Conversion scheme

The [set] transformation scheme (Axiom schema of replacement) is the following statement of set theory :

$\forall x\exists ^{\{1\}}y\ (\phi [x,y])\to \forall a\exists d\forall c\ (c\in d\leftrightarrow \exists b\ (b\in a\ \land \ \phi [b,c])\ )$ ${\ displaystyle \ forall x \ exists ^ {\ {1 \}} y \ (\ phi [x, y]) \ to \ forall a \ exists d \ forall c \ (c \ in d \ leftrightarrow \ exists b \ (b \ in a \ \ land \ \ phi [b, c]) \)}$ ${\ displaystyle \ forall x \ exists ^ {\ {1 \}} y \ (\ phi [x, y]) \ to \ forall a \ exists d \ forall c \ (c \ in d \ leftrightarrow \ exists b \ (b \ in a \ \ land \ \ phi [b, c]) \)}$ where $\forall x\exists ^{\{1\}}y\ (\phi [x,y])\Leftrightarrow \forall x\exists !y\ (\phi [x,y])\Leftrightarrow \forall x\exists y\forall y'(\phi [x,y]\leftrightarrow y=y')$ ${\ displaystyle \ forall x \ exists ^ {\ {1 \}} y \ (\ phi [x, y]) \ Leftrightarrow \ forall x \ exists! y \ (\ phi [x, y]) \ Leftrightarrow \ forall x \ exists y \ forall y '(\ phi [x, y] \ leftrightarrow y = y')}$ ${\ displaystyle \ forall x \ exists ^ {\ {1 \}} y \ (\ phi [x, y]) \ Leftrightarrow \ forall x \ exists! y \ (\ phi [x, y]) \ Leftrightarrow \ forall x \ exists y \ forall y '(\ phi [x, y] \ leftrightarrow y = y')}$

The transformation scheme can be formulated in Russian, namely: "Any set can be transformed into [the same or another] set $d$ ${\ displaystyle d}$ $d$ making a functional judgment $\phi$ ${\ displaystyle \ phi}$ $\ phi$ about all elements $b$ ${\ displaystyle b}$ $b$ given set $a$ ${\ displaystyle a}$ $a$ . "

Example

In the following example, functional judgment

y=x

{\ displaystyle y = x}

y = x

converts every set

a

{\ displaystyle a}

a

in yourself.

\phi [x,y]\leftrightarrow y=x\quad \Rightarrow \quad \forall a\exists d\forall c\ (c\in d\leftrightarrow \exists b\ (b\in a\ \land \ c=b))\quad \Leftrightarrow \quad \forall a\exists d\forall c\ (c\in d\leftrightarrow c\in a)

{\ displaystyle \ phi [x, y] \ leftrightarrow y = x \ quad \ Rightarrow \ quad \ forall a \ exists d \ forall c \ (c \ in d \ leftrightarrow \ exists b \ (b \ in a \ \ land \ c = b)) \ ​​quad \ Leftrightarrow \ quad \ forall a \ exists d \ forall c \ (c \ in d \ leftrightarrow c \ in a)}

\ phi [x, y] \ leftrightarrow y = x \ quad \ Rightarrow \ quad \ forall a \ exists d \ forall c \ (c \ in d \ leftrightarrow \ exists b \ (b \ in a \ \ land \ c = b)) \ ​​quad \ Leftrightarrow \ quad \ forall a \ exists d \ forall c \ (c \ in d \ leftrightarrow c \ in a)

Other wording of the transformation scheme

The conversion scheme is also written as follows:

$\forall a\ (\ \forall b\ (b\in a\to \exists ^{\{1\}}y\ (\phi [b,y])\ )\quad \to \quad \exists d\forall c\ (c\in d\leftrightarrow \exists b\ (b\in a\ \land \ \phi [b,c])\ ))$ ${\ displaystyle \ forall a \ (\ \ forall b \ (b \ in a \ to \ exists ^ {\ {1 \}} y \ (\ phi [b, y]) \) \ quad \ to \ quad \ exists d \ forall c \ (c \ in d \ leftrightarrow \ exists b \ (b \ in a \ \ land \ \ phi [b, c]) \))}$ $\forall a\ (\ \forall b\ (b\in a\to \exists ^{\{1\}}y\ (\phi [b,y])\ )\quad \to \quad \exists d\forall c\ (c\in d\leftrightarrow \exists b\ (b\in a\ \land \ \phi [b,c])\ ))$

