YES We show the termination of the TRS R: app(f(),|0|()) -> true() app(f(),|1|()) -> false() app(f(),app(s(),x)) -> app(f(),x) app(app(app(if(),true()),app(s(),x)),app(s(),y)) -> app(s(),x) app(app(app(if(),false()),app(s(),x)),app(s(),y)) -> app(s(),y) app(app(g(),x),app(c(),y)) -> app(c(),app(app(g(),x),y)) app(app(g(),x),app(c(),y)) -> app(app(g(),x),app(app(app(if(),app(f(),x)),app(c(),app(app(g(),app(s(),x)),y))),app(c(),y))) 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(c_0(),y) -> c_1(y) app(cons_0(),y) -> cons_1(y) app(cons_1(x1),y) -> cons_2(x1,y) app(f_0(),y) -> f_1(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(g_0(),y) -> g_1(y) app(g_1(x1),y) -> g_2(x1,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(map_0(),y) -> map_1(y) app(map_1(x1),y) -> map_2(x1,y) app(s_0(),y) -> s_1(y) f_1(|0|_0()) -> true_0() f_1(|1|_0()) -> false_0() f_1(s_1(x)) -> f_1(x) if_3(true_0(),s_1(x),s_1(y)) -> s_1(x) if_3(false_0(),s_1(x),s_1(y)) -> s_1(y) g_2(x,c_1(y)) -> c_1(g_2(x,y)) g_2(x,c_1(y)) -> g_2(x,if_3(f_1(x),c_1(g_2(s_1(x),y)),c_1(y))) 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_0(),y) -> f_1#(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#(g_1(x1),y) -> g_2#(x1,y) p5: app#(if_2(x1,x2),y) -> if_3#(x1,x2,y) p6: app#(map_1(x1),y) -> map_2#(x1,y) p7: f_1#(s_1(x)) -> f_1#(x) p8: g_2#(x,c_1(y)) -> g_2#(x,y) p9: g_2#(x,c_1(y)) -> g_2#(x,if_3(f_1(x),c_1(g_2(s_1(x),y)),c_1(y))) p10: g_2#(x,c_1(y)) -> if_3#(f_1(x),c_1(g_2(s_1(x),y)),c_1(y)) p11: g_2#(x,c_1(y)) -> f_1#(x) p12: g_2#(x,c_1(y)) -> g_2#(s_1(x),y) p13: map_2#(|fun|,cons_2(x,xs)) -> app#(|fun|,x) p14: map_2#(|fun|,cons_2(x,xs)) -> map_2#(|fun|,xs) p15: filter_2#(|fun|,cons_2(x,xs)) -> filter2_4#(app(|fun|,x),|fun|,x,xs) p16: filter_2#(|fun|,cons_2(x,xs)) -> app#(|fun|,x) p17: filter2_4#(true_0(),|fun|,x,xs) -> filter_2#(|fun|,xs) p18: filter2_4#(false_0(),|fun|,x,xs) -> filter_2#(|fun|,xs) and R consists of: r1: app(c_0(),y) -> c_1(y) r2: app(cons_0(),y) -> cons_1(y) r3: app(cons_1(x1),y) -> cons_2(x1,y) r4: app(f_0(),y) -> f_1(y) r5: app(filter_0(),y) -> filter_1(y) r6: app(filter_1(x1),y) -> filter_2(x1,y) r7: app(filter2_0(),y) -> filter2_1(y) r8: app(filter2_1(x1),y) -> filter2_2(x1,y) r9: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y) r10: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y) r11: app(g_0(),y) -> g_1(y) r12: app(g_1(x1),y) -> g_2(x1,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(map_0(),y) -> map_1(y) r17: app(map_1(x1),y) -> map_2(x1,y) r18: app(s_0(),y) -> s_1(y) r19: f_1(|0|_0()) -> true_0() r20: f_1(|1|_0()) -> false_0() r21: f_1(s_1(x)) -> f_1(x) r22: if_3(true_0(),s_1(x),s_1(y)) -> s_1(x) r23: if_3(false_0(),s_1(x),s_1(y)) -> s_1(y) r24: g_2(x,c_1(y)) -> c_1(g_2(x,y)) r25: g_2(x,c_1(y)) -> g_2(x,if_3(f_1(x),c_1(g_2(s_1(x),y)),c_1(y))) r26: map_2(|fun|,nil_0()) -> nil_0() r27: map_2(|fun|,cons_2(x,xs)) -> cons_2(app(|fun|,x),map_2(|fun|,xs)) r28: filter_2(|fun|,nil_0()) -> nil_0() r29: filter_2(|fun|,cons_2(x,xs)) -> filter2_4(app(|fun|,x),|fun|,x,xs) r30: filter2_4(true_0(),|fun|,x,xs) -> cons_2(x,filter_2(|fun|,xs)) r31: filter2_4(false_0(),|fun|,x,xs) -> filter_2(|fun|,xs) The estimated dependency graph contains the following SCCs: {p2, p3, p6, p13, p14, p15, p16, p17, p18} {p8, p12} {p7} -- 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(c_0(),y) -> c_1(y) r2: app(cons_0(),y) -> cons_1(y) r3: app(cons_1(x1),y) -> cons_2(x1,y) r4: app(f_0(),y) -> f_1(y) r5: app(filter_0(),y) -> filter_1(y) r6: app(filter_1(x1),y) -> filter_2(x1,y) r7: app(filter2_0(),y) -> filter2_1(y) r8: app(filter2_1(x1),y) -> filter2_2(x1,y) r9: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y) r10: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y) r11: app(g_0(),y) -> g_1(y) r12: app(g_1(x1),y) -> g_2(x1,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(map_0(),y) -> map_1(y) r17: app(map_1(x1),y) -> map_2(x1,y) r18: app(s_0(),y) -> s_1(y) r19: f_1(|0|_0()) -> true_0() r20: f_1(|1|_0()) -> false_0() r21: f_1(s_1(x)) -> f_1(x) r22: if_3(true_0(),s_1(x),s_1(y)) -> s_1(x) r23: if_3(false_0(),s_1(x),s_1(y)) -> s_1(y) r24: g_2(x,c_1(y)) -> c_1(g_2(x,y)) r25: g_2(x,c_1(y)) -> g_2(x,if_3(f_1(x),c_1(g_2(s_1(x),y)),c_1(y))) r26: map_2(|fun|,nil_0()) -> nil_0() r27: map_2(|fun|,cons_2(x,xs)) -> cons_2(app(|fun|,x),map_2(|fun|,xs)) r28: filter_2(|fun|,nil_0()) -> nil_0() r29: filter_2(|fun|,cons_2(x,xs)) -> filter2_4(app(|fun|,x),|fun|,x,xs) r30: filter2_4(true_0(),|fun|,x,xs) -> cons_2(x,filter_2(|fun|,xs)) r31: filter2_4(false_0(),|fun|,x,xs) -> filter_2(|fun|,xs) The set of usable rules consists of r1, r5, r7, r8, r13, r18, r19, r20, r26, r28 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{1, x1 - 1, x2 - 27} cons_2_A(x1,x2) = max{26, x1 + 5, x2 + 7} app#_A(x1,x2) = max{1, x1 - 46, x2 - 22} map_1_A(x1) = x1 + 45 filter2_3_A(x1,x2,x3) = max{24, x1 + 12, x2 + 12, x3 + 12} filter2_4#_A(x1,x2,x3,x4) = max{1, x2 - 34, x3 - 34, x4 - 22} false_0_A = 85 filter_2#_A(x1,x2) = max{1, x1 - 34, x2 - 27} filter_1_A(x1) = max{13, x1 + 12} app_A(x1,x2) = max{119, x1 - 49, x2 + 12} true_0_A = 85 c_0_A = 13 c_1_A(x1) = max{14, x1 - 105} filter_0_A = 13 filter2_0_A = 15 filter2_1_A(x1) = max{13, x1 + 12} filter2_2_A(x1,x2) = max{13, x1 - 37, x2 + 12} if_0_A = 13 if_1_A(x1) = max{15, x1 + 12} s_0_A = 45 s_1_A(x1) = max{13, x1 - 1} f_1_A(x1) = max{101, x1 - 1} |0|_0_A = 0 |1|_0_A = 0 map_2_A(x1,x2) = max{13, x1 - 44, x2 - 6} nil_0_A = 8 filter_2_A(x1,x2) = max{13, x1 - 44, x2 - 44} map_2#!_A(x1,x2) = 3 cons_2!_A(x1,x2) = x1 + 8 app#!_A(x1,x2) = max{8, x2 + 4} map_1!_A(x1) = max{33, x1 + 30} filter2_3!_A(x1,x2,x3) = max{x1 - 3, x2 - 1, x3 + 1} filter2_4#!_A(x1,x2,x3,x4) = max{7, x4 + 3} false_0!_A = 3 filter_2#!_A(x1,x2) = max{2, x2 - 1} filter_1!_A(x1) = max{1, x1 - 16} app!_A(x1,x2) = max{x1 - 14, x2 + 29} true_0!