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The professor told me that the solution is trivial, but I must have missed something because I don't even see anyway to start.

Consider an arbitrary language $L$ (which can contains function, constant and relation symbols) and a first-order theory $T$ that can be axiomatized by sentences that have no $\neg$ at all. Prove or find a counterexample that $T$ is always consistent.

user292595
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  • I think this fragment is called "positive first-order logic"; can someone confirm? – goblin GONE Mar 19 '16 at 04:49
  • In general it is non-sense to say what you are forbidding (in this case negation), the important is what you are allowing. Since you have forgotten this part, I find your question not precise enough. For example, what happens if you allow "implicacion (binary)" and "falsum (constant)" as primitive and you don't allow "negation (unary)" as primitive? – boumol Mar 20 '16 at 13:18
  • @boumol: allowing F would make this question even more trivial, while implication have absolutely no effects on the question (WLOG we can also rule out disjunction). – user292595 Mar 21 '16 at 13:46
  • @user292595: You didn't get my point. The important part is that you only allow "conjunction" and "disjunction". – boumol Mar 21 '16 at 16:27

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To show consistency, it is enough to show that there is a model. Consider the $1$-element structure $M$ in which all constant symbols and function symbols have the only possible interpretation, and all relations are true.

We need to show that all the negation-free sentences of our language are true in $M$. In principle this is done by induction on complexity. Start from atomic formulas, and observe that when an atomic formula has its free variables replaced by the one element of $M$, the atomic formula is true in $M$.

André Nicolas
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  • Oh, right. I keep thinking along the proof theory direction. Well I guess the answer is indeed trivial, in more than one sense. – user292595 Mar 19 '16 at 04:56