Theorem of Herbrand

In logical, the theorem of Herbrand establishes a bond between Calcul of the predicates and Calcul of the proposals. Whereas it is possible to determine in an unquestionable way if a proposal of the calculation of the proposals is demonstrable or not, the equivalent question for a formula of the calculation of the predicates is more delicate. The theorem of Herbrand answers this question partially, although one knows since work of Gödel, Tarski, Church, Turing and others, that there does not exist algorithm making it possible to decide if a general formula of the calculation of the predicates is provable or not.

Formulas prénexes

A formula of the calculation of the predicates is prénexe if all the quantifiers which it contains find at the beginning of the formula. Any formula is equivalent to a formula prénexe. For example, the formula $(\backslash \; exists\; X\; \backslash ;\; P\; (X))\; \backslash \; to\; (\backslash \; exists\; there\; \backslash ;\; Q\; (there))$ is equivalent successively to $\backslash \; lnot\; (\backslash \; exists\; X\; \backslash ;\; P\; (X))\; \backslash \; lor\; (\backslash \; exists\; there\; \backslash ;\; Q\; (there))$, $\backslash \; forall\; X\; \backslash ;\; \backslash \; lnot\; P\; (X)\; \backslash \; lor\; \backslash \; exists\; there\; \backslash ;\; Q\; (there)$, and finally $\backslash \; forall\; X\; \backslash ;\; \backslash \; exists\; there\; \backslash ;\; \backslash \; lnot\; P\; (X)\; \backslash \; lor\; Q\; (there)$ (where $\backslash \; to,\; \backslash \; lnot,\; \backslash \; lor,\; \backslash \; land$ indicates respectively the implication, the negation, disjunction, the conjunction. P and Q are unary predicates.

A formula prénexe is universal if it has only universal quantifiers $\backslash \; forall$. It is possible to associate with an unspecified formula a universal formula by applying to him a transformation of Skolem, consisting in introducing new symbols of functions for each existential quantifier $\backslash \; exists$. For example, the form skolemized of $\backslash \; forall\; X\; \backslash ;\; \backslash \; exists\; there\; \backslash ;\; (\backslash \; lnot\; P\; (X)\; \backslash \; lor\; Q\; (there))$ is $\backslash \; forall\; X\; \backslash ;\; \backslash \; lnot\; P\; (X)\; \backslash \; lor\; Q\; (F\; (X))$. Intuitively, so for each X , it exists such as a property R ( X , there ) is checked there, then one can introduce a function F ( X ) = there such as, for all X , R ( X , F ( X )) is checked. It is shown that the initial formula admits a model if and only if its skolemized form admits one of them. In other words, the initial formula is satisfiable if and only if its skolemized form is. In the same way, the initial formula is insatisfiable if and only if its skolemized form is, and thus the negation of the initial formula is provable if and only if the negation of its skolemized form is.

The theorem of Herbrand

Then let us consider a universal formula $F$ (or a whole of formulas), for example of the type $\backslash \; forall\; X,\; \backslash \; forall\; there,\; P\; (X,\; there)$, where $P$ is a predicate, and is a whole of variables $a\_1,\; a\_2,\dots ,\; a\_n,\dots$. Let us consider the three properties:

1. $F$ is consistent , i.e one cannot deduce a contradiction from the assumption $F$. Or $\backslash \; lnot\; F$ is not provable in a system of deduction of the calculation of the predicates, the such natural Déduction for example, which one notes: $\backslash not\backslash vdash\; \backslash lnot\; F$.
2. $F$ is satisfiable , i.e it exists a model in which $F$ is true, which one notes: $\backslash not\backslash vDash\; \backslash lnot\; F$.
3. For entire $n$, there exists a model in which the following proposal that we will note $Q\; (a\_1,\dots ,\; a\_n)$ is true:

$P\; (a\_1,\; a\_1)\; \backslash \; Land\; P\; (a\_1,\; a\_2)\; \backslash \; Land\; P\; (a\_2,\; a\_1)\; \backslash \; Land\; P\; (a\_2,\; a\_2)\; \backslash \; Land\; \backslash \; cdots\; P\; (a\_1,\; a\_n)\; \backslash \; Land\; \backslash \; cdots\; P\; (a\_\; \{n-1\},\; a\_n)\; \backslash \; Land\; P\; (a\_n,\; a\_1)\; \backslash \; Land\; \backslash \; cdots\; P\; (a\_n,\; a\_n)$ what one notes: for all $n$, $\backslash \; not\; \backslash \; vDash\; \backslash \; lnot\; Q\; (a\_1,\dots ,\; a\_n)$

The Théorème of complétude of Gödel states equivalence between 1) and 2). In addition, 2) involves 3), the model in which 2) is checked making it possible to deduce some from the models checking 3). The theorem of Herbrand states that, reciprocally, 3) involves 2), when the $a\_i$ describe a particular field, the field of Herbrand. It is noted that, in the formulation of 3), the universal quantifiers disappeared and that the formulas to be proven then become simple proposals of propositional calculation.

