YES We show the termination of the TRS R: app(app(twice(),f),x) -> app(f,app(f,x)) app(app(map(),f),nil()) -> nil() app(app(map(),f),app(app(cons(),h),t)) -> app(app(cons(),app(f,h)),app(app(map(),f),t)) app(app(fmap(),nil()),x) -> nil() app(app(fmap(),app(app(cons(),f),t_f())),x) -> app(app(cons(),app(f,x)),app(app(fmap(),t_f()),x)) 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(fmap_0(),y) -> fmap_1(y) app(fmap_1(x1),y) -> fmap_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) app(twice_1(x1),y) -> twice_2(x1,y) twice_2(f,x) -> app(f,app(f,x)) map_2(f,nil_0()) -> nil_0() map_2(f,cons_2(h,t)) -> cons_2(app(f,h),map_2(f,t)) fmap_2(nil_0(),x) -> nil_0() fmap_2(cons_2(f,t_f_0()),x) -> cons_2(app(f,x),fmap_2(t_f_0(),x)) -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: app#(fmap_1(x1),y) -> fmap_2#(x1,y) p2: app#(map_1(x1),y) -> map_2#(x1,y) p3: app#(twice_1(x1),y) -> twice_2#(x1,y) p4: twice_2#(f,x) -> app#(f,app(f,x)) p5: twice_2#(f,x) -> app#(f,x) p6: map_2#(f,cons_2(h,t)) -> app#(f,h) p7: map_2#(f,cons_2(h,t)) -> map_2#(f,t) p8: fmap_2#(cons_2(f,t_f_0()),x) -> app#(f,x) p9: fmap_2#(cons_2(f,t_f_0()),x) -> fmap_2#(t_f_0(),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(fmap_0(),y) -> fmap_1(y) r4: app(fmap_1(x1),y) -> fmap_2(x1,y) r5: app(map_0(),y) -> map_1(y) r6: app(map_1(x1),y) -> map_2(x1,y) r7: app(twice_0(),y) -> twice_1(y) r8: app(twice_1(x1),y) -> twice_2(x1,y) r9: twice_2(f,x) -> app(f,app(f,x)) r10: map_2(f,nil_0()) -> nil_0() r11: map_2(f,cons_2(h,t)) -> cons_2(app(f,h),map_2(f,t)) r12: fmap_2(nil_0(),x) -> nil_0() r13: fmap_2(cons_2(f,t_f_0()),x) -> cons_2(app(f,x),fmap_2(t_f_0(),x)) The estimated dependency graph contains the following SCCs: {p1, p2, p3, p4, p5, p6, p7, p8} -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: app#(fmap_1(x1),y) -> fmap_2#(x1,y) p2: fmap_2#(cons_2(f,t_f_0()),x) -> app#(f,x) p3: app#(twice_1(x1),y) -> twice_2#(x1,y) p4: twice_2#(f,x) -> app#(f,x) p5: app#(map_1(x1),y) -> map_2#(x1,y) p6: map_2#(f,cons_2(h,t)) -> map_2#(f,t) p7: map_2#(f,cons_2(h,t)) -> app#(f,h) p8: twice_2#(f,x) -> app#(f,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(fmap_0(),y) -> fmap_1(y) r4: app(fmap_1(x1),y) -> fmap_2(x1,y) r5: app(map_0(),y) -> map_1(y) r6: app(map_1(x1),y) -> map_2(x1,y) r7: app(twice_0(),y) -> twice_1(y) r8: app(twice_1(x1),y) -> twice_2(x1,y) r9: twice_2(f,x) -> app(f,app(f,x)) r10: map_2(f,nil_0()) -> nil_0() r11: map_2(f,cons_2(h,t)) -> cons_2(app(f,h),map_2(f,t)) r12: fmap_2(nil_0(),x) -> nil_0() r13: fmap_2(cons_2(f,t_f_0()),x) -> cons_2(app(f,x),fmap_2(t_f_0(),x)) The set of usable rules consists of r12 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!): max/plus interpretation on natural numbers: app#_A(x1,x2) = 1 fmap_1_A(x1) = x1 + 8 fmap_2#_A(x1,x2) = 1 cons_2_A(x1,x2) = 0 t_f_0_A = 1 twice_1_A(x1) = x1 + 20 twice_2#_A(x1,x2) = 1 map_1_A(x1) = x1 + 20 map_2#_A(x1,x2) = 1 app_A(x1,x2) = 0 fmap_2_A(x1,x2) = max{10, x2 - 19} nil_0_A = 9 app#!_A(x1,x2) = 3 fmap_1!