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 r1, r10, 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!): matrix interpretation with standard point-wise order: carrier: N^1 order: standard order interpretations: app#_A(x1,x2) = x1 fmap_1_A(x1) = x1 + 2 fmap_2#_A(x1,x2) = x1 cons_2_A(x1,x2) = x1 t_f_0_A() = 1 twice_1_A(x1) = x1 + 2 twice_2#_A(x1,x2) = x1 map_1_A(x1) = x1 map_2#_A(x1,x2) = x1 app_A(x1,x2) = 2 cons_0_A() = 1 cons_1_A(x1) = 0 map_2_A(x1,x2) = 0 nil_0_A() = 0 fmap_2_A(x1,x2) = 0 app#!_A(x1,x2) = x1 + 3 fmap_1!_A(x1) = x1 + 5 fmap_2#!_A(x1,x2) = x1 + 4 cons_2!_A(x1,x2) = 2 t_f_0!_A() = 0 twice_1!_A(x1) = 1 twice_2#!_A(x1,x2) = x1 + 4 map_1!_A(x1) = x1 + 5 map_2#!_A(x1,x2) = x1 + 3 app!_A(x1,x2) = 3 cons_0!_A() = 0 cons_1!_A(x1) = 1 map_2!_A(x1,x2) = x2 + 4 nil_0!_A() = 0 fmap_2!_A(x1,x2) = 3 2. marked-algebra reduction triple (>=A, >=A!, >A!): max/plus interpretation on natural numbers: app#_A(x1,x2) = 1 fmap_1_A(x1) = max{1, x1} fmap_2#_A(x1,x2) = 1 cons_2_A(x1,x2) = 3 t_f_0_A = 3 twice_1_A(x1) = max{1, x1} twice_2#_A(x1,x2) = 1 map_1_A(x1) = max{1, x1} map_2#_A(x1,x2) = 1 app_A(x1,x2) = max{6, x2} cons_0_A = 1 cons_1_A(x1) = max{1, x1} map_2_A(x1,x2) = 5 nil_0_A = 3 fmap_2_A(x1,x2) = 6 app#!_A(x1,x2) = x1 + 9 fmap_1!_A(x1) = max{2, x1 - 1} fmap_2#!_A(x1,x2) = max{11, x2 - 1} cons_2!_A(x1,x2) = max{7, x2 + 3} t_f_0!_A = 7 twice_1!_A(x1) = max{0, x1 - 1} twice_2#!_A(x1,x2) = 0 map_1!_A(x1) = x1 + 11 map_2#!_A(x1,x2) = max{10, x1 + 9} app!_A(x1,x2) = x2 + 8 cons_0!_A = 1 cons_1!_A(x1) = max{0, x1 - 1} map_2!_A(x1,x2) = max{x1 + 12, x2} nil_0!_A = 1 fmap_2!_A(x1,x2) = x2 + 8 partial status: pi(app#) = [] pi(fmap_1) = [1] pi(fmap_2#) = [] pi(cons_2) = [1] pi(t_f_0) = [] pi(twice_1) = [1] pi(twice_2#) = [] pi(map_1) = [1] pi(map_2#) = [] pi(app) = [1, 2] pi(cons_0) = [] pi(cons_1) = [1] pi(map_2) = [1, 2] pi(nil_0) = [] pi(fmap_2) = [1, 2] The following rules are removed: p1, p2, p3, p4, p8 -- SCC decomposition. 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(h,t)) -> map_2#(f,t) p3: map_2#(f,cons_2(h,t)) -> app#(f,h) 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} -- 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(h,t)) -> app#(f,h) p3: 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!): matrix interpretation with standard point-wise order: carrier: N^1 order: standard order interpretations: app#_A(x1,x2) = x2 map_1_A(x1) = x1 + 5 map_2#_A(x1,x2) = x2 cons_2_A(x1,x2) = x1 + x2 + 9 app#!_A(x1,x2) = x1 map_1!_A(x1) = x1 + 2 map_2#!_A(x1,x2) = x1 + 1 cons_2!_A(x1,x2) = 4 2. marked-algebra reduction triple (>=A, >=A!, >A!): max/plus interpretation on natural numbers: app#_A(x1,x2) = max{0, x1 - 1, x2 - 4} map_1_A(x1) = x1 map_2#_A(x1,x2) = max{0, x1 - 1, x2 - 11} cons_2_A(x1,x2) = max{10, x1 + 7, x2 + 4} app#!_A(x1,x2) = x2 + 16 map_1!_A(x1) = max{16, x1 - 1} map_2#!_A(x1,x2) = 15 cons_2!_A(x1,x2) = 14 partial status: pi(app#) = [1] pi(map_1) = [1] pi(map_2#) = [1] pi(cons_2) = [1] The following rules are removed: p1, p2 -- 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!): 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) = x1 + x2 + 9 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{7, x1 - 6, x2 - 11} map_2#!_A(x1,x2) = max{19, x2 + 12} cons_2!_A(x1,x2) = max{20, x2 - 6} partial status: pi(map_2#) = [2] pi(cons_2) = [2] The following rules are removed: p1 Then no dependency pair remains.