn | ll= \leadsto
| q = \exists x
| l = x
... | Class Member of Class Builder | https://proofwiki.org/wiki/Class_Member_of_Class_Builder | https://proofwiki.org/wiki/Class_Member_of_Class_Builder | [
"Zermelo-Fraenkel Class Theory"
] | [
"Definition:Class (Class Theory)",
"Definition:Set",
"Definition:Well-Formed Formula",
"Definition:Language of Set Theory",
"Definition:Free Variable",
"Definition:Class (Class Theory)",
"Definition:Class Builder"
] | [
"Substitutivity of Class Equality"
] |
proofwiki-5874 | Class Equality Extension of Set Equality | Let $=_s$ denote set equality.
Let $=_c$ denote class equality.
Let $x$ and $y$ be sets.
Then $x =_s y$ {{iff}} $x =_c y$. | {{begin-eqn}}
{{eqn | l = x
| o = =_s
| r = y
| c =
}}
{{eqn | ll= \leadstoandfrom
| q = \forall z
| l = \leftparen {z \in x}
| o = \iff
| r = \rightparen {z \in y}
| c = {{Defof|Set Equality}}
}}
{{eqn | ll= \leadstoandfrom
| l = x
| o = =_c
| r = y
... | Let $=_s$ denote [[Definition:Set Equality|set equality]].
Let $=_c$ denote [[Definition:Class Equality|class equality]].
Let $x$ and $y$ be [[Definition:Set|sets]].
Then $x =_s y$ {{iff}} $x =_c y$. | {{begin-eqn}}
{{eqn | l = x
| o = =_s
| r = y
| c =
}}
{{eqn | ll= \leadstoandfrom
| q = \forall z
| l = \leftparen {z \in x}
| o = \iff
| r = \rightparen {z \in y}
| c = {{Defof|Set Equality}}
}}
{{eqn | ll= \leadstoandfrom
| l = x
| o = =_c
| r = y
... | Class Equality Extension of Set Equality | https://proofwiki.org/wiki/Class_Equality_Extension_of_Set_Equality | https://proofwiki.org/wiki/Class_Equality_Extension_of_Set_Equality | [
"Zermelo-Fraenkel Class Theory"
] | [
"Definition:Set Equality",
"Definition:Class Equality",
"Definition:Set"
] | [
"Class Membership is Extension of Set Membership",
":Category:Zermelo-Fraenkel Class Theory",
"Definition:Von Neumann-Bernays-Gödel Set Theory",
"Category:Zermelo-Fraenkel Class Theory"
] |
proofwiki-5875 | B-Algebra Induced by Group Induced by B-Algebra | Let $\struct {S, *}$ be a $B$-algebra.
Let $\struct {S, \circ}$ be the group described on $B$-Algebra Induces Group.
Let $\struct {S, *'}$ be the $B$-algebra described on Group Induces $B$-Algebra.
Then $\struct {S, *'} = \struct {S, *}$. | Let $a, b \in S$.
It is required to show that:
:$a *' b = a * b$
To achieve this, recall that $*'$ is defined on Group Induces $B$-Algebra to satisfy:
:$a *' b = a \circ b^{-1}$
which, by the definition of $\circ$ on $B$-Algebra Induces Group comes down to:
{{begin-eqn}}
{{eqn | l = a *' b
| r = a * \paren {0 * b... | Let $\struct {S, *}$ be a [[Definition:B-Algebra|$B$-algebra]].
Let $\struct {S, \circ}$ be the [[Definition:Group|group]] described on [[B-Algebra Induces Group|$B$-Algebra Induces Group]].
Let $\struct {S, *'}$ be the [[Definition:B-Algebra|$B$-algebra]] described on [[Group Induces B-Algebra|Group Induces $B$-Alge... | Let $a, b \in S$.
It is required to show that:
:$a *' b = a * b$
To achieve this, recall that $*'$ is defined on [[Group Induces B-Algebra|Group Induces $B$-Algebra]] to satisfy:
:$a *' b = a \circ b^{-1}$
which, by the definition of $\circ$ on [[B-Algebra Induces Group|$B$-Algebra Induces Group]] comes down to:
... | B-Algebra Induced by Group Induced by B-Algebra | https://proofwiki.org/wiki/B-Algebra_Induced_by_Group_Induced_by_B-Algebra | https://proofwiki.org/wiki/B-Algebra_Induced_by_Group_Induced_by_B-Algebra | [
"Group Theory",
"B-Algebras"
] | [
"Definition:B-Algebra",
"Definition:Group",
"B-Algebra Induces Group",
"Definition:B-Algebra",
"Group Induces B-Algebra"
] | [
"Group Induces B-Algebra",
"B-Algebra Induces Group",
"B-Algebra Induces Group",
"B-Algebra Identity: 0(0x)=x"
] |
proofwiki-5876 | Group Induced by B-Algebra Induced by Group | Let $\struct {S, \circ}$ be a group.
Let $\struct {S, *}$ be the $B$-algebra described on Group Induces $B$-Algebra.
Let $\struct {S, \circ'}$ be the group described on $B$-Algebra Induces Group.
Then $\struct {S, \circ'} = \struct {S, \circ}$. | Let $a, b \in S$.
It is required to show that:
:$a \circ' b = a \circ b$
To achieve this, recall that $\circ'$ is defined on $B$-Algebra Induces Group to satisfy:
:$a \circ' b = a * \paren {e * b}$
which, by the definition of $*$ on Group Induces $B$-Algebra comes down to:
{{begin-eqn}}
{{eqn | l = a \circ' b
| r... | Let $\struct {S, \circ}$ be a [[Definition:Group|group]].
Let $\struct {S, *}$ be the [[Definition:B-Algebra|$B$-algebra]] described on [[Group Induces B-Algebra|Group Induces $B$-Algebra]].
