YES We show the termination of the TRS R: app(app(plus(),|0|()),y) -> 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(app(curry(),g),x),y) -> app(app(g,x),y) inc() -> app(map(),app(app(curry(),plus()),app(s(),|0|()))) 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(curry_0(),y) -> curry_1(y) app(curry_1(x1),y) -> curry_2(x1,y) app(curry_2(x1,x2),y) -> curry_3(x1,x2,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) plus_2(|0|_0(),y) -> 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)) curry_3(g,x,y) -> app(app(g,x),y) inc_0() -> map_1(curry_2(plus_0(),s_1(|0|_0()))) -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: app#(curry_2(x1,x2),y) -> curry_3#(x1,x2,y) p2: app#(map_1(x1),y) -> map_2#(x1,y) p3: app#(plus_1(x1),y) -> plus_2#(x1,y) p4: plus_2#(s_1(x),y) -> plus_2#(x,y) p5: map_2#(f,cons_2(x,xs)) -> app#(f,x) p6: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs) p7: curry_3#(g,x,y) -> app#(app(g,x),y) p8: curry_3#(g,x,y) -> app#(g,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(curry_0(),y) -> curry_1(y) r4: app(curry_1(x1),y) -> curry_2(x1,y) r5: app(curry_2(x1,x2),y) -> curry_3(x1,x2,y) r6: app(map_0(),y) -> map_1(y) r7: app(map_1(x1),y) -> map_2(x1,y) r8: app(plus_0(),y) -> plus_1(y) r9: app(plus_1(x1),y) -> plus_2(x1,y) r10: app(s_0(),y) -> s_1(y) r11: plus_2(|0|_0(),y) -> y r12: plus_2(s_1(x),y) -> s_1(plus_2(x,y)) r13: map_2(f,nil_0()) -> nil_0() r14: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs)) r15: curry_3(g,x,y) -> app(app(g,x),y) r16: inc_0() -> map_1(curry_2(plus_0(),s_1(|0|_0()))) The estimated dependency graph contains the following SCCs: {p1, p2, p5, p6, p7, p8} {p4} -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: app#(curry_2(x1,x2),y) -> curry_3#(x1,x2,y) p2: curry_3#(g,x,y) -> app#(g,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: curry_3#(g,x,y) -> app#(app(g,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(curry_0(),y) -> curry_1(y) r4: app(curry_1(x1),y) -> curry_2(x1,y) r5: app(curry_2(x1,x2),y) -> curry_3(x1,x2,y) r6: app(map_0(),y) -> map_1(y) r7: app(map_1(x1),y) -> map_2(x1,y) r8: app(plus_0(),y) -> plus_1(y) r9: app(plus_1(x1),y) -> plus_2(x1,y) r10: app(s_0(),y) -> s_1(y) r11: plus_2(|0|_0(),y) -> y r12: plus_2(s_1(x),y) -> s_1(plus_2(x,y)) r13: map_2(f,nil_0()) -> nil_0() r14: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs)) r15: curry_3(g,x,y) -> app(app(g,x),y) r16: inc_0() -> map_1(curry_2(plus_0(),s_1(|0|_0()))) The set of usable rules consists of r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15 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: app#_A(x1,x2) = max{9, x1 - 29, x2 - 41} curry_2_A(x1,x2) = max{37, x1 + 36, x2 + 34} curry_3#_A(x1,x2,x3) = max{9, x1 + 1, x2 + 5, x3 - 41} map_1_A(x1) = x1 + 30 