YES We show the termination of the TRS R: is_empty(nil()) -> true() is_empty(cons(x,l)) -> false() hd(cons(x,l)) -> x tl(cons(x,l)) -> l append(l1,l2) -> ifappend(l1,l2,l1) ifappend(l1,l2,nil()) -> l2 ifappend(l1,l2,cons(x,l)) -> cons(x,append(l,l2)) -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: append#(l1,l2) -> ifappend#(l1,l2,l1) p2: ifappend#(l1,l2,cons(x,l)) -> append#(l,l2) and R consists of: r1: is_empty(nil()) -> true() r2: is_empty(cons(x,l)) -> false() r3: hd(cons(x,l)) -> x r4: tl(cons(x,l)) -> l r5: append(l1,l2) -> ifappend(l1,l2,l1) r6: ifappend(l1,l2,nil()) -> l2 r7: ifappend(l1,l2,cons(x,l)) -> cons(x,append(l,l2)) The estimated dependency graph contains the following SCCs: {p1, p2} -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: append#(l1,l2) -> ifappend#(l1,l2,l1) p2: ifappend#(l1,l2,cons(x,l)) -> append#(l,l2) and R consists of: r1: is_empty(nil()) -> true() r2: is_empty(cons(x,l)) -> false() r3: hd(cons(x,l)) -> x r4: tl(cons(x,l)) -> l r5: append(l1,l2) -> ifappend(l1,l2,l1) r6: ifappend(l1,l2,nil()) -> l2 r7: ifappend(l1,l2,cons(x,l)) -> cons(x,append(l,l2)) The set of usable rules consists of r4 Take the reduction pair: generalized weighted path order (>=MG, >G) base order 1: trivial reduction triple (T x T, T x T, Φ) base order 2: lexicographic combination of reduction triples: 1. marked-algebra reduction triple (>=A, >=A!, >A!): matrix interpretation with standard point-wise order: carrier: N^1 order: standard order interpretations: append#_A(x1,x2) = x1 ifappend#_A(x1,x2,x3) = x3 cons_A(x1,x2) = x2 + 2 tl_A(x1) = x1 append#!_A(x1,x2) = x1 + 2 ifappend#!_A(x1,x2,x3) = x3 + 1 cons!_A(x1,x2) = x2 + 1 tl!_A(x1) = x1 2. marked-algebra reduction triple (>=A, >=A!, >A!): matrix interpretation with standard point-wise order: carrier: N^1 order: standard order interpretations: append#_A(x1,x2) = x1 ifappend#_A(x1,x2,x3) = x3 cons_A(x1,x2) = x1 + x2 + 1 tl_A(x1) = x1 append#!_A(x1,x2) = 0 ifappend#!_A(x1,x2,x3) = 1 cons!_A(x1,x2) = 3 tl!_A(x1) = x1 partial status: pi(append#) = [] pi(ifappend#) = [] pi(cons) = [2] pi(tl) = [1] The following rules are removed: p1, p2 Then no dependency pair remains.