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: 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: app#_A(x1,x2) = max{0, x1 - 3, x2 - 63} filter_1_A(x1) = x1 + 2 filter_2#_A(x1,x2) = max{0, x1 - 3, x2 - 63} cons_2_A(x1,x2) = max{61, x1 + 39, x2 + 31} filtersub_2_A(x1,x2) = max{x1 - 70, x2 + 2} filtersub_3#_A(x1,x2,x3) = max{0, x1 - 73, x2 - 3, x3 - 63} false_0_A = 50 app_A(x1,x2) = max{72, x1 - 10, x2 + 49} true_0_A = 72 cons_0_A = 4 cons_1_A(x1) = x1 + 49 filter_0_A = 2 filter_2_A(x1,x2) = max{50, x2 + 20} filtersub_0_A = 8 filtersub_1_A(x1) = max{9, x1 + 2} filtersub_3_A(x1,x2,x3) = max{60, x3 + 20} nil_0_A = 51 app#!_A(x1,x2) = max{x1 + 77, x2 - 31} filter_1!_A(x1) = 77 filter_2#!_A(x1,x2) = max{x1 + 78, x2 - 32} cons_2!_A(x1,x2) = max{0, x1 - 62, x2 - 113} filtersub_2!_A(x1,x2) = max{5, x1 - 187, x2 - 197} filtersub_3#!_A(x1,x2,x3) = max{x2 + 78, x3 - 63} false_0!_A = 79 app!_A(x1,x2) = max{x1 - 3, x2 + 81} true_0!_A = 1 cons_0!_A = 1 cons_1!_A(x1) = max{2, x1 + 1} filter_0!_A = 78 filter_2!_A(x1,x2) = max{80, x1 - 196, x2 - 62} filtersub_0!_A = 1 filtersub_1!_A(x1) = max{4, x1 - 186} filtersub_3!_A(x1,x2,x3) = max{80, x1 - 186, x2 - 196, x3 - 93} nil_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: app#_A(x1,x2) = x1 + x2 filter_1_A(x1) = x1 + 1 filter_2#_A(x1,x2) = x1 + x2 cons_2_A(x1,x2) = x1 + x2 + 4 filtersub_2_A(x1,x2) = x1 + x2 + 1 filtersub_3#_A(x1,x2,x3) = x2 + x3 false_0_A() = 1 app_A(x1,x2) = x1 + x2 + 1 true_0_A() = 1 cons_0_A() = 2 cons_1_A(x1) = x1 + 3 filter_0_A() = 1 filter_2_A(x1,x2) = x1 + x2 + 1 filtersub_0_A() = 1 filtersub_1_A(x1) = x1 + 1 filtersub_3_A(x1,x2,x3) = x2 + x3 + 1 nil_0_A() = 1 app#!_A(x1,x2) = x1 + x2 + 8 filter_1!_A(x1) = 6 filter_2#!_A(x1,x2) = x2 + 3 cons_2!_A(x1,x2) = x2 + 6 filtersub_2!_A(x1,x2) = 0 filtersub_3#!_A(x1,x2,x3) = x3 + 1 false_0!_A() = 0 app!_A(x1,x2) = x1 + 10 true_0!_A() = 0 cons_0!_A() = 0 cons_1!_A(x1) = 13 filter_0!_A() = 0 filter_2!_A(x1,x2) = x2 + 5 filtersub_0!_A() = 0 filtersub_1!_A(x1) = 1 filtersub_3!_A(x1,x2,x3) = x3 + 4 nil_0!_A() = 0 partial status: pi(app#) = [] pi(filter_1) = [1] pi(filter_2#) = [] pi(cons_2) = [] pi(filtersub_2) = [] pi(filtersub_3#) = [] pi(false_0) = [] pi(app) = [2] pi(true_0) = [] pi(cons_0) = [] pi(cons_1) = [1] pi(filter_0) = [] pi(filter_2) = [] pi(filtersub_0) = [] pi(filtersub_1) = [] pi(filtersub_3) = [3] pi(nil_0) = [] The following rules are removed: p1, p2, p3, p4, p5, p6 Then no dependency pair remains.