YES We show the termination of the TRS R: app(app(minus(),x),|0|()) -> x app(app(minus(),app(s(),x)),app(s(),y)) -> app(app(minus(),x),y) app(double(),|0|()) -> |0|() app(double(),app(s(),x)) -> app(s(),app(s(),app(double(),x))) app(app(plus(),|0|()),y) -> y app(app(plus(),app(s(),x)),y) -> app(s(),app(app(plus(),x),y)) app(app(plus(),app(s(),x)),y) -> app(app(plus(),x),app(s(),y)) app(app(plus(),app(s(),x)),y) -> app(s(),app(app(plus(),app(app(minus(),x),y)),app(double(),y))) app(app(map(),f),nil()) -> nil() app(app(map(),f),app(app(cons(),x),xs)) -> app(app(cons(),app(f,x)),app(app(map(),f),xs)) app(app(filter(),f),nil()) -> nil() app(app(filter(),f),app(app(cons(),x),xs)) -> app(app(app(app(filter2(),app(f,x)),f),x),xs) app(app(app(app(filter2(),true()),f),x),xs) -> app(app(cons(),x),app(app(filter(),f),xs)) app(app(app(app(filter2(),false()),f),x),xs) -> app(app(filter(),f),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(double_0(),y) -> double_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(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(plus_0(),y) -> plus_1(y) app(plus_1(x1),y) -> plus_2(x1,y) app(s_0(),y) -> s_1(y) minus_2(x,|0|_0()) -> x minus_2(s_1(x),s_1(y)) -> minus_2(x,y) double_1(|0|_0()) -> |0|_0() double_1(s_1(x)) -> s_1(s_1(double_1(x))) plus_2(|0|_0(),y) -> y plus_2(s_1(x),y) -> s_1(plus_2(x,y)) plus_2(s_1(x),y) -> plus_2(x,s_1(y)) plus_2(s_1(x),y) -> s_1(plus_2(minus_2(x,y),double_1(y))) map_2(f,nil_0()) -> nil_0() map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs)) filter_2(f,nil_0()) -> nil_0() filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs) filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs)) filter2_4(false_0(),f,x,xs) -> filter_2(f,xs) -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: app#(double_0(),y) -> double_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#(map_1(x1),y) -> map_2#(x1,y) p5: app#(minus_1(x1),y) -> minus_2#(x1,y) p6: app#(plus_1(x1),y) -> plus_2#(x1,y) p7: minus_2#(s_1(x),s_1(y)) -> minus_2#(x,y) p8: double_1#(s_1(x)) -> double_1#(x) p9: plus_2#(s_1(x),y) -> plus_2#(x,y) p10: plus_2#(s_1(x),y) -> plus_2#(x,s_1(y)) p11: plus_2#(s_1(x),y) -> plus_2#(minus_2(x,y),double_1(y)) p12: plus_2#(s_1(x),y) -> minus_2#(x,y) p13: plus_2#(s_1(x),y) -> double_1#(y) p14: map_2#(f,cons_2(x,xs)) -> app#(f,x) p15: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs) p16: filter_2#(f,cons_2(x,xs)) -> filter2_4#(app(f,x),f,x,xs) p17: filter_2#(f,cons_2(x,xs)) -> app#(f,x) p18: filter2_4#(true_0(),f,x,xs) -> filter_2#(f,xs) p19: filter2_4#(false_0(),f,x,xs) -> filter_2#(f,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(double_0(),y) -> double_1(y) r4: app(filter_0(),y) -> filter_1(y) r5: app(filter_1(x1),y) -> filter_2(x1,y) r6: app(filter2_0(),y) -> filter2_1(y) r7: app(filter2_1(x1),y) -> filter2_2(x1,y) r8: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y) r9: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y) r10: app(map_0(),y) -> map_1(y) r11: app(map_1(x1),y) -> map_2(x1,y) r12: app(minus_0(),y) -> minus_1(y) r13: app(minus_1(x1),y) -> minus_2(x1,y) r14: app(plus_0(),y) -> plus_1(y) r15: app(plus_1(x1),y) -> plus_2(x1,y) r16: app(s_0(),y) -> s_1(y) r17: minus_2(x,|0|_0()) -> x r18: minus_2(s_1(x),s_1(y)) -> minus_2(x,y) r19: double_1(|0|_0()) -> |0|_0() r20: double_1(s_1(x)) -> s_1(s_1(double_1(x))) r21: plus_2(|0|_0(),y) -> y r22: plus_2(s_1(x),y) -> s_1(plus_2(x,y)) r23: plus_2(s_1(x),y) -> plus_2(x,s_1(y)) r24: plus_2(s_1(x),y) -> s_1(plus_2(minus_2(x,y),double_1(y))) r25: map_2(f,nil_0()) -> nil_0() r26: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs)) r27: filter_2(f,nil_0()) -> nil_0() r28: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs) r29: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs)) r30: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs) The estimated dependency graph contains the following SCCs: {p2, p3, p4, p14, p15, p16, p17, p18, p19} {p9, p10, p11} {p8} {p7} -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: map_2#(f,cons_2(x,xs)) -> app#(f,x) p2: app#(map_1(x1),y) -> map_2#(x1,y) p3: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs) p4: app#(filter2_3(x1,x2,x3),y) -> filter2_4#(x1,x2,x3,y) p5: filter2_4#(false_0(),f,x,xs) -> filter_2#(f,xs) p6: filter_2#(f,cons_2(x,xs)) -> app#(f,x) p7: app#(filter_1(x1),y) -> filter_2#(x1,y) p8: filter_2#(f,cons_2(x,xs)) -> filter2_4#(app(f,x),f,x,xs) p9: filter2_4#(true_0(),f,x,xs) -> filter_2#(f,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(double_0(),y) -> double_1(y) r4: app(filter_0(),y) -> filter_1(y) r5: app(filter_1(x1),y) -> filter_2(x1,y) r6: app(filter2_0(),y) -> filter2_1(y) r7: app(filter2_1(x1),y) -> filter2_2(x1,y) r8: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y) r9: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y) r10: app(map_0(),y) -> map_1(y) r11: app(map_1(x1),y) -> map_2(x1,y) r12: app(minus_0(),y) -> minus_1(y) r13: app(minus_1(x1),y) -> minus_2(x1,y) r14: app(plus_0(),y) -> plus_1(y) r15: app(plus_1(x1),y) -> plus_2(x1,y) r16: app(s_0(),y) -> s_1(y) r17: minus_2(x,|0|_0()) -> x r18: minus_2(s_1(x),s_1(y)) -> minus_2(x,y) r19: double_1(|0|_0()) -> |0|_0() r20: double_1(s_1(x)) -> s_1(s_1(double_1(x))) r21: plus_2(|0|_0(),y) -> y r22: plus_2(s_1(x),y) -> s_1(plus_2(x,y)) r23: plus_2(s_1(x),y) -> plus_2(x,s_1(y)) r24: plus_2(s_1(x),y) -> s_1(plus_2(minus_2(x,y),double_1(y))) r25: map_2(f,nil_0()) -> nil_0() r26: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs)) r27: filter_2(f,nil_0()) -> nil_0() r28: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs) r29: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs)) r30: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs) The set of usable rules consists of r1, r2, r4, r6, r7, r10, r12, r14, r16, r19, r21, r25 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{15, x1 - 1, x2 - 28} cons_2_A(x1,x2) = max{27, x1 + 1, x2 + 1} app#_A(x1,x2) = max{15, x1 - 3, x2 - 27} map_1_A(x1) = max{17, x1 + 9} filter2_3_A(x1,x2,x3) = max{13, x2 + 1} filter2_4#_A(x1,x2,x3,x4) = max{15, x2 - 2, x4 - 27} false_0_A = 43 filter_2#_A(x1,x2) = max{15, x1 - 2, x2 - 28} filter_1_A(x1) = max{17, x1 + 2} app_A(x1,x2) = max{40, x1 - 3, x2 + 16} true_0_A = 43 cons_0_A = 4 cons_1_A(x1) = x1 + 11 filter_0_A = 2 filter2_0_A = 2 filter2_1_A(x1) = x1 + 2 filter2_2_A(x1,x2) = max{x1 - 1, x2 + 16} map_0_A = 2 minus_0_A = 2 minus_1_A(x1) = x1 + 9 plus_0_A = 18 plus_1_A(x1) = max{17, x1 + 7} s_0_A = 2 s_1_A(x1) = max{19, x1 - 6} double_1_A(x1) = max{17, x1 - 20} |0|_0_A = 0 plus_2_A(x1,x2) = max{20, x1 - 21, x2 + 1} map_2_A(x1,x2) = max{17, x1 - 1, x2 + 16} nil_0_A = 20 map_2#!_A(x1,x2) = max{3, x1 + 1} cons_2!_A(x1,x2) = max{7, x1 - 59, x2 - 18} app#!_A(x1,x2) = max{2, x1 - 5} map_1!_A(x1) = x1 + 4 filter2_3!_A(x1,x2,x3) = max{x1 + 9, x2 + 14, x3 + 9} filter2_4#!