YES We show the termination of the TRS R: a(a(append(),nil()),ys) -> ys a(a(append(),a(a(cons(),x),xs)),ys) -> a(a(cons(),x),a(a(append(),xs),ys)) a(a(filter(),f),nil()) -> nil() a(a(filter(),f),a(a(cons(),x),xs)) -> a(a(a(if(),a(f,x)),x),a(a(filter(),f),xs)) a(a(le(),|0|()),y) -> true() a(a(le(),a(s(),x)),|0|()) -> false() a(a(le(),a(s(),x)),a(s(),y)) -> a(a(le(),x),y) a(a(a(if(),true()),x),xs) -> a(a(cons(),x),xs) a(a(a(if(),false()),x),xs) -> xs a(a(not(),f),b) -> a(not2(),a(f,b)) a(not2(),true()) -> false() a(not2(),false()) -> true() a(qs(),nil()) -> nil() a(qs(),a(a(cons(),x),xs)) -> a(a(append(),a(qs(),a(a(filter(),a(le(),x)),xs))),a(a(cons(),x),a(qs(),a(a(filter(),a(not(),a(le(),x))),xs)))) Instead we prove termination of the uncurried system: a(append_0(),y) -> append_1(y) a(append_1(x1),y) -> append_2(x1,y) a(cons_0(),y) -> cons_1(y) a(cons_1(x1),y) -> cons_2(x1,y) a(filter_0(),y) -> filter_1(y) a(filter_1(x1),y) -> filter_2(x1,y) a(if_0(),y) -> if_1(y) a(if_1(x1),y) -> if_2(x1,y) a(if_2(x1,x2),y) -> if_3(x1,x2,y) a(le_0(),y) -> le_1(y) a(le_1(x1),y) -> le_2(x1,y) a(not_0(),y) -> not_1(y) a(not_1(x1),y) -> not_2(x1,y) a(not2_0(),y) -> not2_1(y) a(qs_0(),y) -> qs_1(y) a(s_0(),y) -> s_1(y) append_2(nil_0(),ys) -> ys append_2(cons_2(x,xs),ys) -> cons_2(x,append_2(xs,ys)) filter_2(f,nil_0()) -> nil_0() filter_2(f,cons_2(x,xs)) -> if_3(a(f,x),x,filter_2(f,xs)) le_2(|0|_0(),y) -> true_0() le_2(s_1(x),|0|_0()) -> false_0() le_2(s_1(x),s_1(y)) -> le_2(x,y) if_3(true_0(),x,xs) -> cons_2(x,xs) if_3(false_0(),x,xs) -> xs not_2(f,b) -> not2_1(a(f,b)) not2_1(true_0()) -> false_0() not2_1(false_0()) -> true_0() qs_1(nil_0()) -> nil_0() qs_1(cons_2(x,xs)) -> append_2(qs_1(filter_2(le_1(x),xs)),cons_2(x,qs_1(filter_2(not_1(le_1(x)),xs)))) -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: a#(append_1(x1),y) -> append_2#(x1,y) p2: a#(filter_1(x1),y) -> filter_2#(x1,y) p3: a#(if_2(x1,x2),y) -> if_3#(x1,x2,y) p4: a#(le_1(x1),y) -> le_2#(x1,y) p5: a#(not_1(x1),y) -> not_2#(x1,y) p6: a#(not2_0(),y) -> not2_1#(y) p7: a#(qs_0(),y) -> qs_1#(y) p8: append_2#(cons_2(x,xs),ys) -> append_2#(xs,ys) p9: filter_2#(f,cons_2(x,xs)) -> if_3#(a(f,x),x,filter_2(f,xs)) p10: filter_2#(f,cons_2(x,xs)) -> a#(f,x) p11: filter_2#(f,cons_2(x,xs)) -> filter_2#(f,xs) p12: le_2#(s_1(x),s_1(y)) -> le_2#(x,y) p13: not_2#(f,b) -> not2_1#(a(f,b)) p14: not_2#(f,b) -> a#(f,b) p15: qs_1#(cons_2(x,xs)) -> append_2#(qs_1(filter_2(le_1(x),xs)),cons_2(x,qs_1(filter_2(not_1(le_1(x)),xs)))) p16: qs_1#(cons_2(x,xs)) -> qs_1#(filter_2(le_1(x),xs)) p17: qs_1#(cons_2(x,xs)) -> filter_2#(le_1(x),xs) p18: qs_1#(cons_2(x,xs)) -> qs_1#(filter_2(not_1(le_1(x)),xs)) p19: qs_1#(cons_2(x,xs)) -> filter_2#(not_1(le_1(x)),xs) and R consists of: r1: a(append_0(),y) -> append_1(y) r2: a(append_1(x1),y) -> append_2(x1,y) r3: a(cons_0(),y) -> cons_1(y) r4: a(cons_1(x1),y) -> cons_2(x1,y) r5: a(filter_0(),y) -> filter_1(y) r6: a(filter_1(x1),y) -> filter_2(x1,y) r7: a(if_0(),y) -> if_1(y) r8: a(if_1(x1),y) -> if_2(x1,y) r9: a(if_2(x1,x2),y) -> if_3(x1,x2,y) r10: a(le_0(),y) -> le_1(y) r11: a(le_1(x1),y) -> le_2(x1,y) r12: a(not_0(),y) -> not_1(y) r13: a(not_1(x1),y) -> not_2(x1,y) r14: a(not2_0(),y) -> not2_1(y) r15: a(qs_0(),y) -> qs_1(y) r16: a(s_0(),y) -> s_1(y) r17: append_2(nil_0(),ys) -> ys r18: append_2(cons_2(x,xs),ys) -> cons_2(x,append_2(xs,ys)) r19: filter_2(f,nil_0()) -> nil_0() r20: filter_2(f,cons_2(x,xs)) -> if_3(a(f,x),x,filter_2(f,xs)) r21: le_2(|0|_0(),y) -> true_0() r22: le_2(s_1(x),|0|_0()) -> false_0() r23: le_2(s_1(x),s_1(y)) -> le_2(x,y) r24: if_3(true_0(),x,xs) -> cons_2(x,xs) r25: if_3(false_0(),x,xs) -> xs r26: not_2(f,b) -> not2_1(a(f,b)) r27: not2_1(true_0()) -> false_0() r28: not2_1(false_0()) -> true_0() r29: qs_1(nil_0()) -> nil_0() r30: qs_1(cons_2(x,xs)) -> append_2(qs_1(filter_2(le_1(x),xs)),cons_2(x,qs_1(filter_2(not_1(le_1(x)),xs)))) The estimated dependency graph contains the following SCCs: {p2, p5, p7, p10, p11, p14, p16, p17, p18, p19} {p8} {p12} -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: not_2#(f,b) -> a#(f,b) p2: a#(qs_0(),y) -> qs_1#(y) p3: qs_1#(cons_2(x,xs)) -> filter_2#(not_1(le_1(x)),xs) p4: filter_2#(f,cons_2(x,xs)) -> filter_2#(f,xs) p5: filter_2#(f,cons_2(x,xs)) -> a#(f,x) p6: a#(not_1(x1),y) -> not_2#(x1,y) p7: a#(filter_1(x1),y) -> filter_2#(x1,y) p8: qs_1#(cons_2(x,xs)) -> qs_1#(filter_2(not_1(le_1(x)),xs)) p9: qs_1#(cons_2(x,xs)) -> filter_2#(le_1(x),xs) p10: qs_1#(cons_2(x,xs)) -> qs_1#(filter_2(le_1(x),xs)) and R consists of: r1: a(append_0(),y) -> append_1(y) r2: a(append_1(x1),y) -> append_2(x1,y) r3: a(cons_0(),y) -> cons_1(y) r4: a(cons_1(x1),y) -> cons_2(x1,y) r5: a(filter_0(),y) -> filter_1(y) r6: a(filter_1(x1),y) -> filter_2(x1,y) r7: a(if_0(),y) -> if_1(y) r8: a(if_1(x1),y) -> if_2(x1,y) r9: a(if_2(x1,x2),y) -> if_3(x1,x2,y) r10: a(le_0(),y) -> le_1(y) r11: a(le_1(x1),y) -> le_2(x1,y) r12: a(not_0(),y) -> not_1(y) r13: a(not_1(x1),y) -> not_2(x1,y) r14: a(not2_0(),y) -> not2_1(y) r15: a(qs_0(),y) -> qs_1(y) r16: a(s_0(),y) -> s_1(y) r17: append_2(nil_0(),ys) -> ys r18: append_2(cons_2(x,xs),ys) -> cons_2(x,append_2(xs,ys)) r19: filter_2(f,nil_0()) -> nil_0() r20: filter_2(f,cons_2(x,xs)) -> if_3(a(f,x),x,filter_2(f,xs)) r21: le_2(|0|_0(),y) -> true_0() r22: le_2(s_1(x),|0|_0()) -> false_0() r23: le_2(s_1(x),s_1(y)) -> le_2(x,y) r24: if_3(true_0(),x,xs) -> cons_2(x,xs) r25: if_3(false_0(),x,xs) -> xs r26: not_2(f,b) -> not2_1(a(f,b)) r27: not2_1(true_0()) -> false_0() r28: not2_1(false_0()) -> true_0() r29: qs_1(nil_0()) -> nil_0() r30: qs_1(cons_2(x,xs)) -> append_2(qs_1(filter_2(le_1(x),xs)),cons_2(x,qs_1(filter_2(not_1(le_1(x)),xs)))) The set of usable rules consists of r28 Take the reduction pair: generalized weighted path order (>=MG, >G) base order 1: max/plus interpretation on natural numbers: not_2#_A(x1,x2) = 22 a#_A(x1,x2) = 22 qs_0_A = 23 qs_1#_A(x1) = 22 cons_2_A(x1,x2) = max{x1 + 23, x2 - 27} filter_2#_A(x1,x2) = 22 not_1_A(x1) = max{10, x1 + 8} le_1_A(x1) = 12 filter_1_A(x1) = 23 filter_2_A(x1,x2) = max{x1 + 120, x2 + 28} not2_1_A(x1) = max{10, x1 - 12} false_0_A = 9 true_0_A = 5 base order 2: marked-algebra reduction triple (>=A, >=A!, >A!): matrix interpretation with standard point-wise order: carrier: N^1 order: standard order interpretations: not_2#_A(x1,x2) = x1 a#_A(x1,x2) = x1 qs_0_A() = 9 qs_1#_A(x1) = 7 cons_2_A(x1,x2) = x1 + 2 filter_2#_A(x1,x2) = x1 not_1_A(x1) = x1 + 2 le_1_A(x1) = 5 filter_1_A(x1) = x1 filter_2_A(x1,x2) = x1 + x2 + 2 not2_1_A(x1) = 3 false_0_A() = 8 true_0_A() = 0 not_2#!_A(x1,x2) = x1 + 2 a#!_A(x1,x2) = x1 + 1 qs_0!_A() = 11 qs_1#!_A(x1) = 10 cons_2!_A(x1,x2) = x1 + 13 filter_2#!_A(x1,x2) = x1 + 1 not_1!_A(x1) = 3 le_1!_A(x1) = 0 filter_1!_A(x1) = 0 filter_2!_A(x1,x2) = x1 + x2 + 2 not2_1!_A(x1) = 5 false_0!_A() = 7 true_0!_A() = 6 partial status: pi(not_2#) = [] pi(a#) = [] pi(qs_0) = [] pi(qs_1#) = [] pi(cons_2) = [] pi(filter_2#) = [] pi(not_1) = [] pi(le_1) = [] pi(filter_1) = [] pi(filter_2) = [] pi(not2_1) = [] pi(false_0) = [] pi(true_0) = [] The following rules are removed: p1, p3, p6, p9 -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: a#(qs_0(),y) -> qs_1#(y) p2: filter_2#(f,cons_2(x,xs)) -> filter_2#(f,xs) p3: filter_2#(f,cons_2(x,xs)) -> a#(f,x) p4: a#(filter_1(x1),y) -> filter_2#(x1,y) p5: qs_1#(cons_2(x,xs)) -> qs_1#(filter_2(not_1(le_1(x)),xs)) p6: qs_1#(cons_2(x,xs)) -> qs_1#(filter_2(le_1(x),xs)) and R consists of: r1: a(append_0(),y) -> append_1(y) r2: a(append_1(x1),y) -> append_2(x1,y) r3: a(cons_0(),y) -> cons_1(y) r4: a(cons_1(x1),y) -> cons_2(x1,y) r5: a(filter_0(),y) -> filter_1(y) r6: a(filter_1(x1),y) -> filter_2(x1,y) r7: a(if_0(),y) -> if_1(y) r8: a(if_1(x1),y) -> if_2(x1,y) r9: a(if_2(x1,x2),y) -> if_3(x1,x2,y) r10: a(le_0(),y) -> le_1(y) r11: a(le_1(x1),y) -> le_2(x1,y) r12: a(not_0(),y) -> not_1(y) r13: a(not_1(x1),y) -> not_2(x1,y) r14: a(not2_0(),y) -> not2_1(y) r15: a(qs_0(),y) -> qs_1(y) r16: a(s_0(),y) -> s_1(y) r17: append_2(nil_0(),ys) -> ys r18: append_2(cons_2(x,xs),ys) -> cons_2(x,append_2(xs,ys)) r19: filter_2(f,nil_0()) -> nil_0() r20: filter_2(f,cons_2(x,xs)) -> if_3(a(f,x),x,filter_2(f,xs)) r21: le_2(|0|_0(),y) -> true_0() r22: le_2(s_1(x),|0|_0()) -> false_0() r23: le_2(s_1(x),s_1(y)) -> le_2(x,y) r24: if_3(true_0(),x,xs) -> cons_2(x,xs) r25: if_3(false_0(),x,xs) -> xs r26: not_2(f,b) -> not2_1(a(f,b)) r27: not2_1(true_0()) -> false_0() r28: not2_1(false_0()) -> true_0() r29: qs_1(nil_0()) -> nil_0() r30: qs_1(cons_2(x,xs)) -> append_2(qs_1(filter_2(le_1(x),xs)),cons_2(x,qs_1(filter_2(not_1(le_1(x)),xs)))) The estimated dependency graph contains the following SCCs: {p2, p3, p4} {p5, p6} -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: filter_2#(f,cons_2(x,xs)) -> a#(f,x) p2: a#(filter_1(x1),y) -> filter_2#(x1,y) p3: filter_2#(f,cons_2(x,xs)) -> filter_2#(f,xs) and R consists of: r1: a(append_0(),y) -> append_1(y) r2: a(append_1(x1),y) -> append_2(x1,y) r3: a(cons_0(),y) -> cons_1(y) r4: a(cons_1(x1),y) -> cons_2(x1,y) r5: a(filter_0(),y) -> filter_1(y) r6: a(filter_1(x1),y) -> filter_2(x1,y) r7: a(if_0(),y) -> if_1(y) r8: a(if_1(x1),y) -> if_2(x1,y) r9: a(if_2(x1,x2),y) -> if_3(x1,x2,y) r10: a(le_0(),y) -> le_1(y) r11: a(le_1(x1),y) -> le_2(x1,y) r12: a(not_0(),y) -> not_1(y) r13: a(not_1(x1),y) -> not_2(x1,y) r14: a(not2_0(),y) -> not2_1(y) r15: a(qs_0(),y) -> qs_1(y) r16: a(s_0(),y) -> s_1(y) r17: append_2(nil_0(),ys) -> ys r18: append_2(cons_2(x,xs),ys) -> cons_2(x,append_2(xs,ys)) r19: filter_2(f,nil_0()) -> nil_0() r20: filter_2(f,cons_2(x,xs)) -> if_3(a(f,x),x,filter_2(f,xs)) r21: le_2(|0|_0(),y) -> true_0() r22: le_2(s_1(x),|0|_0()) -> false_0() r23: le_2(s_1(x),s_1(y)) -> le_2(x,y) r24: if_3(true_0(),x,xs) -> cons_2(x,xs) r25: if_3(false_0(),x,xs) -> xs r26: not_2(f,b) -> not2_1(a(f,b)) r27: not2_1(true_0()) -> false_0() r28: not2_1(false_0()) -> true_0() r29: qs_1(nil_0()) -> nil_0() r30: qs_1(cons_2(x,xs)) -> append_2(qs_1(filter_2(le_1(x),xs)),cons_2(x,qs_1(filter_2(not_1(le_1(x)),xs)))) The set of usable rules consists of (no rules) Take the reduction pair: generalized weighted path order (>=MG, >G) base order 1: max/plus interpretation on natural numbers: filter_2#_A(x1,x2) = max{30, x2 + 19} cons_2_A(x1,x2) = max{x1 + 30, x2 + 18} a#_A(x1,x2) = x2 + 30 filter_1_A(x1) = 0 base order 2: marked-algebra reduction triple (>=A, >=A!, >A!): matrix interpretation with standard point-wise order: carrier: N^1 order: standard order interpretations: filter_2#_A(x1,x2) = x1 + x2 cons_2_A(x1,x2) = x1 + x2 + 3 a#_A(x1,x2) = x1 + x2 filter_1_A(x1) = x1 + 2 filter_2#!_A(x1,x2) = x2 cons_2!_A(x1,x2) = x1 + 3 a#!_A(x1,x2) = x2 + 1 filter_1!_A(x1) = 1 partial status: pi(filter_2#) = [2] pi(cons_2) = [1, 2] pi(a#) = [] pi(filter_1) = [] The following rules are removed: p1, p2, p3 Then no dependency pair remains. -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: qs_1#(cons_2(x,xs)) -> qs_1#(filter_2(not_1(le_1(x)),xs)) p2: qs_1#(cons_2(x,xs)) -> qs_1#(filter_2(le_1(x),xs)) and R consists of: r1: a(append_0(),y) -> append_1(y) r2: a(append_1(x1),y) -> append_2(x1,y) r3: a(cons_0(),y) -> cons_1(y) r4: a(cons_1(x1),y) -> cons_2(x1,y) r5: a(filter_0(),y) -> filter_1(y) r6: a(filter_1(x1),y) -> filter_2(x1,y) r7: a(if_0(),y) -> if_1(y) r8: a(if_1(x1),y) -> if_2(x1,y) r9: a(if_2(x1,x2),y) -> if_3(x1,x2,y) r10: a(le_0(),y) -> le_1(y) r11: a(le_1(x1),y) -> le_2(x1,y) r12: a(not_0(),y) -> not_1(y) r13: a(not_1(x1),y) -> not_2(x1,y) r14: a(not2_0(),y) -> not2_1(y) r15: a(qs_0(),y) -> qs_1(y) r16: a(s_0(),y) -> s_1(y) r17: append_2(nil_0(),ys) -> ys r18: append_2(cons_2(x,xs),ys) -> cons_2(x,append_2(xs,ys)) r19: filter_2(f,nil_0()) -> nil_0() r20: filter_2(f,cons_2(x,xs)) -> if_3(a(f,x),x,filter_2(f,xs)) r21: le_2(|0|_0(),y) -> true_0() r22: le_2(s_1(x),|0|_0()) -> false_0() r23: le_2(s_1(x),s_1(y)) -> le_2(x,y) r24: if_3(true_0(),x,xs) -> cons_2(x,xs) r25: if_3(false_0(),x,xs) -> xs r26: not_2(f,b) -> not2_1(a(f,b)) r27: not2_1(true_0()) -> false_0() r28: not2_1(false_0()) -> true_0() r29: qs_1(nil_0()) -> nil_0() r30: qs_1(cons_2(x,xs)) -> append_2(qs_1(filter_2(le_1(x),xs)),cons_2(x,qs_1(filter_2(not_1(le_1(x)),xs)))) The set of usable rules consists of r19, r20, r24, r25 Take the reduction pair: generalized weighted