YES We show the termination of the TRS R: app(not(),app(not(),x)) -> x app(not(),app(app(or(),x),y)) -> app(app(and(),app(not(),x)),app(not(),y)) app(not(),app(app(and(),x),y)) -> app(app(or(),app(not(),x)),app(not(),y)) app(app(and(),x),app(app(or(),y),z)) -> app(app(or(),app(app(and(),x),y)),app(app(and(),x),z)) app(app(and(),app(app(or(),y),z)),x) -> app(app(or(),app(app(and(),x),y)),app(app(and(),x),z)) 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(and_0(),y) -> and_1(y) app(and_1(x1),y) -> and_2(x1,y) 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(not_0(),y) -> not_1(y) app(or_0(),y) -> or_1(y) app(or_1(x1),y) -> or_2(x1,y) not_1(not_1(x)) -> x not_1(or_2(x,y)) -> and_2(not_1(x),not_1(y)) not_1(and_2(x,y)) -> or_2(not_1(x),not_1(y)) and_2(x,or_2(y,z)) -> or_2(and_2(x,y),and_2(x,z)) and_2(or_2(y,z),x) -> or_2(and_2(x,y),and_2(x,z)) 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#(and_1(x1),y) -> and_2#(x1,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#(not_0(),y) -> not_1#(y) p6: not_1#(or_2(x,y)) -> and_2#(not_1(x),not_1(y)) p7: not_1#(or_2(x,y)) -> not_1#(x) p8: not_1#(or_2(x,y)) -> not_1#(y) p9: not_1#(and_2(x,y)) -> not_1#(x) p10: not_1#(and_2(x,y)) -> not_1#(y) p11: and_2#(x,or_2(y,z)) -> and_2#(x,y) p12: and_2#(x,or_2(y,z)) -> and_2#(x,z) p13: and_2#(or_2(y,z),x) -> and_2#(x,y) p14: and_2#(or_2(y,z),x) -> and_2#(x,z) p15: map_2#(f,cons_2(x,xs)) -> app#(f,x) p16: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs) p17: filter_2#(f,cons_2(x,xs)) -> filter2_4#(app(f,x),f,x,xs) p18: filter_2#(f,cons_2(x,xs)) -> app#(f,x) p19: filter2_4#(true_0(),f,x,xs) -> filter_2#(f,xs) p20: filter2_4#(false_0(),f,x,xs) -> filter_2#(f,xs) and R consists of: r1: app(and_0(),y) -> and_1(y) r2: app(and_1(x1),y) -> and_2(x1,y) r3: app(cons_0(),y) -> cons_1(y) r4: app(cons_1(x1),y) -> cons_2(x1,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(map_0(),y) -> map_1(y) r12: app(map_1(x1),y) -> map_2(x1,y) r13: app(not_0(),y) -> not_1(y) r14: app(or_0(),y) -> or_1(y) r15: app(or_1(x1),y) -> or_2(x1,y) r16: not_1(not_1(x)) -> x r17: not_1(or_2(x,y)) -> and_2(not_1(x),not_1(y)) r18: not_1(and_2(x,y)) -> or_2(not_1(x),not_1(y)) r19: and_2(x,or_2(y,z)) -> or_2(and_2(x,y),and_2(x,z)) r20: and_2(or_2(y,z),x) -> or_2(and_2(x,y),and_2(x,z)) r21: map_2(f,nil_0()) -> nil_0() r22: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs)) r23: filter_2(f,nil_0()) -> nil_0() r24: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs) r25: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs)) r26: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs) The estimated dependency graph contains the following SCCs: {p2, p3, p4, p15, p16, p17, p18, p19, p20} {p7, p8, p9, p10} {p11, p12, p13, p14} -- 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(and_0(),y) -> and_1(y) r2: app(and_1(x1),y) -> and_2(x1,y) r3: app(cons_0(),y) -> cons_1(y) r4: app(cons_1(x1),y) -> cons_2(x1,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(map_0(),y) -> map_1(y) r12: app(map_1(x1),y) -> map_2(x1,y) r13: app(not_0(),y) -> not_1(y) r14: app(or_0(),y) -> or_1(y) r15: app(or_1(x1),y) -> or_2(x1,y) r16: not_1(not_1(x)) -> x r17: not_1(or_2(x,y)) -> and_2(not_1(x),not_1(y)) r18: not_1(and_2(x,y)) -> or_2(not_1(x),not_1(y)) r19: and_2(x,or_2(y,z)) -> or_2(and_2(x,y),and_2(x,z)) r20: and_2(or_2(y,z),x) -> or_2(and_2(x,y),and_2(x,z)) r21: map_2(f,nil_0()) -> nil_0() r22: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs)) r23: filter_2(f,nil_0()) -> nil_0() r24: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs) r25: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs)) r26: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs) The set of usable rules consists of r21 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 - 12, x2 - 46} cons_2_A(x1,x2) = max{12, x1 + 1, x2 + 1} app#_A(x1,x2) = max{1, x1 - 17, x2 - 45} map_1_A(x1) = max{7, x1 + 5} filter2_3_A(x1,x2,x3) = max{x1 - 29, x2 + 5, x3 + 16} filter2_4#_A(x1,x2,x3,x4) = max{1, x2 - 16, x3 - 45, x4 - 45} false_0_A = 46 filter_2#_A(x1,x2) = max{1, x1 - 16, x2 - 46} filter_1_A(x1) = max{2, x1 + 1} app_A(x1,x2) = max{46, x1 - 15, x2 + 1} true_0_A = 46 map_2_A(x1,x2) = max{x1 + 47, x2 + 1} nil_0_A = 3 map_2#!