Let $\struct {S, \circ'}$ be the [[Definition:Group|group]] described on [[B-Algebra Induces Group|$B$-Algebra Induces Group]].... | Let $a, b \in S$.
It is required to show that:
:$a \circ' b = a \circ b$
To achieve this, recall that $\circ'$ is defined on [[B-Algebra Induces Group|$B$-Algebra Induces Group]] to satisfy:
:$a \circ' b = a * \paren {e * b}$
which, by the definition of $*$ on [[Group Induces B-Algebra|Group Induces $B$-Algebra]]... | Group Induced by B-Algebra Induced by Group | https://proofwiki.org/wiki/Group_Induced_by_B-Algebra_Induced_by_Group | https://proofwiki.org/wiki/Group_Induced_by_B-Algebra_Induced_by_Group | [
"Group Theory",
"B-Algebras"
] | [
"Definition:Group",
"Definition:B-Algebra",
"Group Induces B-Algebra",
"Definition:Group",
"B-Algebra Induces Group"
] | [
"B-Algebra Induces Group",
"Group Induces B-Algebra",
"Inverse of Group Inverse"
] |
proofwiki-5877 | Union of Doubleton | Let $x$ and $y$ be sets.
Let $\set {x, y}$ be a doubleton.
Then $\bigcup \set {x, y}$ is a set such that:
:$\bigcup \set {x, y} = x \cup y$ | {{begin-eqn}}
{{eqn | o =
| r = z \in \bigcup \set {x, y}
}}
{{eqn | ll= \leadstoandfrom
| o =
| r = \exists w \in \set {x, y}: z \in w
| c = {{Defof|Union of Class}}
}}
{{eqn | ll= \leadstoandfrom
| o =
| r = \exists w: \paren {\paren {w = x \lor w = y} \land z \in w}
| c =... | Let $x$ and $y$ be [[Definition:Set|sets]].
Let $\set {x, y}$ be a [[Definition:Doubleton Class|doubleton]].
Then $\bigcup \set {x, y}$ is a [[Definition:Set|set]] such that:
:$\bigcup \set {x, y} = x \cup y$ | {{begin-eqn}}
{{eqn | o =
| r = z \in \bigcup \set {x, y}
}}
{{eqn | ll= \leadstoandfrom
| o =
| r = \exists w \in \set {x, y}: z \in w
| c = {{Defof|Union of Class}}
}}
{{eqn | ll= \leadstoandfrom
| o =
| r = \exists w: \paren {\paren {w = x \lor w = y} \land z \in w}
| c =... | Union of Doubleton | https://proofwiki.org/wiki/Union_of_Doubleton | https://proofwiki.org/wiki/Union_of_Doubleton | [
"Class Union",
"Doubleton Classes"
] | [
"Definition:Set",
"Definition:Doubleton/Class Theory",
"Definition:Set"
] | [
"Equality implies Substitution",
"Axiom:Axiom of Unions/Class Theory",
"Definition:Set"
] |
proofwiki-5878 | Union of Small Classes is Small | Let $x$ and $y$ be small classes.
Then $x \cup y$ is also small. | Let $\map {\mathscr M} A$ denote that $A$ is small.
{{begin-eqn}}
{{eqn | l = \map {\mathscr M} x \land \map {\mathscr M} y
| o = \leadsto
| r = \map {\mathscr M} {\set {x, y} }
| c = {{Axiom-link|Pairing|Class Theory}}
}}
{{eqn | o = \leadsto
| r = \map {\mathscr M} {\bigcup \set {x, y} }
... | Let $x$ and $y$ be [[Definition:Small Class|small classes]].
Then $x \cup y$ is also [[Definition:Small Class|small]]. | Let $\map {\mathscr M} A$ denote that $A$ is [[Definition:Small Class|small]].
{{begin-eqn}}
{{eqn | l = \map {\mathscr M} x \land \map {\mathscr M} y
| o = \leadsto
| r = \map {\mathscr M} {\set {x, y} }
| c = {{Axiom-link|Pairing|Class Theory}}
}}
{{eqn | o = \leadsto
| r = \map {\mathscr M}... | Union of Small Classes is Small | https://proofwiki.org/wiki/Union_of_Small_Classes_is_Small | https://proofwiki.org/wiki/Union_of_Small_Classes_is_Small | [
"Set Union",
"Zermelo-Fraenkel Class Theory"
] | [
"Definition:Small Class",
"Definition:Small Class"
] | [
"Definition:Small Class",
"Union of Doubleton"
] |
proofwiki-5879 | Category of Finite Sets is Category | Let $\mathbf{Set}$ be the category of finite sets.
Then $\mathbf{Set}$ is a metacategory. | {{mutatis}} the same as that of Category of Sets is Category.
{{qed}}
Category:Category of Finite Sets
r9us00xl8llon0j528uau6br64jkgnt | Let $\mathbf{Set}$ be the [[Definition:Category of Finite Sets|category of finite sets]].
Then $\mathbf{Set}$ is a [[Definition:Metacategory|metacategory]]. | {{mutatis}} the same as that of [[Category of Sets is Category]].
{{qed}}
[[Category:Category of Finite Sets]]
r9us00xl8llon0j528uau6br64jkgnt | Category of Finite Sets is Category | https://proofwiki.org/wiki/Category_of_Finite_Sets_is_Category | https://proofwiki.org/wiki/Category_of_Finite_Sets_is_Category | [
"Category of Finite Sets"
] | [
"Definition:Category of Finite Sets",
"Definition:Metacategory"
] | [
"Category of Sets is Category",
"Category:Category of Finite Sets"
] |
proofwiki-5880 | Category of Ordered Sets is Category | Let $\mathbf{OrdSet}$ be the category of ordered sets.
