Formal definition of truth function
$begingroup$
As Mendelson say in Introduction to Mathematical Logic:
As we have seen, any statement form containing $n$ distinct statement letters determines a corresponding truth function of $n$ arguments.
Thus, it must exist a function $mathcal{F}: {text{statement forms}} to {text{truth function}}$, where the set of statement forms is given by the definition in page 3 and the set of true function is
$$bigcup_{ninmathbb{N}} {T,F}^{{T,F}^n} = bigcup_{ninmathbb{N}} 2^{2^n}.$$
I have an idea of how to define this function, but I want to know if there exists a book or an article where I can find a formal and precise definition of that function?
logic propositional-calculus
$endgroup$
add a comment |
$begingroup$
As Mendelson say in Introduction to Mathematical Logic:
As we have seen, any statement form containing $n$ distinct statement letters determines a corresponding truth function of $n$ arguments.
Thus, it must exist a function $mathcal{F}: {text{statement forms}} to {text{truth function}}$, where the set of statement forms is given by the definition in page 3 and the set of true function is
$$bigcup_{ninmathbb{N}} {T,F}^{{T,F}^n} = bigcup_{ninmathbb{N}} 2^{2^n}.$$
I have an idea of how to define this function, but I want to know if there exists a book or an article where I can find a formal and precise definition of that function?
logic propositional-calculus
$endgroup$
$begingroup$
See similar post: the-number-of-n-variable-preposition-function.
$endgroup$
– Mauro ALLEGRANZA
Oct 30 '17 at 16:14
add a comment |
$begingroup$
As Mendelson say in Introduction to Mathematical Logic:
As we have seen, any statement form containing $n$ distinct statement letters determines a corresponding truth function of $n$ arguments.
Thus, it must exist a function $mathcal{F}: {text{statement forms}} to {text{truth function}}$, where the set of statement forms is given by the definition in page 3 and the set of true function is
$$bigcup_{ninmathbb{N}} {T,F}^{{T,F}^n} = bigcup_{ninmathbb{N}} 2^{2^n}.$$
I have an idea of how to define this function, but I want to know if there exists a book or an article where I can find a formal and precise definition of that function?
logic propositional-calculus
$endgroup$
As Mendelson say in Introduction to Mathematical Logic:
As we have seen, any statement form containing $n$ distinct statement letters determines a corresponding truth function of $n$ arguments.
Thus, it must exist a function $mathcal{F}: {text{statement forms}} to {text{truth function}}$, where the set of statement forms is given by the definition in page 3 and the set of true function is
$$bigcup_{ninmathbb{N}} {T,F}^{{T,F}^n} = bigcup_{ninmathbb{N}} 2^{2^n}.$$
I have an idea of how to define this function, but I want to know if there exists a book or an article where I can find a formal and precise definition of that function?
logic propositional-calculus
logic propositional-calculus
asked Oct 30 '17 at 15:31
andrebantandrebant
12611
12611
$begingroup$
See similar post: the-number-of-n-variable-preposition-function.
$endgroup$
– Mauro ALLEGRANZA
Oct 30 '17 at 16:14
add a comment |
$begingroup$
See similar post: the-number-of-n-variable-preposition-function.
$endgroup$
– Mauro ALLEGRANZA
Oct 30 '17 at 16:14
$begingroup$
See similar post: the-number-of-n-variable-preposition-function.
$endgroup$
– Mauro ALLEGRANZA
Oct 30 '17 at 16:14
$begingroup$
See similar post: the-number-of-n-variable-preposition-function.
$endgroup$
– Mauro ALLEGRANZA
Oct 30 '17 at 16:14
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
But Mendelson's discussion is already entirely precise. He says (2nd edition, but I think this is preserved through later editions):
For every assignment of truth values T or F to the statement letters
occurring in a statement form, there corresponds, by virtue of the
truth tables for the propositional connectives, a truth value for the
statement form. Thus, each statement form determines a truth function.
What could be clearer or more precise than that? Going "formal" wouldn't make it more precise -- it would merely translate Mendelson's crisp description of the map between statement forms and corresponding truth-functions into a much less transparent symbolic description of the same map.