_A = 3 c_0!_A = 1 c_1!_A(x1) = max{0, x1 - 1} filter_0!_A = 2 filter2_0!_A = 1 filter2_1!_A(x1) = max{0, x1 - 4} filter2_2!_A(x1,x2) = max{0, x1 - 4} if_0!_A = 1 if_1!_A(x1) = max{29, x1 + 1} s_0!_A = 32 s_1!_A(x1) = 31 f_1!_A(x1) = 4 |0|_0!_A = 4 |1|_0!_A = 4 map_2!_A(x1,x2) = 32 nil_0!_A = 1 filter_2!_A(x1,x2) = max{27, x1 - 15, x2 - 27} 2. marked-algebra reduction triple (>=A, >=A!, >A!): matrix interpretation with standard point-wise order: carrier: N^1 order: standard order interpretations: map_2#_A(x1,x2) = 0 cons_2_A(x1,x2) = 1 app#_A(x1,x2) = 0 map_1_A(x1) = x1 + 1 filter2_3_A(x1,x2,x3) = x1 + x2 + x3 + 1 filter2_4#_A(x1,x2,x3,x4) = 0 false_0_A() = 1 filter_2#_A(x1,x2) = 0 filter_1_A(x1) = x1 + 4 app_A(x1,x2) = x1 + x2 + 3 true_0_A() = 1 c_0_A() = 4 c_1_A(x1) = 1 filter_0_A() = 1 filter2_0_A() = 1 filter2_1_A(x1) = x1 + 3 filter2_2_A(x1,x2) = x1 + x2 + 1 if_0_A() = 1 if_1_A(x1) = x1 + 1 s_0_A() = 2 s_1_A(x1) = x1 + 1 f_1_A(x1) = x1 + 1 |0|_0_A() = 3 |1|_0_A() = 1 map_2_A(x1,x2) = x2 + 1 nil_0_A() = 1 filter_2_A(x1,x2) = x1 + 1 map_2#!_A(x1,x2) = 1 cons_2!_A(x1,x2) = x1 + 9 app#!_A(x1,x2) = 2 map_1!_A(x1) = 7 filter2_3!_A(x1,x2,x3) = 0 filter2_4#!_A(x1,x2,x3,x4) = 1 false_0!_A() = 6 filter_2#!_A(x1,x2) = x1 + 3 filter_1!_A(x1) = 6 app!_A(x1,x2) = x1 + 4 true_0!_A() = 8 c_0!_A() = 0 c_1!_A(x1) = 7 filter_0!_A() = 5 filter2_0!_A() = 0 filter2_1!_A(x1) = x1 + 6 filter2_2!_A(x1,x2) = 6 if_0!_A() = 1 if_1!_A(x1) = 2 s_0!_A() = 6 s_1!_A(x1) = 5 f_1!_A(x1) = x1 + 6 |0|_0!_A() = 7 |1|_0!_A() = 7 map_2!_A(x1,x2) = 6 nil_0!_A() = 0 filter_2!_A(x1,x2) = 7 partial status: pi(map_2#) = [2] pi(cons_2) = [1, 2] pi(app#) = [2] pi(map_1) = [1] pi(filter2_3) = [] pi(filter2_4#) = [4] pi(false_0) = [] pi(filter_2#) = [2] pi(filter_1) = [1] pi(app) = [1, 2] pi(true_0) = [] pi(c_0) = [] pi(c_1) = [] pi(filter_0) = [] pi(filter2_0) = [] pi(filter2_1) = [1] pi(filter2_2) = [] pi(if_0) = [] pi(if_1) = [1] pi(s_0) = [] pi(s_1) = [1] pi(f_1) = [1] pi(|0|_0) = [] pi(|1|_0) = [] pi(map_2) = [] pi(nil_0) = [] pi(filter_2) = [1] 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: g_2#(x,c_1(y)) -> g_2#(s_1(x),y) p2: g_2#(x,c_1(y)) -> g_2#(x,y) and R consists of: r1: app(c_0(),y) -> c_1(y) r2: app(cons_0(),y) -> cons_1(y) r3: app(cons_1(x1),y) -> cons_2(x1,y) r4: app(f_0(),y) -> f_1(y) r5: app(filter_0(),y) -> filter_1(y) r6: app(filter_1(x1),y) -> filter_2(x1,y) r7: app(filter2_0(),y) -> filter2_1(y) r8: app(filter2_1(x1),y) -> filter2_2(x1,y) r9: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y) r10: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y) r11: app(g_0(),y) -> g_1(y) r12: app(g_1(x1),y) -> g_2(x1,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(map_0(),y) -> map_1(y) r17: app(map_1(x1),y) -> map_2(x1,y) r18: app(s_0(),y) -> s_1(y) r19: f_1(|0|_0()) -> true_0() r20: f_1(|1|_0()) -> false_0() r21: f_1(s_1(x)) -> f_1(x) r22: if_3(true_0(),s_1(x),s_1(y)) -> s_1(x) r23: if_3(false_0(),s_1(x),s_1(y)) -> s_1(y) r24: g_2(x,c_1(y)) -> c_1(g_2(x,y)) r25: g_2(x,c_1(y)) -> g_2(x,if_3(f_1(x),c_1(g_2(s_1(x),y)),c_1(y))) r26: map_2(|fun|,nil_0()) -> nil_0() r27: map_2(|fun|,cons_2(x,xs)) -> cons_2(app(|fun|,x),map_2(|fun|,xs)) r28: filter_2(|fun|,nil_0()) -> nil_0() r29: filter_2(|fun|,cons_2(x,xs)) -> filter2_4(app(|fun|,x),|fun|,x,xs) r30: filter2_4(true_0(),|fun|,x,xs) -> cons_2(x,filter_2(|fun|,xs)) r31: 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: g_2#_A(x1,x2) = max{0, x1 - 41, x2 - 42} c_1_A(x1) = max{40, x1 + 18} s_1_A(x1) = max{40, x1 - 22} g_2#!_A(x1,x2) = max{24, x2 - 15} c_1!_A(x1) = max{24, x1 + 20} s_1!_A(x1) = max{26, x1 - 83} 2. marked-algebra reduction triple (>=A, >=A!, >A!): matrix interpretation with standard point-wise order: carrier: N^1 order: standard order interpretations: g_2#_A(x1,x2) = x1 + x2 c_1_A(x1) = x1 + 3 s_1_A(x1) = x1 + 3 g_2#!_A(x1,x2) = x2 c_1!_A(x1) = x1 + 18 s_1!_A(x1) = 9 partial status: pi(g_2#) = [] pi(c_1) = [] pi(s_1) = [1] The following rules are removed: p1, p2 Then no dependency pair remains. -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: f_1#(s_1(x)) -> f_1#(x) and R consists of: r1: app(c_0(),y) -> c_1(y) r2: app(cons_0(),y) -> cons_1(y) r3: app(cons_1(x1),y) -> cons_2(x1,y) r4: app(f_0(),y) -> f_1(y) r5: app(filter_0(),y) -> filter_1(y) r6: app(filter_1(x1),y) -> filter_2(x1,y) r7: app(filter2_0(),y) -> filter2_1(y) r8: app(filter2_1(x1),y) -> filter2_2(x1,y) r9: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y) r10: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y) r11: app(g_0(),y) -> g_1(y) r12: app(g_1(x1),y) -> g_2(x1,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(map_0(),y) -> map_1(y) r17: app(map_1(x1),y) -> map_2(x1,y) r18: app(s_0(),y) -> s_1(y) r19: f_1(|0|_0()) -> true_0() r20: f_1(|1|_0()) -> false_0() r21: f_1(s_1(x)) -> f_1(x) r22: if_3(true_0(),s_1(x),s_1(y)) -> s_1(x) r23: if_3(false_0(),s_1(x),s_1(y)) -> s_1(y) r24: g_2(x,c_1(y)) -> c_1(g_2(x,y)) r25: g_2(x,c_1(y)) -> g_2(x,if_3(f_1(x),c_1(g_2(s_1(x),y)),c_1(y))) r26: map_2(|fun|,nil_0()) -> nil_0() r27: map_2(|fun|,cons_2(x,xs)) -> cons_2(app(|fun|,x),map_2(|fun|,xs)) r28: filter_2(|fun|,nil_0()) -> nil_0() r29: filter_2(|fun|,cons_2(x,xs)) -> filter2_4(app(|fun|,x),|fun|,x,xs) r30: filter2_4(true_0(),|fun|,x,xs) -> cons_2(x,filter_2(|fun|,xs)) r31: 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_1#_A(x1) = x1 s_1_A(x1) = max{15, x1} f_1#!_A(x1) = max{6, x1 - 16} s_1!_A(x1) = x1 + 5 2. marked-algebra reduction triple (>=A, >=A!, >A!): matrix interpretation with standard point-wise order: carrier: N^1 order: standard order interpretations: f_1#_A(x1) = x1 s_1_A(x1) = x1 + 8 f_1#!_A(x1) = x1 + 27 s_1!_A(x1) = x1 + 26 partial status: pi(f_1#) = [1] pi(s_1) = [1] The following rules are removed: p1 Then no dependency pair remains.