YES We show the termination of the TRS R: app(app(neq(),|0|()),|0|()) -> false() app(app(neq(),|0|()),app(s(),y)) -> true() app(app(neq(),app(s(),x)),|0|()) -> true() app(app(neq(),app(s(),x)),app(s(),y)) -> app(app(neq(),x),y) app(app(filter(),f),nil()) -> nil() app(app(filter(),f),app(app(cons(),y),ys)) -> app(app(app(filtersub(),app(f,y)),f),app(app(cons(),y),ys)) app(app(app(filtersub(),true()),f),app(app(cons(),y),ys)) -> app(app(cons(),y),app(app(filter(),f),ys)) app(app(app(filtersub(),false()),f),app(app(cons(),y),ys)) -> app(app(filter(),f),ys) nonzero() -> app(filter(),app(neq(),|0|())) 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(filtersub_0(),y) -> filtersub_1(y) app(filtersub_1(x1),y) -> filtersub_2(x1,y) app(filtersub_2(x1,x2),y) -> filtersub_3(x1,x2,y) app(neq_0(),y) -> neq_1(y) app(neq_1(x1),y) -> neq_2(x1,y) app(s_0(),y) -> s_1(y) neq_2(|0|_0(),|0|_0()) -> false_0() neq_2(|0|_0(),s_1(y)) -> true_0() neq_2(s_1(x),|0|_0()) -> true_0() neq_2(s_1(x),s_1(y)) -> neq_2(x,y) filter_2(f,nil_0()) -> nil_0() filter_2(f,cons_2(y,ys)) -> filtersub_3(app(f,y),f,cons_2(y,ys)) filtersub_3(true_0(),f,cons_2(y,ys)) -> cons_2(y,filter_2(f,ys)) filtersub_3(false_0(),f,cons_2(y,ys)) -> filter_2(f,ys) nonzero_0() -> filter_1(neq_1(|0|_0())) -- 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#(filtersub_2(x1,x2),y) -> filtersub_3#(x1,x2,y) p3: app#(neq_1(x1),y) -> neq_2#(x1,y) p4: neq_2#(s_1(x),s_1(y)) -> neq_2#(x,y) p5: filter_2#(f,cons_2(y,ys)) -> filtersub_3#(app(f,y),f,cons_2(y,ys)) p6: filter_2#(f,cons_2(y,ys)) -> app#(f,y) p7: filtersub_3#(true_0(),f,cons_2(y,ys)) -> filter_2#(f,ys) p8: filtersub_3#(false_0(),f,cons_2(y,ys)) -> filter_2#(f,ys) 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(filtersub_0(),y) -> filtersub_1(y) r6: app(filtersub_1(x1),y) -> filtersub_2(x1,y) r7: app(filtersub_2(x1,x2),y) -> filtersub_3(x1,x2,y) r8: app(neq_0(),y) -> neq_1(y) r9: app(neq_1(x1),y) -> neq_2(x1,y) r10: app(s_0(),y) -> s_1(y) r11: neq_2(|0|_0(),|0|_0()) -> false_0() r12: neq_2(|0|_0(),s_1(y)) -> true_0() r13: neq_2(s_1(x),|0|_0()) -> true_0() r14: neq_2(s_1(x),s_1(y)) -> neq_2(x,y) r15: filter_2(f,nil_0()) -> nil_0() r16: filter_2(f,cons_2(y,ys)) -> filtersub_3(app(f,y),f,cons_2(y,ys)) r17: filtersub_3(true_0(),f,cons_2(y,ys)) -> cons_2(y,filter_2(f,ys)) r18: filtersub_3(false_0(),f,cons_2(y,ys)) -> filter_2(f,ys) r19: nonzero_0() -> filter_1(neq_1(|0|_0())) The estimated dependency graph