Then $\mathbf{OrdSet}$ is a metacategory. | Let us verify the axioms $(C1)$ up to $(C3)$ for a metacategory.
For any two increasing mappings their composition (in the usual set theoretic sense) is again increasing by Composite of Increasing Mappings is Increasing.
For any set $X$, we have the identity mapping $\operatorname{id}_X$.
By Identity Mapping is Left Id... | Let $\mathbf{OrdSet}$ be the [[Definition:Category of Ordered Sets|category of ordered sets]].
Then $\mathbf{OrdSet}$ is a [[Definition:Metacategory|metacategory]]. | Let us verify the axioms $(C1)$ up to $(C3)$ for a [[Definition:Metacategory|metacategory]].
For any two [[Definition:Increasing Mapping|increasing mappings]] their [[Definition:Composition of Mappings|composition]] (in the usual [[Definition:Set Theory|set theoretic]] sense) is again [[Definition:Increasing Mapping|... | Category of Ordered Sets is Category | https://proofwiki.org/wiki/Category_of_Ordered_Sets_is_Category | https://proofwiki.org/wiki/Category_of_Ordered_Sets_is_Category | [
"Category of Ordered Sets"
] | [
"Definition:Category of Ordered Sets",
"Definition:Metacategory"
] | [
"Definition:Metacategory",
"Definition:Increasing/Mapping",
"Definition:Composition of Mappings",
"Definition:Set Theory",
"Definition:Increasing/Mapping",
"Composite of Increasing Mappings is Increasing",
"Definition:Identity Mapping",
"Identity Mapping is Left Identity",
"Identity Mapping is Right... |
proofwiki-5881 | Set Difference is Set | Let $x$ be a small class.
Let $A$ be a class.
Let $\map \MM B$ denote that $B$ is small.
Then:
:$\map \MM {x \setminus A}$ | Consider $R, S, T \subseteq \mathbb U$, where $\mathbb U$ is considered as the universe.
Then:
{{begin-eqn}}
{{eqn | l = \paren {R \setminus S} \cap T
| r = \paren {R \cap \map \complement S} \cap T
| c = Set Difference as Intersection with Complement
}}
{{eqn | r = \paren {R \cap T} \cap \map \complement S... | Let $x$ be a [[Definition:Small Class|small class]].
Let $A$ be a [[Definition:Class (Class Theory)|class]].
Let $\map \MM B$ denote that $B$ is [[Definition:Small Class|small]].
Then:
:$\map \MM {x \setminus A}$ | Consider $R, S, T \subseteq \mathbb U$, where $\mathbb U$ is considered as the [[Definition:Universal Set|universe]].
Then:
{{begin-eqn}}
{{eqn | l = \paren {R \setminus S} \cap T
| r = \paren {R \cap \map \complement S} \cap T
| c = [[Set Difference as Intersection with Complement]]
}}
{{eqn | r = \pare... | Intersection with Set Difference is Set Difference with Intersection/Proof 1 | https://proofwiki.org/wiki/Set_Difference_is_Set | https://proofwiki.org/wiki/Intersection_with_Set_Difference_is_Set_Difference_with_Intersection/Proof_1 | [
"Class Theory"
] | [
"Definition:Small Class",
"Definition:Class (Class Theory)",
"Definition:Small Class"
] | [
"Definition:Universal Set",
"Set Difference as Intersection with Complement",
"Intersection is Commutative",
"Intersection is Associative",
"Set Difference as Intersection with Complement"
] |
proofwiki-5882 | Set Difference is Set | Let $x$ be a small class.
Let $A$ be a class.
Let $\map \MM B$ denote that $B$ is small.
Then:
:$\map \MM {x \setminus A}$ | {{begin-eqn}}
{{eqn | o =
| r = x \in \paren {R \setminus S} \cap T
| c =
}}
{{eqn | ll= \leadstoandfrom
| o =
| r = \paren {x \in R \land x \notin S} \land x \in T
| c = {{Defof|Set Intersection}} and {{Defof|Set Difference}}
}}
{{eqn | ll= \leadstoandfrom
| o =
| r = \par... | Let $x$ be a [[Definition:Small Class|small class]].
Let $A$ be a [[Definition:Class (Class Theory)|class]].
Let $\map \MM B$ denote that $B$ is [[Definition:Small Class|small]].
Then:
:$\map \MM {x \setminus A}$ | {{begin-eqn}}
{{eqn | o =
| r = x \in \paren {R \setminus S} \cap T
| c =
}}
{{eqn | ll= \leadstoandfrom
| o =
| r = \paren {x \in R \land x \notin S} \land x \in T
| c = {{Defof|Set Intersection}} and {{Defof|Set Difference}}
}}
{{eqn | ll= \leadstoandfrom
| o =
| r = \par... | Intersection with Set Difference is Set Difference with Intersection/Proof 2 | https://proofwiki.org/wiki/Set_Difference_is_Set | https://proofwiki.org/wiki/Intersection_with_Set_Difference_is_Set_Difference_with_Intersection/Proof_2 | [
"Class Theory"
] | [
"Definition:Small Class",
"Definition:Class (Class Theory)",
"Definition:Small Class"
] | [
"Rule of Commutation",
"Rule of Association"
] |
proofwiki-5883 | Set Difference is Set | Let $x$ be a small class.
Let $A$ be a class.
Let $\map \MM B$ denote that $B$ is small.
Then:
:$\map \MM {x \setminus A}$ | {{NotZFC}}
By Set Difference as Intersection with Relative Complement:
:$\paren {x \setminus A} = \paren {x \cap \map \complement A}$
{{wtd|This rule must be proven for classes}}
Next, by Axiom of Subsets Equivalents, $\paren {x \cap \map \complement A}$ is small.