_A(x1,x2,x3,x4) = max{8, x2 - 4} false_0!_A = 1 filter_2#!_A(x1,x2) = max{8, x1 - 4} filter_1!_A(x1) = x1 + 3 app!_A(x1,x2) = max{15, x1 - 3, x2 + 13} true_0!_A = 1 cons_0!_A = 17 cons_1!_A(x1) = 16 filter_0!_A = 2 filter2_0!_A = 10 filter2_1!_A(x1) = max{11, x1 - 3} filter2_2!_A(x1,x2) = max{12, x1 - 2, x2 - 1} map_0!_A = 2 minus_0!_A = 1 minus_1!_A(x1) = max{2, x1 - 1} plus_0!_A = 15 plus_1!_A(x1) = max{14, x1 + 4} s_0!_A = 1 s_1!_A(x1) = 4 double_1!_A(x1) = 2 |0|_0!_A = 1 plus_2!_A(x1,x2) = x2 + 5 map_2!_A(x1,x2) = max{5, x2 - 21} nil_0!_A = 3 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 + 1 app_A(x1,x2) = x1 + x2 + 1 true_0_A() = 1 cons_0_A() = 1 cons_1_A(x1) = x1 + 1 filter_0_A() = 1 filter2_0_A() = 1 filter2_1_A(x1) = x1 + 1 filter2_2_A(x1,x2) = x1 + x2 + 1 map_0_A() = 1 minus_0_A() = 1 minus_1_A(x1) = x1 + 1 plus_0_A() = 1 plus_1_A(x1) = x1 + 1 s_0_A() = 1 s_1_A(x1) = x1 double_1_A(x1) = 0 |0|_0_A() = 0 plus_2_A(x1,x2) = x2 + 1 map_2_A(x1,x2) = x2 + 1 nil_0_A() = 1 map_2#!_A(x1,x2) = 0 cons_2!_A(x1,x2) = 3 app#!_A(x1,x2) = 1 map_1!_A(x1) = x1 + 1 filter2_3!_A(x1,x2,x3) = 0 filter2_4#!_A(x1,x2,x3,x4) = 1 false_0!_A() = 0 filter_2#!_A(x1,x2) = 1 filter_1!_A(x1) = 0 app!_A(x1,x2) = 5 true_0!_A() = 2 cons_0!_A() = 0 cons_1!_A(x1) = x1 + 1 filter_0!_A() = 1 filter2_0!_A() = 0 filter2_1!_A(x1) = 3 filter2_2!_A(x1,x2) = 4 map_0!_A() = 2 minus_0!_A() = 0 minus_1!_A(x1) = 6 plus_0!_A() = 0 plus_1!_A(x1) = 7 s_0!_A() = 3 s_1!_A(x1) = 4 double_1!_A(x1) = x1 + 7 |0|_0!_A() = 0 plus_2!_A(x1,x2) = x1 + 6 map_2!_A(x1,x2) = 2 nil_0!_A() = 0 partial status: pi(map_2#) = [] pi(cons_2) = [] pi(app#) = [] pi(map_1) = [1] pi(filter2_3) = [] pi(filter2_4#) = [] pi(false_0) = [] pi(filter_2#) = [] pi(filter_1) = [1] pi(app) = [1, 2] pi(true_0) = [] pi(cons_0) = [] pi(cons_1) = [] pi(filter_0) = [] pi(filter2_0) = [] pi(filter2_1) = [1] pi(filter2_2) = [2] pi(map_0) = [] pi(minus_0) = [] pi(minus_1) = [1] pi(plus_0) = [] pi(plus_1) = [1] pi(s_0) = [] pi(s_1) = [1] pi(double_1) = [1] pi(|0|_0) = [] pi(plus_2) = [1, 2] pi(map_2) = [] pi(nil_0) = [] The following rules are removed: p1, p2, p6, p7 -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs) p2: app#(filter2_3(x1,x2,x3),y) -> filter2_4#(x1,x2,x3,y) p3: filter2_4#(false_0(),f,x,xs) -> filter_2#(f,xs) p4: filter_2#(f,cons_2(x,xs)) -> filter2_4#(app(f,x),f,x,xs) p5: filter2_4#(true_0(),f,x,xs) -> filter_2#(f,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(double_0(),y) -> double_1(y) r4: app(filter_0(),y) -> filter_1(y) r5: app(filter_1(x1),y) -> filter_2(x1,y) r6: app(filter2_0(),y) -> filter2_1(y) r7: app(filter2_1(x1),y) -> filter2_2(x1,y) r8: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y) r9: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y) r10: app(map_0(),y) -> map_1(y) r11: app(map_1(x1),y) -> map_2(x1,y) r12: app(minus_0(),y) -> minus_1(y) r13: app(minus_1(x1),y) -> minus_2(x1,y) r14: app(plus_0(),y) -> plus_1(y) r15: app(plus_1(x1),y) -> plus_2(x1,y) r16: app(s_0(),y) -> s_1(y) r17: minus_2(x,|0|_0()) -> x r18: minus_2(s_1(x),s_1(y)) -> minus_2(x,y) r19: double_1(|0|_0()) -> |0|_0() r20: double_1(s_1(x)) -> s_1(s_1(double_1(x))) r21: plus_2(|0|_0(),y) -> y r22: plus_2(s_1(x),y) -> s_1(plus_2(x,y)) r23: plus_2(s_1(x),y) -> plus_2(x,s_1(y)) r24: plus_2(s_1(x),y) -> s_1(plus_2(minus_2(x,y),double_1(y))) r25: map_2(f,nil_0()) -> nil_0() r26: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs)) r27: filter_2(f,nil_0()) -> nil_0() r28: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs) r29: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs)) r30: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs) The estimated dependency graph contains the following SCCs: {p1} {p3, p4, p5} -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: map_2#(f,cons_2(x,xs)) -> map_2#(f,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(double_0(),y) -> double_1(y) r4: app(filter_0(),y) -> filter_1(y) r5: app(filter_1(x1),y) -> filter_2(x1,y) r6: app(filter2_0(),y) -> filter2_1(y) r7: app(filter2_1(x1),y) -> filter2_2(x1,y) r8: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y) r9: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y) r10: app(map_0(),y) -> map_1(y) r11: app(map_1(x1),y) -> map_2(x1,y) r12: app(minus_0(),y) -> minus_1(y) r13: app(minus_1(x1),y) -> minus_2(x1,y) r14: app(plus_0(),y) -> plus_1(y) r15: app(plus_1(x1),y) -> plus_2(x1,y) r16: app(s_0(),y) -> s_1(y) r17: minus_2(x,|0|_0()) -> x r18: minus_2(s_1(x),s_1(y)) -> minus_2(x,y) r19: double_1(|0|_0()) -> |0|_0() r20: double_1(s_1(x)) -> s_1(s_1(double_1(x))) r21: plus_2(|0|_0(),y) -> y r22: plus_2(s_1(x),y) -> s_1(plus_2(x,y)) r23: plus_2(s_1(x),y) -> plus_2(x,s_1(y)) r24: plus_2(s_1(x),y) -> s_1(plus_2(minus_2(x,y),double_1(y))) r25: map_2(f,nil_0()) -> nil_0() r26: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs)) r27: filter_2(f,nil_0()) -> nil_0() r28: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs) r29: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs)) r30: filter2_4(false_0(),f,x,xs) -> filter_2(f,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) = 0 cons_2_A(x1,x2) = max{8, x1 - 7, x2 - 8} map_2#!_A(x1,x2) = x1 + 1 cons_2!_A(x1,x2) = max{4, x1 - 32, x2 - 7} 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) = x2 cons_2_A(x1,x2) = x1 + x2 + 5 map_2#!_A(x1,x2) = x2 + 3 cons_2!_A(x1,x2) = x1 + x2 + 2 partial status: pi(map_2#) = [1] pi(cons_2) = [1] The following rules are removed: p1 Then no dependency pair remains. -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: filter2_4#(false_0(),f,x,xs) -> filter_2#(f,xs) p2: filter_2#(f,cons_2(x,xs)) -> filter2_4#(app(f,x),f,x,xs) p3: filter2_4#(true_0(),f,x,xs) -> filter_2#(f,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(double_0(),y) -> double_1(y) r4: app(filter_0(),y) -> filter_1(y) r5: app(filter_1(x1),y) -> filter_2(x1,y) r6: app(filter2_0(),y) -> filter2_1(y) r7: app(filter2_1(x1),y) -> filter2_2(x1,y) r8: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y) r9: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y) r10: app(map_0(),y) -> map_1(y) r11: app(map_1(x1),y) -> map_2(x1,y) r12: app(minus_0(),y) -> minus_1(y) r13: app(minus_1(x1),y) -> minus_2(x1,y) r14: app(plus_0(),y) -> plus_1(y) r15: app(plus_1(x1),y) -> plus_2(x1,y) r16: app(s_0(),y) -> s_1(y) r17: minus_2(x,|0|_0()) -> x r18: minus_2(s_1(x),s_1(y)) -> minus_2(x,y) r19: double_1(|0|_0()) -> |0|_0() r20: double_1(s_1(x)) -> s_1(s_1(double_1(x))) r21: plus_2(|0|_0(),y) -> y r22: plus_2(s_1(x),y) -> s_1(plus_2(x,y)) r23: plus_2(s_1(x),y) -> plus_2(x,s_1(y)) r24: plus_2(s_1(x),y) -> s_1(plus_2(minus_2(x,y),double_1(y))) r25: map_2(f,nil_0()) -> nil_0() r26: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs)) r27: filter_2(f,nil_0()) -> nil_0() r28: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs) r29: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs)) r30: filter2_4(false_0(),f,x,xs) -> filter_2(f,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: filter2_4#_A(x1,x2,x3,x4) = max{1, x2 - 1, x4 - 5} false_0_A = 12 filter_2#_A(x1,x2) = max{1, x1 - 1, x2 - 9} cons_2_A(x1,x2) = max{8, x1 - 2, x2 + 4} app_A(x1,x2) = max{1, x1 - 43} true_0_A = 4 filter2_4#!