path order (>=MG, >G) base order 1: max/plus interpretation on natural numbers: qs_1#_A(x1) = max{52, x1 + 41} cons_2_A(x1,x2) = max{x1 + 8, x2} filter_2_A(x1,x2) = max{7, x1 - 5, x2} not_1_A(x1) = max{4, x1 - 17} le_1_A(x1) = max{16, x1 - 10} nil_0_A = 8 if_3_A(x1,x2,x3) = max{x2 + 8, x3} a_A(x1,x2) = max{14, x1 - 2, x2 + 9} true_0_A = 10 false_0_A = 11 base order 2: marked-algebra reduction triple (>=A, >=A!, >A!): matrix interpretation with standard point-wise order: carrier: N^1 order: standard order interpretations: qs_1#_A(x1) = x1 cons_2_A(x1,x2) = x1 + x2 + 19 filter_2_A(x1,x2) = x1 + x2 + 2 not_1_A(x1) = x1 + 2 le_1_A(x1) = x1 + 2 nil_0_A() = 3 if_3_A(x1,x2,x3) = x2 + x3 + 19 a_A(x1,x2) = 1 true_0_A() = 4 false_0_A() = 6 qs_1#!_A(x1) = x1 cons_2!_A(x1,x2) = 33 filter_2!_A(x1,x2) = x1 + x2 + 18 not_1!_A(x1) = 1 le_1!_A(x1) = 23 nil_0!_A() = 0 if_3!_A(x1,x2,x3) = x3 + 34 a!_A(x1,x2) = 24 true_0!_A() = 32 false_0!_A() = 33 partial status: pi(qs_1#) = [] pi(cons_2) = [] pi(filter_2) = [] pi(not_1) = [] pi(le_1) = [] pi(nil_0) = [] pi(if_3) = [3] pi(a) = [2] pi(true_0) = [] pi(false_0) = [] The following rules are removed: p1, p2 Then no dependency pair remains. -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: append_2#(cons_2(x,xs),ys) -> append_2#(xs,ys) and R consists of: r1: a(append_0(),y) -> append_1(y) r2: a(append_1(x1),y) -> append_2(x1,y) r3: a(cons_0(),y) -> cons_1(y) r4: a(cons_1(x1),y) -> cons_2(x1,y) r5: a(filter_0(),y) -> filter_1(y) r6: a(filter_1(x1),y) -> filter_2(x1,y) r7: a(if_0(),y) -> if_1(y) r8: a(if_1(x1),y) -> if_2(x1,y) r9: a(if_2(x1,x2),y) -> if_3(x1,x2,y) r10: a(le_0(),y) -> le_1(y) r11: a(le_1(x1),y) -> le_2(x1,y) r12: a(not_0(),y) -> not_1(y) r13: a(not_1(x1),y) -> not_2(x1,y) r14: a(not2_0(),y) -> not2_1(y) r15: a(qs_0(),y) -> qs_1(y) r16: a(s_0(),y) -> s_1(y) r17: append_2(nil_0(),ys) -> ys r18: append_2(cons_2(x,xs),ys) -> cons_2(x,append_2(xs,ys)) r19: filter_2(f,nil_0()) -> nil_0() r20: filter_2(f,cons_2(x,xs)) -> if_3(a(f,x),x,filter_2(f,xs)) r21: le_2(|0|_0(),y) -> true_0() r22: le_2(s_1(x),|0|_0()) -> false_0() r23: le_2(s_1(x),s_1(y)) -> le_2(x,y) r24: if_3(true_0(),x,xs) -> cons_2(x,xs) r25: if_3(false_0(),x,xs) -> xs r26: not_2(f,b) -> not2_1(a(f,b)) r27: not2_1(true_0()) -> false_0() r28: not2_1(false_0()) -> true_0() r29: qs_1(nil_0()) -> nil_0() r30: qs_1(cons_2(x,xs)) -> append_2(qs_1(filter_2(le_1(x),xs)),cons_2(x,qs_1(filter_2(not_1(le_1(x)),xs)))) The