Therefore:
:$\map \MM {x \setminus A}$
{{qed}} | Let $x$ be a [[Definition:Small Class|small class]].
Let $A$ be a [[Definition:Class (Class Theory)|class]].
Let $\map \MM B$ denote that $B$ is [[Definition:Small Class|small]].
Then:
:$\map \MM {x \setminus A}$ | {{NotZFC}}
By [[Set Difference as Intersection with Relative Complement]]:
:$\paren {x \setminus A} = \paren {x \cap \map \complement A}$
{{wtd|This rule must be proven for classes}}
Next, by [[Axiom of Subsets Equivalents]], $\paren {x \cap \map \complement A}$ is [[Definition:Small Class|small]].
Therefore:
:$\... | Set Difference is Set | https://proofwiki.org/wiki/Set_Difference_is_Set | https://proofwiki.org/wiki/Set_Difference_is_Set | [
"Class Theory"
] | [
"Definition:Small Class",
"Definition:Class (Class Theory)",
"Definition:Small Class"
] | [
"Set Difference as Intersection with Relative Complement",
"Axiom of Subsets Equivalents",
"Definition:Small Class"
] |
proofwiki-5884 | Nonempty Class has Members | Let $A$ be a class.
Then:
:$A \ne \O \iff \exists x: x \in A$ | {{NotZFC}}
{{begin-eqn}}
{{eqn | l = A \ne \O
| o = \leadstoandfrom
| r = \neg \forall x: \paren {x \in A \iff x \in \O}
| c = {{Defof|Class Equality}}
}}
{{eqn | o = \leadstoandfrom
| r = \neg \forall x: \neg x \in A
| c = {{Defof|Empty Set}}
}}
{{eqn | o = \leadstoandfrom
| r = \ex... | Let $A$ be a [[Definition:Class (Class Theory)|class]].
Then:
:$A \ne \O \iff \exists x: x \in A$ | {{NotZFC}}
{{begin-eqn}}
{{eqn | l = A \ne \O
| o = \leadstoandfrom
| r = \neg \forall x: \paren {x \in A \iff x \in \O}
| c = {{Defof|Class Equality}}
}}
{{eqn | o = \leadstoandfrom
| r = \neg \forall x: \neg x \in A
| c = {{Defof|Empty Set}}
}}
{{eqn | o = \leadstoandfrom
| r = \e... | Nonempty Class has Members | https://proofwiki.org/wiki/Nonempty_Class_has_Members | https://proofwiki.org/wiki/Nonempty_Class_has_Members | [
"Class Theory",
"Empty Set"
] | [
"Definition:Class (Class Theory)"
] | [
"De Morgan's Laws (Predicate Logic)"
] |
proofwiki-5885 | Class is Not Element of Itself | Let $A$ be a class.
Then $A$ is not an element of itself:
:$A \notin A$ | $A \in A$ would create a relational loop for the class membership sign.
By No Membership Loops, this is a contradiction.
{{qed}} | Let $A$ be a [[Definition:Class (Class Theory)|class]].
Then $A$ is not an [[Definition:Element of Class|element]] of itself:
:$A \notin A$ | $A \in A$ would create a [[Definition:Relational Loop|relational loop]] for the [[Definition:Element of Class|class membership]] sign.
By [[No Membership Loops]], this is a contradiction.
{{qed}} | Class is Not Element of Itself | https://proofwiki.org/wiki/Class_is_Not_Element_of_Itself | https://proofwiki.org/wiki/Class_is_Not_Element_of_Itself | [
"Class Theory",
"Axiom of Foundation"
] | [
"Definition:Class (Class Theory)",
"Definition:Element/Class"
] | [
"Definition:Relational Loop",
"Definition:Element/Class",
"No Membership Loops"
] |
proofwiki-5886 | Category of Relations is Category | Let $\mathbf{Rel}$ be the category of relations.
Then $\mathbf{Rel}$ is a metacategory. | Let us verify the axioms $(C1)$ up to $(C3)$ for a metacategory.
We have already defined the composition of morphisms.
For any set $X$, we have the diagonal relation $\operatorname{id}_X$.
By Diagonal Relation is Left Identity and Diagonal Relation is Right Identity it follows that this is the identity morphism for $X$... | Let $\mathbf{Rel}$ be the [[Definition:Category of Relations|category of relations]].
Then $\mathbf{Rel}$ is a [[Definition:Metacategory|metacategory]]. | Let us verify the axioms $(C1)$ up to $(C3)$ for a [[Definition:Metacategory|metacategory]].
We have already defined the [[Definition:Composition of Morphisms|composition of morphisms]].
For any set $X$, we have the [[Definition:Diagonal Relation|diagonal relation]] $\operatorname{id}_X$.
By [[Diagonal Relation is ... | Category of Relations is Category | https://proofwiki.org/wiki/Category_of_Relations_is_Category | https://proofwiki.org/wiki/Category_of_Relations_is_Category | [
"Category of Relations"
] | [
"Definition:Category of Relations",
"Definition:Metacategory"
] | [
"Definition:Metacategory",
"Definition:Composition of Morphisms",
"Definition:Diagonal Relation",
"Diagonal Relation is Left Identity",
"Diagonal Relation is Right Identity",
"Definition:Identity Morphism",
"Composition of Relations is Associative",
"Definition:Metacategory"
] |
proofwiki-5887 | Universal Class is Proper | Let $V$ denote the universal class.
Then $V$ is a proper class. | {{AimForCont}} $V$ is small.
We have that:
:$\operatorname {Ru} \subseteq V$
where $\operatorname {Ru}$ denotes the Russell class.
By Axiom of Subsets Equivalents, $\operatorname {Ru}$ is also small.
This contradicts Russell's Paradox.
{{qed}} | Let $V$ denote the [[Definition:Universal Class|universal class]].
