YES We show the termination of the TRS R: a(a(divides(),|0|()),a(s(),y)) -> true() a(a(divides(),a(s(),x)),a(s(),y)) -> a(a(a(div2(),x),a(s(),y)),y) a(a(a(div2(),x),y),|0|()) -> a(a(divides(),x),y) a(a(a(div2(),|0|()),y),a(s(),z)) -> false() a(a(a(div2(),a(s(),x)),y),a(s(),z)) -> a(a(a(div2(),x),y),z) 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(a(if(),true()),x),xs) -> a(a(cons(),x),xs) a(a(a(if(),false()),x),xs) -> xs a(a(not(),f),x) -> a(not2(),a(f,x)) a(not2(),true()) -> false() a(not2(),false()) -> true() a(sieve(),nil()) -> nil() a(sieve(),a(a(cons(),x),xs)) -> a(a(cons(),x),a(sieve(),a(a(filter(),a(not(),a(divides(),x))),xs))) Instead we prove termination of the uncurried system: a(cons_0(),y) -> cons_1(y) a(cons_1(x1),y) -> cons_2(x1,y) a(div2_0(),y) -> div2_1(y) a(div2_1(x1),y) -> div2_2(x1,y) a(div2_2(x1,x2),y) -> div2_3(x1,x2,y) a(divides_0(),y) -> divides_1(y) a(divides_1(x1),y) -> divides_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(not_0(),y) -> not_1(y) a(not_1(x1),y) -> not_2(x1,y) a(not2_0(),y) -> not2_1(y) a(s_0(),y) -> s_1(y) a(sieve_0(),y) -> sieve_1(y) divides_2(|0|_0(),s_1(y)) -> true_0() divides_2(s_1(x),s_1(y)) -> div2_3(x,s_1(y),y) div2_3(x,y,|0|_0()) -> divides_2(x,y) div2_3(|0|_0(),y,s_1(z)) -> false_0() div2_3(s_1(x),y,s_1(z)) -> div2_3(x,y,z) filter_2(f,nil_0()) -> nil_0() filter_2(f,cons_2(x,xs)) -> if_3(a(f,x),x,filter_2(f,xs)) if_3(true_0(),x,xs) -> cons_2(x,xs) if_3(false_0(),x,xs) -> xs not_2(f,x) -> not2_1(a(f,x)) not2_1(true_0()) -> false_0() not2_1(false_0()) -> true_0() sieve_1(nil_0()) -> nil_0() sieve_1(cons_2(x,xs)) -> cons_2(x,sieve_1(filter_2(not_1(divides_1(x)),xs))) -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: a#(div2_2(x1,x2),y) -> div2_3#(x1,x2,y) p2: a#(divides_1(x1),y) -> divides_2#(x1,y) p3: a#(filter_1(x1),y) -> filter_2#(x1,y) p4: a#(if_2(x1,x2),y) -> if_3#(x1,x2,y) p5: a#(not_1(x1),y) -> not_2#(x1,y) p6: a#(not2_0(),y) -> not2_1#(y) p7: a#(sieve_0(),y) -> sieve_1#(y) p8: divides_2#(s_1(x),s_1(y)) -> div2_3#(x,s_1(y),y) p9: div2_3#(x,y,|0|_0()) -> divides_2#(x,y) p10: div2_3#(s_1(x),y,s_1(z)) -> div2_3#(x,y,z) p11: filter_2#(f,cons_2(x,xs)) -> if_3#(a(f,x),x,filter_2(f,xs)) p12: filter_2#(f,cons_2(x,xs)) -> a#(f,x) p13: filter_2#(f,cons_2(x,xs)) -> filter_2#(f,xs) p14: not_2#(f,x) -> not2_1#(a(f,x)) p15: not_2#(f,x) -> a#(f,x) p16: sieve_1#(cons_2(x,xs)) -> sieve_1#(filter_2(not_1(divides_1(x)),xs)) p17: sieve_1#(cons_2(x,xs)) -> filter_2#(not_1(divides_1(x)),xs) and R consists of: r1: a(cons_0(),y) -> cons_1(y) r2: a(cons_1(x1),y) -> cons_2(x1,y) r3: a(div2_0(),y) -> div2_1(y) r4: a(div2_1(x1),y) -> div2_2(x1,y) r5: a(div2_2(x1,x2),y) -> div2_3(x1,x2,y) r6: a(divides_0(),y) -> divides_1(y) r7: a(divides_1(x1),y) -> divides_2(x1,y) r8: a(filter_0(),y) -> filter_1(y) r9: a(filter_1(x1),y) -> filter_2(x1,y) r10: a(if_0(),y) -> if_1(y) r11: a(if_1(x1),y) -> if_2(x1,y) r12: a(if_2(x1,x2),y) -> if_3(x1,x2,y) r13: a(not_0(),y) -> not_1(y) r14: a(not_1(x1),y) -> not_2(x1,y) r15: a(not2_0(),y) -> not2_1(y) r16: a(s_0(),y) -> s_1(y) r17: a(sieve_0(),y) -> sieve_1(y) r18: divides_2(|0|_0(),s_1(y)) -> true_0() r19: divides_2(s_1(x),s_1(y)) -> div2_3(x,s_1(y),y) r20: div2_3(x,y,|0|_0()) -> divides_2(x,y) r21: div2_3(|0|_0(),y,s_1(z)) -> false_0() r22: div2_3(s_1(x),y,s_1(z)) -> div2_3(x,y,z) r23: filter_2(f,nil_0()) -> nil_0() r24: filter_2(f,cons_2(x,xs)) -> if_3(a(f,x),x,filter_2(f,xs)) r25: if_3(true_0(),x,xs) -> cons_2(x,xs) r26: if_3(false_0(),x,xs) -> xs r27: not_2(f,x) -> not2_1(a(f,x)) r28: not2_1(true_0()) -> false_0() r29: not2_1(false_0()) -> true_0() r30: sieve_1(nil_0()) -> nil_0() r31: sieve_1(cons_2(x,xs)) -> cons_2(x,sieve_1(filter_2(not_1(divides_1(x)),xs))) The estimated dependency graph contains the following SCCs: {p3, p5, p7, p12, p13, p15, p16, p17} {p8, p9, p10} -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: not_2#(f,x) -> a#(f,x) p2: a#(sieve_0(),y) -> sieve_1#(y) p3: sieve_1#(cons_2(x,xs)) -> filter_2#(not_1(divides_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: sieve_1#(cons_2(x,xs)) -> sieve_1#(filter_2(not_1(divides_1(x)),xs)) and R consists of: r1: a(cons_0(),y) -> cons_1(y) r2: a(cons_1(x1),y) -> cons_2(x1,y) r3: a(div2_0(),y) -> div2_1(y) r4: a(div2_1(x1),y) -> div2_2(x1,y) r5: a(div2_2(x1,x2),y) -> div2_3(x1,x2,y) r6: a(divides_0(),y) -> divides_1(y) r7: a(divides_1(x1),y) -> divides_2(x1,y) r8: a(filter_0(),y) -> filter_1(y) r9: a(filter_1(x1),y) -> filter_2(x1,y) r10: a(if_0(),y) -> if_1(y) r11: a(if_1(x1),y) -> if_2(x1,y) r12: a(if_2(x1,x2),y) -> if_3(x1,x2,y) r13: a(not_0(),y) -> not_1(y) r14: a(not_1(x1),y) -> not_2(x1,y) r15: a(not2_0(),y) -> not2_1(y) r16: a(s_0(),y) -> s_1(y) r17: a(sieve_0(),y) -> sieve_1(y) r18: divides_2(|0|_0(),s_1(y)) -> true_0() r19: divides_2(s_1(x),s_1(y)) -> div2_3(x,s_1(y),y) r20: div2_3(x,y,|0|_0()) -> divides_2(x,y) r21: div2_3(|0|_0(),y,s_1(z)) -> false_0() r22: div2_3(s_1(x),y,s_1(z)) -> div2_3(x,y,z) r23: filter_2(f,nil_0()) -> nil_0() r24: filter_2(f,cons_2(x,xs)) -> if_3(a(f,x),x,filter_2(f,xs)) r25: if_3(true_0(),x,xs) -> cons_2(x,xs) r26: if_3(false_0(),x,xs) -> xs r27: not_2(f,x) -> not2_1(a(f,x)) r28: not2_1(true_0()) -> false_0() r29: not2_1(false_0()) -> true_0() r30: sieve_1(nil_0()) -> nil_0() r31: sieve_1(cons_2(x,xs)) -> cons_2(x,sieve_1(filter_2(not_1(divides_1(x)),xs))) The set of usable rules consists of r11, r23, r26 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: not_2#_A(x1,x2) = 1 a#_A(x1,x2) = 1 sieve_0_A = 5 sieve_1#_A(x1) = 1 cons_2_A(x1,x2) = max{4, x1, x2} filter_2#_A(x1,x2) = 1 not_1_A(x1) = max{5, x1 - 16} divides_1_A(x1) = max{26, x1 + 20} filter_1_A(x1) = max{3, x1 - 16} filter_2_A(x1,x2) = max{3, x2} a_A(x1,x2) = max{2, x1 - 27, x2 - 27} if_1_A(x1) = x1 + 5 if_2_A(x1,x2) = max{2, x1 - 22, x2 - 27} nil_0_A = 4 if_3_A(x1,x2,x3) = max{5, x2, x3} false_0_A = 5 not_2#!_A(x1,x2) = 3 a#!_A(x1,x2) = 3 sieve_0!_A = 0 sieve_1#!_A(x1) = 3 cons_2!_A(x1,x2) = x2 + 9 filter_2#!_A(x1,x2) = 3 not_1!_A(x1) = x1 + 4 divides_1!_A(x1) = max{14, x1 - 1} filter_1!_A(x1) = 13 filter_2!_A(x1,x2) = x2 + 11 a!_A(x1,x2) = max{x1 - 6, x2 + 12} if_1!_A(x1) = max{13, x1 - 3} if_2!_A(x1,x2) = max{0, x1 - 2, x2 - 2} nil_0!_A = 0 if_3!_A(x1,x2,x3) = max{10, x2 - 1, x3 - 4} false_0!_A = 3 2. marked-algebra reduction triple (>=A, >=A!, >A!): max/plus interpretation on natural numbers: not_2#_A(x1,x2) = max{21, x1 - 100} a#_A(x1,x2) = max{21, x1 - 100} sieve_0_A = 107 sieve_1#_A(x1) = 21 cons_2_A(x1,x2) = max{x1 + 86, x2} filter_2#_A(x1,x2) = max{21, x1 - 78} not_1_A(x1) = max{99, x1 + 42} divides_1_A(x1) = 37 filter_1_A(x1) = max{50, x1 + 29} filter_2_A(x1,x2) = max{x1 - 12, x2} a_A(x1,x2) = max{3, x1 - 42, x2 - 54} if_1_A(x1) = x1 + 41 if_2_A(x1,x2) = max{3, x1 - 1, x2 - 54} nil_0_A = 84 if_3_A(x1,x2,x3) = max{x2 + 86, x3} false_0_A = 44 not_2#!_A(x1,x2) = max{50, x1 + 5} a#!_A(x1,x2) = max{4, x1 - 1} sieve_0!_A = 106 sieve_1#!_A(x1) = 105 cons_2!_A(x1,x2) = x2 + 87 filter_2#!_A(x1,x2) = x1 + 5 not_1!_A(x1) = max{106, x1 + 49} divides_1!_A(x1) = max{108, x1 - 7} filter_1!_A(x1) = x1 + 6 filter_2!_A(x1,x2) = 0 a!_A(x1,x2) = max{107, x2 + 50} if_1!_A(x1) = max{48, x1 - 2} if_2!_A(x1,x2) = max{49, x1 - 1, x2 - 15} nil_0!_A = 0 if_3!_A(x1,x2,x3) = max{108, x2 - 14, x3 + 88} false_0!