YES We show the termination of the TRS R: app(app(times(),x),|0|()) -> |0|() app(app(times(),x),app(s(),y)) -> app(app(plus(),app(app(times(),x),y)),x) app(app(plus(),x),|0|()) -> x app(app(plus(),|0|()),x) -> x app(app(plus(),x),app(s(),y)) -> app(s(),app(app(plus(),x),y)) app(app(plus(),app(s(),x)),y) -> app(s(),app(app(plus(),x),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(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(plus_0(),y) -> plus_1(y) app(plus_1(x1),y) -> plus_2(x1,y) app(s_0(),y) -> s_1(y) app(times_0(),y) -> times_1(y) app(times_1(x1),y) -> times_2(x1,y) times_2(x,|0|_0()) -> |0|_0() times_2(x,s_1(y)) -> plus_2(times_2(x,y),x) plus_2(x,|0|_0()) -> x plus_2(|0|_0(),x) -> x plus_2(x,s_1(y)) -> s_1(plus_2(x,y)) plus_2(s_1(x),y) -> s_1(plus_2(x,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#(filter_1(x1),y) -> filter_2#(x1,y) p2: app#(filter2_3(x1,x2,x3),y) -> filter2_4#(x1,x2,x3,y) p3: app#(map_1(x1),y) -> map_2#(x1,y) p4: app#(plus_1(x1),y) -> plus_2#(x1,y) p5: app#(times_1(x1),y) -> times_2#(x1,y) p6: times_2#(x,s_1(y)) -> plus_2#(times_2(x,y),x) p7: times_2#(x,s_1(y)) -> times_2#(x,y) p8: plus_2#(x,s_1(y)) -> plus_2#(x,y) p9: plus_2#(s_1(x),y) -> plus_2#(x,y) p10: map_2#(f,cons_2(x,xs)) -> app#(f,x) p11: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs) p12: filter_2#(f,cons_2(x,xs)) -> filter2_4#(app(f,x),f,x,xs) p13: filter_2#(f,cons_2(x,xs)) -> app#(f,x) p14: filter2_4#(true_0(),f,x,xs) -> filter_2#(f,xs) p15: 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(filter_0(),y) -> filter_1(y) r4: app(filter_1(x1),y) -> filter_2(x1,y) r5: app(filter2_0(),y) -> filter2_1(y) r6: app(filter2_1(x1),y) -> filter2_2(x1,y) r7: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y) r8: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y) r9: app(map_0(),y) -> map_1(y) r10: app(map_1(x1),y) -> map_2(x1,y) r11: app(plus_0(),y) -> plus_1(y) r12: app(plus_1(x1),y) -> plus_2(x1,y) r13: app(s_0(),y) -> s_1(y) r14: app(times_0(),y) -> times_1(y) r15: app(times_1(x1),y) -> times_2(x1,y) r16: times_2(x,|0|_0()) -> |0|_0() r17: times_2(x,s_1(y)) -> plus_2(times_2(x,y),x) r18: plus_2(x,|0|_0()) -> x r19: plus_2(|0|_0(),x) -> x r20: plus_2(x,s_1(y)) -> s_1(plus_2(x,y)) r21: plus_2(s_1(x),y) -> s_1(plus_2(x,y)) r22: map_2(f,nil_0()) -> nil_0() r23: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs)) r24: filter_2(f,nil_0()) -> nil_0() r25: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs) r26: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs)) r27: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs) The estimated dependency graph contains the following SCCs: {p1, p2, p3, p10, p11, p12, p13, p14, p15} {p7} {p8, p9} -- 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(x,xs)) -> app#(f,x) p3: app#(map_1(x1),y) -> map_2#(x1,y) p4: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs) p5: map_2#(f,cons_2(x,xs)) -> app#(f,x) p6: app#(filter2_3(x1,x2,x3),y) -> filter2_4#(x1,x2,x3,y) p7: filter2_4#(false_0(),f,x,xs) -> filter_2#(f,xs) 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(filter_0(),y) -> filter_1(y) r4: app(filter_1(x1),y) -> filter_2(x1,y) r5: app(filter2_0(),y) -> filter2_1(y) r6: app(filter2_1(x1),y) -> filter2_2(x1,y) r7: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y) r8: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y) r9: app(map_0(),y) -> map_1(y) r10: app(map_1(x1),y) -> map_2(x1,y) r11: app(plus_0(),y) -> plus_1(y) r12: app(plus_1(x1),y) -> plus_2(x1,y) r13: app(s_0(),y) -> s_1(y) r14: app(times_0(),y) -> times_1(y) r15: app(times_1(x1),y) -> times_2(x1,y) r16: times_2(x,|0|_0()) -> |0|_0() r17: times_2(x,s_1(y)) -> plus_2(times_2(x,y),x) r18: plus_2(x,|0|_0()) -> x r19: plus_2(|0|_0(),x) -> x r20: plus_2(x,s_1(y)) -> s_1(plus_2(x,y)) r21: plus_2(s_1(x),y) -> s_1(plus_2(x,y)) r22: map_2(f,nil_0()) -> nil_0() r23: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs)) r24: filter_2(f,nil_0()) -> nil_0() r25: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs) r26: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs)) r27: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs) The set of usable rules consists of r16, r18, r19, r20, r21, r22 Take the reduction pair: generalized weighted path order (>=MG, >G) base order 1: matrix interpretation with standard point-wise order: carrier: N^1 order: standard order interpretations: app#_A(x1,x2) = 3 filter_1_A(x1) = x1 + 2 filter_2#_A(x1,x2) = 3 cons_2_A(x1,x2) = 5 map_1_A(x1) = 4 map_2#_A(x1,x2) = 3 filter2_3_A(x1,x2,x3) = x2 + x3 + 1 filter2_4#_A(x1,x2,x3,x4) = 3 false_0_A() = 4 app_A(x1,x2) = x1 + x2 + 6 true_0_A() = 4 times_2_A(x1,x2) = x1 + x2 + 3 |0|_0_A() = 1 plus_2_A(x1,x2) = x1 + x2 + 3 s_1_A(x1) = x1 + 7 map_2_A(x1,x2) = 2 nil_0_A() = 1 base order 2: marked-algebra reduction triple (>=A, >=A!, >A!): max/plus interpretation on natural numbers: app#_A(x1,x2) = max{1, x1 - 6, x2 - 3} filter_1_A(x1) = max{2, x1 + 1} filter_2#_A(x1,x2) = max{1, x1 - 5, x2 - 22} cons_2_A(x1,x2) = max{21, x1 + 19, x2 + 6} map_1_A(x1) = max{2, x1 + 1} map_2#_A(x1,x2) = max{1, x1 - 5, x2 - 22} filter2_3_A(x1,x2,x3) = max{x1 + 5, x2 + 4, x3 + 1} filter2_4#_A(x1,x2,x3,x4) = max{1, x2 - 5, x3 - 5, x4 - 16} false_0_A = 15 app_A(x1,x2) = max{7, x1 - 6} true_0_A = 15 times_2_A(x1,x2) = max{7, x1 - 4, x2 - 3} |0|_0_A = 0 plus_2_A(x1,x2) = max{7, x1, x2 + 1} s_1_A(x1) = max{2, x1 - 2} map_2_A(x1,x2) = max{26, x1 - 11, x2 - 27} nil_0_A = 24 app#!_A(x1,x2) = max{19, x2 + 14} filter_1!_A(x1) = max{1, x1 - 24} filter_2#!_A(x1,x2) = max{15, x2 - 1} cons_2!_A(x1,x2) = max{12, x1 - 12, x2 + 10} map_1!_A(x1) = 0 map_2#!_A(x1,x2) = x2 + 14 filter2_3!_A(x1,x2,x3) = max{0, x2 - 24} filter2_4#!_A(x1,x2,x3,x4) = max{15, x4 - 1} false_0!_A = 1 app!_A(x1,x2) = max{11, x1 - 3, x2 - 23} true_0!_A = 13 times_2!_A(x1,x2) = max{4, x1 - 28, x2 - 25} |0|_0!_A = 0 plus_2!_A(x1,x2) = max{3, x1 - 23, x2 - 24} s_1!_A(x1) = max{3, x1 - 27} map_2!_A(x1,x2) = max{0, x1 - 4, x2 - 22} nil_0!_A = 1 partial status: pi(app#) = [] pi(filter_1) = [1] pi(filter_2#) = [] pi(cons_2) = [] pi(map_1) = [] pi(map_2#) = [] pi(filter2_3) = [] pi(filter2_4#) = [] pi(false_0) = [] pi(app) = [1] pi(true_0) = [] pi(times_2) = [2] pi(|0|_0) = [] pi(plus_2) = [1] pi(s_1) = [] pi(map_2) = [] pi(nil_0) = [] The following rules are removed: p1, p2, p4, p5, p6, p8 -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: app#(map_1(x1),y) -> map_2#(x1,y) p2: filter2_4#(false_0(),f,x,xs) -> filter_2#(f,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(filter_0(),y) -> filter_1(y) r4: app(filter_1(x1),y) -> filter_2(x1,y) r5: app(filter2_0(),y) -> filter2_1(y) r6: app(filter2_1(x1),y) -> filter2_2(x1,y) r7: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y) r8: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y) r9: app(map_0(),y) -> map_1(y) r10: app(map_1(x1),y) -> map_2(x1,y) r11: app(plus_0(),y) -> plus_1(y) r12: app(plus_1(x1),y) -> plus_2(x1,y) r13: app(s_0(),y) -> s_1(y) r14: app(times_0(),y) -> times_1(y) r15: app(times_1(x1),y) -> times_2(x1,y) r16: times_2(x,|0|_0()) -> |0|_0() r17: times_2(x,s_1(y)) -> plus_2(times_2(x,y),x) r18: plus_2(x,|0|_0()) -> x r19: plus_2(|0|_0(),x) -> x r20: plus_2(x,s_1(y)) -> s_1(plus_2(x,y)) r21: plus_2(s_1(x),y) -> s_1(plus_2(x,y)) r22: map_2(f,nil_0()) -> nil_0() r23: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs)) r24: filter_2(f,nil_0()) -> nil_0() r25: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs) r26: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs)) r27: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs) The estimated dependency graph contains the following SCCs: (no SCCs) -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: times_2#(x,s_1(y)) -> times_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(filter_0(),y) -> filter_1(y) r4: app(filter_1(x1),y) -> filter_2(x1,y) r5: app(filter2_0(),y) -> filter2_1(y) r6: app(filter2_1(x1),y) -> filter2_2(x1,y) r7: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y) r8: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y) r9: app(map_0(),y) -> map_1(y) r10: app(map_1(x1),y) -> map_2(x1,y) r11: app(plus_0(),y) -> plus_1(y) r12: app(plus_1(x1),y) -> plus_2(x1,y) r13: app(s_0(),y) -> s_1(y) r14: app(times_0(),y) -> times_1(y) r15: app(times_1(x1),y) -> times_2(x1,y) r16: times_2(x,|0|_0()) -> |0|_0() r17: times_2(x,s_1(y)) -> plus_2(times_2(x,y),x) r18: plus_2(x,|0|_0()) -> x r19: plus_2(|0|_0(),x) -> x r20: plus_2(x,s_1(y)) -> s_1(plus_2(x,y)) r21: plus_2(s_1(x),y) -> s_1(plus_2(x,y)) r22: map_2(f,nil_0()) -> nil_0() r23: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs)) r24: filter_2(f,nil_0()) -> nil_0() r25: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs) r26: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs)) r27: 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: matrix interpretation with standard point-wise order: carrier: N^1 order: standard order interpretations: times_2#_A(x1,x2) = x1 + x2 + 4 s_1_A(x1) = x1 + 3 base order 2: marked-algebra reduction triple (>=A, >=A!, >A!): max/plus interpretation on natural numbers: times_2#_A(x1,x2) = 0 s_1_A(x1) = max{12, x1 - 12} times_2#!_A(x1,x2) = 0 s_1!_A(x1) = 0 partial status: pi(times_2#) = [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: plus_2#(x,s_1(y)) -> plus_2#(x,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(filter_0(),y) -> filter_1(y) r4: app(filter_1(x1),y) -> filter_2(x1,y) r5: app(filter2_0(),y) -> filter2_1(y) r6: app(filter2_1(x1),y) -> filter2_2(x1,y) r7: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y) r8: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y) r9: app(map_0(),y) -> map_1(y) r10: app(map_1(x1),y) -> map_2(x1,y) r11: app(plus_0(),y) -> plus_1(y) r12: app(plus_1(x1),y) -> plus_2(x1,y) r13: app(s_0(),y) -> s_1(y) r14: app(times_0(),y) -> times_1(y) r15: app(times_1(x1),y) -> times_2(x1,y) r16: times_2(x,|0|_0()) -> |0|_0() r17: times_2(x,s_1(y)) -> plus_2(times_2(x,y),x) r18: plus_2(x,|0|_0()) -> x r19: plus_2(|0|_0(),x) -> x r20: plus_2(x,s_1(y)) -> s_1(plus_2(x,y)) r21: plus_2(s_1(x),y) -> s_1(plus_2(x,y)) r22: map_2(f,nil_0()) -> nil_0() r23: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs)) r24: filter_2(f,nil_0()) -> nil_0() r25: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs) r26: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs)) r27: 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: matrix interpretation with standard point-wise order: carrier: N^1 order: standard order interpretations: plus_2#_A(x1,x2) = x1 + x2 + 3 s_1_A(x1) = x1 + 2 base order 2: marked-algebra reduction triple (>=A, >=A!, >A!): max/plus interpretation on natural numbers: plus_2#_A(x1,x2) = 0 s_1_A(x1) = max{4, x1 - 4} plus_2#!_A(x1,x2) = max{0, x1 - 5} s_1!_A(x1) = max{2, x1 - 15} partial status: pi(plus_2#) = [1, 2] pi(s_1) = [] The following rules are removed: p1, p2 Then no dependency pair remains.