 Solution

A fragment has the finite model property if $\boxed{\text{every satisfiable sentence in the fragment has a finite model}}$. This does not mean that every model is finite, nor that there is one uniform bound on model size for the entire fragment.

ALC has the finite model property, including satisfiability with respect to finite TBoxes. A tableau algorithm for ALC uses finitely many possible labels and direct blocking in description logic to obtain a finite representation of a model. The resulting finite-model claim is a property of the logic; termination alone, for an arbitrary procedure, would not establish it.

Full first-order logic does not have the finite model property. For example, the first-order sentencerequires an irreflexive transitive relation with a successor for every element. It is satisfied by the usual strict order on the natural numbers. In a nonempty finite domain, repeatedly taking successors eventually repeats an element; transitivity would then force $R(x,x)$, contradicting irreflexivity. Hence this satisfiable first-order sentence has no finite model.