Herbrand’s theorem and applications

Aman| Darsh | Mihir

Reminder

In first-order logic, there is no algorithm to decide whether a formula is valid or non valid.

Why?While number of predicates and functions on the domain can be finite, the domain can be infinite. Brute force is not guaranteed to halt in finite time.

Semi-decidabilityA problem is said to be semi-decidable if there exists a program such that:

1. If it ends, then it correctly decides whether the formula is valid or not.2. If the formula is not valid, the termination of this program is not guaranteed.

No program exists that is guaranteed to terminate with the right answer.

Introduction

An interpretation is an assignment of meaning to the symbols of a formal language.

In Propositional calculus

A function that maps each symbol to true or false.

This function is known as a truth assignment or valuation function.

There are 2n interpretations for a language with n propositions.

Interpretations

The number of relational sentences (that can be assigned a truth value) are infinite.

An infinite number of interpretations.

For example,

r(x,y) => (x>y)

Domain is the set of naturals

Interpretations (Set of statements assigned TRUE)

{r(1,1)}

{r(1,2)}

{r(1,3),r(1,4)}

…

Number of interpretations is uncountable!

Relational logic

Satisfiability

● A formula F is unsatisfiable iff there is no interpretation I such that I(F) = true

● Hence, if one could check all interpretations, one could confirm (un)satisfiability of F

● This is not feasible in First Order Logic

● Given any set of sentences, there is a specially defined subset of interpretations called Herbrand interpretations

● Under certain conditions, checking just the Herbrand interpretations suffices to determine satisfiability.

● Much easier than checking all interpretations.

Motivation

● The Herbrand universe for a set of sentences in Relational Logic (with at least one object constant) is the set of all ground terms that can be formed from just the constants used in those sentences

● If there are no object constants, then we add an arbitrary object constant, say a.● This determines the domain of interpretation of a formula for Herbrand interpretations

For example, given the set of sentences :

1.{P(x), R(x)}

Herbrand universe = {a}

2. {∀x.(r(a,x) ⇒ r(x,b)) , ∀x.∀y.∀z.(r(x,y) ∧r(x,y) ⇒ r(x,z))}

Herbrand universe = {a,b}

3. {∀x.(f(a,x)⇒ g(x))}

Herbrand universe = {a,f(a,a), g(a),f(a,g(a)), f(g(a),a), f(g(a),g(a)), g(g(a)) ...}

The Herbrand universe

Herbrand Base

● The Herbrand base for a set of sentences is the set of all ground atomic sentences that can be formed using just the constants in the Herbrand universe.

F = {¬q(x) ∨ r(x), p(x) ∨ ¬r(x), ¬p(x)}

H(F) = {a}

HB(F) = {p(a), q(a),r(a)}

● Uses predicates, while Herbrand universe uses functions.

● A Herbrand model for a set of sentences is any subset of the Herbrand base for those sentences

A Herbrand interpretation has

● D = the Herbrand universe,● R = a subset of the Herbrand base that is TRUE.

R defines a mapping from the Herbrand base to {TRUE,FALSE}.

Essentially, we assign a truth value to each sentence-- similar to a valuation function.

Herbrand interpretation

Example

Formula

F = {p(y), q(a) ∨ ¬p(f (x)), ¬q(x)}

Herbrand Base

HB(F) = {p(t) | t ∈ H(F)} ∪ {q(t) | t ∈ H(F)}

Herbrand Universe

H(F) = {f n (a) | n ≥ 0}

Herbrand Interpretation

Any subset of HB(F)

Herbrand theorem

A set of quantifier-free sentences is satisfiable if and only if it has a Herbrand model that satisfies it

● F is unsatisfiable iff there is no interpretation I such that I(F) = true● in order to check this, we should consider all models:

○ if F is propositional with n different propositions, then there are 2^n models○ in a first order formula, the number of interpretations can be non-countable

● it would be useful to have a subset of interpretations of F such that○ it contains a smaller (finite or countable) number of interpretations○ analyzing it is sufficient to decide the satisfiability of F

● such interpretations exist for every formula, and are called Herbrand interpretations

Satisfiability and Interpretation

THE PROBLEM

● Herbrand’s theorem is the basis for most proof techniques in automatic theorem proving

● in order to decide the (un)satisfiability of a formula F, it is enough to study its Herbrand interpretations

● it is necessary to have an ordered and exhaustive way to produce the Herbrand interpretations● this can be done by means of semantic trees

Herbrand Theorem

How is it useful?