Then $V$ is a [[Definition:Proper Class|proper class]]. | {{AimForCont}} $V$ is [[Definition:Small Class|small]].
We have that:
:$\operatorname {Ru} \subseteq V$
where $\operatorname {Ru}$ denotes the [[Definition:Russell Class|Russell class]].
By [[Axiom of Subsets Equivalents]], $\operatorname {Ru}$ is also [[Definition:Small Class|small]].
This contradicts [[Russell's... | Universal Class is Proper/Proof 1 | https://proofwiki.org/wiki/Universal_Class_is_Proper | https://proofwiki.org/wiki/Universal_Class_is_Proper/Proof_1 | [
"Universal Class",
"Universal Class is Proper"
] | [
"Definition:Universal Class",
"Definition:Class (Class Theory)/Proper Class"
] | [
"Definition:Small Class",
"Definition:Russell Class",
"Axiom of Subsets Equivalents",
"Definition:Small Class",
"Russell's Paradox"
] |
proofwiki-5888 | Universal Class is Proper | Let $V$ denote the universal class.
Then $V$ is a proper class. | {{NotZFC}}
Let $I_V$ be the identity mapping on $V$.
By Identity Mapping is Bijection it follows that $I_V$ is a bijection.
Therefore, by the Axiom of Limitation of Size, $V$ is proper.
{{qed}} | Let $V$ denote the [[Definition:Universal Class|universal class]].
Then $V$ is a [[Definition:Proper Class|proper class]]. | {{NotZFC}}
Let $I_V$ be the [[Definition:Identity Mapping|identity mapping]] on $V$.
By [[Identity Mapping is Bijection]] it follows that $I_V$ is a [[Definition:Class Bijection|bijection]].
Therefore, by the [[Axiom:Axiom of Limitation of Size|Axiom of Limitation of Size]], $V$ is [[Definition:Proper Class|proper... | Universal Class is Proper/Proof 2 | https://proofwiki.org/wiki/Universal_Class_is_Proper | https://proofwiki.org/wiki/Universal_Class_is_Proper/Proof_2 | [
"Universal Class",
"Universal Class is Proper"
] | [
"Definition:Universal Class",
"Definition:Class (Class Theory)/Proper Class"
] | [
"Definition:Identity Mapping",
"Identity Mapping is Bijection",
"Definition:Bijection/Class Theory",
"Axiom:Axiom of Limitation of Size",
"Definition:Class (Class Theory)/Proper Class"
] |
proofwiki-5889 | Universal Class is Proper | Let $V$ denote the universal class.
Then $V$ is a proper class. | {{AimForCont}} $V$ is small.
By Cantor's Theorem, there is no surjection from $V$ to $\powerset V$.
By definition of universal class:
:$\powerset V \subseteq V$
By Injection from Subset to Superset, there exists an injection from $\powerset V$ to $V$.
By Injection has Surjective Left Inverse Mapping, there is a surject... | Let $V$ denote the [[Definition:Universal Class|universal class]].
Then $V$ is a [[Definition:Proper Class|proper class]]. | {{AimForCont}} $V$ is [[Definition:Small Class|small]].
By [[Cantor's Theorem]], there is no [[Definition:Surjection|surjection]] from $V$ to $\powerset V$.
By definition of [[Definition:Universal Class|universal class]]:
:$\powerset V \subseteq V$
By [[Injection from Subset to Superset]], there exists an [[Definiti... | Universal Class is Proper/Proof 3 | https://proofwiki.org/wiki/Universal_Class_is_Proper | https://proofwiki.org/wiki/Universal_Class_is_Proper/Proof_3 | [
"Universal Class",
"Universal Class is Proper"
] | [
"Definition:Universal Class",
"Definition:Class (Class Theory)/Proper Class"
] | [
"Definition:Small Class",
"Cantor's Theorem",
"Definition:Surjection",
"Definition:Universal Class",
"Injection from Subset to Superset",
"Definition:Injection",
"Injection has Surjective Left Inverse Mapping",
"Definition:Surjection",
"Cantor's Theorem",
"Definition:Class (Class Theory)/Proper Cl... |
proofwiki-5890 | Composite Functor is Functor | Let $\mathbf C, \mathbf D$ and $\mathbf E$ be metacategories.
Let $F: \mathbf C \to \mathbf D$ and $G: \mathbf D \to \mathbf E$ be covariant functors.
Let $GF: \mathbf C \to \mathbf E$ be the composition of $G$ with $F$.
Then $G F$ is also a covariant functor. | Let $f, g$ be morphisms of $\mathbf C$ such that $g \circ f$ is defined.
Then:
{{begin-eqn}}
{{eqn | l = \map {G F} {g \circ f}
| r = \map G {\map F {g \circ f} }
| c = {{Defof|Composition of Functors}}
}}
{{eqn | r = \map G {F g \circ F f}
| c = $F$ is a Covariant Functor
}}
{{eqn | r = \map G {F g} ... | Let $\mathbf C, \mathbf D$ and $\mathbf E$ be [[Definition:Metacategory|metacategories]].
Let $F: \mathbf C \to \mathbf D$ and $G: \mathbf D \to \mathbf E$ be [[Definition:Covariant Functor|covariant functors]].
Let $GF: \mathbf C \to \mathbf E$ be the [[Definition:Composition of Functors|composition of $G$ with $F$]... | Let $f, g$ be [[Definition:Morphism (Category Theory)|morphisms]] of $\mathbf C$ such that $g \circ f$ is defined.