_A = 57 partial status: pi(not_2#) = [] pi(a#) = [] pi(sieve_0) = [] pi(sieve_1#) = [] pi(cons_2) = [1, 2] pi(filter_2#) = [] pi(not_1) = [1] pi(divides_1) = [1] pi(filter_1) = [] pi(filter_2) = [2] pi(a) = [2] pi(if_1) = [1] pi(if_2) = [2] pi(nil_0) = [] pi(if_3) = [2, 3] pi(false_0) = [] The following rules are removed: p1, p2, p3, p5, p6, p7 -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: filter_2#(f,cons_2(x,xs)) -> filter_2#(f,xs) p2: sieve_1#(cons_2(x,xs)) -> sieve_1#(filter_2(not_1(divides_1(x)),xs)) and R consists of: r1: a(cons_0(),y) -> cons_1(y) r2: a(cons_1(x1),y) -> cons_2(x1,y) r3: a(div2_0(),y) -> div2_1(y) r4: a(div2_1(x1),y) -> div2_2(x1,y) r5: a(div2_2(x1,x2),y) -> div2_3(x1,x2,y) r6: a(divides_0(),y) -> divides_1(y) r7: a(divides_1(x1),y) -> divides_2(x1,y) r8: a(filter_0(),y) -> filter_1(y) r9: a(filter_1(x1),y) -> filter_2(x1,y) r10: a(if_0(),y) -> if_1(y) r11: a(if_1(x1),y) -> if_2(x1,y) r12: a(if_2(x1,x2),y) -> if_3(x1,x2,y) r13: a(not_0(),y) -> not_1(y) r14: a(not_1(x1),y) -> not_2(x1,y) r15: a(not2_0(),y) -> not2_1(y) r16: a(s_0(),y) -> s_1(y) r17: a(sieve_0(),y) -> sieve_1(y) r18: divides_2(|0|_0(),s_1(y)) -> true_0() r19: divides_2(s_1(x),s_1(y)) -> div2_3(x,s_1(y),y) r20: div2_3(x,y,|0|_0()) -> divides_2(x,y) r21: div2_3(|0|_0(),y,s_1(z)) -> false_0() r22: div2_3(s_1(x),y,s_1(z)) -> div2_3(x,y,z) r23: filter_2(f,nil_0()) -> nil_0() r24: filter_2(f,cons_2(x,xs)) -> if_3(a(f,x),x,filter_2(f,xs)) r25: if_3(true_0(),x,xs) -> cons_2(x,xs) r26: if_3(false_0(),x,xs) -> xs r27: not_2(f,x) -> not2_1(a(f,x)) r28: not2_1(true_0()) -> false_0() r29: not2_1(false_0()) -> true_0() r30: sieve_1(nil_0()) -> nil_0() r31: sieve_1(cons_2(x,xs)) -> cons_2(x,sieve_1(filter_2(not_1(divides_1(x)),xs))) The estimated dependency graph contains the following SCCs: {p1} {p2} -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: filter_2#(f,cons_2(x,xs)) -> filter_2#(f,xs) and R consists of: r1: a(cons_0(),y) -> cons_1(y) r2: a(cons_1(x1),y) -> cons_2(x1,y) r3: a(div2_0(),y) -> div2_1(y) r4: a(div2_1(x1),y) -> div2_2(x1,y) r5: a(div2_2(x1,x2),y) -> div2_3(x1,x2,y) r6: a(divides_0(),y) -> divides_1(y) r7: a(divides_1(x1),y) -> divides_2(x1,y) r8: a(filter_0(),y) -> filter_1(y) r9: a(filter_1(x1),y) -> filter_2(x1,y) r10: a(if_0(),y) -> if_1(y) r11: a(if_1(x1),y) -> if_2(x1,y) r12: a(if_2(x1,x2),y) -> if_3(x1,x2,y) r13: a(not_0(),y) -> not_1(y) r14: a(not_1(x1),y) -> not_2(x1,y) r15: a(not2_0(),y) -> not2_1(y) r16: a(s_0(),y) -> s_1(y) r17: a(sieve_0(),y) -> sieve_1(y) r18: divides_2(|0|_0(),s_1(y)) -> true_0() r19: divides_2(s_1(x),s_1(y)) -> div2_3(x,s_1(y),y) r20: div2_3(x,y,|0|_0()) -> divides_2(x,y) r21: div2_3(|0|_0(),y,s_1(z)) -> false_0() r22: div2_3(s_1(x),y,s_1(z)) -> div2_3(x,y,z) r23: filter_2(f,nil_0()) -> nil_0() r24: filter_2(f,cons_2(x,xs)) -> if_3(a(f,x),x,filter_2(f,xs)) r25: if_3(true_0(),x,xs) -> cons_2(x,xs) r26: if_3(false_0(),x,xs) -> xs r27: not_2(f,x) -> not2_1(a(f,x)) r28: not2_1(true_0()) -> false_0() r29: not2_1(false_0()) -> true_0() r30: sieve_1(nil_0()) -> nil_0() r31: sieve_1(cons_2(x,xs)) -> cons_2(x,sieve_1(filter_2(not_1(divides_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: 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: filter_2#_A(x1,x2) = 0 cons_2_A(x1,x2) = max{66, x1 + 65, x2 - 75} filter_2#!_A(x1,x2) = max{1, x2 - 67} cons_2!_A(x1,x2) = x2 + 3 2. marked-algebra reduction triple (>=A, >=A!, >A!): max/plus interpretation on natural numbers: filter_2#_A(x1,x2) = 0 cons_2_A(x1,x2) = max{26, x1 - 141, x2 - 2} filter_2#!_A(x1,x2) = max{3, x1 - 1, x2 - 27} cons_2!_A(x1,x2) = max{2, x1 - 226, x2 - 118} partial status: pi(filter_2#) = [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: sieve_1#(cons_2(x,xs)) -> sieve_1#(filter_2(not_1(divides_1(x)),xs)) and R consists of: r1: a(cons_0(),y) -> cons_1(y) r2: a(cons_1(x1),y) -> cons_2(x1,y) r3: a(div2_0(),y) -> div2_1(y) r4: a(div2_1(x1),y) -> div2_2(x1,y) r5: a(div2_2(x1,x2),y) -> div2_3(x1,x2,y) r6: a(divides_0(),y) -> divides_1(y) r7: a(divides_1(x1),y) -> divides_2(x1,y) r8: a(filter_0(),y) -> filter_1(y) r9: a(filter_1(x1),y) -> filter_2(x1,y) r10: a(if_0(),y) -> if_1(y) r11: a(if_1(x1),y) -> if_2(x1,y) r12: a(if_2(x1,x2),y) -> if_3(x1,x2,y) r13: a(not_0(),y) -> not_1(y) r14: a(not_1(x1),y) -> not_2(x1,y) r15: a(not2_0(),y) -> not2_1(y) r16: a(s_0(),y) -> s_1(y) r17: a(sieve_0(),y) -> sieve_1(y) r18: divides_2(|0|_0(),s_1(y)) -> true_0() r19: divides_2(s_1(x),s_1(y)) -> div2_3(x,s_1(y),y) r20: div2_3(x,y,|0|_0()) -> divides_2(x,y) r21: div2_3(|0|_0(),y,s_1(z)) -> false_0() r22: div2_3(s_1(x),y,s_1(z)) -> div2_3(x,y,z) r23: filter_2(f,nil_0()) -> nil_0() r24: filter_2(f,cons_2(x,xs)) -> if_3(a(f,x),x,filter_2(f,xs)) r25: if_3(true_0(),x,xs) -> cons_2(x,xs) r26: if_3(false_0(),x,xs) -> xs r27: not_2(f,x) -> not2_1(a(f,x)) r28: not2_1(true_0()) -> false_0() r29: not2_1(false_0()) -> true_0() r30: sieve_1(nil_0()) -> nil_0() r31: sieve_1(cons_2(x,xs)) -> cons_2(x,sieve_1(filter_2(not_1(divides_1(x)),xs))) The set of usable rules consists of r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17, r18, r19, r20, r21, r22, r23, r24, r25, r26, r27, r28, r29, r30, r31 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: sieve_1#_A(x1) = max{0, x1 - 