Let HB(F) = {A1, A2, A3, ..} be the Herbrand base of a formula F in clause form: a semantic tree for F is a binary tree where

● every level of the tree corresponds to a ground atom of HB(F)● the two links from a node at level i − 1 to nodes at level i are labeled, resp., with Ai and ¬Ai

Semantic trees (Robinson 1968, Kowalski-Hayes 1969)

Definition:

1. F = {p(y), q(a) ∨ ¬p(f (x)), ¬q(x)}

H(F) = {f n (a) | n ≥ 0} HB(F) = {p(t) | t ∈ H(F)} ∪ {q(t) | t ∈ H(F)}

● every Herbrand interpretation falsifies some instance of some clause, so that F is unsatisfiable

Semantic trees (Robinson 1968, Kowalski-Hayes 1969)

Examples:

● a semantic tree is complete if every path from the root to a leaf contains Ai or ¬Ai for all Ai ∈ HB(F)

○ a complete tree for F contains all Herbrand interpretations of F

● given a node N, I(N) is the set of all literals which label the path from the root to N○ I(N) partially represents a Herbrand interpretation

● a node N is a failure node (denoted by z) if I(N) makes some ground instance of some clause false, and I(N1 ) for any predecessor N1 of N does not

● a tree is closed iff all paths from the root to a leaf contain a failure node

Semantic trees (Robinson 1968, Kowalski-Hayes 1969)

Completeness, failure nodes and closed trees

A set S of clauses is unsatisfiable iff corresponding to every complete semantic tree of S, there is a finite closed semantic tree.

Herbrand Theorem

Version 1

Alternatively

A set of quantifier-free sentences is satisfiable if and only if it has a Herbrand model that satisfies it

Let T be the complete semantic tree for S

● Consider each branch B of T: for each branch B there is a complete Herbrand interpretation Ib

● S is unsatisfiable: Ib must falsify a ground instance C1 of some clause C of S● Since C1 is finite, then there must be a failure node Nb which is a finite number of

links away from the root● Since for every branch B we can find a failure node Nb there is a corresponding

closed semantic tree T1● Since only a finite number of links are connected to each node then T1 is finite

Herbrand Theorem: Version 1

S is unsatisfiable => there is a finite closed semantic tree

T1 finite closed semantic tree corresponding to every complete semantic tree T of S

● For every branch of T there must be an interpretation that falsify S● Every possible interpretation falsifies S, thus S is unsatisfiable.

Herbrand Theorem: Version 1

If there is a finite closed semantic tree then S is unsatisfiable

● Method:Add negation of conclusion to the premises to form the satisfaction set. Loop over Herbrand interpretations.

● Cross out each interpretation that does not satisfy the sentences in the satisfaction set. If all Herbrand interpretations are crossed out, by the Herbrand Theorem, the set is unsatisfiable.

● Termination: Since there are only finitely many Herbrand interpretations, the process halts.

Application:Herbrand Method

Definition:

Ground Relational LogicNo variables, no functions, no quantifiers

Universal Relational LogicNo functions, no quantifiers

Free variables implicitly universally quantified

Functional Relational LogicNo quantifiers, with function constants

Cases of Relational Logic

Ground relational logic

Premises:p(a) ⇒ q(a)p(a) ∨ q(a)

Conclusionq(a)

Herbrand Interpretations{}{p(a)}{q(a)}{p(a),q(a)}

4 such interpretations

Satisfaction Setp(a) ⇒ q(a)p(a)∨ q(a)¬q(a)

Universal Relational Logic

Premisesp(x) ⇒ q(x)p(a) ∨ q(a)

Conclusionq(a)

Herbrand Interpretations{}{p(a)}{q(a)}{p(a),q(a)}

Satisfaction Setp(x) ⇒ q(x)p(a) ∨ q(a)¬q(a)

Universal Relational Logic

Premisesp(x) ⇒ q(x)p(x) ∨ q(x)

Conclusionq(x)

Herbrand Interpretations?