In a dual way, by posing $G\; =\; \backslash \; lnot\; F$ and $R\; =\; \backslash \; lnot\; Q$, one obtains equivalences between the three properties:

1. $G$ is provable , i.e it exists a demonstration of $G$ in a system of deduction of propositional calculation. what one notes: $\backslash vdash\; G$.
2. $G$ is valid , i.e. $G$ is true in very model, which one notes: $\backslash vDash\; G$.
3. There exists an entirety $n$ for which $R\; (a\_1,\dots ,\; a\_n)$ is valid: $\backslash \; vDash\; R\; (a\_1,\dots ,\; a\_n)$

$G$ is then a theorem.

If $F$ is not satisfiable (what occurs if and only if $G\; =\; \backslash \; lnot\; F$ is a theorem), then by checking $Q\; (a\_1,\dots ,\; a\_n)$ for $n=1$, then $n=2$, etc, one will end up falling on an entirety $n$ such as $Q\; (a\_1,\dots ,\; a\_n)$ is false, and reciprocally.

On the other hand, if $F$ is satisfiable (and thus if $G$ is not a theorem), then for all $n$, $Q\; (a\_1,\dots ,\; a\_n)$ will be true, and the computing process will not finish, except in the particular case where the variables $a\_i$ are of finished number.

That illustrates the fact that the whole of the formulas not satisfiables or the whole of the theorems is units recursively énumérables, but that the calculation of the predicates is not Décidable.

Examples

The field of Herbrand of the model is defined at least by a constant element (in order to be nonempty) and consists of all the terms which one can form starting from the constants and of the functions used in the formulas considered. One defines a model of Herbrand by allotting the true value to certain predicates defined on these terms.

Example 1 : let us consider the formula $F\; =\; \backslash \; forall\; X,\; P\; (X)\; \backslash \; Land\; \backslash \; lnot\; P\; (a)$, where $a$ is a constant. The field of Herbrand is consisted of the singleton $\backslash \; \{has\; \backslash \}$. One forms the formula $Q\_1\; then\; =\; P\; (a)\; \backslash \; Land\; \backslash \; lnot\; P\; (a)$ which is false, therefore $F$ is not satisfiable.

Example 2 : One considers the formula $F\; =\; \backslash \; forall\; X,\; (P\; (X)\; \backslash \; lor\; Q\; (X))\; \backslash \; Land\; \backslash \; lnot\; P\; (a)\; \backslash \; Land\; \backslash \; lnot\; Q\; (b)$. The field of Herbrand is $\backslash \; \{has,\; B\; \backslash \}$. One forms $Q\_1\; then\; =\; (P\; (a)\; \backslash \; lor\; Q\; (a))\; \backslash \; Land\; \backslash \; lnot\; P\; (a)\; \backslash \; Land\; \backslash \; lnot\; Q\; (b)$ which is true in a model where one gives the true value to $Q\; (a)$, then one forms $Q\_2\; =\; Q\_1\; \backslash \; Land\; (P\; (b)\; \backslash \; lor\; Q\; (b))\; \backslash \; Land\; \backslash \; lnot\; P\; (a)\; \backslash \; Land\; \backslash \; lnot\; Q\; (b)$ which is true in a model where one gives the true value to $\backslash \; \{Q\; (a),\; P\; (b)\; \backslash \}$. Having exhausted the field of Herbrand, calculation finishes and $F$ is satisfiable in the model previously definite.

Example 3 : One considers the formula $F\; =\; \backslash \; forall\; X,\; \backslash \; forall\; there,\; P\; (X)\; \backslash \; Land\; \backslash \; lnot\; P\; (F\; (there))$. The field of Herbrand consists of $\backslash \; \{has,\; F\; (a),\; f^2\; (a),\dots ,\; f^n\; (a),\dots \; \backslash \}$. One forms $Q\_1\; then\; =\; P\; (a)\; \backslash \; Land\; \backslash \; lnot\; P\; (F\; (a))$ which has a model by allotting the value true to $P\; (a)$ but false with $P\; (F\; (a))$. Then, by taking the first two elements of the $a$ field and $f\; (a)$, one forms $Q\_2\; =\; P\; (a)\; \backslash \; Land\; \backslash \; lnot\; P\; (F\; (a))\; \backslash \; Land\; P\; (a)\; \backslash \; Land\; \backslash \; lnot\; P\; (F\; (F\; (a)))\; \backslash \; Land\; P\; (F\; (a))\; \backslash \; Land\; \backslash \; lnot\; P\; (F\; (a))\; \backslash \; Land\; P\; (F\; (a))\; \backslash \; Land\; \backslash \; lnot\; P\; (F\; (F\; (a)))$ which cannot have of model because of the conjunction distorts $P\; (F\; (a))\; \backslash \; Land\; \backslash \; lnot\; P\; (F\; (a))$. Thus $F$ does not have a model.

