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(app(comp(),f),g),x) -> app(f,app(g,x)) app(twice(),f) -> app(app(comp(),f),f) Instead we prove termination of the uncurried system: app(comp_0(),y) -> comp_1(y) app(comp_1(x1),y) -> comp_2(x1,y) app(comp_2(x1,x2),y) -> comp_3(x1,x2,y) 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(twice_0(),y) -> twice_1(y) map_2(f,nil_0()) -> nil_0() map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs)) comp_3(f,g,x) -> app(f,app(g,x)) twice_1(f) -> comp_2(f,f) -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: app#(comp_2(x1,x2),y) -> comp_3#(x1,x2,y) p2: app#(map_1(x1),y) -> map_2#(x1,y) p3: app#(twice_0(),y) -> twice_1#(y) p4: map_2#(f,cons_2(x,xs)) -> app#(f,x) p5: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs) p6: comp_3#(f,g,x) -> app#(f,app(g,x)) p7: comp_3#(f,g,x) -> app#(g,x) and R consists of: r1: app(comp_0(),y) -> comp_1(y) r2: app(comp_1(x1),y) -> comp_2(x1,y) r3: app(comp_2(x1,x2),y) -> comp_3(x1,x2,y) r4: app(cons_0(),y) -> cons_1(y) r5: app(cons_1(x1),y) -> cons_2(x1,y) r6: app(map_0(),y) -> map_1(y) r7: app(map_1(x1),y) -> map_2(x1,y) r8: app(twice_0(),y) -> twice_1(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: comp_3(f,g,x) -> app(f,app(g,x)) r12: twice_1(f) -> comp_2(f,f) The estimated dependency graph contains the following SCCs: {p1, p2, p4, p5, p6, p7} -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: app#(comp_2(x1,x2),y) -> comp_3#(x1,x2,y) p2: comp_3#(f,g,x) -> app#(g,x) p3: app#(map_1(x1),y) -> map_2#(x1,y) p4: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs) p5: map_2#(f,cons_2(x,xs)) -> app#(f,x) p6: comp_3#(f,g,x) -> app#(f,app(g,x)) and R consists of: r1: app(comp_0(),y) -> comp_1(y) r2: app(comp_1(x1),y) -> comp_2(x1,y) r3: app(comp_2(x1,x2),y) -> comp_3(x1,x2,y) r4: app(cons_0(),y) -> cons_1(y) r5: app(cons_1(x1),y) -> cons_2(x1,y) r6: app(map_0(),y) -> map_1(y) r7: app(map_1(x1),y) -> map_2(x1,y) r8: app(twice_0(),y) -> twice_1(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: comp_3(f,g,x) -> app(f,app(g,x)) r12: twice_1(f) -> comp_2(f,f) The set of usable rules consists of r2, r5 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: app#_A(x1,x2) = x1 comp_2_A(x1,x2) = x1 + x2 + 4 comp_3#_A(x1,x2,x3) = x1 + x2 map_1_A(x1) = x1 + 3 map_2#_A(x1,x2) = x1 cons_2_A(x1,x2) = 6 app_A(x1,x2) = x1 + x2 + 1 comp_1_A(x1) = x1 + 3 cons_1_A(x1) = x1 + 5 app#!_A(x1,x2) = x1 + 8 comp_2!_A(x1,x2) = 2 comp_3#!_A(x1,x2,x3) = x1 + x2 + 11 map_1!_A(x1) = x1 + 9 map_2#!_A(x1,x2) = x1 + 9 cons_2!_A(x1,x2) = 7 app!_A(x1,x2) = 10 comp_1!_A(x1) = 3 cons_1!_A(x1) = 6 2. marked-algebra reduction triple (>=A, >=A!, >A!): max/plus interpretation on natural numbers: app#_A(x1,x2) = 1 comp_2_A(x1,x2) = 7 comp_3#_A(x1,x2,x3) = 1 map_1_A(x1) = max{10, x1 - 2} map_2#_A(x1,x2) = 1 cons_2_A(x1,x2) = 3 app_A(x1,x2) = 7 comp_1_A(x1) = max{8, x1 - 5} cons_1_A(x1) = max{8, x1 - 2} app#!_A(x1,x2) = max{9, x1 + 4} comp_2!_A(x1,x2) = max{4, x1 - 15, x2 - 1} comp_3#!_A(x1,x2,x3) = 6 map_1!_A(x1) = 10 map_2#!_A(x1,x2) = 10 cons_2!_A(x1,x2) = max{8, x1 + 6, x2 - 6} app!_A(x1,x2) = max{x1 - 9, x2 + 6} comp_1!_A(x1) = max{3, x1 - 16} cons_1!_A(x1) = 7 partial status: pi(app#) = [] pi(comp_2) = [1] pi(comp_3#) = [] pi(map_1) = [] pi(map_2#) = [] pi(cons_2) = [2] pi(app) = [1, 2] pi(comp_1) = [1] pi(cons_1) = [1] The following rules are removed: p1, p2, p3, p5, 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) and R consists of: r1: app(comp_0(),y) -> comp_1(y) r2: app(comp_1(x1),y) -> comp_2(x1,y) r3: app(comp_2(x1,x2),y) -> comp_3(x1,x2,y) r4: app(cons_0(),y) -> cons_1(y) r5: app(cons_1(x1),y) -> cons_2(x1,y) r6: app(map_0(),y) -> map_1(y) r7: app(map_1(x1),y) -> map_2(x1,y) r8: app(twice_0(),y) -> twice_1(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: comp_3(f,g,x) -> app(f,app(g,x)) r12: twice_1(f) -> comp_2(f,f) The estimated dependency graph contains the following SCCs: {p1} -- 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) and R consists of: r1: app(comp_0(),y) -> comp_1(y) r2: app(comp_1(x1),y) -> comp_2(x1,y) r3: app(comp_2(x1,x2),y) -> comp_3(x1,x2,y) r4: app(cons_0(),y) -> cons_1(y) r5: app(cons_1(x1),y) -> cons_2(x1,y) r6: app(map_0(),y) -> map_1(y) r7: app(map_1(x1),y) -> map_2(x1,y) r8: app(twice_0(),y) -> twice_1(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: comp_3(f,g,x) -> app(f,app(g,x)) r12: twice_1(f) -> comp_2(f,f) The set of usable rules consists of (no rules) 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: map_2#_A(x1,x2) = x2 cons_2_A(x1,x2) = x2 + 2 map_2#!_A(x1,x2) = x2 + 4 cons_2!_A(x1,x2) = 3 2. marked-algebra reduction triple (>=A, >=A!, >A!): max/plus interpretation on natural numbers: map_2#_A(x1,x2) = 0 cons_2_A(x1,x2) = max{4, x1 - 4, x2 - 4} map_2#!_A(x1,x2) = x2 + 8 cons_2!_A(x1,x2) = x1 + 2 partial status: pi(map_2#) = [] pi(cons_2) = [] The following rules are removed: p1 Then no dependency pair remains.