YES We show the termination of the TRS R: app(app(map(),f),nil()) -> nil() app(app(map(),f),app(app(cons(),x),xs)) -> app(app(cons(),app(f,x)),app(app(map(),f),xs)) app(app(treemap(),f),app(app(node(),x),xs)) -> app(app(node(),app(f,x)),app(app(map(),app(treemap(),f)),xs)) Instead we prove termination of the uncurried system: app(cons_0(),y) -> cons_1(y) app(cons_1(x1),y) -> cons_2(x1,y) app(map_0(),y) -> map_1(y) app(map_1(x1),y) -> map_2(x1,y) app(node_0(),y) -> node_1(y) app(node_1(x1),y) -> node_2(x1,y) app(treemap_0(),y) -> treemap_1(y) app(treemap_1(x1),y) -> treemap_2(x1,y) map_2(f,nil_0()) -> nil_0() map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs)) treemap_2(f,node_2(x,xs)) -> node_2(app(f,x),map_2(treemap_1(f),xs)) -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: app#(map_1(x1),y) -> map_2#(x1,y) p2: app#(treemap_1(x1),y) -> treemap_2#(x1,y) p3: map_2#(f,cons_2(x,xs)) -> app#(f,x) p4: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs) p5: treemap_2#(f,node_2(x,xs)) -> app#(f,x) p6: treemap_2#(f,node_2(x,xs)) -> map_2#(treemap_1(f),xs) and R consists of: r1: app(cons_0(),y) -> cons_1(y) r2: app(cons_1(x1),y) -> cons_2(x1,y) r3: app(map_0(),y) -> map_1(y) r4: app(map_1(x1),y) -> map_2(x1,y) r5: app(node_0(),y) -> node_1(y) r6: app(node_1(x1),y) -> node_2(x1,y) r7: app(treemap_0(),y) -> treemap_1(y) r8: app(treemap_1(x1),y) -> treemap_2(x1,y) r9: map_2(f,nil_0()) -> nil_0() r10: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs)) r11: treemap_2(f,node_2(x,xs)) -> node_2(app(f,x),map_2(treemap_1(f),xs)) The estimated dependency graph contains the following SCCs: {p1, p2, p3, p4, p5, p6} -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: app#(map_1(x1),y) -> map_2#(x1,y) p2: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs) p3: map_2#(f,cons_2(x,xs)) -> app#(f,x) p4: app#(treemap_1(x1),y) -> treemap_2#(x1,y) p5: treemap_2#(f,node_2(x,xs)) -> map_2#(treemap_1(f),xs) p6: treemap_2#(f,node_2(x,xs)) -> app#(f,x) and R consists of: r1: app(cons_0(),y) -> cons_1(y) r2: app(cons_1(x1),y) -> cons_2(x1,y) r3: app(map_0(),y) -> map_1(y) r4: app(map_1(x1),y) -> map_2(x1,y) r5: app(node_0(),y) -> node_1(y) r6: app(node_1(x1),y) -> node_2(x1,y) r7: app(treemap_0(),y) -> treemap_1(y) r8: app(treemap_1(x1),y) -> treemap_2(x1,y) r9: map_2(f,nil_0()) -> nil_0() r10: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs)) r11: treemap_2(f,node_2(x,xs)) -> node_2(app(f,x),map_2(treemap_1(f),xs)) The set of usable rules consists of (no rules) Take the reduction pair: generalized weighted path order (>=MG, >G) base order 1: max/plus interpretation on natural numbers: app#_A(x1,x2) = max{3, x1 - 12} map_1_A(x1) = max{2, x1 + 1} map_2#_A(x1,x2) = max{3, x1 - 12} cons_2_A(x1,x2) = max{0, x1 - 17, x2 - 17} treemap_1_A(x1) = x1 + 16 treemap_2#_A(x1,x2) = x1 + 4 node_2_A(x1,x2) = max{8, x1 - 14, x2 + 7} base order 2: marked-algebra reduction triple (>=A, >=A!, >A!): max/plus interpretation on natural numbers: app#_A(x1,x2) = max{0, x1 - 19, x2 - 20} map_1_A(x1) = x1 + 18 map_2#_A(x1,x2) = max{0, x1 - 19, x2 - 23} cons_2_A(x1,x2) = max{x1 + 22, x2 + 14} treemap_1_A(x1) = max{9, x1} treemap_2#_A(x1,x2) = max{0, x1 - 19, x2 - 20} node_2_A(x1,x2) = max{9, x1 + 2, x2 - 3} app#!