What does individual variables mean ? Can be propositional variables?
$begingroup$
We know that any particular first-order language is
determined by its symbols. These consist of ;
$ 1-) $ A denumerable list of symbols called individual variables.
- A denumerable list of symbols $(text{not in }1)$ called individual parameters.
- the connectives ; $land ,lor ,lnot , rightarrow $
- for each natural number $n$, a set of $n$-ary relation symbols (also
called predicate symbols).
- for each natural number $n$, a set of $n$-ary function symbols.
- the quantifiers ; $ forall , exists$
- parentheses and the comma. $( , )$
My qeustion :
$1-) $ What does individual variables mean ? Can be propositional variables?
$2-) $ In first-order language ; Can be definition symbol?
logic first-order-logic
edited Dec 17 '18 at 14:04
Mauro ALLEGRANZA
66.5k449115
asked Dec 17 '18 at 13:47
Almot1960Almot1960
2,312823
$endgroup$
add a comment |
$begingroup$
We know that any particular first-order language is
determined by its symbols. These consist of ;
$ 1-) $ A denumerable list of symbols called individual variables.
- A denumerable list of symbols $(text{not in }1)$ called individual parameters.
- the connectives ; $land ,lor ,lnot , rightarrow $
- for each natural number $n$, a set of $n$-ary relation symbols (also
called predicate symbols).
- for each natural number $n$, a set of $n$-ary function symbols.
- the quantifiers ; $ forall , exists$
- parentheses and the comma. $( , )$
My qeustion :
$1-) $ What does individual variables mean ? Can be propositional variables?
$2-) $ In first-order language ; Can be definition symbol?
logic first-order-logic
edited Dec 17 '18 at 14:04
Mauro ALLEGRANZA
66.5k449115
asked Dec 17 '18 at 13:47
Almot1960Almot1960
2,312823
$endgroup$
add a comment |
$begingroup$
We know that any particular first-order language is
determined by its symbols. These consist of ;
$ 1-) $ A denumerable list of symbols called individual variables.
- A denumerable list of symbols $(text{not in }1)$ called individual parameters.
- the connectives ; $land ,lor ,lnot , rightarrow $
- for each natural number $n$, a set of $n$-ary relation symbols (also
called predicate symbols).
- for each natural number $n$, a set of $n$-ary function symbols.
- the quantifiers ; $ forall , exists$
- parentheses and the comma. $( , )$
My qeustion :
$1-) $ What does individual variables mean ? Can be propositional variables?
$2-) $ In first-order language ; Can be definition symbol?
logic first-order-logic
edited Dec 17 '18 at 14:04
Mauro ALLEGRANZA
66.5k449115
asked Dec 17 '18 at 13:47
Almot1960Almot1960
2,312823
$endgroup$
We know that any particular first-order language is
determined by its symbols. These consist of ;
$ 1-) $ A denumerable list of symbols called individual variables.
- A denumerable list of symbols $(text{not in }1)$ called individual parameters.
- the connectives ; $land ,lor ,lnot , rightarrow $
- for each natural number $n$, a set of $n$-ary relation symbols (also
called predicate symbols).
- for each natural number $n$, a set of $n$-ary function symbols.
- the quantifiers ; $ forall , exists$
- parentheses and the comma. $( , )$
My qeustion :
$1-) $ What does individual variables mean ? Can be propositional variables?
$2-) $ In first-order language ; Can be definition symbol?
logic first-order-logic
logic first-order-logic
edited Dec 17 '18 at 14:04
Mauro ALLEGRANZA
66.5k449115
asked Dec 17 '18 at 13:47
Almot1960Almot1960
2,312823
edited Dec 17 '18 at 14:04
Mauro ALLEGRANZA
66.5k449115
asked Dec 17 '18 at 13:47
Almot1960Almot1960
2,312823
edited Dec 17 '18 at 14:04
Mauro ALLEGRANZA
66.5k449115
edited Dec 17 '18 at 14:04
Mauro ALLEGRANZA
66.5k449115
edited Dec 17 '18 at 14:04
Mauro ALLEGRANZA
66.5k449115
66.5k449115
asked Dec 17 '18 at 13:47
Almot1960Almot1960
2,312823
asked Dec 17 '18 at 13:47
Almot1960Almot1960
2,312823
asked Dec 17 '18 at 13:47
Almot1960Almot1960
2,312823
2,312823
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
1) What does individual variable mean ?
It is a term, i.e. a symbol that acts as a name for an object.
Thus, it cannot be a propositional variables, i.e. a symbols that stands for a sentence.
Consider the simple example from first-order language of arithmetic : $(x=0)$.
In this formula $x$ must be replaced by a number in order to give an arithmetical meaning to the formula.
2) In first-order language, can be definition symbol ?
A definition must either introduce a term, i.e. a symbol acting as a name for an object, or a predicate letter, i.e. a symbol naming a property.
Again, examples from first order arithmetic : we start from the basic symbols of the language : $0$ (an individual constant denoting the number $text {zero}$), the unary function $s(x)$ (the $text {successor}$ function) and the binary function $+(x,y)$ (the $text {sum}$ operation, abbreviated with : $(x+y)$).
With them we define the new constant $1$ as $s(0)$.
And we define the new binary predicate $<(n,m)$ (the relation $text {less than}$, abbreviated with $(n < m)$) as follows :
$(n < m) text { iff } exists z (m=n+s(z))$.
answered Dec 17 '18 at 13:56
Mauro ALLEGRANZAMauro ALLEGRANZA
66.5k449115
$endgroup$
add a comment |
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1 Answer
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1 Answer
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active
oldest
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oldest
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$begingroup$
1) What does individual variable mean ?