_A(x1,x2) = max{1, x1 - 2} cons_2!_A(x1,x2) = max{0, x1 - 50, x2 - 64} app#!_A(x1,x2) = max{0, x1 - 3} map_1!_A(x1) = max{2, x1 - 1} filter2_3!_A(x1,x2,x3) = x3 + 1 filter2_4#!_A(x1,x2,x3,x4) = max{0, x2 - 3} false_0!_A = 1 filter_2#!_A(x1,x2) = max{0, x1 - 3} filter_1!_A(x1) = x1 + 3 app!_A(x1,x2) = max{6, x1 - 6, x2 + 4} true_0!_A = 1 map_2!_A(x1,x2) = max{7, x2 - 4} nil_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: map_2#_A(x1,x2) = 0 cons_2_A(x1,x2) = x2 + 7 app#_A(x1,x2) = 0 map_1_A(x1) = x1 filter2_3_A(x1,x2,x3) = x2 + x3 filter2_4#_A(x1,x2,x3,x4) = 0 false_0_A() = 9 filter_2#_A(x1,x2) = 0 filter_1_A(x1) = x1 + 1 app_A(x1,x2) = 7 true_0_A() = 9 map_2_A(x1,x2) = x2 + 1 nil_0_A() = 1 map_2#!_A(x1,x2) = x1 + 1 cons_2!_A(x1,x2) = x1 + x2 + 18 app#!_A(x1,x2) = 0 map_1!_A(x1) = x1 + 2 filter2_3!_A(x1,x2,x3) = 3 filter2_4#!_A(x1,x2,x3,x4) = x2 + x4 + 2 false_0!_A() = 13 filter_2#!_A(x1,x2) = x1 + x2 + 2 filter_1!_A(x1) = 17 app!_A(x1,x2) = x1 + 15 true_0!_A() = 1 map_2!_A(x1,x2) = 14 nil_0!_A() = 13 partial status: pi(map_2#) = [1] pi(cons_2) = [2] pi(app#) = [1] pi(map_1) = [1] pi(filter2_3) = [2] pi(filter2_4#) = [2] pi(false_0) = [] pi(filter_2#) = [1] pi(filter_1) = [1] pi(app) = [1, 2] pi(true_0) = [] pi(map_2) = [1] pi(nil_0) = [] The following rules are removed: p1, p2, p4, p6, p7, p8 -- 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: 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(and_0(),y) -> and_1(y) r2: app(and_1(x1),y) -> and_2(x1,y) r3: app(cons_0(),y) -> cons_1(y) r4: app(cons_1(x1),y) -> cons_2(x1,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(map_0(),y) -> map_1(y) r12: app(map_1(x1),y) -> map_2(x1,y) r13: app(not_0(),y) -> not_1(y) r14: app(or_0(),y) -> or_1(y) r15: app(or_1(x1),y) -> or_2(x1,y) r16: not_1(not_1(x)) -> x r17: not_1(or_2(x,y)) -> and_2(not_1(x),not_1(y)) r18: not_1(and_2(x,y)) -> or_2(not_1(x),not_1(y)) r19: and_2(x,or_2(y,z)) -> or_2(and_2(x,y),and_2(x,z)) r20: and_2(or_2(y,z),x) -> or_2(and_2(x,y),and_2(x,z)) r21: map_2(f,nil_0()) -> nil_0() r22: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs)) r23: filter_2(f,nil_0()) -> nil_0() r24: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs) r25: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs)) r26: 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: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs) and R consists of: r1: app(and_0(),y) -> and_1(y) r2: app(and_1(x1),y) -> and_2(x1,y) r3: app(cons_0(),y) -> cons_1(y) r4: app(cons_1(x1),y) -> cons_2(x1,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(map_0(),y) -> map_1(y) r12: app(map_1(x1),y) -> map_2(x1,y) r13: app(not_0(),y) -> not_1(y) r14: app(or_0(),y) -> or_1(y) r15: app(or_1(x1),y) -> or_2(x1,y) r16: not_1(not_1(x)) -> x r17: not_1(or_2(x,y)) -> and_2(not_1(x),not_1(y)) r18: not_1(and_2(x,y)) -> or_2(not_1(x),not_1(y)) r19: and_2(x,or_2(y,z)) -> or_2(and_2(x,y),and_2(x,z)) r20: and_2(or_2(y,z),x) -> or_2(and_2(x,y),and_2(x,z)) r21: map_2(f,nil_0()) -> nil_0() r22: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs)) r23: filter_2(f,nil_0()) -> nil_0() r24: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs) r25: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs)) r26: 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{3, x1 - 4, x2 - 5} map_2#!