Questions: Decidable Fragments of First-Order Logic

5 questions to test your understanding

Score: 0 / 5
Question 1 Multiple Choice

First-order logic with only two variable names (x and y, reused freely) is decidable. What happens if you add a third variable name?

AThe three-variable fragment is still decidable, just harder — it becomes EXPTIME-complete instead of NEXPTIME-complete
BAdding a third variable restores full FOL expressiveness and undecidability
CThe three-variable fragment is decidable but only for finite models
DThree variables make the fragment decidable for satisfiability but undecidable for validity
Question 2 Multiple Choice

Modal logic is decidable. What is the connection between modal logic's decidability and the theory of decidable fragments of first-order logic?

AModal logic is unrelated to FOL — its decidability comes from different proof-theoretic grounds
BModal logic translates into a restricted fragment of FOL via the standard translation, and decidability of modal logic corresponds to decidability of that FOL fragment
CModal logic is decidable only for propositional (zero-order) formulas, not when quantifiers are present
DModal logic's Kripke semantics is inherently finite, making all satisfiability problems trivially decidable
Question 3 True / False

Monadic first-order logic — where all predicates take exactly one argument — is decidable.

TTrue
FFalse
Question 4 True / False

Since full first-order logic is undecidable, most fragments of first-order logic obtained by restricting the set of allowed formulas are also undecidable.

TTrue
FFalse
Question 5 Short Answer

What determines whether a fragment of first-order logic is decidable or undecidable, and why does the ability to encode Turing machine computations matter?

Think about your answer, then reveal below.