$endgroup$
add a comment |
$begingroup$
This is my own idea:
Notation:
$S$ - the set of statement letters
$W$ the set of statement forms
$mathrm{Sub}(A)$ - the set of all subformulas of $Ain W$.
It is a known fact that each function $nu_0:Srightarrow{T,F}$ can be uniquely extended to a function $nu:Wrightarrow{T,F}$ in such way that the vaules on negation, conjuction etc. depend on values on constituents.
Another thing we need is that for any $Ain W$, the set $mathrm{Sub}(A)cap S$ is nonempty and finite. This is proved by structural induction.
From the definition, $S$ is countable. Let $S={a_1,a_2,a_3,ldots}$, where $(a_n)$ is injective. We define as well an order on $S$ by condition $a_ile a_j :iff ile j$
Now let's go to defining $mathcal{F}$. Fix $Ain W$. The set $mathrm{Sub}(A)cap S$ is a finite,nonempty subset of $S$. Therefore it can be enumerated: $mathrm{Sub}(A)cap S={b_1,ldots,b_n}$ for some $ninmathbb{N}$ and injective $(b_k)_{kle n}$ (to avoid axiom of choice $(b_k)_{kle n}$ can be uniquely chosen to be monotonic). We will define $xi:{T,F}^nrightarrow{T,F}$ and then set $mathcal{F}(A):=xi$. Fix $xin{T,F}^n$ and set $nu_0:Srightarrow{T,F}$ by conditions: $nu_0(a):=F$ for $ain Ssetminus{b_1,ldots,b_n}$ and if $ain{b_1,ldots,b_n}$ then $a=b_i$ for some $i$ and we set $nu_0(a):=x_i$. $v_0$ can be uniquely extended to $nu:Wrightarrow{T,F}$ and we set $xi(x):=nu(A)$.
An important thing to note is that $mathcal{F}$ depends on the choice of enumeration $(a_n)$.
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
But Mendelson's discussion is already entirely precise. He says (2nd edition, but I think this is preserved through later editions):
For every assignment of truth values T or F to the statement letters
occurring in a statement form, there corresponds, by virtue of the
truth tables for the propositional connectives, a truth value for the
statement form. Thus, each statement form determines a truth function.
What could be clearer or more precise than that? Going "formal" wouldn't make it more precise -- it would merely translate Mendelson's crisp description of the map between statement forms and corresponding truth-functions into a much less transparent symbolic description of the same map.
$endgroup$
add a comment |
$begingroup$
But Mendelson's discussion is already entirely precise. He says (2nd edition, but I think this is preserved through later editions):
For every assignment of truth values T or F to the statement letters
occurring in a statement form, there corresponds, by virtue of the
truth tables for the propositional connectives, a truth value for the
statement form. Thus, each statement form determines a truth function.
What could be clearer or more precise than that? Going "formal" wouldn't make it more precise -- it would merely translate Mendelson's crisp description of the map between statement forms and corresponding truth-functions into a much less transparent symbolic description of the same map.
$endgroup$
add a comment |
$begingroup$
But Mendelson's discussion is already entirely precise. He says (2nd edition, but I think this is preserved through later editions):
For every assignment of truth values T or F to the statement letters
occurring in a statement form, there corresponds, by virtue of the
truth tables for the propositional connectives, a truth value for the
statement form. Thus, each statement form determines a truth function.
What could be clearer or more precise than that? Going "formal" wouldn't make it more precise -- it would merely translate Mendelson's crisp description of the map between statement forms and corresponding truth-functions into a much less transparent symbolic description of the same map.
$endgroup$
But Mendelson's discussion is already entirely precise. He says (2nd edition, but I think this is preserved through later editions):
For every assignment of truth values T or F to the statement letters
occurring in a statement form, there corresponds, by virtue of the
truth tables for the propositional connectives, a truth value for the
statement form. Thus, each statement form determines a truth function.
