Talk:Herbrand universe

I can't tell the difference between a term algebra, and a Herbrand universe. These seem to be the same thing; maybe "term algebra" is more general, because it does not assume that the terms are first-order logic constructs?? Can someone clarify? linas (talk) 15:53, 9 April 2008 (UTC)


 * Err, maybe I've got it .. a Herbrand universe cannot contain any free variables, by definition. A term algebra does not even raise the issue of free variables, and so one can define a term algebra to consist of a set of constants, variables, and other terms. linas (talk) 15:57, 9 April 2008 (UTC)

Finiteness of a Herbrand Universe
Is it true that if the vocabulary contains at least one function symbol which has arity greater than zero and at least one constant symbol, then the Herbrand Universe of the Vocabulary is infinite?

Let g/1 be a unary function symbol, and c a constant -- one can also say a nullary function symbol. Then one could construct the ground terms g(c), g(g(c)), g(g(g(c))) ... i.e. infinitely many terms within the Herbrand Universe, correct? We could add this to the article. —Preceding unsigned comment added by 141.84.23.241 (talk) 09:50, 18 April 2008 (UTC)


 * Yes, that is correct. The Herbrand universe is only finite if all function symbols are constants. The sub-language of clausal logic with this property is decidable (but the decision problem is NEXPTIME-complete, if I remember correctly). --Stephan Schulz (talk) 09:57, 18 April 2008 (UTC)


 * It may become finite only after an interpretation (logic) is given. After an interpretation is given, one might find out that g(c) is just c, in which case, the whole infinity collapses down. But this is no longer the Herbrand universe, it is the interpretation of the universe.linas (talk) 17:54, 31 August 2008 (UTC)


 * However, (and I am confused on this point, as I do not have an adequate reference on the topic), the Herbrand universe still assumes that reduction has been done, right? i.e. reducible expressions have been eliminated? So, if I have unary functions f and g, and the language has that f(g(c)) is c for all constants c, then the Herbrand universe does not contain f(g(c)), but only c, because f(g(c)) reduces to c. Right? The Herebrand universe is essentially a free object, with the equivalence relation given by the reduction of the language? Right? linas (talk) 18:20, 31 August 2008 (UTC)


 * Guhh, this is confusing. Maybe I'm supposed to just consider formal languages, which are defined extensively (i.e. as a set of sentences). In that case, there are no re-writing rules, since rewriting rules are equivalence relations, and thus predicates (so, for the previous example, it is true that for all c, f(g(c)) = c is the equivalence relation, the sentence "forall x, f(g(x))=x" lies outside of the language, and P(x,y) is the predicate f(g(x))=y, which is true whenever x=y (for all groundings of x and y). Sheesh. I really need a good review. linas (talk) 21:23, 31 August 2008 (UTC)