set of usable rules consists of (no rules) Take the reduction pair: generalized weighted path order (>=MG, >G) base order 1: max/plus interpretation on natural numbers: append_2#_A(x1,x2) = max{x1 + 21, x2 + 20} cons_2_A(x1,x2) = max{x1 + 22, x2 + 20} base order 2: marked-algebra reduction triple (>=A, >=A!, >A!): matrix interpretation with standard point-wise order: carrier: N^1 order: standard order interpretations: append_2#_A(x1,x2) = x1 cons_2_A(x1,x2) = x1 + x2 + 3 append_2#!_A(x1,x2) = 4 cons_2!_A(x1,x2) = x2 + 3 partial status: pi(append_2#) = [1] pi(cons_2) = [1, 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: le_2#(s_1(x),s_1(y)) -> le_2#(x,y) and R consists of: r1: a(append_0(),y) -> append_1(y) r2: a(append_1(x1),y) -> append_2(x1,y) r3: a(cons_0(),y) -> cons_1(y) r4: a(cons_1(x1),y) -> cons_2(x1,y) r5: a(filter_0(),y) -> filter_1(y) r6: a(filter_1(x1),y) -> filter_2(x1,y) r7: a(if_0(),y) -> if_1(y) r8: a(if_1(x1),y) -> if_2(x1,y) r9: a(if_2(x1,x2),y) -> if_3(x1,x2,y) r10: a(le_0(),y) -> le_1(y) r11: a(le_1(x1),y) -> le_2(x1,y) r12: a(not_0(),y) -> not_1(y) r13: a(not_1(x1),y) -> not_2(x1,y) r14: a(not2_0(),y) -> not2_1(y) r15: a(qs_0(),y) -> qs_1(y) r16: a(s_0(),y) -> s_1(y) r17: append_2(nil_0(),ys) -> ys r18: append_2(cons_2(x,xs),ys) -> cons_2(x,append_2(xs,ys)) r19: filter_2(f,nil_0()) -> nil_0() r20: filter_2(f,cons_2(x,xs)) -> if_3(a(f,x),x,filter_2(f,xs)) r21: le_2(|0|_0(),y) -> true_0() r22: le_2(s_1(x),|0|_0()) -> false_0() r23: le_2(s_1(x),s_1(y)) -> le_2(x,y) r24: if_3(true_0(),x,xs) -> cons_2(x,xs) r25: if_3(false_0(),x,xs) -> xs r26: not_2(f,b) -> not2_1(a(f,b)) r27: not2_1(true_0()) -> false_0() r28: not2_1(false_0()) -> true_0() r29: qs_1(nil_0()) -> nil_0() r30: qs_1(cons_2(x,xs)) -> append_2(qs_1(filter_2(le_1(x),xs)),cons_2(x,qs_1(filter_2(not_1(le_1(x)),xs)))) The set of usable rules consists of (no rules) Take the reduction pair: generalized weighted path order (>=MG, >G) base order 1: max/plus interpretation on natural numbers: le_2#_A(x1,x2) = max{x1 + 113, x2 + 113} s_1_A(x1) = max{120, x1 + 112} base order 2: marked-algebra reduction triple (>=A, >=A!, >A!): matrix interpretation with standard point-wise order: carrier: N^1 order: standard order interpretations: le_2#_A(x1,x2) = x1 + x2 s_1_A(x1) = x1 + 2 le_2#!_A(x1,x2) = x1 + x2 + 3 s_1!_A(x1) = 2 partial status: pi(le_2#) = [1, 2] pi(s_1) = [1] The following rules are removed: p1 Then no dependency pair remains.