What could be clearer or more precise than that? Going "formal" wouldn't make it more precise -- it would merely translate Mendelson's crisp description of the map between statement forms and corresponding truth-functions into a much less transparent symbolic description of the same map.
answered Oct 31 '17 at 22:33
Peter SmithPeter Smith
40.9k340120
40.9k340120
add a comment |
add a comment |
$begingroup$
This is my own idea:
Notation:
$S$ - the set of statement letters
$W$ the set of statement forms
$mathrm{Sub}(A)$ - the set of all subformulas of $Ain W$.
It is a known fact that each function $nu_0:Srightarrow{T,F}$ can be uniquely extended to a function $nu:Wrightarrow{T,F}$ in such way that the vaules on negation, conjuction etc. depend on values on constituents.
Another thing we need is that for any $Ain W$, the set $mathrm{Sub}(A)cap S$ is nonempty and finite. This is proved by structural induction.
From the definition, $S$ is countable. Let $S={a_1,a_2,a_3,ldots}$, where $(a_n)$ is injective. We define as well an order on $S$ by condition $a_ile a_j :iff ile j$
Now let's go to defining $mathcal{F}$. Fix $Ain W$. The set $mathrm{Sub}(A)cap S$ is a finite,nonempty subset of $S$. Therefore it can be enumerated: $mathrm{Sub}(A)cap S={b_1,ldots,b_n}$ for some $ninmathbb{N}$ and injective $(b_k)_{kle n}$ (to avoid axiom of choice $(b_k)_{kle n}$ can be uniquely chosen to be monotonic). We will define $xi:{T,F}^nrightarrow{T,F}$ and then set $mathcal{F}(A):=xi$. Fix $xin{T,F}^n$ and set $nu_0:Srightarrow{T,F}$ by conditions: $nu_0(a):=F$ for $ain Ssetminus{b_1,ldots,b_n}$ and if $ain{b_1,ldots,b_n}$ then $a=b_i$ for some $i$ and we set $nu_0(a):=x_i$. $v_0$ can be uniquely extended to $nu:Wrightarrow{T,F}$ and we set $xi(x):=nu(A)$.
An important thing to note is that $mathcal{F}$ depends on the choice of enumeration $(a_n)$.
$endgroup$
add a comment |
$begingroup$
This is my own idea:
Notation:
$S$ - the set of statement letters
$W$ the set of statement forms
$mathrm{Sub}(A)$ - the set of all subformulas of $Ain W$.
It is a known fact that each function $nu_0:Srightarrow{T,F}$ can be uniquely extended to a function $nu:Wrightarrow{T,F}$ in such way that the vaules on negation, conjuction etc. depend on values on constituents.
Another thing we need is that for any $Ain W$, the set $mathrm{Sub}(A)cap S$ is nonempty and finite. This is proved by structural induction.
From the definition, $S$ is countable. Let $S={a_1,a_2,a_3,ldots}$, where $(a_n)$ is injective. We define as well an order on $S$ by condition $a_ile a_j :iff ile j$
Now let's go to defining $mathcal{F}$. Fix $Ain W$. The set $mathrm{Sub}(A)cap S$ is a finite,nonempty subset of $S$. Therefore it can be enumerated: $mathrm{Sub}(A)cap S={b_1,ldots,b_n}$ for some $ninmathbb{N}$ and injective $(b_k)_{kle n}$ (to avoid axiom of choice $(b_k)_{kle n}$ can be uniquely chosen to be monotonic). We will define $xi:{T,F}^nrightarrow{T,F}$ and then set $mathcal{F}(A):=xi$. Fix $xin{T,F}^n$ and set $nu_0:Srightarrow{T,F}$ by conditions: $nu_0(a):=F$ for $ain Ssetminus{b_1,ldots,b_n}$ and if $ain{b_1,ldots,b_n}$ then $a=b_i$ for some $i$ and we set $nu_0(a):=x_i$. $v_0$ can be uniquely extended to $nu:Wrightarrow{T,F}$ and we set $xi(x):=nu(A)$.
An important thing to note is that $mathcal{F}$ depends on the choice of enumeration $(a_n)$.