Then:
{{begin-eqn}}
{{eqn | l = \map {G F} {g \circ f}
| r = \map G {\map F {g \circ f} }
| c = {{Defof|Composition of Functors}}
}}
{{eqn | r = \map G {F g \circ F f}
| c = $F$ is a [[... | Composite Functor is Functor | https://proofwiki.org/wiki/Composite_Functor_is_Functor | https://proofwiki.org/wiki/Composite_Functor_is_Functor | [
"Functors"
] | [
"Definition:Metacategory",
"Definition:Functor/Covariant",
"Definition:Composition of Functors",
"Definition:Functor/Covariant"
] | [
"Definition:Morphism",
"Definition:Functor/Covariant",
"Definition:Functor/Covariant",
"Definition:Object (Category Theory)",
"Definition:Functor/Covariant",
"Definition:Functor/Covariant",
"Definition:Functor/Covariant",
"Category:Functors"
] |
proofwiki-5891 | Identity Functor is Functor | Let $\mathbf C$ be a metacategory.
Let $\operatorname{id}_{\mathbf C}: \mathbf C \to \mathbf C$ be the identity functor on $\mathbf C$.
Then $\operatorname{id}_{\mathbf C}$ is a functor. | Let $f, g$ be morphisms of $\mathbf C$ such that $g \circ f$ is defined.
Then:
{{begin-eqn}}
{{eqn | l = \operatorname{id}_{\mathbf C} \left({g \circ f}\right)
| r = g \circ f
| c = {{Defof|Identity Functor}}
}}
{{eqn | r = \operatorname{id}_{\mathbf C} g \circ \operatorname{id}_{\mathbf C} f
| c = {{... | Let $\mathbf C$ be a [[Definition:Metacategory|metacategory]].
Let $\operatorname{id}_{\mathbf C}: \mathbf C \to \mathbf C$ be the [[Definition:Identity Functor|identity functor]] on $\mathbf C$.
Then $\operatorname{id}_{\mathbf C}$ is a [[Definition:Covariant Functor|functor]]. | Let $f, g$ be [[Definition:Morphism (Category Theory)|morphisms]] of $\mathbf C$ such that $g \circ f$ is defined.
Then:
{{begin-eqn}}
{{eqn | l = \operatorname{id}_{\mathbf C} \left({g \circ f}\right)
| r = g \circ f
| c = {{Defof|Identity Functor}}
}}
{{eqn | r = \operatorname{id}_{\mathbf C} g \circ \o... | Identity Functor is Functor | https://proofwiki.org/wiki/Identity_Functor_is_Functor | https://proofwiki.org/wiki/Identity_Functor_is_Functor | [
"Functors"
] | [
"Definition:Metacategory",
"Definition:Identity Functor",
"Definition:Functor/Covariant"
] | [
"Definition:Morphism",
"Definition:Object",
"Definition:Functor/Covariant",
"Category:Functors"
] |
proofwiki-5892 | Category of Categories is Category | Let $\mathbf{Cat}$ be the category of categories.
Then $\mathbf{Cat}$ is a metacategory. | Let us verify the axioms $(\text C 1)$ up to $(\text C 3)$ for a metacategory.
For any two functors their composition is again a functor by Composite Functor is Functor.
For any small category $\mathbf C$, we have the identity functor $\operatorname {id}_{\mathbf C}$.
By Identity Functor is Left Identity and Identity F... | Let $\mathbf{Cat}$ be the [[Definition:Category of Categories|category of categories]].
Then $\mathbf{Cat}$ is a [[Definition:Metacategory|metacategory]]. | Let us verify the axioms $(\text C 1)$ up to $(\text C 3)$ for a [[Definition:Metacategory|metacategory]].
For any two [[Definition:Covariant Functor|functors]] their [[Definition:Composition of Functors|composition]] is again a [[Definition:Covariant Functor|functor]] by [[Composite Functor is Functor]].
For any [... | Category of Categories is Category | https://proofwiki.org/wiki/Category_of_Categories_is_Category | https://proofwiki.org/wiki/Category_of_Categories_is_Category | [
"Category of Categories"
] | [
"Definition:Category of Categories",
"Definition:Metacategory"
] | [
"Definition:Metacategory",
"Definition:Functor/Covariant",
"Definition:Composition of Functors",
"Definition:Functor/Covariant",
"Composite Functor is Functor",
"Definition:Small Category",
"Definition:Identity Functor",
"Identity Functor is Left Identity",
"Identity Functor is Right Identity",
"D... |
proofwiki-5893 | Identity Functor is Left Identity | Let $\mathbf C$ and $\mathbf D$ be metacategories.
Let $F: \mathbf C \to \mathbf D$ be a functor, and let $\operatorname{id}_{\mathbf D}$ be the identity functor on $\mathbf D$.
Then the composite functor $\operatorname{id}_{\mathbf D} F$ satisfies:
:$\operatorname{id}_{\mathbf D} F = F$ | We have, for all objects $C$ of $\mathbf C$:
:$\operatorname{id}_{\mathbf D} F C = \map {\operatorname{id}_{\mathbf D} } {F C} = F C$
by definition of composition of functors and of identity functor.
Similarly, we have, for a morphism $f$ of $\mathbf C$:
:$\operatorname{id}_{\mathbf D} F f = \map {\operatorname{id}_{\m... | Let $\mathbf C$ and $\mathbf D$ be [[Definition:Metacategory|metacategories]].
Let $F: \mathbf C \to \mathbf D$ be a [[Definition:Covariant Functor|functor]], and let $\operatorname{id}_{\mathbf D}$ be the [[Definition:Identity Functor|identity functor]] on $\mathbf D$.
Then the [[Definition:Composite Functor|compos... | We have, for all [[Definition:Object|objects]] $C$ of $\mathbf C$:
:$\operatorname{id}_{\mathbf D} F C = \map {\operatorname{id}_{\mathbf D} } {F C} = F C$
by definition of [[Definition:Composition of Functors|composition of functors]] and of [[Definition:Identity Functor|identity functor]].
