 We have so far shown that with the given ingredients our Liar sentence \(L\) implies a contradiction (thus formalizing the reasoning in ULiar). From here, it is one short step to all-out absurdity—if the lone contradiction weren’t already absurd enough. We invoke EFQ to finish the proof. (Well, we also assume that \(A \wedge B\) implies \(A\) and \(B\), i.e., that simplification is valid in \(\mathcal{L}\); but in fact this assumption is not really necessary.)