8} cons_2_A(x1,x2) = max{42, x1 + 7, x2 + 23} filter_2_A(x1,x2) = x2 not_1_A(x1) = max{26, x1 - 1} divides_1_A(x1) = x1 + 27 a_A(x1,x2) = max{63, x1 - 27, x2 + 34} cons_0_A = 0 cons_1_A(x1) = x1 + 34 div2_0_A = 17 div2_1_A(x1) = x1 div2_2_A(x1,x2) = max{x1 - 35, x2} div2_3_A(x1,x2,x3) = max{63, x1 - 62, x2 - 49, x3 - 49} divides_0_A = 0 divides_2_A(x1,x2) = max{63, x1 - 62, x2 - 49} filter_0_A = 0 filter_1_A(x1) = x1 if_0_A = 0 if_1_A(x1) = x1 + 14 if_2_A(x1,x2) = max{x1 - 35, x2 + 34} if_3_A(x1,x2,x3) = max{42, x2 + 7, x3 + 23} not_0_A = 0 not_2_A(x1,x2) = max{63, x1 - 91, x2 - 30} not2_0_A = 0 not2_1_A(x1) = max{63, x1 - 64} s_0_A = 13 s_1_A(x1) = max{35, x1} sieve_0_A = 0 sieve_1_A(x1) = x1 + 18 |0|_0_A = 48 true_0_A = 63 false_0_A = 63 nil_0_A = 2 sieve_1#!_A(x1) = 4 cons_2!_A(x1,x2) = max{17, x2} filter_2!_A(x1,x2) = max{3, x2 - 3} not_1!_A(x1) = max{15, x1 - 3} divides_1!_A(x1) = x1 + 8 a!_A(x1,x2) = max{16, x1, x2 + 13} cons_0!_A = 1 cons_1!_A(x1) = max{0, x1 - 2} div2_0!_A = 18 div2_1!_A(x1) = max{17, x1 - 1} div2_2!_A(x1,x2) = max{10, x1 - 35, x2 - 71} div2_3!_A(x1,x2,x3) = max{9, x1 - 36, x2 - 72, x3 - 73} divides_0!_A = 6 divides_2!_A(x1,x2) = max{9, x1 - 36, x2 - 72} filter_0!_A = 1 filter_1!_A(x1) = max{0, x1 - 4} if_0!_A = 15 if_1!_A(x1) = max{14, x1 - 36} if_2!_A(x1,x2) = max{15, x1 - 35, x2} if_3!_A(x1,x2,x3) = max{21, x3} not_0!_A = 16 not_2!_A(x1,x2) = max{15, x1 - 2, x2} not2_0!_A = 15 not2_1!_A(x1) = max{14, x1 - 64} s_0!_A = 13 s_1!_A(x1) = max{12, x1 - 38} sieve_0!_A = 1 sieve_1!_A(x1) = max{0, x1 - 5} |0|_0!_A = 12 true_0!_A = 11 false_0!_A = 10 nil_0!_A = 1 2. marked-algebra reduction triple (>=A, >=A!, >A!): max/plus interpretation on natural numbers: sieve_1#_A(x1) = max{1, x1 - 3} cons_2_A(x1,x2) = 2 filter_2_A(x1,x2) = max{4, x1 - 64} not_1_A(x1) = max{3, x1 - 1} divides_1_A(x1) = 4 a_A(x1,x2) = max{x1 - 119, x2 + 236} cons_0_A = 90 cons_1_A(x1) = x1 + 90 div2_0_A = 95 div2_1_A(x1) = x1 + 90 div2_2_A(x1,x2) = max{x1 - 29, x2 + 90} div2_3_A(x1,x2,x3) = max{32, x2 - 39, x3 - 39} divides_0_A = 118 divides_2_A(x1,x2) = max{32, x2 - 39} filter_0_A = 90 filter_1_A(x1) = x1 + 90 if_0_A = 92 if_1_A(x1) = x1 + 236 if_2_A(x1,x2) = max{x1 + 117, x2 + 92} if_3_A(x1,x2,x3) = max{4, x3} not_0_A = 117 not_2_A(x1,x2) = max{x1 - 202, x2 + 32} not2_0_A = 0 not2_1_A(x1) = max{32, x1 - 204} s_0_A = 90 s_1_A(x1) = x1 + 38 sieve_0_A = 0 sieve_1_A(x1) = max{4, x1 - 3} |0|_0_A = 38 true_0_A = 32 false_0_A = 1 nil_0_A = 1 sieve_1#!_A(x1) = 1 cons_2!_A(x1,x2) = x2 + 20 filter_2!