Satisfaction Setp(x) ⇒ q(x)p(x) ∨ q(x)¬q(x) ? Incorrect¬∀x.q(x) ? Has universal quantifer

Negation distributed over universal quantification by flipping

the universal quantifier to an existential quantifier.

¬∀x1...∀xn.ϕ↓∃x1... ∃xn.¬ϕ

If a sentence is purely universal, then this distribution leads to

a purely existential sentence.

Universal and Existential Sentences

The Skolemization of a purely existential sentence is the sentence obtained by dropping the existential quantifiers and replacing all variables systematically by brand new constants.

Example:

∃x.¬q(x)

↓¬q(c)

∃x.∃y.(p(x,y) ∧ q(x,b))

↓p(c,d) ∧ q(c,b)

Skolemization

Skolemization Theorem A set of sentences in Relational Logic is satisfiable if and only if its Skolemization is satisfiable.

A modification of the Herbrand Method can be used.

Significance of Skolemization

Original Definition● Add negation of conclusion to the premises to form the satisfaction set.● Loop over Herbrand interpretations…

New Definition

● Negate the conclusion.● Add its Skolemization to the premises to form the satisfaction set.● Loop over Herbrand interpretations…

Satisfaction set after Skolemization

Premisesp(x) ⇒ q(x)p(x) ∨ q(x)

Conclusionq(x)

Negation¬∀x.q(x)

Existentialization∃x.¬q(x)

Skolemization¬q(c)

Satisfaction Setp(x) ⇒ q(x)p(x) ∨ q(x)¬q(c)

Herbrand universe

OldIn the absence of function constants, this is exactly the set of

constants in the set of sentences.

{a,b}

NewIn the presence of function constants, all ground functional terms are included.

{a, b, f(a), f(b), g(a), g(b), f(f(a)), f(f(b)), f(g(a)), f(g(b)), ...}

Functional relational logic

The Herbrand theorem applies

Herbrand Theorem

A set of quantifier-free sentences has a model if and only if it has a Herbrand model.

No quantifiers; so we are okay.The Modified Herbrand Method works.

Unfortunately, the size of the Herbrand universe for a functional language is infinite.

Finite time checks for logical entailment are not feasible.

● The Herbrand method works for Ground logic.● The Modified Herbrand method works for Universal Logic.● The Herbrand Method does not work for Existential Logic. These● cases can be handled by Skolemizing to form sentences in Functional Logic and

then using Modified Herbrand Method.● Modified Herbrand Method works for Functional Logic, but there are infinitely

many interpretations.● In any case, the number of Herbrand interpretations can be very large.

Summary

About Herbrand (Read)

Jacques Herbrand ● (Paris, France, February 12, 1908 - La B´erarde, Is`ere, France, July 27,

1931)● PhD at Ecole Normale Superieure, Paris, in 1929 ● joined the army in October 1929 ● H. universe, H. base, H. interpretation, H. structure, H. quotient ● Herbrand’s Theorem:

○ actually, two different results have this name introduced the notion of recursive function

○ worked with John von Neumann and Emmy Noether ● died falling from a mountain in the Alps while climbing

References

● Herbrand Theorem proof: http://profs.sci.univr.it/~farinelli/courses/ar/slides/herbrand.pdf● Herbrand Theorem and Interpretation: [Chang-Lee 4] ● Semantic Tree:

○ http://costa.ls.fi.upm.es/~damiano/teaching/emcl/cl_08_09/slides/05herbrand.pdf○ http://www.cs.miami.edu/~geoff/Courses/TPTPSYS/FirstOrder/Herbrand.shtml

● http://logic.stanford.edu/classes/cs157/2004/lectures/lecture07.pdf

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