map_2#_A(x1,x2) = max{9, x1 + 1, x2 - 48} cons_2_A(x1,x2) = max{47, x1 + 7, x2} app_A(x1,x2) = max{x1 + 26, x2 + 34} cons_0_A = 0 cons_1_A(x1) = max{21, x1 - 5} curry_0_A = 21 curry_1_A(x1) = max{21, x1 + 10} curry_3_A(x1,x2,x3) = max{63, x1 + 52, x2 + 60, x3 + 34} map_0_A = 35 map_2_A(x1,x2) = max{x1 + 33, x2 + 34} plus_0_A = 37 plus_1_A(x1) = 63 plus_2_A(x1,x2) = max{49, x2} s_0_A = 0 s_1_A(x1) = max{29, x1 - 50} |0|_0_A = 13 nil_0_A = 34 app#!_A(x1,x2) = max{x1 - 25, x2 + 8} curry_2!_A(x1,x2) = max{9, x1 - 38, x2 - 1} curry_3#!_A(x1,x2,x3) = max{10, x1 + 2, x2 + 9, x3 + 8} map_1!_A(x1) = 13 map_2#!_A(x1,x2) = max{x1 + 4, x2 + 7} cons_2!_A(x1,x2) = max{12, x2 - 70} app!_A(x1,x2) = max{x1 - 22, x2 + 12} cons_0!_A = 1 cons_1!_A(x1) = 0 curry_0!_A = 2 curry_1!_A(x1) = 1 curry_3!_A(x1,x2,x3) = max{x1 + 4, x2 + 12, x3 + 12} map_0!_A = 1 map_2!_A(x1,x2) = max{1, x1 - 37, x2 - 35} plus_0!_A = 1 plus_1!_A(x1) = max{14, x1 - 1} plus_2!_A(x1,x2) = max{13, x2 + 12} s_0!_A = 11 s_1!_A(x1) = max{12, x1 - 50} |0|_0!_A = 29 nil_0!_A = 0 2. marked-algebra reduction triple (>=A, >=A!, >A!): matrix interpretation with standard point-wise order: carrier: N^1 order: standard order interpretations: app#_A(x1,x2) = 0 curry_2_A(x1,x2) = 1 curry_3#_A(x1,x2,x3) = 0 map_1_A(x1) = 1 map_2#_A(x1,x2) = 0 cons_2_A(x1,x2) = 1 app_A(x1,x2) = x2 + 1 cons_0_A() = 1 cons_1_A(x1) = x1 + 1 curry_0_A() = 1 curry_1_A(x1) = 1 curry_3_A(x1,x2,x3) = x3 + 1 map_0_A() = 1 map_2_A(x1,x2) = 1 plus_0_A() = 1 plus_1_A(x1) = 0 plus_2_A(x1,x2) = x2 + 1 s_0_A() = 1 s_1_A(x1) = 1 |0|_0_A() = 1 nil_0_A() = 1 app#!_A(x1,x2) = 6 curry_2!_A(x1,x2) = 5 curry_3#!_A(x1,x2,x3) = 6 map_1!_A(x1) = 3 map_2#!_A(x1,x2) = 5 cons_2!_A(x1,x2) = 1 app!_A(x1,x2) = x2 + 4 cons_0!_A() = 1 cons_1!_A(x1) = 0 curry_0!_A() = 5 curry_1!_A(x1) = 4 curry_3!_A(x1,x2,x3) = x3 + 4 map_0!_A() = 2 map_2!_A(x1,x2) = 2 plus_0!_A() = 0 plus_1!_A(x1) = 5 plus_2!_A(x1,x2) = x2 + 3 s_0!_A() = 0 s_1!_A(x1) = 2 |0|_0!_A() = 5 nil_0!_A() = 0 partial status: pi(app#) = [] pi(curry_2) = [2] pi(curry_3#) = [] pi(map_1) = [] pi(map_2#) = [] pi(cons_2) = [] pi(app) = [2] pi(cons_0) = [] pi(cons_1) = [] pi(curry_0) = [] pi(curry_1) = [] pi(curry_3) = [3] pi(map_0) = [] pi(map_2) = [2] pi(plus_0) = [] pi(plus_1) = [] pi(plus_2) = [2] pi(s_0) = [] pi(s_1) = [] pi(|0|_0) = [] pi(nil_0) = [] The following rules are removed: p2, p3, p5 -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: app#(curry_2(x1,x2),y) -> curry_3#(x1,x2,y) p2: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs) p3: curry_3#(g,x,y) -> app#(app(g,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(curry_0(),y) -> curry_1(y) r4: app(curry_1(x1),y) -> curry_2(x1,y) r5: app(curry_2(x1,x2),y) -> curry_3(x1,x2,y) r6: app(map_0(),y) -> map_1(y) r7: app(map_1(x1),y) -> map_2(x1,y) r8: app(plus_0(),y) -> plus_1(y) r9: app(plus_1(x1),y) -> plus_2(x1,y) r10: app(s_0(),y) -> s_1(y) r11: plus_2(|0|_0(),y) -> y r12: plus_2(s_1(x),y) -> s_1(plus_2(x,y)) r13: map_2(f,nil_0()) -> nil_0() r14: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs)) r15: curry_3(g,x,y) -> app(app(g,x),y) r16: inc_0() -> map_1(curry_2(plus_0(),s_1(|0|_0()))) The estimated dependency graph contains the following SCCs: {p1, p3} {p2} -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: app#(curry_2(x1,x2),y) -> curry_3#(x1,x2,y) p2: curry_3#(g,x,y) -> app#(app(g,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(curry_0(),y) -> curry_1(y) r4: app(curry_1(x1),y) -> curry_2(x1,y) r5: app(curry_2(x1,x2),y) -> curry_3(x1,x2,y) r6: app(map_0(),y) -> map_1(y) r7: app(map_1(x1),y) -> map_2(x1,y) r8: app(plus_0(),y) -> plus_1(y) r9: app(plus_1(x1),y) -> plus_2(x1,y) r10: app(s_0(),y) -> s_1(y) r11: plus_2(|0|_0(),y) -> y r12: plus_2(s_1(x),y) -> s_1(plus_2(x,y)) r13: map_2(f,nil_0()) -> nil_0() r14: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs)) r15: curry_3(g,x,y) -> app(app(g,x),y) r16: inc_0() -> map_1(curry_2(plus_0(),s_1(|0|_0()))) The set of usable rules consists of r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15 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: app#_A(x1,x2) = max{0, x1 - 24, x2 - 2} curry_2_A(x1,x2) = max{x1 - 14, x2 + 13} curry_3#_A(x1,x2,x3) = max{0, x1 - 38, x2 - 11, x3 - 2} app_A(x1,x2) = max{23, x1 - 24, x2 + 13} cons_0_A = 0 cons_1_A(x1) = max{14, x1 + 13} cons_2_A(x1,x2) = x2 + 13 curry_0_A = 16 curry_1_A(x1) = max{16, x1 + 13} curry_3_A(x1,x2,x3) = max{23, x1 - 38, x2 - 11, x3 + 13} map_0_A = 16 map_1_A(x1) = max{18, x1 - 17} map_2_A(x1,x2) = x2 + 11 plus_0_A = 0 plus_1_A(x1) = max{19, x1 + 13} plus_2_A(x1,x2) = x2 + 11 s_0_A = 22 s_1_A(x1) = max{3, x1 - 3} |0|_0_A = 1 nil_0_A = 2 app#!_A(x1,x2) = 6 curry_2!_A(x1,x2) = max{6, x2 + 1} curry_3#!_A(x1,x2,x3) = 6 app!_A(x1,x2) = max{x1 - 15, x2 + 13} cons_0!_A = 1 cons_1!_A(x1) = max{0, x1 - 1} cons_2!_A(x1,x2) = max{1, x2 - 12} curry_0!_A = 3 curry_1!_A(x1) = 2 curry_3!_A(x1,x2,x3) = max{x1 - 29, x2 - 2, x3 + 13} map_0!_A = 1 map_1!_A(x1) = 0 map_2!_A(x1,x2) = x2 + 12 plus_0!_A = 1 plus_1!_A(x1) = x1 + 4 plus_2!_A(x1,x2) = x2 + 7 s_0!_A = 7 s_1!_A(x1) = 6 |0|_0!_A = 6 nil_0!_A = 13 2. marked-algebra reduction triple (>=A, >=A!, >A!): matrix interpretation with standard point-wise order: carrier: N^1 order: standard order interpretations: app#_A(x1,x2) = x1 curry_2_A(x1,x2) = x1 + x2 + 3 curry_3#_A(x1,x2,x3) = x1 + x2 + 1 app_A(x1,x2) = x1 + x2 + 1 cons_0_A() = 1 cons_1_A(x1) = x1 + 1 cons_2_A(x1,x2) = 1 curry_0_A() = 1 curry_1_A(x1) = x1 + 2 curry_3_A(x1,x2,x3) = x1 + x2 + x3 + 2 map_0_A() = 1 map_1_A(x1) = 1 map_2_A(x1,x2) = 1 plus_0_A() = 1 plus_1_A(x1) = x1 + 1 plus_2_A(x1,x2) = x2 + 1 s_0_A() = 1 s_1_A(x1) = 1 |0|_0_A() = 1 nil_0_A() = 1 app#!