_A(x1,x2,x3,x4) = max{6, x3 - 1, x4 + 4} false_0!_A = 6 filter_2#!_A(x1,x2) = max{5, x2 + 3} cons_2!_A(x1,x2) = max{x1, x2 - 14} app!_A(x1,x2) = max{12, x1 - 3, x2 + 9} true_0!_A = 1 2. marked-algebra reduction triple (>=A, >=A!, >A!): matrix interpretation with standard point-wise order: carrier: N^1 order: standard order interpretations: filter2_4#_A(x1,x2,x3,x4) = x4 false_0_A() = 1 filter_2#_A(x1,x2) = x2 cons_2_A(x1,x2) = x1 + x2 + 7 app_A(x1,x2) = x1 + x2 true_0_A() = 1 filter2_4#!_A(x1,x2,x3,x4) = x2 + x4 + 3 false_0!_A() = 0 filter_2#!_A(x1,x2) = x1 + x2 + 2 cons_2!_A(x1,x2) = 6 app!_A(x1,x2) = x2 + 8 true_0!_A() = 3 partial status: pi(filter2_4#) = [] pi(false_0) = [] pi(filter_2#) = [] pi(cons_2) = [2] pi(app) = [2] pi(true_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: plus_2#(s_1(x),y) -> plus_2#(minus_2(x,y),double_1(y)) p2: plus_2#(s_1(x),y) -> plus_2#(x,s_1(y)) p3: plus_2#(s_1(x),y) -> plus_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(double_0(),y) -> double_1(y) r4: app(filter_0(),y) -> filter_1(y) r5: app(filter_1(x1),y) -> filter_2(x1,y) r6: app(filter2_0(),y) -> filter2_1(y) r7: app(filter2_1(x1),y) -> filter2_2(x1,y) r8: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y) r9: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y) r10: app(map_0(),y) -> map_1(y) r11: app(map_1(x1),y) -> map_2(x1,y) r12: app(minus_0(),y) -> minus_1(y) r13: app(minus_1(x1),y) -> minus_2(x1,y) r14: app(plus_0(),y) -> plus_1(y) r15: app(plus_1(x1),y) -> plus_2(x1,y) r16: app(s_0(),y) -> s_1(y) r17: minus_2(x,|0|_0()) -> x r18: minus_2(s_1(x),s_1(y)) -> minus_2(x,y) r19: double_1(|0|_0()) -> |0|_0() r20: double_1(s_1(x)) -> s_1(s_1(double_1(x))) r21: plus_2(|0|_0(),y) -> y r22: plus_2(s_1(x),y) -> s_1(plus_2(x,y)) r23: plus_2(s_1(x),y) -> plus_2(x,s_1(y)) r24: plus_2(s_1(x),y) -> s_1(plus_2(minus_2(x,y),double_1(y))) r25: map_2(f,nil_0()) -> nil_0() r26: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs)) r27: filter_2(f,nil_0()) -> nil_0() r28: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs) r29: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs)) r30: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs) The set of usable rules consists of r17, r18, r19, r20 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: plus_2#_A(x1,x2) = max{0, x2 - 1} s_1_A(x1) = x1 minus_2_A(x1,x2) = max{8, x1, x2 + 1} double_1_A(x1) = max{0, x1 - 18} |0|_0_A = 0 plus_2#!_A(x1,x2) = x2 + 10 s_1!_A(x1) = x1 + 13 minus_2!_A(x1,x2) = 9 double_1!_A(x1) = max{14, x1 + 11} |0|_0!_A = 15 2. marked-algebra reduction triple (>=A, >=A!, >A!): matrix interpretation with standard point-wise order: carrier: N^1 order: standard order interpretations: plus_2#_A(x1,x2) = 0 s_1_A(x1) = x1 minus_2_A(x1,x2) = x1 + 5 double_1_A(x1) = 1 |0|_0_A() = 1 plus_2#!_A(x1,x2) = 10 s_1!_A(x1) = 12 minus_2!_A(x1,x2) = 9 double_1!_A(x1) = x1 + 11 |0|_0!