YES Proof: This system is quasi-decreasing. By \cite{O02}, p. 214, Proposition 7.2.50. This system is of type 3 or smaller. This system is deterministic. System R transformed to U(R). Call external tool: ttt2 - trs 30 Input: lte(0, 0) -> tt lte(s(x), 0) -> ff lte(s(x), s(y)) -> lte(x, y) ?1(tt, x, y) -> tt lte(x, s(y)) -> ?1(lte(x, y), x, y) DP Processor: DPs: lte#(s(x),s(y)) -> lte#(x,y) lte#(x,s(y)) -> lte#(x,y) lte#(x,s(y)) -> ?1#(lte(x,y),x,y) TRS: lte(0(),0()) -> tt() lte(s(x),0()) -> ff() lte(s(x),s(y)) -> lte(x,y) ?1(tt(),x,y) -> tt() lte(x,s(y)) -> ?1(lte(x,y),x,y) TDG Processor: DPs: lte#(s(x),s(y)) -> lte#(x,y) lte#(x,s(y)) -> lte#(x,y) lte#(x,s(y)) -> ?1#(lte(x,y),x,y) TRS: lte(0(),0()) -> tt() lte(s(x),0()) -> ff() lte(s(x),s(y)) -> lte(x,y) ?1(tt(),x,y) -> tt() lte(x,s(y)) -> ?1(lte(x,y),x,y) graph: lte#(s(x),s(y)) -> lte#(x,y) -> lte#(x,s(y)) -> ?1#(lte(x,y),x,y) lte#(s(x),s(y)) -> lte#(x,y) -> lte#(x,s(y)) -> lte#(x,y) lte#(s(x),s(y)) -> lte#(x,y) -> lte#(s(x),s(y)) -> lte#(x,y) lte#(x,s(y)) -> lte#(x,y) -> lte#(x,s(y)) -> ?1#(lte(x,y),x,y) lte#(x,s(y)) -> lte#(x,y) -> lte#(x,s(y)) -> lte#(x,y) lte#(x,s(y)) -> lte#(x,y) -> lte#(s(x),s(y)) -> lte#(x,y) SCC Processor: #sccs: 1 #rules: 2 #arcs: 6/9 DPs: lte#(s(x),s(y)) -> lte#(x,y) lte#(x,s(y)) -> lte#(x,y) TRS: lte(0(),0()) -> tt() lte(s(x),0()) -> ff() lte(s(x),s(y)) -> lte(x,y) ?1(tt(),x,y) -> tt() lte(x,s(y)) -> ?1(lte(x,y),x,y) Subterm Criterion Processor: simple projection: pi(lte#) = 0 problem: DPs: lte#(x,s(y)) -> lte#(x,y) TRS: lte(0(),0()) -> tt() lte(s(x),0()) -> ff() lte(s(x),s(y)) -> lte(x,y) ?1(tt(),x,y) -> tt() lte(x,s(y)) -> ?1(lte(x,y),x,y) Subterm Criterion Processor: simple projection: pi(lte#) = 1 problem: DPs: TRS: lte(0(),0()) -> tt() lte(s(x),0()) -> ff() lte(s(x),s(y)) -> lte(x,y) ?1(tt(),x,y) -> tt() lte(x,s(y)) -> ?1(lte(x,y),x,y) Qed