contains the following SCCs: {p1, p2, p5, p6, p7, p8} {p4} -- 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(y,ys)) -> app#(f,y) p3: app#(filtersub_2(x1,x2),y) -> filtersub_3#(x1,x2,y) p4: filtersub_3#(false_0(),f,cons_2(y,ys)) -> filter_2#(f,ys) p5: filter_2#(f,cons_2(y,ys)) -> filtersub_3#(app(f,y),f,cons_2(y,ys)) p6: filtersub_3#(true_0(),f,cons_2(y,ys)) -> filter_2#(f,ys) 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(filtersub_0(),y) -> filtersub_1(y) r6: app(filtersub_1(x1),y) -> filtersub_2(x1,y) r7: app(filtersub_2(x1,x2),y) -> filtersub_3(x1,x2,y) r8: app(neq_0(),y) -> neq_1(y) r9: app(neq_1(x1),y) -> neq_2(x1,y) r10: app(s_0(),y) -> s_1(y) r11: neq_2(|0|_0(),|0|_0()) -> false_0() r12: neq_2(|0|_0(),s_1(y)) -> true_0() r13: neq_2(s_1(x),|0|_0()) -> true_0() r14: neq_2(s_1(x),s_1(y)) -> neq_2(x,y) r15: filter_2(f,nil_0()) -> nil_0() r16: filter_2(f,cons_2(y,ys)) -> filtersub_3(app(f,y),f,cons_2(y,ys)) r17: filtersub_3(true_0(),f,cons_2(y,ys)) -> cons_2(y,filter_2(f,ys)) r18: filtersub_3(false_0(),f,cons_2(y,ys)) -> filter_2(f,ys) r19: nonzero_0() -> filter_1(neq_1(|0|_0())) 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 Take the reduction pair: generalized weighted path order (>=MG, >G) base order 1: max/plus interpretation on natural numbers: app#_A(x1,x2) = 3 filter_1_A(x1) = max{6, x1 - 3} filter_2#_A(x1,x2) = 3 cons_2_A(x1,x2) = max{2, x2 - 6} filtersub_2_A(x1,x2) = 6 filtersub_3#_A(x1,x2,x3) = 3 false_0_A = 6 app_A(x1,x2) = max{9, x2 + 3} true_0_A = 4 cons_0_A = 2 cons_1_A(x1) = max{1, x1 - 3} filter_0_A = 1 filter_2_A(x1,x2) = 5 filtersub_0_A = 1 filtersub_1_A(x1) = max{7, x1 - 3} filtersub_3_A(x1,x2,x3) = 5 neq_0_A = 1 neq_1_A(x1) = max{0, x1 - 3} neq_2_A(x1,x2) = max{7, x2 - 9} s_0_A = 1 s_1_A(x1) = max{8, x1 + 1} |0|_0_A = 2 nil_0_A = 1 base order 2: marked-algebra reduction triple (>=A, >=A!, >A!): max/plus interpretation on natural numbers: app#_A(x1,x2) = max{0, x1 - 5, x2 - 2} filter_1_A(x1) = max{4, x1 + 2} filter_2#_A(x1,x2) = max{0, x1 - 5, x2 - 2} cons_2_A(x1,x2) = max{x1, x2} filtersub_2_A(x1,x2) = max{4, x1 - 1, x2 + 2} filtersub_3#_A(x1,x2,x3) = max{0, x1 - 16, x2 - 5, x3 - 2} false_0_A = 9 app_A(x1,x2) = max{15, x1 - 5, x2 + 8} true_0_A = 9 cons_0_A = 4 cons_1_A(x1) = x1 + 6 filter_0_A = 4 filter_2_A(x1,x2) = max{9, x1 - 3, x2 + 4} filtersub_0_A = 4 filtersub_1_A(x1) = x1 + 4 filtersub_3_A(x1,x2,x3) = max{9, x1 - 10, x2 - 3, x3 + 4} neq_0_A = 4 neq_1_A(x1) = max{0, x1 - 3} neq_2_A(x1,x2) = max{9, x1 - 10, x2 - 10} s_0_A = 4 s_1_A(x1) = max{9, x1} |0|_0_A = 0 nil_0_A = 0 app#!