_A() = 13 partial status: pi(plus_2#) = [1] pi(s_1) = [1] pi(minus_2) = [1] pi(double_1) = [1] pi(|0|_0) = [] The following rules are removed: p1 -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: plus_2#(s_1(x),y) -> plus_2#(x,s_1(y)) p2: plus_2#(s_1(x),y) -> plus_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(double_0(),y) -> double_1(y) r4: app(filter_0(),y) -> filter_1(y) r5: app(filter_1(x1),y) -> filter_2(x1,y) r6: app(filter2_0(),y) -> filter2_1(y) r7: app(filter2_1(x1),y) -> filter2_2(x1,y) r8: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y) r9: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y) r10: app(map_0(),y) -> map_1(y) r11: app(map_1(x1),y) -> map_2(x1,y) r12: app(minus_0(),y) -> minus_1(y) r13: app(minus_1(x1),y) -> minus_2(x1,y) r14: app(plus_0(),y) -> plus_1(y) r15: app(plus_1(x1),y) -> plus_2(x1,y) r16: app(s_0(),y) -> s_1(y) r17: minus_2(x,|0|_0()) -> x r18: minus_2(s_1(x),s_1(y)) -> minus_2(x,y) r19: double_1(|0|_0()) -> |0|_0() r20: double_1(s_1(x)) -> s_1(s_1(double_1(x))) r21: plus_2(|0|_0(),y) -> y r22: plus_2(s_1(x),y) -> s_1(plus_2(x,y)) r23: plus_2(s_1(x),y) -> plus_2(x,s_1(y)) r24: plus_2(s_1(x),y) -> s_1(plus_2(minus_2(x,y),double_1(y))) r25: map_2(f,nil_0()) -> nil_0() r26: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs)) r27: filter_2(f,nil_0()) -> nil_0() r28: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs) r29: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs)) r30: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs) The estimated dependency graph contains the following SCCs: {p1, p2} -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: plus_2#(s_1(x),y) -> plus_2#(x,s_1(y)) p2: plus_2#(s_1(x),y) -> plus_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(double_0(),y) -> double_1(y) r4: app(filter_0(),y) -> filter_1(y) r5: app(filter_1(x1),y) -> filter_2(x1,y) r6: app(filter2_0(),y) -> filter2_1(y) r7: app(filter2_1(x1),y) -> filter2_2(x1,y) r8: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y) r9: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y) r10: app(map_0(),y) -> map_1(y) r11: app(map_1(x1),y) -> map_2(x1,y) r12: app(minus_0(),y) -> minus_1(y) r13: app(minus_1(x1),y) -> minus_2(x1,y) r14: app(plus_0(),y) -> plus_1(y) r15: app(plus_1(x1),y) -> plus_2(x1,y) r16: app(s_0(),y) -> s_1(y) r17: minus_2(x,|0|_0()) -> x r18: minus_2(s_1(x),s_1(y)) -> minus_2(x,y) r19: double_1(|0|_0()) -> |0|_0() r20: double_1(s_1(x)) -> s_1(s_1(double_1(x))) r21: plus_2(|0|_0(),y) -> y r22: plus_2(s_1(x),y) -> s_1(plus_2(x,y)) r23: plus_2(s_1(x),y) -> plus_2(x,s_1(y)) r24: plus_2(s_1(x),y) -> s_1(plus_2(minus_2(x,y),double_1(y))) r25: map_2(f,nil_0()) -> nil_0() r26: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs)) r27: filter_2(f,nil_0()) -> nil_0() r28: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs) r29: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs)) r30: filter2_4(false_0(),f,x,xs) -> filter_2(f,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: plus_2#_A(x1,x2) = max{0, x2 - 28} s_1_A(x1) = 27 plus_2#!_A(x1,x2) = max{86, x2 + 58} s_1!_A(x1) = max{86, x1 + 57} 2. marked-algebra reduction triple (>=A, >=A!, >A!): matrix interpretation with standard point-wise order: carrier: N^1 order: standard order interpretations: plus_2#_A(x1,x2) = x1 + x2 s_1_A(x1) = x1 + 3 plus_2#!_A(x1,x2) = x1 + x2 + 10 s_1!