Examples

1. In the following example, functional judgment

y=2b'

{\ displaystyle y = 2b '}

y=2b'

converts many natural numbers

\mathbb {N}

{\ displaystyle \ mathbb {N}}

\mathbb {N}

into many even numbers

\{0,2,4,...\}

{\ displaystyle \ {0,2,4, ... \}}

\{0,2,4,...\}

.

{\begin{aligned}a=\mathbb {N} \ \land \ (\phi [b',y]\leftrightarrow y=2b')\quad \Rightarrow \quad \exists d\forall c\ (c\in d\leftrightarrow \exists b\ (b\in \mathbb {N} \ \land \ c=2b))\\\ \Leftrightarrow \exists d\forall c\ (c\in d\leftrightarrow c\in \{0,2,4,...\})\end{aligned}}

{\ displaystyle {\ begin {aligned} a = \ mathbb {N} \ \ land \ (\ phi [b ', y] \ leftrightarrow y = 2b') \ quad \ Rightarrow \ quad \ exists d \ forall c \ ( c \ in d \ leftrightarrow \ exists b \ (b \ in \ mathbb {N} \ \ land \ c = 2b)) \\\ \ Leftrightarrow \ exists d \ forall c \ (c \ in d \ leftrightarrow c \ in \ {0,2,4, ... \}) \ end {aligned}}}

{\begin{aligned}a={\mathbb {N}}\ \land \ (\phi [b',y]\leftrightarrow y=2b')\quad \Rightarrow \quad \exists d\forall c\ (c\in d\leftrightarrow \exists b\ (b\in {\mathbb {N}}\ \land \ c=2b))\\\ \Leftrightarrow \exists d\forall c\ (c\in d\leftrightarrow c\in \{0,2,4,...\})\end{aligned}}

2. In the following example, functional judgment

(b'=0\to y=a_{1})\ \land \ (b'\neq 0\to y=a_{2})

{\ displaystyle (b '= 0 \ to y = a_ {1}) \ \ land \ (b' \ neq 0 \ to y = a_ {2})}

(b'=0\to y=a_{1})\ \land \ (b'\neq 0\to y=a_{2})

converts a set of real numbers

\mathbb {R}

{\ displaystyle \ mathbb {R}}

\mathbb {R}

into a [unordered] pair

\{a_{1},\ a_{2}\}

{\ displaystyle \ {a_ {1}, \ a_ {2} \}}

\{a_{1},\ a_{2}\}

.

{\begin{aligned}a=\mathbb {R} \quad \land \quad (\phi [b',y]\leftrightarrow (b'=0\to y=a_{1})\ \land \ (b'\neq 0\to y=a_{2}))\quad \Rightarrow \\\ \exists d\forall c\ (c\in d\leftrightarrow \exists b\ (b\in \mathbb {R} \ \land \ (b=0\to c=a_{1})\land (b\neq 0\to c=a_{2})\ ))\\\ \Leftrightarrow \exists d\forall c\ (c\in d\leftrightarrow c=a_{1}\ \lor \ c=a_{2})\end{aligned}}

{\ displaystyle {\ begin {aligned} a = \ mathbb {R} \ quad \ land \ quad (\ phi [b ', y] \ leftrightarrow (b' = 0 \ to y = a_ {1}) \ \ land \ (b '\ neq 0 \ to y = a_ {2})) \ quad \ Rightarrow \\\ \ exists d \ forall c \ (c \ in d \ leftrightarrow \ exists b \ (b \ in \ mathbb {R } \ \ land \ (b = 0 \ to c = a_ {1}) \ land (b \ neq 0 \ to c = a_ {2}) \)) \\\ \ Leftrightarrow \ exists d \ forall c \ ( c \ in d \ leftrightarrow c = a_ {1} \ \ lor \ c = a_ {2}) \ end {aligned}}}