_A(x1,x2) = max{8, x1 + 6} filter_1!_A(x1) = max{0, x1 - 1} filter_2#!_A(x1,x2) = max{9, x1 + 7} cons_2!_A(x1,x2) = 0 filtersub_2!_A(x1,x2) = max{0, x2 - 1} filtersub_3#!_A(x1,x2,x3) = max{9, x2 + 7} false_0!_A = 0 app!_A(x1,x2) = max{15, x1 - 7, x2 - 1} true_0!_A = 3 cons_0!_A = 1 cons_1!_A(x1) = max{0, x1 - 2} filter_0!_A = 1 filter_2!_A(x1,x2) = 0 filtersub_0!_A = 1 filtersub_1!_A(x1) = max{0, x1 - 2} filtersub_3!_A(x1,x2,x3) = 0 neq_0!_A = 1 neq_1!_A(x1) = max{2, x1 - 2} neq_2!_A(x1,x2) = max{14, x1 + 4, x2 + 5} s_0!_A = 1 s_1!_A(x1) = x1 + 4 |0|_0!_A = 0 nil_0!_A = 0 partial status: pi(app#) = [] pi(filter_1) = [] pi(filter_2#) = [] pi(cons_2) = [] pi(filtersub_2) = [] pi(filtersub_3#) = [] pi(false_0) = [] pi(app) = [] pi(true_0) = [] pi(cons_0) = [] pi(cons_1) = [] pi(filter_0) = [] pi(filter_2) = [] pi(filtersub_0) = [] pi(filtersub_1) = [] pi(filtersub_3) = [] pi(neq_0) = [] pi(neq_1) = [] pi(neq_2) = [] pi(s_0) = [] pi(s_1) = [1] pi(|0|_0) = [] pi(nil_0) = [] The following rules are removed: p1, p2, p3 -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: filtersub_3#(false_0(),f,cons_2(y,ys)) -> filter_2#(f,ys) p2: filter_2#(f,cons_2(y,ys)) -> filtersub_3#(app(f,y),f,cons_2(y,ys)) p3: filtersub_3#(true_0(),f,cons_2(y,ys)) -> filter_2#(f,ys) 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(filtersub_0(),y) -> filtersub_1(y) r6: app(filtersub_1(x1),y) -> filtersub_2(x1,y) r7: app(filtersub_2(x1,x2),y) -> filtersub_3(x1,x2,y) r8: app(neq_0(),y) -> neq_1(y) r9: app(neq_1(x1),y) -> neq_2(x1,y) r10: app(s_0(),y) -> s_1(y) r11: neq_2(|0|_0(),|0|_0()) -> false_0() r12: neq_2(|0|_0(),s_1(y)) -> true_0() r13: neq_2(s_1(x),|0|_0()) -> true_0() r14: neq_2(s_1(x),s_1(y)) -> neq_2(x,y) r15: filter_2(f,nil_0()) -> nil_0() r16: filter_2(f,cons_2(y,ys)) -> filtersub_3(app(f,y),f,cons_2(y,ys)) r17: filtersub_3(true_0(),f,cons_2(y,ys)) -> cons_2(y,filter_2(f,ys)) r18: filtersub_3(false_0(),f,cons_2(y,ys)) -> filter_2(f,ys) r19: nonzero_0() -> filter_1(neq_1(|0|_0())) 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: filtersub_3#(false_0(),f,cons_2(y,ys)) -> filter_2#(f,ys) p2: filter_2#(f,cons_2(y,ys)) -> filtersub_3#(app(f,y),f,cons_2(y,ys)) p3: filtersub_3#(true_0(),f,cons_2(y,ys)) -> filter_2#(f,ys) 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(filtersub_0(),y) -> filtersub_1(y) r6: app(filtersub_1(x1),y) -> filtersub_2(x1,y) r7: app(filtersub_2(x1,x2),y) -> filtersub_3(x1,x2,y) r8: app(neq_0(),y) -> neq_1(y) r9: app(neq_1(x1),y) -> neq_2(x1,y) r10: app(s_0(),y) -> s_1(y) r11: neq_2(|0|_0(),|0|_0()) -> false_0() r12: neq_2(|0|_0(),s_1(y)) -> true_0() r13: neq_2(s_1(x),|0|_0()) -> true_0() r14: neq_2(s_1(x),s_1(y)) -> neq_2(x,y) r15: filter_2(f,nil_0()) -> nil_0() r16: filter_2(f,cons_2(y,ys)) -> filtersub_3(app(f,y),f,cons_2(y,ys)) r17: filtersub_3(true_0(),f,cons_2(y,ys)) -> cons_2(y,filter_2(f,ys)) r18: filtersub_3(false_0(),f,cons_2(y,ys)) -> filter_2(f,ys) r19: nonzero_0() -> filter_1(neq_1(|0|_0())) 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 Take the reduction pair: generalized weighted path order (>=MG, >G) base order 1: max/plus interpretation on natural numbers: filtersub_3#_A(x1,x2,x3) = max{1, x3 - 1} false_0_A = 10 cons_2_A(x1,x2) = x2 + 5 filter_2#_A(x1,x2) = max{2, x2} app_A(x1,x2) = max{10, x1 + 2, x2 + 6} true_0_A = 7 cons_0_A = 1 cons_1_A(x1) = max{4, x1 - 4} filter_0_A = 1 filter_1_A(x1) = max{10, x1 - 8} filter_2_A(x1,x2) = x2 + 3 filtersub_0_A = 1 filtersub_1_A(x1) = max{10, x1 - 4} filtersub_2_A(x1,x2) = max{0, x1 - 3, x2 - 3} filtersub_3_A(x1,x2,x3) = max{4, x3 + 3} neq_0_A = 4 neq_1_A(x1) = max{5, x1 - 22} neq_2_A(x1,x2) = max{9, x1 - 21, x2 - 9} s_0_A = 1 s_1_A(x1) = max{8, x1 + 5} |0|_0_A = 20 nil_0_A = 4 base order 2: marked-algebra reduction triple (>=A, >=A!, >A!): max/plus interpretation on natural numbers: filtersub_3#_A(x1,x2,x3) = max{0, x1 - 49, x2 - 3, x3 - 30} false_0_A = 8 cons_2_A(x1,x2) = max{28, x1 - 7, x2} filter_2#_A(x1,x2) = max{0, x1 - 3, x2 - 30} app_A(x1,x2) = max{48, x1 - 8, x2 + 7} true_0_A = 7 cons_0_A = 1 cons_1_A(x1) = x1 + 1 filter_0_A = 1 filter_1_A(x1) = max{1, x1 - 1} filter_2_A(x1,x2) = max{8, x1 - 9, x2 + 1} filtersub_0_A = 3 filtersub_1_A(x1) = x1 + 7 filtersub_2_A(x1,x2) = max{x1 - 1, x2 + 1} filtersub_3_A(x1,x2,x3) = max{8, x1 - 49, x2 - 9, x3 + 1} neq_0_A = 1 neq_1_A(x1) = max{0, x1 - 6} neq_2_A(x1,x2) = max{48, x1 - 14, x2 - 9} s_0_A = 1 s_1_A(x1) = max{8, x1} |0|_0_A = 6 nil_0_A = 9 filtersub_3#!_A(x1,x2,x3) = max{0, x1 - 8} false_0!_A = 14 cons_2!_A(x1,x2) = max{12, x1 - 54, x2 - 105} filter_2#!_A(x1,x2) = max{13, x2 + 11} app!_A(x1,x2) = max{17, x1 - 2, x2 + 3} true_0!_A = 4 cons_0!_A = 1 cons_1!_A(x1) = max{4, x1 - 55} filter_0!_A = 17 filter_1!_A(x1) = max{16, x1 - 1} filter_2!_A(x1,x2) = max{15, x1 - 4, x2 - 10} filtersub_0!_A = 1 filtersub_1!_A(x1) = 17 filtersub_2!_A(x1,x2) = max{4, x2 - 97} filtersub_3!_A(x1,x2,x3) = max{14, x1 - 103, x2 - 96, x3 - 46} neq_0!_A = 1 neq_1!_A(x1) = max{2, x1 - 10} neq_2!_A(x1,x2) = max{4, x1 - 9, x2 - 11} s_0!_A = 1 s_1!_A(x1) = max{5, x1 - 10} |0|_0!_A = 15 nil_0!_A = 0 partial status: pi(filtersub_3#) = [] pi(false_0) = [] pi(cons_2) = [2] pi(filter_2#) = [2] pi(app) = [2] pi(true_0) = [] pi(cons_0) = [] pi(cons_1) = [] pi(filter_0) = [] pi(filter_1) = [] pi(filter_2) = [] pi(filtersub_0) = [] pi(filtersub_1) = [] pi(filtersub_2) = [] pi(filtersub_3) = [] pi(neq_0) = [] pi(neq_1) = [] pi(neq_2) = [] pi(s_0) = [] pi(s_1) = [1] pi(|0|_0) = [] pi(nil_0) = [] 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: neq_2#(s_1(x),s_1(y)) -> neq_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(filtersub_0(),y) -> filtersub_1(y) r6: app(filtersub_1(x1),y) -> filtersub_2(x1,y) r7: app(filtersub_2(x1,x2),y) -> filtersub_3(x1,x2,y) r8: app(neq_0(),y) -> neq_1(y) r9: app(neq_1(x1),y) -> neq_2(x1,y) r10: app(s_0(),y) -> s_1(y) r11: neq_2(|0|_0(),|0|_0()) -> false_0() r12: neq_2(|0|_0(),s_1(y)) -> true_0() r13: neq_2(s_1(x),|0|_0()) -> true_0() r14: neq_2(s_1(x),s_1(y)) -> neq_2(x,y) r15: filter_2(f,nil_0()) -> nil_0() r16: filter_2(f,cons_2(y,ys)) -> filtersub_3(app(f,y),f,cons_2(y,ys)) r17: filtersub_3(true_0(),f,cons_2(y,ys)) -> cons_2(y,filter_2(f,ys)) r18: filtersub_3(false_0(),f,cons_2(y,ys)) -> filter_2(f,ys) r19: nonzero_0() -> filter_1(neq_1(|0|_0())) The set of usable rules consists of r19 Take the reduction pair: generalized weighted path order (>=MG, >G) base order 1: max/plus interpretation on natural numbers: neq_2#_A(x1,x2) = 1 s_1_A(x1) = max{4, x1 - 38} nonzero_0_A = 17 filter_1_A(x1) = max{3, x1 - 17} neq_1_A(x1) = max{2, x1 - 17} |0|_0_A = 16 base order 2: marked-algebra reduction triple (>=A, >=A!, >A!): max/plus interpretation on natural numbers: neq_2#_A(x1,x2) = max{x1, x2} s_1_A(x1) = x1 + 12 nonzero_0_A = 17 filter_1_A(x1) = max{17, x1 - 13} neq_1_A(x1) = max{12, x1 + 4} |0|_0_A = 8 neq_2#!_A(x1,x2) = max{3, x1 - 47, x2 - 8} s_1!_A(x1) = max{2, x1 - 47} nonzero_0!_A = 3 filter_1!_A(x1) = max{2, x1 - 46} neq_1!_A(x1) = max{2, x1 - 46} |0|_0!_A = 2 partial status: pi(neq_2#) = [] pi(s_1) = [] pi(nonzero_0) = [] pi(filter_1) = [] pi(neq_1) = [] pi(|0|_0) = [] The following rules are removed: p1 Then no dependency pair remains.