_A(x1) = 14 partial status: pi(plus_2#) = [] pi(s_1) = [1] The following rules are removed: p2 -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: plus_2#(s_1(x),y) -> plus_2#(x,s_1(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(double_0(),y) -> double_1(y) r4: app(filter_0(),y) -> filter_1(y) r5: app(filter_1(x1),y) -> filter_2(x1,y) r6: app(filter2_0(),y) -> filter2_1(y) r7: app(filter2_1(x1),y) -> filter2_2(x1,y) r8: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y) r9: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y) r10: app(map_0(),y) -> map_1(y) r11: app(map_1(x1),y) -> map_2(x1,y) r12: app(minus_0(),y) -> minus_1(y) r13: app(minus_1(x1),y) -> minus_2(x1,y) r14: app(plus_0(),y) -> plus_1(y) r15: app(plus_1(x1),y) -> plus_2(x1,y) r16: app(s_0(),y) -> s_1(y) r17: minus_2(x,|0|_0()) -> x r18: minus_2(s_1(x),s_1(y)) -> minus_2(x,y) r19: double_1(|0|_0()) -> |0|_0() r20: double_1(s_1(x)) -> s_1(s_1(double_1(x))) r21: plus_2(|0|_0(),y) -> y r22: plus_2(s_1(x),y) -> s_1(plus_2(x,y)) r23: plus_2(s_1(x),y) -> plus_2(x,s_1(y)) r24: plus_2(s_1(x),y) -> s_1(plus_2(minus_2(x,y),double_1(y))) r25: map_2(f,nil_0()) -> nil_0() r26: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs)) r27: filter_2(f,nil_0()) -> nil_0() r28: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs) r29: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs)) r30: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs) The estimated dependency graph contains the following SCCs: {p1} -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: plus_2#(s_1(x),y) -> plus_2#(x,s_1(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(double_0(),y) -> double_1(y) r4: app(filter_0(),y) -> filter_1(y) r5: app(filter_1(x1),y) -> filter_2(x1,y) r6: app(filter2_0(),y) -> filter2_1(y) r7: app(filter2_1(x1),y) -> filter2_2(x1,y) r8: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y) r9: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y) r10: app(map_0(),y) -> map_1(y) r11: app(map_1(x1),y) -> map_2(x1,y) r12: app(minus_0(),y) -> minus_1(y) r13: app(minus_1(x1),y) -> minus_2(x1,y) r14: app(plus_0(),y) -> plus_1(y) r15: app(plus_1(x1),y) -> plus_2(x1,y) r16: app(s_0(),y) -> s_1(y) r17: minus_2(x,|0|_0()) -> x r18: minus_2(s_1(x),s_1(y)) -> minus_2(x,y) r19: double_1(|0|_0()) -> |0|_0() r20: double_1(s_1(x)) -> s_1(s_1(double_1(x))) r21: plus_2(|0|_0(),y) -> y r22: plus_2(s_1(x),y) -> s_1(plus_2(x,y)) r23: plus_2(s_1(x),y) -> plus_2(x,s_1(y)) r24: plus_2(s_1(x),y) -> s_1(plus_2(minus_2(x,y),double_1(y))) r25: map_2(f,nil_0()) -> nil_0() r26: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs)) r27: filter_2(f,nil_0()) -> nil_0() r28: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs) r29: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs)) r30: filter2_4(false_0(),f,x,xs) -> filter_2(f,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: plus_2#_A(x1,x2) = max{0, x1 - 19, x2 - 18} s_1_A(x1) = max{17, x1} plus_2#!_A(x1,x2) = max{26, x1 - 18, x2 + 8} s_1!_A(x1) = x1 + 7 2. marked-algebra reduction triple (>=A, >=A!, >A!): matrix interpretation with standard point-wise order: carrier: N^1 order: standard order interpretations: plus_2#_A(x1,x2) = x1 + x2 s_1_A(x1) = x1 + 7 plus_2#!_A(x1,x2) = x1 s_1!