Example 4 : The formula $G\; is\; considered\; =\; (\backslash \; exists\; X,\; \backslash \; forall\; there,\; P\; (X,\; there))\; \backslash \; to\; (\backslash \; forall\; there,\; \backslash \; exists\; X,\; P\; (X,\; there))$ which one wants to show that it is about a theorem. After having famous the variables X and in the second part of G in order to avoid there having dependant variables of the same name, one obtains a form prénexe equivalent to G:

$(\backslash \; exists\; X,\; \backslash \; forall\; there,\; P\; (X,\; there))\; \backslash \; to\; (\backslash \; forall\; Z,\; \backslash \; exists\; T,\; P\; (T,\; Z))$

$\backslash \; lnot\; (\backslash \; exists\; X,\; \backslash \; forall\; there,\; P\; (X,\; there))\; \backslash \; lor\; (\backslash \; forall\; Z,\; \backslash \; exists\; T,\; P\; (T,\; Z))$

$(\backslash \; forall\; X,\; \backslash \; exists\; there,\; \backslash \; lnot\; P\; (X,\; there))\; \backslash \; lor\; (\backslash \; forall\; Z,\; \backslash \; exists\; T,\; P\; (T,\; Z))$

$\backslash \; forall\; X,\; \backslash \; exists\; there,\; \backslash \; forall\; Z,\; \backslash \; exists\; T,\; \backslash \; lnot\; P\; (X,\; there)\; \backslash \; lor\; P\; (T,\; Z)$

The formula prénexe $F$ equivalent to $\backslash \; lnot\; G$ is:

$\backslash \; exists\; X,\; \backslash \; forall\; there,\; \backslash \; exists\; Z,\; \backslash \; forall\; T,\; P\; (X,\; there)\; \backslash \; Land\; \backslash \; lnot\; P\; (T,\; Z)$

whose skolemized form is:

$\backslash \; forall\; there,\; \backslash \; forall\; T,\; P\; (has,\; there)\; \backslash \; Land\; \backslash \; lnot\; P\; (T,\; F\; (there))$

Formation of the formulas $Q\_1$ by taking for $(there,\; T)$ the couple $(has,\; a)$ then $Q\_2$ by taking $(there,\; T)\; =\; (F\; (a),\; a)$ leads to $P\; (has,\; a)\; \backslash \; Land\; \backslash \; lnot\; P\; (has,\; F\; (a))\; \backslash \; Land\; P\; (has,\; F\; (a))\; \backslash \; Land\; \backslash \; lnot\; P\; (has,\; F\; (F\; (a)))$ which is false in very model. $F$ is thus not satisfiable and $G$ is a theorem.

Example 5 : The formula $G\; is\; considered\; =\; (\backslash \; forall\; X,\; \backslash \; exists\; there,\; P\; (X,\; there))\; \backslash \; to\; (\backslash \; exists\; there,\; \backslash \; forall\; X,\; P\; (X,\; there))$. By a process comparable with the preceding example, one obtains for form prénexe equivalent to $G$: $\backslash \; exists\; X,\; \backslash \; forall\; there,\; \backslash \; exists\; Z,\; \backslash \; forall\; T,\; \backslash \; lnot\; P\; (X,\; there)\; \backslash \; lor\; P\; (T,\; Z)$. The formula prénexe $F$ equivalent to $\backslash \; lnot\; G$ is:

$\backslash \; forall\; X,\; \backslash \; exists\; there,\; \backslash \; forall\; Z,\; \backslash \; exists\; T,\; P\; (X,\; there)\; \backslash \; Land\; \backslash \; lnot\; P\; (T,\; Z)$

whose skolemized form is:

$\backslash \; forall\; X,\; \backslash \; forall\; Z,\; P\; (X,\; F\; (X))\; \backslash \; Land\; \backslash \; lnot\; P\; (G\; (X,\; Z),\; Z)$

It is satisfiable by giving the value truth to all $P\; (U,\; F\; (U))$ and forgery with all $P\; (G\; (U,\; v),\; v)$, where $u$ and $v$ describe the field of Herbrand $\backslash \; \{has,\; F\; (a),\; G\; (has,\; a),\; F\; (F\; (a)),\; F\; (G\; (has,\; a)),\; G\; (has,\; F\; (a)),\dots \; \backslash \}$, and by giving an arbitrary value to the other predicates. However, the successive calculation of the corresponding formulas $Q\_1$, $Q\_2$,… does not finish. $F$ is satisfiable and thus $G$ is not a theorem, but the preceding step does not make it possible to show it by a calculation in a finished number of stages.

Random links:

Shining Forces CD | Franz of Paula Adam von Waldstein | Zoltán Czibor | The Bus | Raoul Maltravers | Planète_de_trésor

© 2007-2008 speedlook.com; article text available under the terms of GFDL, from fr.wikipedia.org