Similarly, we have, for... | Identity Functor is Left Identity | https://proofwiki.org/wiki/Identity_Functor_is_Left_Identity | https://proofwiki.org/wiki/Identity_Functor_is_Left_Identity | [
"Functors"
] | [
"Definition:Metacategory",
"Definition:Functor/Covariant",
"Definition:Identity Functor",
"Definition:Composition of Functors"
] | [
"Definition:Object",
"Definition:Composition of Functors",
"Definition:Identity Functor",
"Definition:Morphism",
"Category:Functors"
] |
proofwiki-5894 | Identity Functor is Right Identity | Let $\mathbf C$ and $\mathbf D$ be metacategories.
Let $F: \mathbf C \to \mathbf D$ be a functor, and let $\operatorname{id}_{\mathbf C}$ be the identity functor on $\mathbf C$.
Then the composite functor $F \operatorname{id}_{\mathbf C}$ satisfies:
:$F \operatorname{id}_{\mathbf C} = F$ | We have, for all objects $C$ of $\mathbf C$:
:$F \operatorname{id}_{\mathbf C} C = F \left({\operatorname{id}_{\mathbf C} C}\right) = F C$
by definition of composition of functors and of identity functor.
Similarly, we have, for a morphism $f$ of $\mathbf C$:
:$F \operatorname{id}_{\mathbf C} f = F \left({\operatorname... | Let $\mathbf C$ and $\mathbf D$ be [[Definition:Metacategory|metacategories]].
Let $F: \mathbf C \to \mathbf D$ be a [[Definition:Covariant Functor|functor]], and let $\operatorname{id}_{\mathbf C}$ be the [[Definition:Identity Functor|identity functor]] on $\mathbf C$.
Then the [[Definition:Composite Functor|compos... | We have, for all [[Definition:Object|objects]] $C$ of $\mathbf C$:
:$F \operatorname{id}_{\mathbf C} C = F \left({\operatorname{id}_{\mathbf C} C}\right) = F C$
by definition of [[Definition:Composition of Functors|composition of functors]] and of [[Definition:Identity Functor|identity functor]].
Similarly, we have... | Identity Functor is Right Identity | https://proofwiki.org/wiki/Identity_Functor_is_Right_Identity | https://proofwiki.org/wiki/Identity_Functor_is_Right_Identity | [
"Functors"
] | [
"Definition:Metacategory",
"Definition:Functor/Covariant",
"Definition:Identity Functor",
"Definition:Composition of Functors"
] | [
"Definition:Object",
"Definition:Composition of Functors",
"Definition:Identity Functor",
"Definition:Morphism",
"Category:Functors"
] |
proofwiki-5895 | Composition of Functors is Associative | Let $\mathbf A$, $\mathbf B$, $\mathbf C$ and $\mathbf D$ be metacategories.
Let $F: \mathbf A \to \mathbf B$, $G: \mathbf B \to \mathbf C$ and $H: \mathbf C \to \mathbf D$ be functors.
Then composition of functors is associative:
:$H \paren {G F} = \paren {H G} F$ | Let $A$ be an object of $\mathbf A$.
Then, solely by the definition of composite functor:
{{begin-eqn}}
{{eqn | l = H \paren {G F} A
| r = H \paren {G F A}
}}
{{eqn | r = H \paren {G \paren {F A} }
}}
{{eqn | r = H G \paren {F A}
}}
{{eqn | r = \paren {H G} F A
}}
{{end-eqn}}
Then, {{mutatis}}, the same proof wor... | Let $\mathbf A$, $\mathbf B$, $\mathbf C$ and $\mathbf D$ be [[Definition:Metacategory|metacategories]].
Let $F: \mathbf A \to \mathbf B$, $G: \mathbf B \to \mathbf C$ and $H: \mathbf C \to \mathbf D$ be [[Definition:Covariant Functor|functors]].
Then [[Definition:Composition of Functors|composition of functors]] is... | Let $A$ be an [[Definition:Object (Category Theory)|object]] of $\mathbf A$.
Then, solely by the definition of [[Definition:Composite Functor|composite functor]]:
{{begin-eqn}}
{{eqn | l = H \paren {G F} A
| r = H \paren {G F A}
}}
{{eqn | r = H \paren {G \paren {F A} }
}}
{{eqn | r = H G \paren {F A}
}}
{{eqn ... | Composition of Functors is Associative | https://proofwiki.org/wiki/Composition_of_Functors_is_Associative | https://proofwiki.org/wiki/Composition_of_Functors_is_Associative | [
"Functors"
] | [
"Definition:Metacategory",
"Definition:Functor/Covariant",
"Definition:Composition of Functors",
"Definition:Associative Operation"
] | [
"Definition:Object (Category Theory)",
"Definition:Composition of Functors",
"Definition:Morphism",
"Category:Functors"
] |
proofwiki-5896 | Abel-Ruffini Theorem | There is no general algebraic solution for determining all the roots of a polynomial of degree $5$ or higher. | {{ProofWanted|Galois theory seems the best choice; we probably also want the actual proof by Abel}}
{{tidy}}
{{MissingLinks}}
The general polynomial:
:$\map f x = x^n - s_1 x^{n - 1} + \cdots + \paren {-1}^n s_n$
has Galois group $S_n$.
From Symmetric Group on Greater than 4 Letters is Not Solvable:
:$S_n$ is not solva... | There is no general [[Definition:Algebraic Solution|algebraic solution]] for determining all the [[Definition:Root of Polynomial|roots]] of a [[Definition:Polynomial over Field|polynomial]] of [[Definition:Degree (Polynomial)|degree]] $5$ or higher. | {{ProofWanted|Galois theory seems the best choice; we probably also want the actual proof by Abel}}
{{tidy}}
{{MissingLinks}}
The general polynomial:
:$\map f x = x^n - s_1 x^{n - 1} + \cdots + \paren {-1}^n s_n$
has [[Definition:Galois Group of Polynomial|Galois group]] $S_n$.