_A(x1,x2) = x1 + 4 curry_2!_A(x1,x2) = 3 curry_3#!_A(x1,x2,x3) = x1 + x2 + 6 app!_A(x1,x2) = x2 + 5 cons_0!_A() = 0 cons_1!_A(x1) = 1 cons_2!_A(x1,x2) = 2 curry_0!_A() = 7 curry_1!_A(x1) = 6 curry_3!_A(x1,x2,x3) = x3 + 5 map_0!_A() = 5 map_1!_A(x1) = 4 map_2!_A(x1,x2) = 3 plus_0!_A() = 1 plus_1!_A(x1) = x1 + 6 plus_2!_A(x1,x2) = 4 s_0!_A() = 0 s_1!_A(x1) = 3 |0|_0!_A() = 3 nil_0!_A() = 0 partial status: pi(app#) = [] pi(curry_2) = [] pi(curry_3#) = [] pi(app) = [2] pi(cons_0) = [] pi(cons_1) = [] pi(cons_2) = [] pi(curry_0) = [] pi(curry_1) = [] pi(curry_3) = [3] pi(map_0) = [] pi(map_1) = [] pi(map_2) = [] pi(plus_0) = [] pi(plus_1) = [1] pi(plus_2) = [2] pi(s_0) = [] pi(s_1) = [] pi(|0|_0) = [] pi(nil_0) = [] 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: 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(curry_0(),y) -> curry_1(y) r4: app(curry_1(x1),y) -> curry_2(x1,y) r5: app(curry_2(x1,x2),y) -> curry_3(x1,x2,y) r6: app(map_0(),y) -> map_1(y) r7: app(map_1(x1),y) -> map_2(x1,y) r8: app(plus_0(),y) -> plus_1(y) r9: app(plus_1(x1),y) -> plus_2(x1,y) r10: app(s_0(),y) -> s_1(y) r11: plus_2(|0|_0(),y) -> y r12: plus_2(s_1(x),y) -> s_1(plus_2(x,y)) r13: map_2(f,nil_0()) -> nil_0() r14: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs)) r15: curry_3(g,x,y) -> app(app(g,x),y) r16: inc_0() -> map_1(curry_2(plus_0(),s_1(|0|_0()))) 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 - 13, x2 - 14} map_2#!_A(x1,x2) = max{21, x2 + 15} cons_2!_A(x1,x2) = max{21, x1 - 6, x2 + 20} 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) = x2 + 2 map_2#!_A(x1,x2) = x2 + 6 cons_2!_A(x1,x2) = x2 + 6 partial status: pi(map_2#) = [2] pi(cons_2) = [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: 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(curry_0(),y) -> curry_1(y) r4: app(curry_1(x1),y) -> curry_2(x1,y) r5: app(curry_2(x1,x2),y) -> curry_3(x1,x2,y) r6: app(map_0(),y) -> map_1(y) r7: app(map_1(x1),y) -> map_2(x1,y) r8: app(plus_0(),y) -> plus_1(y) r9: app(plus_1(x1),y) -> plus_2(x1,y) r10: app(s_0(),y) -> s_1(y) r11: plus_2(|0|_0(),y) -> y r12: plus_2(s_1(x),y) -> s_1(plus_2(x,y)) r13: map_2(f,nil_0()) -> nil_0() r14: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs)) r15: curry_3(g,x,y) -> app(app(g,x),y) r16: inc_0() -> map_1(curry_2(plus_0(),s_1(|0|_0()))) 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) = 0 s_1_A(x1) = max{21, x1 + 11} plus_2#!_A(x1,x2) = max{9, x1 - 1} s_1!_A(x1) = 8 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 s_1_A(x1) = x1 + 2 plus_2#!_A(x1,x2) = 0 s_1!_A(x1) = 2 partial status: pi(plus_2#) = [] pi(s_1) = [] The following rules are removed: p1 Then no dependency pair remains.