It is a term, i.e. a symbol that acts as a name for an object.
Thus, it cannot be a propositional variables, i.e. a symbols that stands for a sentence.
Consider the simple example from first-order language of arithmetic : $(x=0)$.
In this formula $x$ must be replaced by a number in order to give an arithmetical meaning to the formula.
2) In first-order language, can be definition symbol ?
A definition must either introduce a term, i.e. a symbol acting as a name for an object, or a predicate letter, i.e. a symbol naming a property.
Again, examples from first order arithmetic : we start from the basic symbols of the language : $0$ (an individual constant denoting the number $text {zero}$), the unary function $s(x)$ (the $text {successor}$ function) and the binary function $+(x,y)$ (the $text {sum}$ operation, abbreviated with : $(x+y)$).
With them we define the new constant $1$ as $s(0)$.
And we define the new binary predicate $<(n,m)$ (the relation $text {less than}$, abbreviated with $(n < m)$) as follows :
$(n < m) text { iff } exists z (m=n+s(z))$.
answered Dec 17 '18 at 13:56
Mauro ALLEGRANZAMauro ALLEGRANZA
66.5k449115
$endgroup$
add a comment |
$begingroup$
1) What does individual variable mean ?
It is a term, i.e. a symbol that acts as a name for an object.
Thus, it cannot be a propositional variables, i.e. a symbols that stands for a sentence.
Consider the simple example from first-order language of arithmetic : $(x=0)$.
In this formula $x$ must be replaced by a number in order to give an arithmetical meaning to the formula.
2) In first-order language, can be definition symbol ?
A definition must either introduce a term, i.e. a symbol acting as a name for an object, or a predicate letter, i.e. a symbol naming a property.
Again, examples from first order arithmetic : we start from the basic symbols of the language : $0$ (an individual constant denoting the number $text {zero}$), the unary function $s(x)$ (the $text {successor}$ function) and the binary function $+(x,y)$ (the $text {sum}$ operation, abbreviated with : $(x+y)$).
With them we define the new constant $1$ as $s(0)$.
And we define the new binary predicate $<(n,m)$ (the relation $text {less than}$, abbreviated with $(n < m)$) as follows :
$(n < m) text { iff } exists z (m=n+s(z))$.
answered Dec 17 '18 at 13:56
Mauro ALLEGRANZAMauro ALLEGRANZA
66.5k449115
$endgroup$
add a comment |
$begingroup$
1) What does individual variable mean ?
It is a term, i.e. a symbol that acts as a name for an object.
Thus, it cannot be a propositional variables, i.e. a symbols that stands for a sentence.
Consider the simple example from first-order language of arithmetic : $(x=0)$.
In this formula $x$ must be replaced by a number in order to give an arithmetical meaning to the formula.
2) In first-order language, can be definition symbol ?
A definition must either introduce a term, i.e. a symbol acting as a name for an object, or a predicate letter, i.e. a symbol naming a property.
Again, examples from first order arithmetic : we start from the basic symbols of the language : $0$ (an individual constant denoting the number $text {zero}$), the unary function $s(x)$ (the $text {successor}$ function) and the binary function $+(x,y)$ (the $text {sum}$ operation, abbreviated with : $(x+y)$).
With them we define the new constant $1$ as $s(0)$.
And we define the new binary predicate $<(n,m)$ (the relation $text {less than}$, abbreviated with $(n < m)$) as follows :
$(n < m) text { iff } exists z (m=n+s(z))$.
answered Dec 17 '18 at 13:56
Mauro ALLEGRANZAMauro ALLEGRANZA
66.5k449115
$endgroup$
1) What does individual variable mean ?
It is a term, i.e. a symbol that acts as a name for an object.
Thus, it cannot be a propositional variables, i.e. a symbols that stands for a sentence.
Consider the simple example from first-order language of arithmetic : $(x=0)$.
In this formula $x$ must be replaced by a number in order to give an arithmetical meaning to the formula.
2) In first-order language, can be definition symbol ?
A definition must either introduce a term, i.e. a symbol acting as a name for an object, or a predicate letter, i.e. a symbol naming a property.
Again, examples from first order arithmetic : we start from the basic symbols of the language : $0$ (an individual constant denoting the number $text {zero}$), the unary function $s(x)$ (the $text {successor}$ function) and the binary function $+(x,y)$ (the $text {sum}$ operation, abbreviated with : $(x+y)$).
With them we define the new constant $1$ as $s(0)$.
And we define the new binary predicate $<(n,m)$ (the relation $text {less than}$, abbreviated with $(n < m)$) as follows :
$(n < m) text { iff } exists z (m=n+s(z))$.
answered Dec 17 '18 at 13:56
Mauro ALLEGRANZAMauro ALLEGRANZA
66.5k449115
edited Dec 17 '18 at 14:03
answered Dec 17 '18 at 13:56
Mauro ALLEGRANZAMauro ALLEGRANZA
66.5k449115
answered Dec 17 '18 at 13:56
Mauro ALLEGRANZAMauro ALLEGRANZA
66.5k449115
answered Dec 17 '18 at 13:56
Mauro ALLEGRANZAMauro ALLEGRANZA
66.5k449115
66.5k449115
add a comment |
add a comment |
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