_A(x1) = 8 partial status: pi(plus_2#) = [2] pi(s_1) = [1] The following rules are removed: p1 Then no dependency pair remains. -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: double_1#(s_1(x)) -> double_1#(x) and R consists of: r1: app(cons_0(),y) -> cons_1(y) r2: app(cons_1(x1),y) -> cons_2(x1,y) r3: app(double_0(),y) -> double_1(y) r4: app(filter_0(),y) -> filter_1(y) r5: app(filter_1(x1),y) -> filter_2(x1,y) r6: app(filter2_0(),y) -> filter2_1(y) r7: app(filter2_1(x1),y) -> filter2_2(x1,y) r8: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y) r9: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y) r10: app(map_0(),y) -> map_1(y) r11: app(map_1(x1),y) -> map_2(x1,y) r12: app(minus_0(),y) -> minus_1(y) r13: app(minus_1(x1),y) -> minus_2(x1,y) r14: app(plus_0(),y) -> plus_1(y) r15: app(plus_1(x1),y) -> plus_2(x1,y) r16: app(s_0(),y) -> s_1(y) r17: minus_2(x,|0|_0()) -> x r18: minus_2(s_1(x),s_1(y)) -> minus_2(x,y) r19: double_1(|0|_0()) -> |0|_0() r20: double_1(s_1(x)) -> s_1(s_1(double_1(x))) r21: plus_2(|0|_0(),y) -> y r22: plus_2(s_1(x),y) -> s_1(plus_2(x,y)) r23: plus_2(s_1(x),y) -> plus_2(x,s_1(y)) r24: plus_2(s_1(x),y) -> s_1(plus_2(minus_2(x,y),double_1(y))) r25: map_2(f,nil_0()) -> nil_0() r26: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs)) r27: filter_2(f,nil_0()) -> nil_0() r28: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs) r29: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs)) r30: filter2_4(false_0(),f,x,xs) -> filter_2(f,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: double_1#_A(x1) = 0 s_1_A(x1) = x1 + 9 double_1#!_A(x1) = x1 + 34 s_1!_A(x1) = x1 + 33 2. marked-algebra reduction triple (>=A, >=A!, >A!): matrix interpretation with standard point-wise order: carrier: N^1 order: standard order interpretations: double_1#_A(x1) = 0 s_1_A(x1) = 3 double_1#!_A(x1) = 0 s_1!_A(x1) = 2 partial status: pi(double_1#) = [] pi(s_1) = [] The following rules are removed: p1 Then no dependency pair remains. -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: minus_2#(s_1(x),s_1(y)) -> minus_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(double_0(),y) -> double_1(y) r4: app(filter_0(),y) -> filter_1(y) r5: app(filter_1(x1),y) -> filter_2(x1,y) r6: app(filter2_0(),y) -> filter2_1(y) r7: app(filter2_1(x1),y) -> filter2_2(x1,y) r8: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y) r9: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y) r10: app(map_0(),y) -> map_1(y) r11: app(map_1(x1),y) -> map_2(x1,y) r12: app(minus_0(),y) -> minus_1(y) r13: app(minus_1(x1),y) -> minus_2(x1,y) r14: app(plus_0(),y) -> plus_1(y) r15: app(plus_1(x1),y) -> plus_2(x1,y) r16: app(s_0(),y) -> s_1(y) r17: minus_2(x,|0|_0()) -> x r18: minus_2(s_1(x),s_1(y)) -> minus_2(x,y) r19: double_1(|0|_0()) -> |0|_0() r20: double_1(s_1(x)) -> s_1(s_1(double_1(x))) r21: plus_2(|0|_0(),y) -> y r22: plus_2(s_1(x),y) -> s_1(plus_2(x,y)) r23: plus_2(s_1(x),y) -> plus_2(x,s_1(y)) r24: plus_2(s_1(x),y) -> s_1(plus_2(minus_2(x,y),double_1(y))) r25: map_2(f,nil_0()) -> nil_0() r26: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs)) r27: filter_2(f,nil_0()) -> nil_0() r28: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs) r29: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs)) r30: filter2_4(false_0(),f,x,xs) -> filter_2(f,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: minus_2#_A(x1,x2) = 0 s_1_A(x1) = max{13, x1 + 10} minus_2#!_A(x1,x2) = 1 s_1!_A(x1) = 6 2. marked-algebra reduction triple (>=A, >=A!, >A!): matrix interpretation with standard point-wise order: carrier: N^1 order: standard order interpretations: minus_2#_A(x1,x2) = 0 s_1_A(x1) = 5 minus_2#!_A(x1,x2) = 2 s_1!_A(x1) = 1 partial status: pi(minus_2#) = [1] pi(s_1) = [1] The following rules are removed: p1 Then no dependency pair remains.