Is First-order logic Turing complete?

Question

Is First-order logic also called predicate logic Turing complete?

EDIT: If we do computation by checking if premises logically entail a conclusion. And the method is resolution .

Answer 1

You seem to be asking whether you can use the deductive rules of first-order logic to simulate a Turing machine. One possible answer is yes, in the following sense: You can use the elements $s$ of a structure to represent steps in a computation, a series of predicates $P(s)$ to represent the state at that step, and a group of first-order axioms such as $P(s) \rightarrow Q(s+1)$ to initialize the state and govern the transitions. If you do this correctly, your Turing machine halts iff you can use these axioms to prove $(\exists s)[s$ is configured in a halting state$]$. But there are also other ways to interpret your question.

Answer 2

[Some Turing-Complete Extensions of First-Order Logic], by Antti Kuusisto, contains the following remark in the second paragraph:

A crucial weakness in the expressivity of k-th order predicate logic is that only a finite amount of information can be encoded by a finite number of quantified relations over a finite domain. Intuitively, there is no infinitely expandable memory available. Thus k-th order logic is not Turing-complete.