_A(x1) = x1 + 8 fmap_2#!_A(x1,x2) = 3 cons_2!_A(x1,x2) = x1 + 7 t_f_0!_A = 5 twice_1!_A(x1) = max{5, x1 - 1} twice_2#!_A(x1,x2) = max{4, x1 - 2} map_1!_A(x1) = x1 + 23 map_2#!_A(x1,x2) = x1 + 22 app!_A(x1,x2) = max{6, x1 + 4} fmap_2!_A(x1,x2) = x1 + 8 nil_0!_A = 18 2. marked-algebra reduction triple (>=A, >=A!, >A!): max/plus interpretation on natural numbers: app#_A(x1,x2) = 36 fmap_1_A(x1) = max{2, x1} fmap_2#_A(x1,x2) = 36 cons_2_A(x1,x2) = max{0, x1 - 12} t_f_0_A = 23 twice_1_A(x1) = max{13, x1 - 12} twice_2#_A(x1,x2) = 36 map_1_A(x1) = 2 map_2#_A(x1,x2) = 36 app_A(x1,x2) = max{13, x1 - 23} fmap_2_A(x1,x2) = 9 nil_0_A = 1 app#!_A(x1,x2) = 4 fmap_1!_A(x1) = x1 + 21 fmap_2#!_A(x1,x2) = max{x1 + 11, x2 - 1} cons_2!_A(x1,x2) = max{x1 + 6, x2 - 2} t_f_0!_A = 5 twice_1!_A(x1) = max{18, x1 - 1} twice_2#!_A(x1,x2) = max{19, x1 - 1, x2 - 1} map_1!_A(x1) = x1 + 8 map_2#!_A(x1,x2) = 7 app!_A(x1,x2) = 20 fmap_2!_A(x1,x2) = max{20, x1} nil_0!_A = 7 partial status: pi(app#) = [1] pi(fmap_1) = [1] pi(fmap_2#) = [1] pi(cons_2) = [1] pi(t_f_0) = [] pi(twice_1) = [1] pi(twice_2#) = [1] pi(map_1) = [1] pi(map_2#) = [1] pi(app) = [1] pi(fmap_2) = [1] pi(nil_0) = [] The following rules are removed: p1, p2, p3, p4, p5, p7, p8 -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: map_2#(f,cons_2(h,t)) -> map_2#(f,t) and R consists of: r1: app(cons_0(),y) -> cons_1(y) r2: app(cons_1(x1),y) -> cons_2(x1,y) r3: app(fmap_0(),y) -> fmap_1(y) r4: app(fmap_1(x1),y) -> fmap_2(x1,y) r5: app(map_0(),y) -> map_1(y) r6: app(map_1(x1),y) -> map_2(x1,y) r7: app(twice_0(),y) -> twice_1(y) r8: app(twice_1(x1),y) -> twice_2(x1,y) r9: twice_2(f,x) -> app(f,app(f,x)) r10: map_2(f,nil_0()) -> nil_0() r11: map_2(f,cons_2(h,t)) -> cons_2(app(f,h),map_2(f,t)) r12: fmap_2(nil_0(),x) -> nil_0() r13: fmap_2(cons_2(f,t_f_0()),x) -> cons_2(app(f,x),fmap_2(t_f_0(),x)) 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(h,t)) -> map_2#(f,t) and R consists of: r1: app(cons_0(),y) -> cons_1(y) r2: app(cons_1(x1),y) -> cons_2(x1,y) r3: app(fmap_0(),y) -> fmap_1(y) r4: app(fmap_1(x1),y) -> fmap_2(x1,y) r5: app(map_0(),y) -> map_1(y) r6: app(map_1(x1),y) -> map_2(x1,y) r7: app(twice_0(),y) -> twice_1(y) r8: app(twice_1(x1),y) -> twice_2(x1,y) r9: twice_2(f,x) -> app(f,app(f,x)) r10: map_2(f,nil_0()) -> nil_0() r11: map_2(f,cons_2(h,t)) -> cons_2(app(f,h),map_2(f,t)) r12: fmap_2(nil_0(),x) -> nil_0() r13: fmap_2(cons_2(f,t_f_0()),x) -> cons_2(app(f,x),fmap_2(t_f_0(),x)) 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!): max/plus interpretation on natural numbers: map_2#_A(x1,x2) = 0 cons_2_A(x1,x2) = max{16, x1 - 2, x2 - 3} map_2#!_A(x1,x2) = max{1, x2 - 17} cons_2!_A(x1,x2) = max{2, x1 - 8, x2 - 11} 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{13, x1 - 3, x2 - 3} map_2#!_A(x1,x2) = max{6, x1 - 1, x2 - 33} cons_2!_A(x1,x2) = max{5, x1 - 22, x2 - 32} partial status: pi(map_2#) = [2] pi(cons_2) = [2] The following rules are removed: p1 Then no dependency pair remains.