YES We show the termination of the TRS R: app(app(*(),x),app(app(+(),y),z)) -> app(app(+(),app(app(*(),x),y)),app(app(*(),x),z)) 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(filter(),f),nil()) -> nil() app(app(filter(),f),app(app(cons(),x),xs)) -> app(app(app(app(filter2(),app(f,x)),f),x),xs) app(app(app(app(filter2(),true()),f),x),xs) -> app(app(cons(),x),app(app(filter(),f),xs)) app(app(app(app(filter2(),false()),f),x),xs) -> app(app(filter(),f),xs) Instead we prove termination of the uncurried system: app(*_0(),y) -> *_1(y) app(*_1(x1),y) -> *_2(x1,y) app(+_0(),y) -> +_1(y) app(+_1(x1),y) -> +_2(x1,y) app(cons_0(),y) -> cons_1(y) app(cons_1(x1),y) -> cons_2(x1,y) app(filter_0(),y) -> filter_1(y) app(filter_1(x1),y) -> filter_2(x1,y) app(filter2_0(),y) -> filter2_1(y) app(filter2_1(x1),y) -> filter2_2(x1,y) app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y) app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y) app(map_0(),y) -> map_1(y) app(map_1(x1),y) -> map_2(x1,y) *_2(x,+_2(y,z)) -> +_2(*_2(x,y),*_2(x,z)) map_2(f,nil_0()) -> nil_0() map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs)) filter_2(f,nil_0()) -> nil_0() filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs) filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs)) filter2_4(false_0(),f,x,xs) -> filter_2(f,xs) -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: app#(*_1(x1),y) -> *_2#(x1,y) p2: app#(filter_1(x1),y) -> filter_2#(x1,y) p3: app#(filter2_3(x1,x2,x3),y) -> filter2_4#(x1,x2,x3,y) p4: app#(map_1(x1),y) -> map_2#(x1,y) p5: *_2#(x,+_2(y,z)) -> *_2#(x,y) p6: *_2#(x,+_2(y,z)) -> *_2#(x,z) p7: map_2#(f,cons_2(x,xs)) -> app#(f,x) p8: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs) p9: filter_2#(f,cons_2(x,xs)) -> filter2_4#(app(f,x),f,x,xs) p10: filter_2#(f,cons_2(x,xs)) -> app#(f,x) p11: filter2_4#(true_0(),f,x,xs) -> filter_2#(f,xs) p12: filter2_4#(false_0(),f,x,xs) -> filter_2#(f,xs) and R consists of: r1: app(*_0(),y) -> *_1(y) r2: app(*_1(x1),y) -> *_2(x1,y) r3: app(+_0(),y) -> +_1(y) r4: app(+_1(x1),y) -> +_2(x1,y) r5: app(cons_0(),y) -> cons_1(y) r6: app(cons_1(x1),y) -> cons_2(x1,y) r7: app(filter_0(),y) -> filter_1(y) r8: app(filter_1(x1),y) -> filter_2(x1,y) r9: app(filter2_0(),y) -> filter2_1(y) r10: app(filter2_1(x1),y) -> filter2_2(x1,y) r11: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y) r12: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y) r13: app(map_0(),y) -> map_1(y) r14: app(map_1(x1),y) -> map_2(x1,y) r15: *_2(x,+_2(y,z)) -> +_2(*_2(x,y),*_2(x,z)) r16: map_2(f,nil_0()) -> nil_0() r17: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs)) r18: filter_2(f,nil_0()) -> nil_0() r19: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs) r20: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs)) r21: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs) The estimated dependency graph contains the following SCCs: {p2, p3, p4, p7, p8, p9, p10, p11, p12} {p5, p6} -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: map_2#(f,cons_2(x,xs)) -> app#(f,x) p2: app#(map_1(x1),y) -> map_2#(x1,y) p3: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs) p4: app#(filter2_3(x1,x2,x3),y) -> filter2_4#(x1,x2,x3,y) p5: filter2_4#(false_0(),f,x,xs) -> filter_2#(f,xs) p6: filter_2#(f,cons_2(x,xs)) -> app#(f,x) p7: app#(filter_1(x1),y) -> filter_2#(x1,y) p8: filter_2#(f,cons_2(x,xs)) -> filter2_4#(app(f,x),f,x,xs) p9: filter2_4#(true_0(),f,x,xs) -> filter_2#(f,xs) and R consists of: r1: app(*_0(),y) -> *_1(y) r2: app(*_1(x1),y) -> *_2(x1,y) r3: app(+_0(),y) -> +_1(y) r4: app(+_1(x1),y) -> +_2(x1,y) r5: app(cons_0(),y) -> cons_1(y) r6: app(cons_1(x1),y) -> cons_2(x1,y) r7: app(filter_0(),y) -> filter_1(y) r8: app(filter_1(x1),y) -> filter_2(x1,y) r9: app(filter2_0(),y) -> filter2_1(y) r10: app(filter2_1(x1),y) -> filter2_2(x1,y) r11: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y) r12: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y) r13: app(map_0(),y) -> map_1(y) r14: app(map_1(x1),y) -> map_2(x1,y) r15: *_2(x,+_2(y,z)) -> +_2(*_2(x,y),*_2(x,z)) r16: map_2(f,nil_0()) -> nil_0() r17: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs)) r18: filter_2(f,nil_0()) -> nil_0() r19: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs) r20: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs)) r21: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs) 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) = max{2, x1 - 16, x2 - 38} cons_2_A(x1,x2) = max{x1 + 37, x2 + 7} app#_A(x1,x2) = max{2, x1 - 26, x2 - 1} map_1_A(x1) = x1 + 10 filter2_3_A(x1,x2,x3) = max{x1 + 23, x2 + 1, x3 + 25} filter2_4#_A(x1,x2,x3,x4) = max{2, x2 - 25, x3 - 1, x4 - 29} false_0_A = 15 filter_2#_A(x1,x2) = max{2, x1 - 25, x2 - 36} filter_1_A(x1) = x1 + 8 app_A(x1,x2) = max{x1 - 24, x2 + 7} true_0_A = 15 map_2#!_A(x1,x2) = max{x1 + 17, x2 - 19} cons_2!_A(x1,x2) = max{0, x2 - 80} app#!_A(x1,x2) = max{x1 + 7, x2 + 17} map_1!_A(x1) = max{2, x1 - 32} filter2_3!_A(x1,x2,x3) = max{5, x2 - 1} filter2_4#!_A(x1,x2,x3,x4) = max{31, x2 + 7, x4 - 4} false_0!_A = 1 filter_2#!_A(x1,x2) = max{x1 + 7, x2 - 4} filter_1!_A(x1) = max{5, x1 - 1} app!_A(x1,x2) = max{4, x2 - 6} true_0!_A = 1 2. 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) = 0 cons_2_A(x1,x2) = x2 + 6 app#_A(x1,x2) = 0 map_1_A(x1) = x1 + 3 filter2_3_A(x1,x2,x3) = x2 + x3 + 7 filter2_4#_A(x1,x2,x3,x4) = 0 false_0_A() = 1 filter_2#_A(x1,x2) = 0 filter_1_A(x1) = 3 app_A(x1,x2) = x1 + x2 + 1 true_0_A() = 1 map_2#!_A(x1,x2) = x1 + 6 cons_2!_A(x1,x2) = 4 app#!_A(x1,x2) = x1 + 5 map_1!_A(x1) = 2 filter2_3!_A(x1,x2,x3) = x1 + x2 + 13 filter2_4#!_A(x1,x2,x3,x4) = x2 + x4 + 3 false_0!_A() = 0 filter_2#!_A(x1,x2) = x1 + x2 + 2 filter_1!_A(x1) = 1 app!_A(x1,x2) = x1 + 7 true_0!_A() = 3 partial status: pi(map_2#) = [] pi(cons_2) = [] pi(app#) = [] pi(map_1) = [] pi(filter2_3) = [] pi(filter2_4#) = [] pi(false_0) = [] pi(filter_2#) = [] pi(filter_1) = [] pi(app) = [] pi(true_0) = [] The following rules are removed: p1, p2, p4, p5, p6, p7, p8, p9 -- 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(*_0(),y) -> *_1(y) r2: app(*_1(x1),y) -> *_2(x1,y) r3: app(+_0(),y) -> +_1(y) r4: app(+_1(x1),y) -> +_2(x1,y) r5: app(cons_0(),y) -> cons_1(y) r6: app(cons_1(x1),y) -> cons_2(x1,y) r7: app(filter_0(),y) -> filter_1(y) r8: app(filter_1(x1),y) -> filter_2(x1,y) r9: app(filter2_0(),y) -> filter2_1(y) r10: app(filter2_1(x1),y) -> filter2_2(x1,y) r11: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y) r12: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y) r13: app(map_0(),y) -> map_1(y) r14: app(map_1(x1),y) -> map_2(x1,y) r15: *_2(x,+_2(y,z)) -> +_2(*_2(x,y),*_2(x,z)) r16: map_2(f,nil_0()) -> nil_0() r17: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs)) r18: filter_2(f,nil_0()) -> nil_0() r19: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs) r20: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs)) r21: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs) 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(*_0(),y) -> *_1(y) r2: app(*_1(x1),y) -> *_2(x1,y) r3: app(+_0(),y) -> +_1(y) r4: app(+_1(x1),y) -> +_2(x1,y) r5: app(cons_0(),y) -> cons_1(y) r6: app(cons_1(x1),y) -> cons_2(x1,y) r7: app(filter_0(),y) -> filter_1(y) r8: app(filter_1(x1),y) -> filter_2(x1,y) r9: app(filter2_0(),y) -> filter2_1(y) r10: app(filter2_1(x1),y) -> filter2_2(x1,y) r11: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y) r12: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y) r13: app(map_0(),y) -> map_1(y) r14: app(map_1(x1),y) -> map_2(x1,y) r15: *_2(x,+_2(y,z)) -> +_2(*_2(x,y),*_2(x,z)) r16: map_2(f,nil_0()) -> nil_0() r17: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs)) r18: filter_2(f,nil_0()) -> nil_0() r19: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs) r20: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs)) r21: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs) 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{3, x1 - 4, x2 - 6} map_2#!_A(x1,x2) = 0 cons_2!_A(x1,x2) = max{x1 + 1, x2 + 7} 2. 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 + 5 map_2#!_A(x1,x2) = x2 cons_2!_A(x1,x2) = x1 + x2 + 5 partial status: pi(map_2#) = [] pi(cons_2) = [2] The following rules are removed: p1 Then no dependency pair remains. -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: *_2#(x,+_2(y,z)) -> *_2#(x,y) p2: *_2#(x,+_2(y,z)) -> *_2#(x,z) and R consists of: r1: app(*_0(),y) -> *_1(y) r2: app(*_1(x1),y) -> *_2(x1,y) r3: app(+_0(),y) -> +_1(y) r4: app(+_1(x1),y) -> +_2(x1,y) r5: app(cons_0(),y) -> cons_1(y) r6: app(cons_1(x1),y) -> cons_2(x1,y) r7: app(filter_0(),y) -> filter_1(y) r8: app(filter_1(x1),y) -> filter_2(x1,y) r9: app(filter2_0(),y) -> filter2_1(y) r10: app(filter2_1(x1),y) -> filter2_2(x1,y) r11: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y) r12: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y) r13: app(map_0(),y) -> map_1(y) r14: app(map_1(x1),y) -> map_2(x1,y) r15: *_2(x,+_2(y,z)) -> +_2(*_2(x,y),*_2(x,z)) r16: map_2(f,nil_0()) -> nil_0() r17: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs)) r18: filter_2(f,nil_0()) -> nil_0() r19: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs) r20: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs)) r21: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs) 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: *_2#_A(x1,x2) = 0 +_2_A(x1,x2) = max{3, x1 - 3, x2 - 3} *_2#!_A(x1,x2) = 2 +_2!_A(x1,x2) = max{11, x1 + 1, x2 - 3} 2. marked-algebra reduction triple (>=A, >=A!, >A!): matrix interpretation with standard point-wise order: carrier: N^1 order: standard order interpretations: *_2#_A(x1,x2) = x2 +_2_A(x1,x2) = x1 + x2 + 7 *_2#!_A(x1,x2) = x2 + 32 +_2!_A(x1,x2) = x2 + 33 partial status: pi(*_2#) = [1] pi(+_2) = [] The following rules are removed: p1, p2 Then no dependency pair remains.