1. Herbrand Logic Semantics
Computational Logic Lecture 17
Herbrand Logic Semantics
Michael Genesereth Autumn 2011

2. Propositional Interpretations
For a language with n constants, there are 2n interpretations.
9/19/2019

3. Relational Interpretations
|i| a b r
{,}   {}
{,}   {}
{,}   {}
{,}   {, }
{,}   {}
{,}   {}
{,}   {}
{,}   {, }
{,}   {}
{,}   {}
{,}   {}
{,}   {, }
. . .
Infinitely many interpretations.
9/19/2019

4. Logical Entailment
A set of premises logically entails a conclusion if and only if every interpretation that satisfies the premises also satisfies the conclusion.
In the case of Propositional Logic, the number of interpretations is finite, and so it is possible to check logical entailment directly in finite time.
In the case of Relational Logic, the number of interpretations is infinite, and so a direct check of logical entailment is not feasible.
9/19/2019

5. Good News
Given any set of sentences, there is a special subset of interpretations called Herbrand interpretations.
Under certain conditions, checking just the Herbrand interpretations suffices to determine logical entailment.
Since there are fewer Herbrand interpretations than interpretations in general, checking just the Herbrand interpretations is less work than checking all interpretations.
9/19/2019

6. Herbrand
In HLT, the Herbrand universe for a relational language is the set of all object constants in that language.
In HLF, the Herbrand universe for is the set of all ground terms that can be formed from the constants in the language.
Difference: In HLF, functional terms map to themselves rather than mapping to object constants.
9/19/2019

7. Ground Relational Sentences
A dataset Delta satisfies a ground relational sentence phi if and only if phi is contained in Delta.
9/19/2019

8. Logical Sentences |= iff    |=(  ) iff |= and |=
|=(  ) iff |= or |=
|=(  ) iff |# or |=
|=(  ) iff |= or |#
|=(  ) iff |= and |= or |# and |#
9/19/2019

9. Quantified Sentences
A dataset satisfies a universally quantified sentence if and only if it satisfies every instance of the scope of that sentence.
A dataset satisfies an existentially quantified sentence if and only if it satisfies some instance of the scope of that sentence.
9/19/2019

10. 9/19/2019

11. Example Interpretation |i|={, } i(a) =  i(b) = 
i(r) = {,, ,}
Herbrand Base
{r(a,a), r(a,b), r(b,a), r(b,b)}
Herbrand Interpretation
|i|={a,b}
i(a) = a
i(b) = b
i(r) = {a,b, b,b}
9/19/2019

12. Example Interpretation |i|={, } i(a) =  i(b) = 
i(r) = {,, ,}
Herbrand Base
{r(a,a), r(a,b), r(b,a), r(b,b)}
Herbrand Interpretation
|i|={a,b}
i(a) = a
i(b) = b
i(r) = {a,b, b,b}
9/19/2019

13. Example Interpretation |i|={, } i(a) =  i(b) = 
i(r) = {,, ,}
Herbrand Base
{r(a,a), r(a,b), r(b,a), r(b,b)}
Herbrand Interpretation
|i|={a,b}
i(a) = a
i(b) = b
i(r) = {a,b, b,b}
9/19/2019

14. 9/19/2019

15. Transitivity Theorem
It is not possible in first-order logic to define transitive closure in first-order logic.
More precisely, it is not possible characterize the set of structures consisting of an arbitrary universe, an arbitrary binary relation, and the transitive closure of that relation.
9/19/2019

16. Expressive Completeness Theorem
Metatheorem: In HLF, the theory of arithmetic is finitely axiomatizable.
Proof: If there were a finite axiomatization, then the theory would be decidable. However, arithmetic is not decidable. Therefore, there is no finite axiomatization.
9/19/2019

17. Resolution Incompleteness
Metatheorem: Resolution is not complete for HL.
Proof: If there were a finite axiomatization, then the theory would be decidable. However, arithmetic is not decidable. Therefore, there is no finite axiomatization.
9/19/2019

18. Deductive Incompleteness
Metatheorem: There is no complete proof procedure for HL.
Proof: If there were a finite axiomatization, then the theory would be decidable. However, arithmetic is not decidable. Therefore, there is no finite axiomatization.
9/19/2019

19. 9/19/2019

20. Special Cases
9/19/2019

21. Herbrand Theorem
Herbrand Theorem: A set of quantifier-free sentences in Relational Logic is satisfiable if and only if it has a Herbrand model.
9/19/2019

22. Herbrand Theorem Does Not Apply
Sentences
r(a,a)
r(b,b)
x.r(x,x)
Model Herbrand Construction:
|i|={, , } |h|={a,b}
i(a) =  h(a) = a
i(b) =  h(b) = b
i(r) = {,, ,} h(r) = {a,a, b,b}
Herbrand construction does not satisfy and no Herbrand model satisfies.
9/19/2019

23. Good News (Sort of…)
Solution: Skolemize the sentences to get rid of explicit quantifiers.
Small hitch #1 is that we are now dealing with existential sentences but not purely existential sentences; so we need to extend the technique we just looked at. See upcoming lecture.
Small hitch #2 is that there are some difficulties using the Modified Herbrand Method with Functional Logic. See next few slides.
9/19/2019

24. Special Cases Ground Relational Logic
no variables, no functions, no quantifiers
Universal Relational Logic
no functions, no quantifiers
free variables implicitly universally quantified
Existential Relational Logic
no functions
 Functional Relational Logic
no quantifiers
9/19/2019

25. Herbrand Universe for Functional Logic
Old Definition: The Herbrand universe for a set of sentences in Relational Logic is the set of all ground terms that can be formed from just the constants used in those sentences.
In the absence of function constants, this is exactly the set of constants in the set of sentences.
{a,b}
In 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)), ...}
9/19/2019

26. Herbrand Theorem for Functional Logic
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. Hooray!!
9/19/2019

27. Sad Theorem
The size of the Herbrand universe for a functional language is infinite.
Upshot: Checking the Herbrand interpretations for a language to determine logical entailment is not feasible in finite time.
9/19/2019

28. General Problem The number of Herbrand models can be very large.
n object constants
m k-ary relation constants
Number of k-ary tuples: nk
Number of k-ary relations: 2^nk
Number of Herbrand Interpretations: (2^nk)^m
10 object constants
3 2-ary relation constants
Number of k-ary tuples: 100
Number of k-ary relations: 2100
Number of Herbrand Interpretations: 2300
9/19/2019

29. Summary Herbrand Method works for Ground Logic.
Modified Method works for Universal Logic.
Herbrand Method does not work for Existential Logic. These cases can be handled by Skolemizing to form sentences in Functional Logic.
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.
Solution: Use formal proofs!
9/19/2019

30. 9/19/2019