_A(x1,x2) = max{2, x1 - 4} not_1!_A(x1) = max{26, x1 - 50} divides_1!_A(x1) = max{26, x1 - 39} a!_A(x1,x2) = max{2, x1 - 91, x2} cons_0!_A = 4 cons_1!_A(x1) = max{3, x1 - 1} div2_0!_A = 1 div2_1!_A(x1) = max{3, x1 - 1} div2_2!_A(x1,x2) = max{1, x1 - 39, x2 - 8} div2_3!_A(x1,x2,x3) = max{31, x1 - 39, x2 - 7, x3 - 41} divides_0!_A = 1 divides_2!_A(x1,x2) = max{29, x1 - 40, x2 - 7} filter_0!_A = 4 filter_1!_A(x1) = max{3, x1 - 4} if_0!_A = 148 if_1!_A(x1) = max{147, x1 + 1} if_2!_A(x1,x2) = max{146, x2} if_3!_A(x1,x2,x3) = x3 + 59 not_0!_A = 1 not_2!_A(x1,x2) = max{27, x2 - 1} not2_0!_A = 1 not2_1!_A(x1) = max{28, x1 - 2} s_0!_A = 33 s_1!_A(x1) = max{32, x1 - 40} sieve_0!_A = 1 sieve_1!_A(x1) = max{25, x1 + 15} |0|_0!_A = 31 true_0!_A = 30 false_0!_A = 29 nil_0!_A = 26 partial status: pi(sieve_1#) = [1] pi(cons_2) = [2] pi(filter_2) = [] pi(not_1) = [] pi(divides_1) = [] pi(a) = [2] pi(cons_0) = [] pi(cons_1) = [1] pi(div2_0) = [] pi(div2_1) = [1] pi(div2_2) = [2] pi(div2_3) = [] pi(divides_0) = [] pi(divides_2) = [] pi(filter_0) = [] pi(filter_1) = [1] pi(if_0) = [] pi(if_1) = [] pi(if_2) = [2] pi(if_3) = [3] pi(not_0) = [] pi(not_2) = [2] pi(not2_0) = [] pi(not2_1) = [] pi(s_0) = [] pi(s_1) = [1] pi(sieve_0) = [] pi(sieve_1) = [1] pi(|0|_0) = [] pi(true_0) = [] pi(false_0) = [] pi(nil_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: div2_3#(x,y,|0|_0()) -> divides_2#(x,y) p2: divides_2#(s_1(x),s_1(y)) -> div2_3#(x,s_1(y),y) p3: div2_3#(s_1(x),y,s_1(z)) -> div2_3#(x,y,z) and R consists of: r1: a(cons_0(),y) -> cons_1(y) r2: a(cons_1(x1),y) -> cons_2(x1,y) r3: a(div2_0(),y) -> div2_1(y) r4: a(div2_1(x1),y) -> div2_2(x1,y) r5: a(div2_2(x1,x2),y) -> div2_3(x1,x2,y) r6: a(divides_0(),y) -> divides_1(y) r7: a(divides_1(x1),y) -> divides_2(x1,y) r8: a(filter_0(),y) -> filter_1(y) r9: a(filter_1(x1),y) -> filter_2(x1,y) r10: a(if_0(),y) -> if_1(y) r11: a(if_1(x1),y) -> if_2(x1,y) r12: a(if_2(x1,x2),y) -> if_3(x1,x2,y) r13: a(not_0(),y) -> not_1(y) r14: a(not_1(x1),y) -> not_2(x1,y) r15: a(not2_0(),y) -> not2_1(y) r16: a(s_0(),y) -> s_1(y) r17: a(sieve_0(),y) -> sieve_1(y) r18: divides_2(|0|_0(),s_1(y)) -> true_0() r19: divides_2(s_1(x),s_1(y)) -> div2_3(x,s_1(y),y) r20: div2_3(x,y,|0|_0()) -> divides_2(x,y) r21: div2_3(|0|_0(),y,s_1(z)) -> false_0() r22: div2_3(s_1(x),y,s_1(z)) -> div2_3(x,y,z) r23: filter_2(f,nil_0()) -> nil_0() r24: filter_2(f,cons_2(x,xs)) -> if_3(a(f,x),x,filter_2(f,xs)) r25: if_3(true_0(),x,xs) -> cons_2(x,xs) r26: if_3(false_0(),x,xs) -> xs r27: not_2(f,x) -> not2_1(a(f,x)) r28: not2_1(true_0()) -> false_0() r29: not2_1(false_0()) -> true_0() r30: sieve_1(nil_0()) -> nil_0() r31: sieve_1(cons_2(x,xs)) -> cons_2(x,sieve_1(filter_2(not_1(divides_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: 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: div2_3#_A(x1,x2,x3) = max{0, x1 - 51, x2 - 51, x3 - 50} |0|_0_A = 0 divides_2#_A(x1,x2) = max{0, x1 - 52, x2 - 51} s_1_A(x1) = max{50, x1 + 1} div2_3#!_A(x1,x2,x3) = max{7, x1 - 56, x2 - 55, x3 - 54} |0|_0!_A = 8 divides_2#!_A(x1,x2) = max{7, x1 - 56, x2 - 55} s_1!_A(x1) = max{8, x1 - 53} 2. marked-algebra reduction triple (>=A, >=A!, >A!): max/plus interpretation on natural numbers: div2_3#_A(x1,x2,x3) = max{0, x1 - 4, x2 - 10, x3 - 10} |0|_0_A = 9 divides_2#_A(x1,x2) = max{0, x1 - 11, x2 - 10} s_1_A(x1) = max{9, x1 + 7} div2_3#!_A(x1,x2,x3) = max{1, x1 - 12, x2 - 16, x3 - 10} |0|_0!_A = 12 divides_2#!_A(x1,x2) = max{1, x1 - 13, x2 - 16} s_1!_A(x1) = max{12, x1 - 11} partial status: pi(div2_3#) = [1] pi(|0|_0) = [] pi(divides_2#) = [1] pi(s_1) = [1] The following rules are removed: p2, p3 -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: div2_3#(x,y,|0|_0()) -> divides_2#(x,y) and R consists of: r1: a(cons_0(),y) -> cons_1(y) r2: a(cons_1(x1),y) -> cons_2(x1,y) r3: a(div2_0(),y) -> div2_1(y) r4: a(div2_1(x1),y) -> div2_2(x1,y) r5: a(div2_2(x1,x2),y) -> div2_3(x1,x2,y) r6: a(divides_0(),y) -> divides_1(y) r7: a(divides_1(x1),y) -> divides_2(x1,y) r8: a(filter_0(),y) -> filter_1(y) r9: a(filter_1(x1),y) -> filter_2(x1,y) r10: a(if_0(),y) -> if_1(y) r11: a(if_1(x1),y) -> if_2(x1,y) r12: a(if_2(x1,x2),y) -> if_3(x1,x2,y) r13: a(not_0(),y) -> not_1(y) r14: a(not_1(x1),y) -> not_2(x1,y) r15: a(not2_0(),y) -> not2_1(y) r16: a(s_0(),y) -> s_1(y) r17: a(sieve_0(),y) -> sieve_1(y) r18: divides_2(|0|_0(),s_1(y)) -> true_0() r19: divides_2(s_1(x),s_1(y)) -> div2_3(x,s_1(y),y) r20: div2_3(x,y,|0|_0()) -> divides_2(x,y) r21: div2_3(|0|_0(),y,s_1(z)) -> false_0() r22: div2_3(s_1(x),y,s_1(z)) -> div2_3(x,y,z) r23: filter_2(f,nil_0()) -> nil_0() r24: filter_2(f,cons_2(x,xs)) -> if_3(a(f,x),x,filter_2(f,xs)) r25: if_3(true_0(),x,xs) -> cons_2(x,xs) r26: if_3(false_0(),x,xs) -> xs r27: not_2(f,x) -> not2_1(a(f,x)) r28: not2_1(true_0()) -> false_0() r29: not2_1(false_0()) -> true_0() r30: sieve_1(nil_0()) -> nil_0() r31: sieve_1(cons_2(x,xs)) -> cons_2(x,sieve_1(filter_2(not_1(divides_1(x)),xs))) The estimated dependency graph contains the following SCCs: (no SCCs)