 This is certainly possible. The special deduction theorem holds trivially; if we define\(\{p\} \dproves q\) as \(\varnothing \vDash(p \Rightarrow q)\); that is, \(\{p\}\) defeasibly entails \(q\) if and only if (by definition) \((p \Rightarrow q)\) is a theorem of the classical conditional  logic.[9]