$endgroup$
add a comment |
$begingroup$
This is my own idea:
Notation:
$S$ - the set of statement letters
$W$ the set of statement forms
$mathrm{Sub}(A)$ - the set of all subformulas of $Ain W$.
It is a known fact that each function $nu_0:Srightarrow{T,F}$ can be uniquely extended to a function $nu:Wrightarrow{T,F}$ in such way that the vaules on negation, conjuction etc. depend on values on constituents.
Another thing we need is that for any $Ain W$, the set $mathrm{Sub}(A)cap S$ is nonempty and finite. This is proved by structural induction.
From the definition, $S$ is countable. Let $S={a_1,a_2,a_3,ldots}$, where $(a_n)$ is injective. We define as well an order on $S$ by condition $a_ile a_j :iff ile j$
Now let's go to defining $mathcal{F}$. Fix $Ain W$. The set $mathrm{Sub}(A)cap S$ is a finite,nonempty subset of $S$. Therefore it can be enumerated: $mathrm{Sub}(A)cap S={b_1,ldots,b_n}$ for some $ninmathbb{N}$ and injective $(b_k)_{kle n}$ (to avoid axiom of choice $(b_k)_{kle n}$ can be uniquely chosen to be monotonic). We will define $xi:{T,F}^nrightarrow{T,F}$ and then set $mathcal{F}(A):=xi$. Fix $xin{T,F}^n$ and set $nu_0:Srightarrow{T,F}$ by conditions: $nu_0(a):=F$ for $ain Ssetminus{b_1,ldots,b_n}$ and if $ain{b_1,ldots,b_n}$ then $a=b_i$ for some $i$ and we set $nu_0(a):=x_i$. $v_0$ can be uniquely extended to $nu:Wrightarrow{T,F}$ and we set $xi(x):=nu(A)$.
An important thing to note is that $mathcal{F}$ depends on the choice of enumeration $(a_n)$.
$endgroup$
This is my own idea:
Notation:
$S$ - the set of statement letters
$W$ the set of statement forms
$mathrm{Sub}(A)$ - the set of all subformulas of $Ain W$.
It is a known fact that each function $nu_0:Srightarrow{T,F}$ can be uniquely extended to a function $nu:Wrightarrow{T,F}$ in such way that the vaules on negation, conjuction etc. depend on values on constituents.
Another thing we need is that for any $Ain W$, the set $mathrm{Sub}(A)cap S$ is nonempty and finite. This is proved by structural induction.
From the definition, $S$ is countable. Let $S={a_1,a_2,a_3,ldots}$, where $(a_n)$ is injective. We define as well an order on $S$ by condition $a_ile a_j :iff ile j$
Now let's go to defining $mathcal{F}$. Fix $Ain W$. The set $mathrm{Sub}(A)cap S$ is a finite,nonempty subset of $S$. Therefore it can be enumerated: $mathrm{Sub}(A)cap S={b_1,ldots,b_n}$ for some $ninmathbb{N}$ and injective $(b_k)_{kle n}$ (to avoid axiom of choice $(b_k)_{kle n}$ can be uniquely chosen to be monotonic). We will define $xi:{T,F}^nrightarrow{T,F}$ and then set $mathcal{F}(A):=xi$. Fix $xin{T,F}^n$ and set $nu_0:Srightarrow{T,F}$ by conditions: $nu_0(a):=F$ for $ain Ssetminus{b_1,ldots,b_n}$ and if $ain{b_1,ldots,b_n}$ then $a=b_i$ for some $i$ and we set $nu_0(a):=x_i$. $v_0$ can be uniquely extended to $nu:Wrightarrow{T,F}$ and we set $xi(x):=nu(A)$.
An important thing to note is that $mathcal{F}$ depends on the choice of enumeration $(a_n)$.
edited Jan 23 at 19:06
answered Jan 23 at 18:45
KulistyKulisty
419216
419216
add a comment |
add a comment |
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$begingroup$
See similar post: the-number-of-n-variable-preposition-function.
$endgroup$
– Mauro ALLEGRANZA
Oct 30 '17 at 16:14