_A(x1,x2) = max{x1 - 37, x2 + 8} map_1!_A(x1) = max{2, x1 - 12} map_2#!_A(x1,x2) = max{7, x1 - 37, x2 - 14} cons_2!_A(x1,x2) = max{0, x1 - 14} treemap_1!_A(x1) = max{7, x1 - 1} treemap_2#!_A(x1,x2) = max{8, x1 - 37, x2 - 10} node_2!_A(x1,x2) = max{5, x1 - 4, x2 - 27} partial status: pi(app#) = [] pi(map_1) = [1] pi(map_2#) = [] pi(cons_2) = [] pi(treemap_1) = [1] pi(treemap_2#) = [] pi(node_2) = [2] The following rules are removed: p1, p6 -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs) p2: map_2#(f,cons_2(x,xs)) -> app#(f,x) p3: app#(treemap_1(x1),y) -> treemap_2#(x1,y) p4: treemap_2#(f,node_2(x,xs)) -> map_2#(treemap_1(f),xs) and R consists of: r1: app(cons_0(),y) -> cons_1(y) r2: app(cons_1(x1),y) -> cons_2(x1,y) r3: app(map_0(),y) -> map_1(y) r4: app(map_1(x1),y) -> map_2(x1,y) r5: app(node_0(),y) -> node_1(y) r6: app(node_1(x1),y) -> node_2(x1,y) r7: app(treemap_0(),y) -> treemap_1(y) r8: app(treemap_1(x1),y) -> treemap_2(x1,y) r9: map_2(f,nil_0()) -> nil_0() r10: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs)) r11: treemap_2(f,node_2(x,xs)) -> node_2(app(f,x),map_2(treemap_1(f),xs)) The estimated dependency graph contains the following SCCs: {p1, p2, p3, p4} -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs) p2: map_2#(f,cons_2(x,xs)) -> app#(f,x) p3: app#(treemap_1(x1),y) -> treemap_2#(x1,y) p4: treemap_2#(f,node_2(x,xs)) -> map_2#(treemap_1(f),xs) and R consists of: r1: app(cons_0(),y) -> cons_1(y) r2: app(cons_1(x1),y) -> cons_2(x1,y) r3: app(map_0(),y) -> map_1(y) r4: app(map_1(x1),y) -> map_2(x1,y) r5: app(node_0(),y) -> node_1(y) r6: app(node_1(x1),y) -> node_2(x1,y) r7: app(treemap_0(),y) -> treemap_1(y) r8: app(treemap_1(x1),y) -> treemap_2(x1,y) r9: map_2(f,nil_0()) -> nil_0() r10: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs)) r11: treemap_2(f,node_2(x,xs)) -> node_2(app(f,x),map_2(treemap_1(f),xs)) The set of usable rules consists of (no rules) Take the reduction pair: generalized weighted path order (>=MG, >G) base order 1: max/plus interpretation on natural numbers: map_2#_A(x1,x2) = max{15, x1 + 1} cons_2_A(x1,x2) = max{13, x1 - 28, x2 - 43} app#_A(x1,x2) = max{14, x1 + 1} treemap_1_A(x1) = max{14, x1 - 68} treemap_2#_A(x1,x2) = max{15, x1 - 67} node_2_A(x1,x2) = max{2, x1 - 69, x2 - 17} base order 2: marked-algebra reduction triple (>=A, >=A!, >A!): max/plus interpretation on natural numbers: map_2#_A(x1,x2) = max{0, x1 - 3, x2 - 41} cons_2_A(x1,x2) = max{x1 + 27, x2 + 40} app#_A(x1,x2) = max{0, x1 - 3, x2 - 14} treemap_1_A(x1) = max{2, x1 - 9} treemap_2#_A(x1,x2) = max{0, x1 - 12, x2 - 14} node_2_A(x1,x2) = max{5, x1 - 7, x2 - 3} map_2#!_A(x1,x2) = max{12, x2 - 17} cons_2!_A(x1,x2) = max{13, x1 - 3, x2 - 42} app#!_A(x1,x2) = max{11, x2 + 9} treemap_1!_A(x1) = 14 treemap_2#!_A(x1,x2) = max{10, x2 + 8} node_2!_A(x1,x2) = max{14, x1 - 19, x2 - 16} partial status: pi(map_2#) = [] pi(cons_2) = [] pi(app#) = [1] pi(treemap_1) = [] pi(treemap_2#) = [] pi(node_2) = [] The following rules are removed: p1, p2, p3, p4 Then no dependency pair remains.