{\begin{aligned}a={\mathbb {R}}\quad \land \quad (\phi [b',y]\leftrightarrow (b'=0\to y=a_{1})\ \land \ (b'\neq 0\to y=a_{2}))\quad \Rightarrow \\\ \exists d\forall c\ (c\in d\leftrightarrow \exists b\ (b\in {\mathbb {R}}\ \land \ (b=0\to c=a_{1})\land (b\neq 0\to c=a_{2})\ ))\\\ \Leftrightarrow \exists d\forall c\ (c\in d\leftrightarrow c=a_{1}\ \lor \ c=a_{2})\end{aligned}}

3. In the following example, functional judgment

(0\leq b'\leq 1\to y=b')\ \land \ (\neg (0\leq b'\leq 1)\to y=1)

{\ displaystyle (0 \ leq b '\ leq 1 \ to y = b') \ \ land \ (\ neg (0 \ leq b '\ leq 1) \ to y = 1)}

(0\leq b'\leq 1\to y=b')\ \land \ (\neg (0\leq b'\leq 1)\to y=1)

converts a set of integers

\mathbb {Z}

{\ displaystyle \ mathbb {Z}}

\mathbb {Z}

into a subset of natural numbers

\{n:\ n\in \mathbb {N} \ \land \ n<2\}

{\ displaystyle \ {n: \ n \ in \ mathbb {N} \ \ land \ n <2 \}}

\{n:\ n\in \mathbb {N} \ \land \ n<2\}

.

{\begin{aligned}a=\mathbb {Z} \quad \land \quad (\phi [b',y]\leftrightarrow (0\leq b'\leq 1\to y=b')\land (\neg (0\leq b'\leq 1)\to y=1))\quad \Rightarrow \\\ \exists d\forall c\ (c\in d\leftrightarrow \exists b\ (b\in \mathbb {Z} \land (0\leq b\leq 1\to c=b)\land (b<0\lor b>1\to c=1)))\\\ \Leftrightarrow \exists d\forall c\ (c\in d\leftrightarrow c\in \{n:\ n\in \mathbb {N} \ \land \ n<2\}\ )\end{aligned}}

{\ displaystyle {\ begin {aligned} a = \ mathbb {Z} \ quad \ land \ quad (\ phi [b ', y] \ leftrightarrow (0 \ leq b' \ leq 1 \ to y = b ') \ land (\ neg (0 \ leq b '\ leq 1) \ to y = 1)) \ quad \ Rightarrow \\\ \ exists d \ forall c \ (c \ in d \ leftrightarrow \ exists b \ (b \ in \ mathbb {Z} \ land (0 \ leq b \ leq 1 \ to c = b) \ land (b <0 \ lor b> 1 \ to c = 1))) \\\ \ Leftrightarrow \ exists d \ forall c \ (c \ in d \ leftrightarrow c \ in \ {n: \ n \ in \ mathbb {N} \ \ land \ n <2 \} \) \ end {aligned}}}

{\begin{aligned}a={\mathbb {Z}}\quad \land \quad (\phi [b',y]\leftrightarrow (0\leq b'\leq 1\to y=b')\land (\neg (0\leq b'\leq 1)\to y=1))\quad \Rightarrow \\\ \exists d\forall c\ (c\in d\leftrightarrow \exists b\ (b\in {\mathbb {Z}}\land (0\leq b\leq 1\to c=b)\land (b<0\lor b>1\to c=1)))\\\ \Leftrightarrow \exists d\forall c\ (c\in d\leftrightarrow c\in \{n:\ n\in {\mathbb {N}}\ \land \ n<2\}\ )\end{aligned}}

The conversion scheme is also written as follows:

$\forall a\ (\ \forall b\ (b\in a\to \exists ^{\{0,1\}}y\ (\phi [b,y]))\quad \to \quad \exists d\forall c\ (c\in d\leftrightarrow \exists b\ (b\in a\ \land \ \phi [b,c])\ ))$ ${\ displaystyle \ forall a \ (\ \ forall b \ (b \ in a \ to \ exists ^ {\ {0,1 \}} y \ (\ phi [b, y])) \ quad \ to \ quad \ exists d \ forall c \ (c \ in d \ leftrightarrow \ exists b \ (b \ in a \ \ land \ \ phi [b, c]) \))}$ where $\exists ^{\{0,1\}}y\ (\phi [b,y])\Leftrightarrow \forall y\forall y'\ (\phi [b,y]\ \land \ \phi [b,y']\to y=y')$ ${\ displaystyle \ exists ^ {\ {0,1 \}} y \ (\ phi [b, y]) \ Leftrightarrow \ forall y \ forall y '\ (\ phi [b, y] \ \ land \ \ phi [b, y '] \ to y = y')}$

Notes

1. The relationship between the transformation scheme and the axiom of a pair is expressed by the following statement:

${\begin{aligned}\forall a_{1}\forall a_{2}\ (a={\mathcal {P}}({\mathcal {P}}(\varnothing ))\quad \land \quad (\phi [b',y]\ \leftrightarrow \ (b'=\varnothing \to y=a_{1})\land (b'\neq \varnothing \to y=a_{2})\ )\\\ \rightarrow \quad (\exists d\forall c\ (c\in d\ \leftrightarrow \ \exists b\ (b\in a\land \phi [b,c]))\ \rightarrow \ \exists c\forall b\ (b\in c\leftrightarrow b=a_{1}\lor b=a_{2})\ )),\end{aligned}}$ ${\ displaystyle {\ begin {aligned} \ forall a_ {1} \ forall a_ {2} \ (a = {\ mathcal {P}} ({\ mathcal {P}} (\ varnothing)) \ quad \ land \ quad (\ phi [b ', y] \ \ leftrightarrow \ (b' = \ varnothing \ to y = a_ {1}) \ land (b '\ neq \ varnothing \ to y = a_ {2}) \) \ \\ \ rightarrow \ quad (\ exists d \ forall c \ (c \ in d \ \ leftrightarrow \ \ exists b \ (b \ in a \ land \ phi [b, c])) \ \ rightarrow \ \ exists c \ forall b \ (b \ in c \ leftrightarrow b = a_ {1} \ lor b = a_ {2}) \)), \ end {aligned}}}$

Where

{\mathcal {P}}({\mathcal {P}}(\varnothing ))

{\ displaystyle {\ mathcal {P}} ({\ mathcal {P}} (\ varnothing))}

- Boolean Boolean empty set.

2. The relationship between the conversion scheme and the allocation scheme is expressed by the following statement:

${\begin{aligned}\forall a\ (\ x\in \{b:b\in a\land \Phi [b]\}\quad \land \quad (\phi [b',y]\ \leftrightarrow \ (\Phi [b']\to y=b')\land (\neg \Phi [b']\to y=x)\ )\\\ \to \quad (\exists d\forall c\ (c\in d\leftrightarrow \exists b\ (b\in a\land \phi [b,c]))\ \leftrightarrow \ \exists c\forall b\ (b\in c\leftrightarrow b\in a\land \Phi [b]))\ )\end{aligned}}$ ${\ displaystyle {\ begin {aligned} \ forall a \ (\ x \ in \ {b: b \ in a \ land \ Phi [b] \} \ quad \ land \ quad (\ phi [b ', y] \ \ leftrightarrow \ (\ Phi [b '] \ to y = b') \ land (\ neg \ Phi [b '] \ to y = x) \) \\\ \ to \ quad (\ exists d \ forall c \ (c \ in d \ leftrightarrow \ exists b \ (b \ in a \ land \ phi [b, c])) \ \ leftrightarrow \ \ exists c \ forall b \ (b \ in c \ leftrightarrow b \ in a \ land \ Phi [b])) \) \ end {aligned}}}$

Historical background

The transformation scheme was not included in the set of axioms of set theory formulated by the German mathematician Ernst Zermelo in 1908.

The transformation scheme was proposed by Adolf Frenkel in 1922 , a little later and independently of it, the scheme was proposed by the Norwegian mathematician Turalf Skulem .

Conversion scheme

Other wording of the transformation scheme

Notes

Historical background

See also

Literature

More articles: