YES We show the termination of the TRS R: app(perfectp(),|0|()) -> false() app(perfectp(),app(s(),x)) -> app(app(app(app(f(),x),app(s(),|0|())),app(s(),x)),app(s(),x)) app(app(app(app(f(),|0|()),y),|0|()),u) -> true() app(app(app(app(f(),|0|()),y),app(s(),z)),u) -> false() app(app(app(app(f(),app(s(),x)),|0|()),z),u) -> app(app(app(app(f(),x),u),app(app(minus(),z),app(s(),x))),u) app(app(app(app(f(),app(s(),x)),app(s(),y)),z),u) -> app(app(app(if(),app(app(le(),x),y)),app(app(app(app(f(),app(s(),x)),app(app(minus(),y),x)),z),u)),app(app(app(app(f(),x),u),z),u)) app(app(map(),|fun|),nil()) -> nil() app(app(map(),|fun|),app(app(cons(),x),xs)) -> app(app(cons(),app(|fun|,x)),app(app(map(),|fun|),xs)) app(app(filter(),|fun|),nil()) -> nil() app(app(filter(),|fun|),app(app(cons(),x),xs)) -> app(app(app(app(filter2(),app(|fun|,x)),|fun|),x),xs) app(app(app(app(filter2(),true()),|fun|),x),xs) -> app(app(cons(),x),app(app(filter(),|fun|),xs)) app(app(app(app(filter2(),false()),|fun|),x),xs) -> app(app(filter(),|fun|),xs) 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(f_0(),y) -> f_1(y) app(f_1(x1),y) -> f_2(x1,y) app(f_2(x1,x2),y) -> f_3(x1,x2,y) app(f_3(x1,x2,x3),y) -> f_4(x1,x2,x3,y) app(filter_0(),y) -> filter_1(y) app(filter_1(x1),y) -> filter_2(x1,y) app(filter2_0(),y) -> filter2_1(y) app(filter2_1(x1),y) -> filter2_2(x1,y) app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y) app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y) app(if_0(),y) -> if_1(y) app(if_1(x1),y) -> if_2(x1,y) app(if_2(x1,x2),y) -> if_3(x1,x2,y) app(le_0(),y) -> le_1(y) app(le_1(x1),y) -> le_2(x1,y) app(map_0(),y) -> map_1(y) app(map_1(x1),y) -> map_2(x1,y) app(minus_0(),y) -> minus_1(y) app(minus_1(x1),y) -> minus_2(x1,y) app(perfectp_0(),y) -> perfectp_1(y) app(s_0(),y) -> s_1(y) perfectp_1(|0|_0()) -> false_0() perfectp_1(s_1(x)) -> f_4(x,s_1(|0|_0()),s_1(x),s_1(x)) f_4(|0|_0(),y,|0|_0(),u) -> true_0() f_4(|0|_0(),y,s_1(z),u) -> false_0() f_4(s_1(x),|0|_0(),z,u) -> f_4(x,u,minus_2(z,s_1(x)),u) f_4(s_1(x),s_1(y),z,u) -> if_3(le_2(x,y),f_4(s_1(x),minus_2(y,x),z,u),f_4(x,u,z,u)) map_2(|fun|,nil_0()) -> nil_0() map_2(|fun|,cons_2(x,xs)) -> cons_2(app(|fun|,x),map_2(|fun|,xs)) filter_2(|fun|,nil_0()) -> nil_0() filter_2(|fun|,cons_2(x,xs)) -> filter2_4(app(|fun|,x),|fun|,x,xs) filter2_4(true_0(),|fun|,x,xs) -> cons_2(x,filter_2(|fun|,xs)) filter2_4(false_0(),|fun|,x,xs) -> filter_2(|fun|,xs) -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: app#(f_3(x1,x2,x3),y) -> f_4#(x1,x2,x3,y) p2: app#(filter_1(x1),y) -> filter_2#(x1,y) p3: app#(filter2_3(x1,x2,x3),y) -> filter2_4#(x1,x2,x3,y) p4: app#(map_1(x1),y) -> map_2#(x1,y) p5: app#(perfectp_0(),y) -> perfectp_1#(y) p6: perfectp_1#(s_1(x)) -> f_4#(x,s_1(|0|_0()),s_1(x),s_1(x)) p7: f_4#(s_1(x),|0|_0(),z,u) -> f_4#(x,u,minus_2(z,s_1(x)),u) p8: f_4#(s_1(x),s_1(y),z,u) -> f_4#(s_1(x),minus_2(y,x),z,u) p9: f_4#(s_1(x),s_1(y),z,u) -> f_4#(x,u,z,u) p10: map_2#(|fun|,cons_2(x,xs)) -> app#(|fun|,x) p11: map_2#(|fun|,cons_2(x,xs)) -> map_2#(|fun|,xs) p12: filter_2#(|fun|,cons_2(x,xs)) -> filter2_4#(app(|fun|,x),|fun|,x,xs) p13: filter_2#(|fun|,cons_2(x,xs)) -> app#(|fun|,x) p14: filter2_4#(true_0(),|fun|,x,xs) -> filter_2#(|fun|,xs) p15: filter2_4#(false_0(),|fun|,x,xs) -> filter_2#(|fun|,xs) and R consists of: r1: app(cons_0(),y) -> cons_1(y) r2: app(cons_1(x1),y) -> cons_2(x1,y) r3: app(f_0(),y) -> f_1(y) r4: app(f_1(x1),y) -> f_2(x1,y) r5: app(f_2(x1,x2),y) -> f_3(x1,x2,y) r6: app(f_3(x1,x2,x3),y) -> f_4(x1,x2,x3,y) r7: app(filter_0(),y) -> filter_1(y) r8: app(filter_1(x1),y) -> filter_2(x1,y) r9: app(filter2_0(),y) -> filter2_1(y) r10: app(filter2_1(x1),y) -> filter2_2(x1,y) r11: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y) r12: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y) r13: app(if_0(),y) -> if_1(y) r14: app(if_1(x1),y) -> if_2(x1,y) r15: app(if_2(x1,x2),y) -> if_3(x1,x2,y) r16: app(le_0(),y) -> le_1(y) r17: app(le_1(x1),y) -> le_2(x1,y) r18: app(map_0(),y) -> map_1(y) r19: app(map_1(x1),y) -> map_2(x1,y) r20: app(minus_0(),y) -> minus_1(y) r21: app(minus_1(x1),y) -> minus_2(x1,y) r22: app(perfectp_0(),y) -> perfectp_1(y) r23: app(s_0(),y) -> s_1(y) r24: perfectp_1(|0|_0()) -> false_0() r25: perfectp_1(s_1(x)) -> f_4(x,s_1(|0|_0()),s_1(x),s_1(x)) r26: f_4(|0|_0(),y,|0|_0(),u) -> true_0() r27: f_4(|0|_0(),y,s_1(z),u) -> false_0() r28: f_4(s_1(x),|0|_0(),z,u) -> f_4(x,u,minus_2(z,s_1(x)),u) r29: f_4(s_1(x),s_1(y),z,u) -> if_3(le_2(x,y),f_4(s_1(x),minus_2(y,x),z,u),f_4(x,u,z,u)) r30: map_2(|fun|,nil_0()) -> nil_0() r31: map_2(|fun|,cons_2(x,xs)) -> cons_2(app(|fun|,x),map_2(|fun|,xs)) r32: filter_2(|fun|,nil_0()) -> nil_0() r33: filter_2(|fun|,cons_2(x,xs)) -> filter2_4(app(|fun|,x),|fun|,x,xs) r34: filter2_4(true_0(),|fun|,x,xs) -> cons_2(x,filter_2(|fun|,xs)) r35: filter2_4(false_0(),|fun|,x,xs) -> filter_2(|fun|,xs) The estimated dependency graph contains the following SCCs: {p2, p3, p4, p10, p11, p12, p13, p14, p15} {p7, p9} -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: map_2#(|fun|,cons_2(x,xs)) -> app#(|fun|,x) p2: app#(map_1(x1),y) -> map_2#(x1,y) p3: map_2#(|fun|,cons_2(x,xs)) -> map_2#(|fun|,xs) p4: app#(filter2_3(x1,x2,x3),y) -> filter2_4#(x1,x2,x3,y) p5: filter2_4#(false_0(),|fun|,x,xs) -> filter_2#(|fun|,xs) p6: filter_2#(|fun|,cons_2(x,xs)) -> app#(|fun|,x) p7: app#(filter_1(x1),y) -> filter_2#(x1,y) p8: filter_2#(|fun|,cons_2(x,xs)) -> filter2_4#(app(|fun|,x),|fun|,x,xs) p9: filter2_4#(true_0(),|fun|,x,xs) -> filter_2#(|fun|,xs) and R consists of: r1: app(cons_0(),y) -> cons_1(y) r2: app(cons_1(x1),y) -> cons_2(x1,y) r3: app(f_0(),y) -> f_1(y) r4: app(f_1(x1),y) -> f_2(x1,y) r5: app(f_2(x1,x2),y) -> f_3(x1,x2,y) r6: app(f_3(x1,x2,x3),y) -> f_4(x1,x2,x3,y) r7: app(filter_0(),y) -> filter_1(y) r8: app(filter_1(x1),y) -> filter_2(x1,y) r9: app(filter2_0(),y) -> filter2_1(y) r10: app(filter2_1(x1),y) -> filter2_2(x1,y) r11: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y) r12: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y) r13: app(if_0(),y) -> if_1(y) r14: app(if_1(x1),y) -> if_2(x1,y) r15: app(if_2(x1,x2),y) -> if_3(x1,x2,y) r16: app(le_0(),y) -> le_1(y) r17: app(le_1(x1),y) -> le_2(x1,y) r18: app(map_0(),y) -> map_1(y) r19: app(map_1(x1),y) -> map_2(x1,y) r20: app(minus_0(),y) -> minus_1(y) r21: app(minus_1(x1),y) -> minus_2(x1,y) r22: app(perfectp_0(),y) -> perfectp_1(y) r23: app(s_0(),y) -> s_1(y) r24: perfectp_1(|0|_0()) -> false_0() r25: perfectp_1(s_1(x)) -> f_4(x,s_1(|0|_0()),s_1(x),s_1(x)) r26: f_4(|0|_0(),y,|0|_0(),u) -> true_0() r27: f_4(|0|_0(),y,s_1(z),u) -> false_0() r28: f_4(s_1(x),|0|_0(),z,u) -> f_4(x,u,minus_2(z,s_1(x)),u) r29: f_4(s_1(x),s_1(y),z,u) -> if_3(le_2(x,y),f_4(s_1(x),minus_2(y,x),z,u),f_4(x,u,z,u)) r30: map_2(|fun|,nil_0()) -> nil_0() r31: map_2(|fun|,cons_2(x,xs)) -> cons_2(app(|fun|,x),map_2(|fun|,xs)) r32: filter_2(|fun|,nil_0()) -> nil_0() r33: filter_2(|fun|,cons_2(x,xs)) -> filter2_4(app(|fun|,x),|fun|,x,xs) r34: filter2_4(true_0(),|fun|,x,xs) -> cons_2(x,filter_2(|fun|,xs)) r35: filter2_4(false_0(),|fun|,x,xs) -> filter_2(|fun|,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: map_2#_A(x1,x2) = max{3, x1 - 10} cons_2_A(x1,x2) = max{60, x1 + 58, x2 + 49} app#_A(x1,x2) = max{3, x1 - 82} map_1_A(x1) = max{82, x1 + 72} filter2_3_A(x1,x2,x3) = max{x1 + 81, x2 + 74} filter2_4#_A(x1,x2,x3,x4) = max{3, x2 - 8} false_0_A = 19 filter_2#_A(x1,x2) = max{3, x1 - 8} filter_1_A(x1) = x1 + 96 app_A(x1,x2) = max{19, x2 - 12} true_0_A = 17 map_2#!_A(x1,x2) = max{88, x2 + 37} cons_2!_A(x1,x2) = max{86, x2 + 60} app#!_A(x1,x2) = x2 + 94 map_1!_A(x1) = max{56, x1 + 1} filter2_3!_A(x1,x2,x3) = max{103, x2 + 89} filter2_4#!_A(x1,x2,x3,x4) = x4 + 93 false_0!_A = 93 filter_2#!_A(x1,x2) = max{92, x2 + 45} filter_1!_A(x1) = x1 app!_A(x1,x2) = max{104, x1 + 10, x2 + 57} true_0!_A = 91 2. marked-algebra reduction triple (>=A, >=A!, >A!): max/plus interpretation on natural numbers: map_2#_A(x1,x2) = max{1, x2 - 216} cons_2_A(x1,x2) = max{215, x1 + 52, x2 + 56} app#_A(x1,x2) = max{1, x2 - 164} map_1_A(x1) = 53 filter2_3_A(x1,x2,x3) = max{x1 + 2, x2 + 144, x3} filter2_4#_A(x1,x2,x3,x4) = max{1, x4 - 164} false_0_A = 15 filter_2#_A(x1,x2) = max{1, x2 - 216} filter_1_A(x1) = 51 app_A(x1,x2) = max{49, x1 - 52, x2 - 112} true_0_A = 13 map_2#!_A(x1,x2) = max{x1 + 158, x2 + 60} cons_2!_A(x1,x2) = 155 app#!_A(x1,x2) = max{156, x1 + 39, x2 + 59} map_1!_A(x1) = x1 + 159 filter2_3!_A(x1,x2,x3) = max{2, x3 + 1} filter2_4#!_A(x1,x2,x3,x4) = max{182, x2 + 94, x3 + 1, x4 + 58} false_0!_A = 179 filter_2#!_A(x1,x2) = max{181, x1 + 93} filter_1!_A(x1) = 180 app!_A(x1,x2) = x2 + 179 true_0!_A = 173 partial status: pi(map_2#) = [] pi(cons_2) = [] pi(app#) = [] pi(map_1) = [1] pi(filter2_3) = [1, 3] pi(filter2_4#) = [] pi(false_0) = [] pi(filter_2#) = [] pi(filter_1) = [] pi(app) = [1, 2] pi(true_0) = [] The following rules are removed: p1, p2, p3, p4, p5, p6, p7, p8, p9 Then no dependency pair remains. -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: f_4#(s_1(x),|0|_0(),z,u) -> f_4#(x,u,minus_2(z,s_1(x)),u) p2: f_4#(s_1(x),s_1(y),z,u) -> f_4#(x,u,z,u) and R consists of: r1: app(cons_0(),y) -> cons_1(y) r2: app(cons_1(x1),y) -> cons_2(x1,y) r3: app(f_0(),y) -> f_1(y) r4: app(f_1(x1),y) -> f_2(x1,y) r5: app(f_2(x1,x2),y) -> f_3(x1,x2,y) r6: app(f_3(x1,x2,x3),y) -> f_4(x1,x2,x3,y) r7: app(filter_0(),y) -> filter_1(y) r8: app(filter_1(x1),y) -> filter_2(x1,y) r9: app(filter2_0(),y) -> filter2_1(y) r10: app(filter2_1(x1),y) -> filter2_2(x1,y) r11: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y) r12: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y) r13: app(if_0(),y) -> if_1(y) r14: app(if_1(x1),y) -> if_2(x1,y) r15: app(if_2(x1,x2),y) -> if_3(x1,x2,y) r16: app(le_0(),y) -> le_1(y) r17: app(le_1(x1),y) -> le_2(x1,y) r18: app(map_0(),y) -> map_1(y) r19: app(map_1(x1),y) -> map_2(x1,y) r20: app(minus_0(),y) -> minus_1(y) r21: app(minus_1(x1),y) -> minus_2(x1,y) r22: app(perfectp_0(),y) -> perfectp_1(y) r23: app(s_0(),y) -> s_1(y) r24: perfectp_1(|0|_0()) -> false_0() r25: perfectp_1(s_1(x)) -> f_4(x,s_1(|0|_0()),s_1(x),s_1(x)) r26: f_4(|0|_0(),y,|0|_0(),u) -> true_0() r27: f_4(|0|_0(),y,s_1(z),u) -> false_0() r28: f_4(s_1(x),|0|_0(),z,u) -> f_4(x,u,minus_2(z,s_1(x)),u) r29: f_4(s_1(x),s_1(y),z,u) -> if_3(le_2(x,y),f_4(s_1(x),minus_2(y,x),z,u),f_4(x,u,z,u)) r30: map_2(|fun|,nil_0()) -> nil_0() r31: map_2(|fun|,cons_2(x,xs)) -> cons_2(app(|fun|,x),map_2(|fun|,xs)) r32: filter_2(|fun|,nil_0()) -> nil_0() r33: filter_2(|fun|,cons_2(x,xs)) -> filter2_4(app(|fun|,x),|fun|,x,xs) r34: filter2_4(true_0(),|fun|,x,xs) -> cons_2(x,filter_2(|fun|,xs)) r35: filter2_4(false_0(),|fun|,x,xs) -> filter_2(|fun|,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: f_4#_A(x1,x2,x3,x4) = max{0, x1 - 11, x2 - 13, x3 - 8, x4 - 12} s_1_A(x1) = x1 |0|_0_A = 1 minus_2_A(x1,x2) = max{7, x1 - 3, x2 - 3} f_4#!_A(x1,x2,x3,x4) = x1 + 1 s_1!_A(x1) = max{10, x1 + 2} |0|_0!_A = 9 minus_2!_A(x1,x2) = max{11, x1 + 3, x2 + 8} 2. marked-algebra reduction triple (>=A, >=A!, >A!): max/plus interpretation on natural numbers: f_4#_A(x1,x2,x3,x4) = 0 s_1_A(x1) = x1 |0|_0_A = 0 minus_2_A(x1,x2) = max{x1 + 12, x2 + 12} f_4#!_A(x1,x2,x3,x4) = max{9, x1 - 73, x2 - 74, x4 - 73} s_1!_A(x1) = max{0, x1 - 2} |0|_0!_A = 8 minus_2!_A(x1,x2) = max{10, x1 - 1, x2 - 72} partial status: pi(f_4#) = [1] pi(s_1) = [1] pi(|0|_0) = [] pi(minus_2) = [] The following rules are removed: p1, p2 Then no dependency pair remains.