YES We show the termination of the TRS R: app(pred(),app(s(),x)) -> x app(app(minus(),x),|0|()) -> x app(app(minus(),x),app(s(),y)) -> app(pred(),app(app(minus(),x),y)) app(app(quot(),|0|()),app(s(),y)) -> |0|() app(app(quot(),app(s(),x)),app(s(),y)) -> app(s(),app(app(quot(),app(app(minus(),x),y)),app(s(),y))) 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(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) app(minus_0(),y) -> minus_1(y) app(minus_1(x1),y) -> minus_2(x1,y) app(pred_0(),y) -> pred_1(y) app(quot_0(),y) -> quot_1(y) app(quot_1(x1),y) -> quot_2(x1,y) app(s_0(),y) -> s_1(y) pred_1(s_1(x)) -> x minus_2(x,|0|_0()) -> x minus_2(x,s_1(y)) -> pred_1(minus_2(x,y)) quot_2(|0|_0(),s_1(y)) -> |0|_0() quot_2(s_1(x),s_1(y)) -> s_1(quot_2(minus_2(x,y),s_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)) 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#(filter_1(x1),y) -> filter_2#(x1,y) p2: app#(filter2_3(x1,x2,x3),y) -> filter2_4#(x1,x2,x3,y) p3: app#(map_1(x1),y) -> map_2#(x1,y) p4: app#(minus_1(x1),y) -> minus_2#(x1,y) p5: app#(pred_0(),y) -> pred_1#(y) p6: app#(quot_1(x1),y) -> quot_2#(x1,y) p7: minus_2#(x,s_1(y)) -> pred_1#(minus_2(x,y)) p8: minus_2#(x,s_1(y)) -> minus_2#(x,y) p9: quot_2#(s_1(x),s_1(y)) -> quot_2#(minus_2(x,y),s_1(y)) p10: quot_2#(s_1(x),s_1(y)) -> minus_2#(x,y) p11: map_2#(f,cons_2(x,xs)) -> app#(f,x) p12: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs) p13: filter_2#(f,cons_2(x,xs)) -> filter2_4#(app(f,x),f,x,xs) p14: filter_2#(f,cons_2(x,xs)) -> app#(f,x) p15: filter2_4#(true_0(),f,x,xs) -> filter_2#(f,xs) p16: filter2_4#(false_0(),f,x,xs) -> filter_2#(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(filter_0(),y) -> filter_1(y) r4: app(filter_1(x1),y) -> filter_2(x1,y) r5: app(filter2_0(),y) -> filter2_1(y) r6: app(filter2_1(x1),y) -> filter2_2(x1,y) r7: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y) r8: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y) r9: app(map_0(),y) -> map_1(y) r10: app(map_1(x1),y) -> map_2(x1,y) r11: app(minus_0(),y) -> minus_1(y) r12: app(minus_1(x1),y) -> minus_2(x1,y) r13: app(pred_0(),y) -> pred_1(y) r14: app(quot_0(),y) -> quot_1(y) r15: app(quot_1(x1),y) -> quot_2(x1,y) r16: app(s_0(),y) -> s_1(y) r17: pred_1(s_1(x)) -> x r18: minus_2(x,|0|_0()) -> x r19: minus_2(x,s_1(y)) -> pred_1(minus_2(x,y)) r20: quot_2(|0|_0(),s_1(y)) -> |0|_0() r21: quot_2(s_1(x),s_1(y)) -> s_1(quot_2(minus_2(x,y),s_1(y))) r22: map_2(f,nil_0()) -> nil_0() r23: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs)) r24: filter_2(f,nil_0()) -> nil_0() r25: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs) r26: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs)) r27: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs) The estimated dependency graph contains the following SCCs: {p1, p2, p3, p11, p12, p13, p14, p15, p16} {p9} {p8} -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: app#(filter_1(x1),y) -> filter_2#(x1,y) p2: filter_2#(f,cons_2(x,xs)) -> app#(f,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: app#(filter2_3(x1,x2,x3),y) -> filter2_4#(x1,x2,x3,y) p7: filter2_4#(false_0(),f,x,xs) -> filter_2#(f,xs) 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(cons_0(),y) -> cons_1(y) r2: app(cons_1(x1),y) -> cons_2(x1,y) r3: app(filter_0(),y) -> filter_1(y) r4: app(filter_1(x1),y) -> filter_2(x1,y) r5: app(filter2_0(),y) -> filter2_1(y) r6: app(filter2_1(x1),y) -> filter2_2(x1,y) r7: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y) r8: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y) r9: app(map_0(),y) -> map_1(y) r10: app(map_1(x1),y) -> map_2(x1,y) r11: app(minus_0(),y) -> minus_1(y) r12: app(minus_1(x1),y) -> minus_2(x1,y) r13: app(pred_0(),y) -> pred_1(y) r14: app(quot_0(),y) -> quot_1(y) r15: app(quot_1(x1),y) -> quot_2(x1,y) r16: app(s_0(),y) -> s_1(y) r17: pred_1(s_1(x)) -> x r18: minus_2(x,|0|_0()) -> x r19: minus_2(x,s_1(y)) -> pred_1(minus_2(x,y)) r20: quot_2(|0|_0(),s_1(y)) -> |0|_0() r21: quot_2(s_1(x),s_1(y)) -> s_1(quot_2(minus_2(x,y),s_1(y))) r22: map_2(f,nil_0()) -> nil_0() r23: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs)) r24: filter_2(f,nil_0()) -> nil_0() r25: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs) r26: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs)) r27: 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!): matrix interpretation with standard point-wise order: carrier: N^1 order: standard order interpretations: app#_A(x1,x2) = x1 + x2 filter_1_A(x1) = x1 + 3 filter_2#_A(x1,x2) = x1 + x2 + 3 cons_2_A(x1,x2) = x1 + x2 + 15 map_1_A(x1) = x1 + 3 map_2#_A(x1,x2) = x1 + x2 filter2_3_A(x1,x2,x3) = x1 + x2 + x3 + 7 filter2_4#_A(x1,x2,x3,x4) = x2 + x4 + 3 false_0_A() = 3 app_A(x1,x2) = x1 + x2 + 1 true_0_A() = 3 app#!_A(x1,x2) = 2 filter_1!_A(x1) = x1 + 15 filter_2#!_A(x1,x2) = 3 cons_2!_A(x1,x2) = x2 + 1 map_1!_A(x1) = x1 + 15 map_2#!_A(x1,x2) = x1 + 3 filter2_3!_A(x1,x2,x3) = x1 + x2 + 13 filter2_4#!_A(x1,x2,x3,x4) = 3 false_0!_A() = 4 app!_A(x1,x2) = x2 + 14 true_0!_A() = 12 2. 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 filter_1_A(x1) = x1 filter_2#_A(x1,x2) = x2 cons_2_A(x1,x2) = x1 + x2 + 17 map_1_A(x1) = x1 map_2#_A(x1,x2) = x2 filter2_3_A(x1,x2,x3) = x1 + x3 filter2_4#_A(x1,x2,x3,x4) = x4 false_0_A() = 1 app_A(x1,x2) = x1 + x2 + 7 true_0_A() = 1 app#!_A(x1,x2) = x1 + 31 filter_1!_A(x1) = 28 filter_2#!_A(x1,x2) = x2 + 13 cons_2!_A(x1,x2) = x2 + 2 map_1!_A(x1) = 33 map_2#!_A(x1,x2) = x1 + x2 + 32 filter2_3!_A(x1,x2,x3) = x1 + 28 filter2_4#!_A(x1,x2,x3,x4) = x4 + 29 false_0!_A() = 14 app!_A(x1,x2) = 27 true_0!_A() = 12 partial status: pi(app#) = [1] pi(filter_1) = [1] pi(filter_2#) = [1] pi(cons_2) = [2] pi(map_1) = [1] pi(map_2#) = [1] pi(filter2_3) = [1, 2] pi(filter2_4#) = [2] pi(false_0) = [] pi(app) = [] pi(true_0) = [] The following rules are removed: p1, p2, p3, p4, p5, p6, p7, p8, p9 Then no dependency pair remains. -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: quot_2#(s_1(x),s_1(y)) -> quot_2#(minus_2(x,y),s_1(y)) and R consists of: r1: app(cons_0(),y) -> cons_1(y) r2: app(cons_1(x1),y) -> cons_2(x1,y) r3: app(filter_0(),y) -> filter_1(y) r4: app(filter_1(x1),y) -> filter_2(x1,y) r5: app(filter2_0(),y) -> filter2_1(y) r6: app(filter2_1(x1),y) -> filter2_2(x1,y) r7: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y) r8: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y) r9: app(map_0(),y) -> map_1(y) r10: app(map_1(x1),y) -> map_2(x1,y) r11: app(minus_0(),y) -> minus_1(y) r12: app(minus_1(x1),y) -> minus_2(x1,y) r13: app(pred_0(),y) -> pred_1(y) r14: app(quot_0(),y) -> quot_1(y) r15: app(quot_1(x1),y) -> quot_2(x1,y) r16: app(s_0(),y) -> s_1(y) r17: pred_1(s_1(x)) -> x r18: minus_2(x,|0|_0()) -> x r19: minus_2(x,s_1(y)) -> pred_1(minus_2(x,y)) r20: quot_2(|0|_0(),s_1(y)) -> |0|_0() r21: quot_2(s_1(x),s_1(y)) -> s_1(quot_2(minus_2(x,y),s_1(y))) r22: map_2(f,nil_0()) -> nil_0() r23: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs)) r24: filter_2(f,nil_0()) -> nil_0() r25: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs) r26: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs)) r27: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs) The set of usable rules consists of r17, r18, r19 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: quot_2#_A(x1,x2) = x1 s_1_A(x1) = x1 + 3 minus_2_A(x1,x2) = x1 + 2 pred_1_A(x1) = x1 |0|_0_A() = 3 quot_2#!_A(x1,x2) = x1 + 14 s_1!_A(x1) = 15 minus_2!_A(x1,x2) = 14 pred_1!_A(x1) = 13 |0|_0!_A() = 0 2. marked-algebra reduction triple (>=A, >=A!, >A!): matrix interpretation with standard point-wise order: carrier: N^1 order: standard order interpretations: quot_2#_A(x1,x2) = x1 s_1_A(x1) = x1 + 4 minus_2_A(x1,x2) = x1 + 3 pred_1_A(x1) = x1 |0|_0_A() = 4 quot_2#!_A(x1,x2) = x1 + x2 + 7 s_1!_A(x1) = 2 minus_2!_A(x1,x2) = x2 + 14 pred_1!_A(x1) = x1 + 16 |0|_0!_A() = 0 partial status: pi(quot_2#) = [1] pi(s_1) = [1] pi(minus_2) = [1] pi(pred_1) = [1] pi(|0|_0) = [] 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: minus_2#(x,s_1(y)) -> minus_2#(x,y) and R consists of: r1: app(cons_0(),y) -> cons_1(y) r2: app(cons_1(x1),y) -> cons_2(x1,y) r3: app(filter_0(),y) -> filter_1(y) r4: app(filter_1(x1),y) -> filter_2(x1,y) r5: app(filter2_0(),y) -> filter2_1(y) r6: app(filter2_1(x1),y) -> filter2_2(x1,y) r7: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y) r8: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y) r9: app(map_0(),y) -> map_1(y) r10: app(map_1(x1),y) -> map_2(x1,y) r11: app(minus_0(),y) -> minus_1(y) r12: app(minus_1(x1),y) -> minus_2(x1,y) r13: app(pred_0(),y) -> pred_1(y) r14: app(quot_0(),y) -> quot_1(y) r15: app(quot_1(x1),y) -> quot_2(x1,y) r16: app(s_0(),y) -> s_1(y) r17: pred_1(s_1(x)) -> x r18: minus_2(x,|0|_0()) -> x r19: minus_2(x,s_1(y)) -> pred_1(minus_2(x,y)) r20: quot_2(|0|_0(),s_1(y)) -> |0|_0() r21: quot_2(s_1(x),s_1(y)) -> s_1(quot_2(minus_2(x,y),s_1(y))) r22: map_2(f,nil_0()) -> nil_0() r23: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs)) r24: filter_2(f,nil_0()) -> nil_0() r25: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs) r26: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs)) r27: 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!): matrix interpretation with standard point-wise order: carrier: N^1 order: standard order interpretations: minus_2#_A(x1,x2) = 0 s_1_A(x1) = 3 minus_2#!_A(x1,x2) = 8 s_1!_A(x1) = 9 2. marked-algebra reduction triple (>=A, >=A!, >A!): matrix interpretation with standard point-wise order: carrier: N^1 order: standard order interpretations: minus_2#_A(x1,x2) = x2 s_1_A(x1) = x1 + 2 minus_2#!_A(x1,x2) = x2 s_1!_A(x1) = x1 + 3 partial status: pi(minus_2#) = [1, 2] pi(s_1) = [1] The following rules are removed: p1 Then no dependency pair remains.