_A(x1,x2) = 0 cons_2!_A(x1,x2) = max{4, x2 - 12} 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 cons_2!_A(x1,x2) = x1 + 9 partial status: pi(map_2#) = [] pi(cons_2) = [] 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: not_1#(and_2(x,y)) -> not_1#(y) p2: not_1#(and_2(x,y)) -> not_1#(x) p3: not_1#(or_2(x,y)) -> not_1#(y) p4: not_1#(or_2(x,y)) -> not_1#(x) and R consists of: r1: app(and_0(),y) -> and_1(y) r2: app(and_1(x1),y) -> and_2(x1,y) r3: app(cons_0(),y) -> cons_1(y) r4: app(cons_1(x1),y) -> cons_2(x1,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(map_0(),y) -> map_1(y) r12: app(map_1(x1),y) -> map_2(x1,y) r13: app(not_0(),y) -> not_1(y) r14: app(or_0(),y) -> or_1(y) r15: app(or_1(x1),y) -> or_2(x1,y) r16: not_1(not_1(x)) -> x r17: not_1(or_2(x,y)) -> and_2(not_1(x),not_1(y)) r18: not_1(and_2(x,y)) -> or_2(not_1(x),not_1(y)) r19: and_2(x,or_2(y,z)) -> or_2(and_2(x,y),and_2(x,z)) r20: and_2(or_2(y,z),x) -> or_2(and_2(x,y),and_2(x,z)) r21: map_2(f,nil_0()) -> nil_0() r22: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs)) r23: filter_2(f,nil_0()) -> nil_0() r24: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs) r25: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs)) r26: 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: not_1#_A(x1) = 0 and_2_A(x1,x2) = max{257, x1 + 35, x2 - 258} or_2_A(x1,x2) = max{71, x1 - 258} not_1#!_A(x1) = 1 and_2!_A(x1,x2) = max{37, x1 - 34, x2 - 107} or_2!_A(x1,x2) = max{40, x1 - 256, x2 - 292} 2. marked-algebra reduction triple (>=A, >=A!, >A!): matrix interpretation with standard point-wise order: carrier: N^1 order: standard order interpretations: not_1#_A(x1) = x1 and_2_A(x1,x2) = x1 + x2 + 15 or_2_A(x1,x2) = x1 + x2 + 8 not_1#!_A(x1) = x1 + 13 and_2!_A(x1,x2) = x1 + x2 + 14 or_2!_A(x1,x2) = x1 + x2 + 21 partial status: pi(not_1#) = [] pi(and_2) = [] pi(or_2) = [1] The following rules are removed: p1, p2, p3, p4 Then no dependency pair remains. -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: and_2#(x,or_2(y,z)) -> and_2#(x,y) p2: and_2#(or_2(y,z),x) -> and_2#(x,z) p3: and_2#(or_2(y,z),x) -> and_2#(x,y) p4: and_2#(x,or_2(y,z)) -> and_2#(x,z) and R consists of: r1: app(and_0(),y) -> and_1(y) r2: app(and_1(x1),y) -> and_2(x1,y) r3: app(cons_0(),y) -> cons_1(y) r4: app(cons_1(x1),y) -> cons_2(x1,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(map_0(),y) -> map_1(y) r12: app(map_1(x1),y) -> map_2(x1,y) r13: app(not_0(),y) -> not_1(y) r14: app(or_0(),y) -> or_1(y) r15: app(or_1(x1),y) -> or_2(x1,y) r16: not_1(not_1(x)) -> x r17: not_1(or_2(x,y)) -> and_2(not_1(x),not_1(y)) r18: not_1(and_2(x,y)) -> or_2(not_1(x),not_1(y)) r19: and_2(x,or_2(y,z)) -> or_2(and_2(x,y),and_2(x,z)) r20: and_2(or_2(y,z),x) -> or_2(and_2(x,y),and_2(x,z)) r21: map_2(f,nil_0()) -> nil_0() r22: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs)) r23: filter_2(f,nil_0()) -> nil_0() r24: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs) r25: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs)) r26: 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: and_2#_A(x1,x2) = 0 or_2_A(x1,x2) = max{6, x1 - 26, x2 - 58} and_2#!_A(x1,x2) = 0 or_2!_A(x1,x2) = max{8, x2 - 9} 2. marked-algebra reduction triple (>=A, >=A!, >A!): matrix interpretation with standard point-wise order: carrier: N^1 order: standard order interpretations: and_2#_A(x1,x2) = x1 + x2 or_2_A(x1,x2) = x1 + x2 + 29 and_2#!_A(x1,x2) = x1 + x2 or_2!_A(x1,x2) = x1 + x2 + 19 partial status: pi(and_2#) = [] pi(or_2) = [1, 2] The following rules are removed: p1, p2, p3, p4 Then no dependency pair remains.