A finite language with an infinite model but no finite ones

Question

So I have this question from Logic for Applications by Nerode and Shore. The chapter on Predicate Logic.

Find a finite language $L$ and a finite set of sentences $S$ that has an infinite model but no finite ones.

Proposition 7.5 says that a tableau is finite if every path is contradictory. So am I just finding a sentence with at least one contradictory path? I'm a bit confused here.

score 0 · Answer 1 · answered Jan 07 '18 at 22:16

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I have no idea what the relevant background is supposed to be for this problem as presented in your textbook, but here's a complicated abstract algebra example. There are groups, such as the Higman group, which are finitely presented but have no nontrivial finite quotients. So you can consider $L$ the language of groups and the sentences given by the existence of elements satisfying the presentation of the Higman group together with the axiom that one of these elements is not the identity.

answered Jan 07 '18 at 22:16

Qiaochu Yuan

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A total dense order also works, I guess, or, in the same spirit, an ordered field. – Mariano Suárez-Álvarez Jan 07 '18 at 22:31
Thanks, I ended up using a Herbrand base. – Tim Jan 08 '18 at 15:39

A finite language with an infinite model but no finite ones

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