YES We show the termination of the TRS R: 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) 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) 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) -- 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: filter_2#(f,cons_2(y,ys)) -> filtersub_3#(app(f,y),f,cons_2(y,ys)) p4: filter_2#(f,cons_2(y,ys)) -> app#(f,y) p5: filtersub_3#(true_0(),f,cons_2(y,ys)) -> filter_2#(f,ys) p6: 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: filter_2(f,nil_0()) -> nil_0() r9: filter_2(f,cons_2(y,ys)) -> filtersub_3(app(f,y),f,cons_2(y,ys)) r10: filtersub_3(true_0(),f,cons_2(y,ys)) -> cons_2(y,filter_2(f,ys)) r11: filtersub_3(false_0(),f,cons_2(y,ys)) -> filter_2(f,ys) The estimated dependency graph contains the following SCCs: {p1, p2, p3, p4, p5, p6} -- 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: filter_2(f,nil_0()) -> nil_0() r9: filter_2(f,cons_2(y,ys)) -> filtersub_3(app(f,y),f,cons_2(y,ys)) r10: filtersub_3(true_0(),f,cons_2(y,ys)) -> cons_2(y,filter_2(f,ys)) r11: filtersub_3(false_0(),f,cons_2(y,ys)) -> filter_2(f,ys) The set of usable rules consists of r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11 Take the reduction pair: generalized weighted path order (>=MG, >G) base order 1: max/plus interpretation on natural numbers: app#_A(x1,x2) = 4 filter_1_A(x1) = max{10, x1 - 6} filter_2#_A(x1,x2) = 4 cons_2_A(x1,x2) = max{9, x2 + 5} filtersub_2_A(x1,x2) = max{7, x2 - 6} filtersub_3#_A(x1,x2,x3) = 4 false_0_A = 7 app_A(x1,x2) = max{10, x2 + 6} true_0_A = 7 cons_0_A = 1 cons_1_A(x1) = max{8, x1 - 6} filter_0_A = 1 filter_2_A(x1,x2) = max{4, x2} filtersub_0_A = 1 filtersub_1_A(x1) = max{0, x1 - 6} filtersub_3_A(x1,x2,x3) = max{8, x3} nil_0_A = 5 base order 2: marked-algebra reduction triple (>=A, >=A!, >A!): max/plus interpretation on natural numbers: app#_A(x1,x2) = max{24, x1 - 43, x2 - 11} filter_1_A(x1) = max{42, x1 + 23} filter_2#_A(x1,x2) = max{0, x1 - 43, x2 - 11} cons_2_A(x1,x2) = max{35, x1 + 10, x2} filtersub_2_A(x1,x2) = max{35, x1 - 14, x2 + 23} filtersub_3#_A(x1,x2,x3) = max{24, x1 - 57, x2 - 43, x3 - 11} false_0_A = 56 app_A(x1,x2) = max{58, x1 - 24, x2 + 34} true_0_A = 56 cons_0_A = 23 cons_1_A(x1) = max{58, x1 + 34} filter_0_A = 23 filter_2_A(x1,x2) = max{58, x2} filtersub_0_A = 23 filtersub_1_A(x1) = x1 + 23 filtersub_3_A(x1,x2,x3) = max{58, x3} nil_0_A = 1 app#!_A(x1,x2) = max{5, x1 - 36, x2 - 41} filter_1!_A(x1) = 2 filter_2#!_A(x1,x2) = max{5, x1 - 14, x2 - 50} cons_2!_A(x1,x2) = max{7, x1 + 4, x2 - 51} filtersub_2!_A(x1,x2) = max{2, x1 - 111, x2 - 57} filtersub_3#!_A(x1,x2,x3) = max{5, x1 - 74, x2 - 14, x3 - 50} false_0!_A = 1 app!_A(x1,x2) = max{34, x1 - 57, x2 - 52} true_0!_A = 6 cons_0!_A = 1 cons_1!_A(x1) = max{8, x1 - 53} filter_0!_A = 1 filter_2!_A(x1,x2) = max{34, x1 - 57, x2 - 2} filtersub_0!_A = 2 filtersub_1!_A(x1) = max{1, x1 - 112} filtersub_3!_A(x1,x2,x3) = max{3, x1 - 110, x2 - 58} 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) = [2] pi(filtersub_0) = [] pi(filtersub_1) = [] pi(filtersub_3) = [3] pi(nil_0) = [] The following rules are removed: p1 -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: filter_2#(f,cons_2(y,ys)) -> app#(f,y) p2: app#(filtersub_2(x1,x2),y) -> filtersub_3#(x1,x2,y) p3: filtersub_3#(false_0(),f,cons_2(y,ys)) -> filter_2#(f,ys) p4: filter_2#(f,cons_2(y,ys)) -> filtersub_3#(app(f,y),f,cons_2(y,ys)) p5: 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: filter_2(f,nil_0()) -> nil_0() r9: filter_2(f,cons_2(y,ys)) -> filtersub_3(app(f,y),f,cons_2(y,ys)) r10: filtersub_3(true_0(),f,cons_2(y,ys)) -> cons_2(y,filter_2(f,ys)) r11: filtersub_3(false_0(),f,cons_2(y,ys)) -> filter_2(f,ys) The estimated dependency graph contains the following SCCs: {p1, p2, p3, p4, p5} -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: filter_2#(f,cons_2(y,ys)) -> app#(f,y) p2: app#(filtersub_2(x1,x2),y) -> filtersub_3#(x1,x2,y) p3: filtersub_3#(true_0(),f,cons_2(y,ys)) -> filter_2#(f,ys) p4: filter_2#(f,cons_2(y,ys)) -> filtersub_3#(app(f,y),f,cons_2(y,ys)) p5: 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: filter_2(f,nil_0()) -> nil_0() r9: filter_2(f,cons_2(y,ys)) -> filtersub_3(app(f,y),f,cons_2(y,ys)) r10: filtersub_3(true_0(),f,cons_2(y,ys)) -> cons_2(y,filter_2(f,ys)) r11: filtersub_3(false_0(),f,cons_2(y,ys)) -> filter_2(f,ys) The set of usable rules consists of r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11 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{19, x1 - 7} cons_2_A(x1,x2) = 0 app#_A(x1,x2) = max{19, x1 - 8} filtersub_2_A(x1,x2) = x2 + 13 filtersub_3#_A(x1,x2,x3) = max{19, x2 - 7} true_0_A = 2 app_A(x1,x2) = max{20, x1 - 8, x2 + 17} false_0_A = 18 cons_0_A = 19 cons_1_A(x1) = max{18, x1 - 1} filter_0_A = 7 filter_1_A(x1) = x1 + 12 filter_2_A(x1,x2) = max{6, x1 + 4} filtersub_0_A = 7 filtersub_1_A(x1) = x1 + 16 filtersub_3_A(x1,x2,x3) = max{6, x2 + 4} nil_0_A = 5 base order 2: marked-algebra reduction triple (>=A, >=A!, >A!): max/plus interpretation on natural numbers: filter_2#_A(x1,x2) = max{x1 + 3, x2 - 11} cons_2_A(x1,x2) = max{12, x1 + 2, x2} app#_A(x1,x2) = max{2, x1 - 5, x2 - 11} filtersub_2_A(x1,x2) = max{x1 - 9, x2 + 8} filtersub_3#_A(x1,x2,x3) = max{x1 - 29, x2 + 3, x3 - 11} true_0_A = 17 app_A(x1,x2) = max{32, x1 - 9, x2 + 11} false_0_A = 0 cons_0_A = 0 cons_1_A(x1) = max{12, x1 + 11} filter_0_A = 8 filter_1_A(x1) = x1 + 8 filter_2_A(x1,x2) = max{32, x1 - 27, x2 + 1} filtersub_0_A = 8 filtersub_1_A(x1) = max{12, x1 + 8} filtersub_3_A(x1,x2,x3) = max{32, x1 - 18, x2 - 27, x3 + 1} nil_0_A = 33 filter_2#!_A(x1,x2) = max{x1 + 15, x2 - 24} cons_2!_A(x1,x2) = x1 + 12 app#!_A(x1,x2) = max{15, x1 + 8, x2 - 23} filtersub_2!_A(x1,x2) = max{0, x1 - 34, x2 - 2} filtersub_3#!_A(x1,x2,x3) = max{x1 - 33, x2 + 15, x3 - 24} true_0!_A = 0 app!_A(x1,x2) = max{11, x1 - 13, x2 - 1} false_0!_A = 0 cons_0!_A = 12 cons_1!_A(x1) = max{11, x1 - 2} filter_0!_A = 1 filter_1!_A(x1) = max{0, x1 - 2} filter_2!_A(x1,x2) = 0 filtersub_0!_A = 1 filtersub_1!_A(x1) = max{0, x1 - 35} filtersub_3!_A(x1,x2,x3) = 0 nil_0!_A = 0 partial status: pi(filter_2#) = [] pi(cons_2) = [] pi(app#) = [] pi(filtersub_2) = [2] pi(filtersub_3#) = [] pi(true_0) = [] pi(app) = [] pi(false_0) = [] pi(cons_0) = [] pi(cons_1) = [] pi(filter_0) = [] pi(filter_1) = [1] pi(filter_2) = [1] pi(filtersub_0) = [] pi(filtersub_1) = [1] pi(filtersub_3) = [2] pi(nil_0) = [] The following rules are removed: p2 -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: filter_2#(f,cons_2(y,ys)) -> app#(f,y) p2: filtersub_3#(true_0(),f,cons_2(y,ys)) -> filter_2#(f,ys) p3: filter_2#(f,cons_2(y,ys)) -> filtersub_3#(app(f,y),f,cons_2(y,ys)) p4: 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: filter_2(f,nil_0()) -> nil_0() r9: filter_2(f,cons_2(y,ys)) -> filtersub_3(app(f,y),f,cons_2(y,ys)) r10: filtersub_3(true_0(),f,cons_2(y,ys)) -> cons_2(y,filter_2(f,ys)) r11: filtersub_3(false_0(),f,cons_2(y,ys)) -> filter_2(f,ys) The estimated dependency graph contains the following SCCs: {p2, p3, p4} -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: filtersub_3#(true_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#(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: filter_2(f,nil_0()) -> nil_0() r9: filter_2(f,cons_2(y,ys)) -> filtersub_3(app(f,y),f,cons_2(y,ys)) r10: filtersub_3(true_0(),f,cons_2(y,ys)) -> cons_2(y,filter_2(f,ys)) r11: filtersub_3(false_0(),f,cons_2(y,ys)) -> filter_2(f,ys) The set of usable rules consists of r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11 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{5, x1} true_0_A = 6 cons_2_A(x1,x2) = 2 filter_2#_A(x1,x2) = 5 app_A(x1,x2) = 4 false_0_A = 6 cons_0_A = 0 cons_1_A(x1) = 0 filter_0_A = 1 filter_1_A(x1) = 0 filter_2_A(x1,x2) = 3 filtersub_0_A = 0 filtersub_1_A(x1) = 0 filtersub_2_A(x1,x2) = 0 filtersub_3_A(x1,x2,x3) = 3 nil_0_A = 1 base order 2: marked-algebra reduction triple (>=A, >=A!, >A!): max/plus interpretation on natural numbers: filtersub_3#_A(x1,x2,x3) = max{1, x2 - 1} true_0_A = 7 cons_2_A(x1,x2) = max{0, x2 - 7} filter_2#_A(x1,x2) = max{1, x1 - 1} app_A(x1,x2) = max{5, x2 - 5} false_0_A = 5 cons_0_A = 0 cons_1_A(x1) = 1 filter_0_A = 0 filter_1_A(x1) = max{1, x1 - 5} filter_2_A(x1,x2) = 1 filtersub_0_A = 0 filtersub_1_A(x1) = max{1, x1 - 6} filtersub_2_A(x1,x2) = max{1, x2 - 5} filtersub_3_A(x1,x2,x3) = 1 nil_0_A = 1 filtersub_3#!_A(x1,x2,x3) = 0 true_0!_A = 0 cons_2!_A(x1,x2) = max{x1 + 2, x2 + 1} filter_2#!_A(x1,x2) = max{9, x1 + 2} app!_A(x1,x2) = max{8, x1 + 4, x2 + 6} false_0!_A = 1 cons_0!_A = 1 cons_1!_A(x1) = max{8, x1 - 1} filter_0!_A = 6 filter_1!_A(x1) = max{5, x1 - 1} filter_2!_A(x1,x2) = x1 + 4 filtersub_0!_A = 1 filtersub_1!_A(x1) = 7 filtersub_2!_A(x1,x2) = max{0, x1 - 3} filtersub_3!_A(x1,x2,x3) = x2 + 4 nil_0!_A = 5 partial status: pi(filtersub_3#) = [1] pi(true_0) = [] pi(cons_2) = [] pi(filter_2#) = [] pi(app) = [] pi(false_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(nil_0) = [] The following rules are removed: p1, p2, p3 Then no dependency pair remains.