From [[Symmetric Group on Greater than... | Abel-Ruffini Theorem | https://proofwiki.org/wiki/Abel-Ruffini_Theorem | https://proofwiki.org/wiki/Abel-Ruffini_Theorem | [
"Polynomial Theory",
"Polynomial Equations",
"5"
] | [
"Definition:Algebraic Solution",
"Definition:Root of Polynomial",
"Definition:Polynomial over Ring",
"Definition:Degree of Polynomial"
] | [
"Definition:Galois Group of Polynomial",
"Symmetric Group on Greater than 4 Letters is Not Solvable",
"Definition:Solvable Group",
"Polynomial is Solvable by Radicals iff Galois Group is Solvable"
] |
proofwiki-5897 | Cartesian Product is Small | Let $a$ and $b$ be small classes.
Then their Cartesian product $a \times b$ is small:
:$\map {\MM} {a \times b}$ | By Binary Cartesian Product in Kuratowski Formalization contained in Power Set of Power Set of Union:
:$a \times b \subseteq \powerset {\powerset {a \cup b} }$
By Union of Small Classes is Small, $a \cup b$ is small.
By the Axiom of Powers, $\powerset {\powerset {a \cup b} }$ is small.
By Axiom of Subsets Equivalents, ... | Let $a$ and $b$ be [[Definition:Small Class|small classes]].
Then their [[Definition:Cartesian Product|Cartesian product]] $a \times b$ is [[Definition:Small Class|small]]:
:$\map {\MM} {a \times b}$ | By [[Binary Cartesian Product in Kuratowski Formalization contained in Power Set of Power Set of Union]]:
:$a \times b \subseteq \powerset {\powerset {a \cup b} }$
By [[Union of Small Classes is Small]], $a \cup b$ is [[Definition:Small Class|small]].
By the [[Axiom:Axiom of Powers (Set Theory)|Axiom of Powers]], $... | Cartesian Product is Small | https://proofwiki.org/wiki/Cartesian_Product_is_Small | https://proofwiki.org/wiki/Cartesian_Product_is_Small | [
"Zermelo-Fraenkel Class Theory"
] | [
"Definition:Small Class",
"Definition:Cartesian Product",
"Definition:Small Class"
] | [
"Binary Cartesian Product in Kuratowski Formalization contained in Power Set of Power Set of Union",
"Union of Small Classes is Small",
"Definition:Small Class",
"Axiom:Axiom of Powers/Set Theory",
"Definition:Small Class",
"Axiom of Subsets Equivalents",
"Definition:Small Class"
] |
proofwiki-5898 | Preorder Category is Category | Let $\struct {S, \precsim}$ be a preordered set.
Let $\mathbf S$ be its associated preorder category.
Then $\mathbf S$ is a category. | Let us verify the axioms $(\text C 1)$ up to $(\text C 3)$ for a metacategory.
Note that for objects $a, b$ of $\mathbf S$, there is at most ''one'' morphism $a \to b$, by definition of $\mathbf S$.
Suppose that $a \to b$ and $b \to c$ are morphisms of $\mathbf S$.
Then we have $a \precsim b$ and $b \precsim c$, and as... | Let $\struct {S, \precsim}$ be a [[Definition:Preordered Set|preordered set]].
Let $\mathbf S$ be its associated [[Definition:Preorder Category|preorder category]].
Then $\mathbf S$ is a [[Definition:Category|category]]. | Let us verify the axioms $(\text C 1)$ up to $(\text C 3)$ for a [[Definition:Metacategory|metacategory]].
Note that for [[Definition:Object (Category Theory)|objects]] $a, b$ of $\mathbf S$, there is at most ''one'' [[Definition:Morphism (Category Theory)|morphism]] $a \to b$, by [[Definition:Preorder Category|defini... | Preorder Category is Category | https://proofwiki.org/wiki/Preorder_Category_is_Category | https://proofwiki.org/wiki/Preorder_Category_is_Category | [
"Preorder Categories"
] | [
"Definition:Preordering/Preordered Set",
"Definition:Preorder Category",
"Definition:Category"
] | [
"Definition:Metacategory",
"Definition:Object (Category Theory)",
"Definition:Morphism",
"Definition:Preorder Category",
"Definition:Morphism",
"Definition:Preordering",
"Definition:Morphism",
"Definition:Morphism",
"Definition:Composition of Morphisms",
"Definition:Morphism",
"Definition:Preord... |
proofwiki-5899 | Category Induces Preorder | Let $\mathbf S$ be a category with set of objects $S$.
Then the binary relation $\precsim$ defined by:
:$\forall a, b \in S: a \precsim b \iff \exists f: a \to b$
is a preorder on $S$. | It suffices to establish $\precsim$ is reflexive and transitive. | Let $\mathbf S$ be a [[Definition:Category|category]] with set of [[Definition:Object (Category Theory)|objects]] $S$.
Then the [[Definition:Binary Relation|binary relation]] $\precsim$ defined by:
:$\forall a, b \in S: a \precsim b \iff \exists f: a \to b$
is a [[Definition:Preordering|preorder]] on $S$. | It suffices to establish $\precsim$ is [[Definition:Reflexive Relation|reflexive]] and [[Definition:Transitive Relation|transitive]]. | Category Induces Preorder | https://proofwiki.org/wiki/Category_Induces_Preorder | https://proofwiki.org/wiki/Category_Induces_Preorder | [
"Category Theory",
"Preorder Theory"
] | [
"Definition:Category",
"Definition:Object (Category Theory)",
"Definition:Relation",
"Definition:Preordering"
] | [
"Definition:Reflexive Relation",
"Definition:Transitive Relation"
] |
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