consistent Henkin theory with language that has only one constant symbol

as the title says, I am looking for a consistent Henking theory (either a complete or an incomplete one, or both of them) whose language has only one constant symbol (but may have more predicate- and/or function symbols).

I'd be thankful for any help or hint!

• What is your definition of Henkin theory? How about the theory of the structure with one element, in the language with equality and a single constant symbol? – spaceisdarkgreen Dec 1 '18 at 23:58
• @spaceisdarkgreen The definition of Henkin theory is the following: Let $\Sigma$ be a theory in the language L. Then $\Sigma$ is called a Henkin theory, if for every closed formula of the form $\exists x \phi(x)$ there's a constant symbol $c \in L$ so that $\Sigma \vdash (\exists \phi(x) \rightarrow \phi(c))$. – Studentu Dec 2 '18 at 15:52
• @spaceisdarkgreen So I think the example you suggested would work? (At least I can't find any counterexample for that language.) – Studentu Dec 2 '18 at 15:55

The example spaceisdarkgreen gave is in fact essentially the only one: take $$\varphi(x)\equiv (\exists y(y\not=c)\implies x\not=c).$$ Then $$\Sigma\vdash\exists x\varphi(x)$$, and so if $$\Sigma$$ is to have the Henkin property we must have $$\Sigma\vdash\varphi(c)$$. But $$\Sigma\vdash \varphi(c)\iff \forall y(y=c)$$. So $$\Sigma$$ describes the one-element structure (precisely: every model of $$\Sigma$$ has a single element, named by $$c$$).
A less trivial (in this context) notion of Henkin theory is gotten by replacing "constant symbol" by "term" in the definition: that is, requiring that for each formula $$\varphi$$, there be some (variable-free, or "closed") term $$t$$ such $$\Sigma\vdash\exists x\varphi(x)\iff\varphi(t).$$ Here, having a single constant symbol isn't too much of a problem, since we can still produce lots of terms by using function symbols.
• Great, I understand the counterexample, thank you so much! However, I've got a further question and a question to your second example: 1) Do you know if this Henkin theory is complete or incomplete (and how do you know)? 2) Do I understand it correctly, that for your alternative definition of Henkin theory, there would be also structures with more than one element so that $\Sigma$ is a Henkin theory? – Studentu Dec 2 '18 at 21:55
• (2) Yes. For a concrete example, take $\{-1,1\}$ with multiplication and a constant naming $-1$. (1) It depends. $\Sigma$ definitely decides every sentence with no non-logical relation symbols: just get rid of all quantifiers and replace all terms (open or closed) by "$c$," and then check that the result is a tautology. For example, the sentence $$(*)\quad\forall x,y\exists z(F(G(x), y)=y\implies F(x,z)=x)$$ boils down under this process to just "$c=c\implies c=c$," which is true; so $\Sigma$ proves $(*)$. (cont'd) – Noah Schweber Dec 3 '18 at 0:10
• Basically, the point is that since $\Sigma$ proves $\forall x,y(x=y)$, $\Sigma$ proves that every equality is true. In the absence of non-logical relation symbols, this means that $\Sigma$ decides every sentence, since all sentences are built up from equalities (via Booleans and quantifiers). If, however, there are relation symbols in the language, then $\Sigma$ need not be complete. For example, take $\Sigma$ to be precisely $\{\forall x(x=c)\}$; then $\Sigma$ is a Henkin theory in the language $\{c,R\}$ with $R$ a unary relation symbol, but $\Sigma$ does not prove or disprove $R(c)$. – Noah Schweber Dec 3 '18 at 0:11