YES We show the termination of the TRS R: |0|(|#|()) -> |#|() +(x,|#|()) -> x +(|#|(),x) -> x +(|0|(x),|0|(y)) -> |0|(+(x,y)) +(|0|(x),|1|(y)) -> |1|(+(x,y)) +(|1|(x),|0|(y)) -> |1|(+(x,y)) +(|1|(x),|1|(y)) -> |0|(+(+(x,y),|1|(|#|()))) +(+(x,y),z) -> +(x,+(y,z)) *(|#|(),x) -> |#|() *(|0|(x),y) -> |0|(*(x,y)) *(|1|(x),y) -> +(|0|(*(x,y)),y) *(*(x,y),z) -> *(x,*(y,z)) *(x,+(y,z)) -> +(*(x,y),*(x,z)) app(nil(),l) -> l app(cons(x,l1),l2) -> cons(x,app(l1,l2)) sum(nil()) -> |0|(|#|()) sum(cons(x,l)) -> +(x,sum(l)) sum(app(l1,l2)) -> +(sum(l1),sum(l2)) prod(nil()) -> |1|(|#|()) prod(cons(x,l)) -> *(x,prod(l)) prod(app(l1,l2)) -> *(prod(l1),prod(l2)) -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: +#(|0|(x),|0|(y)) -> |0|#(+(x,y)) p2: +#(|0|(x),|0|(y)) -> +#(x,y) p3: +#(|0|(x),|1|(y)) -> +#(x,y) p4: +#(|1|(x),|0|(y)) -> +#(x,y) p5: +#(|1|(x),|1|(y)) -> |0|#(+(+(x,y),|1|(|#|()))) p6: +#(|1|(x),|1|(y)) -> +#(+(x,y),|1|(|#|())) p7: +#(|1|(x),|1|(y)) -> +#(x,y) p8: +#(+(x,y),z) -> +#(x,+(y,z)) p9: +#(+(x,y),z) -> +#(y,z) p10: *#(|0|(x),y) -> |0|#(*(x,y)) p11: *#(|0|(x),y) -> *#(x,y) p12: *#(|1|(x),y) -> +#(|0|(*(x,y)),y) p13: *#(|1|(x),y) -> |0|#(*(x,y)) p14: *#(|1|(x),y) -> *#(x,y) p15: *#(*(x,y),z) -> *#(x,*(y,z)) p16: *#(*(x,y),z) -> *#(y,z) p17: *#(x,+(y,z)) -> +#(*(x,y),*(x,z)) p18: *#(x,+(y,z)) -> *#(x,y) p19: *#(x,+(y,z)) -> *#(x,z) p20: app#(cons(x,l1),l2) -> app#(l1,l2) p21: sum#(nil()) -> |0|#(|#|()) p22: sum#(cons(x,l)) -> +#(x,sum(l)) p23: sum#(cons(x,l)) -> sum#(l) p24: sum#(app(l1,l2)) -> +#(sum(l1),sum(l2)) p25: sum#(app(l1,l2)) -> sum#(l1) p26: sum#(app(l1,l2)) -> sum#(l2) p27: prod#(cons(x,l)) -> *#(x,prod(l)) p28: prod#(cons(x,l)) -> prod#(l) p29: prod#(app(l1,l2)) -> *#(prod(l1),prod(l2)) p30: prod#(app(l1,l2)) -> prod#(l1) p31: prod#(app(l1,l2)) -> prod#(l2) and R consists of: r1: |0|(|#|()) -> |#|() r2: +(x,|#|()) -> x r3: +(|#|(),x) -> x r4: +(|0|(x),|0|(y)) -> |0|(+(x,y)) r5: +(|0|(x),|1|(y)) -> |1|(+(x,y)) r6: +(|1|(x),|0|(y)) -> |1|(+(x,y)) r7: +(|1|(x),|1|(y)) -> |0|(+(+(x,y),|1|(|#|()))) r8: +(+(x,y),z) -> +(x,+(y,z)) r9: *(|#|(),x) -> |#|() r10: *(|0|(x),y) -> |0|(*(x,y)) r11: *(|1|(x),y) -> +(|0|(*(x,y)),y) r12: *(*(x,y),z) -> *(x,*(y,z)) r13: *(x,+(y,z)) -> +(*(x,y),*(x,z)) r14: app(nil(),l) -> l r15: app(cons(x,l1),l2) -> cons(x,app(l1,l2)) r16: sum(nil()) -> |0|(|#|()) r17: sum(cons(x,l)) -> +(x,sum(l)) r18: sum(app(l1,l2)) -> +(sum(l1),sum(l2)) r19: prod(nil()) -> |1|(|#|()) r20: prod(cons(x,l)) -> *(x,prod(l)) r21: prod(app(l1,l2)) -> *(prod(l1),prod(l2)) The estimated dependency graph contains the following SCCs: {p23, p25, p26} {p28, p30, p31} {p11, p14, p15, p16, p18, p19} {p2, p3, p4, p6, p7, p8, p9} {p20} -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: sum#(app(l1,l2)) -> sum#(l2) p2: sum#(app(l1,l2)) -> sum#(l1) p3: sum#(cons(x,l)) -> sum#(l) and R consists of: r1: |0|(|#|()) -> |#|() r2: +(x,|#|()) -> x r3: +(|#|(),x) -> x r4: +(|0|(x),|0|(y)) -> |0|(+(x,y)) r5: +(|0|(x),|1|(y)) -> |1|(+(x,y)) r6: +(|1|(x),|0|(y)) -> |1|(+(x,y)) r7: +(|1|(x),|1|(y)) -> |0|(+(+(x,y),|1|(|#|()))) r8: +(+(x,y),z) -> +(x,+(y,z)) r9: *(|#|(),x) -> |#|() r10: *(|0|(x),y) -> |0|(*(x,y)) r11: *(|1|(x),y) -> +(|0|(*(x,y)),y) r12: *(*(x,y),z) -> *(x,*(y,z)) r13: *(x,+(y,z)) -> +(*(x,y),*(x,z)) r14: app(nil(),l) -> l r15: app(cons(x,l1),l2) -> cons(x,app(l1,l2)) r16: sum(nil()) -> |0|(|#|()) r17: sum(cons(x,l)) -> +(x,sum(l)) r18: sum(app(l1,l2)) -> +(sum(l1),sum(l2)) r19: prod(nil()) -> |1|(|#|()) r20: prod(cons(x,l)) -> *(x,prod(l)) r21: prod(app(l1,l2)) -> *(prod(l1),prod(l2)) 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: sum#_A(x1) = x1 app_A(x1,x2) = x1 + x2 + 9 cons_A(x1,x2) = x1 + x2 + 8 base order 2: marked-algebra reduction triple (>=A, >=A!, >A!): max/plus interpretation on natural numbers: sum#_A(x1) = max{0, x1 - 2} app_A(x1,x2) = max{61, x1 + 1, x2 + 1} cons_A(x1,x2) = max{54, x1 - 2, x2 + 1} sum#!_A(x1) = 6 app!_A(x1,x2) = max{7, x1 - 99} cons!_A(x1,x2) = max{x1 + 6, x2 - 98} partial status: pi(sum#) = [1] pi(app) = [1, 2] pi(cons) = [2] 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: prod#(app(l1,l2)) -> prod#(l2) p2: prod#(app(l1,l2)) -> prod#(l1) p3: prod#(cons(x,l)) -> prod#(l) and R consists of: r1: |0|(|#|()) -> |#|() r2: +(x,|#|()) -> x r3: +(|#|(),x) -> x r4: +(|0|(x),|0|(y)) -> |0|(+(x,y)) r5: +(|0|(x),|1|(y)) -> |1|(+(x,y)) r6: +(|1|(x),|0|(y)) -> |1|(+(x,y)) r7: +(|1|(x),|1|(y)) -> |0|(+(+(x,y),|1|(|#|()))) r8: +(+(x,y),z) -> +(x,+(y,z)) r9: *(|#|(),x) -> |#|() r10: *(|0|(x),y) -> |0|(*(x,y)) r11: *(|1|(x),y) -> +(|0|(*(x,y)),y) r12: *(*(x,y),z) -> *(x,*(y,z)) r13: *(x,+(y,z)) -> +(*(x,y),*(x,z)) r14: app(nil(),l) -> l r15: app(cons(x,l1),l2) -> cons(x,app(l1,l2)) r16: sum(nil()) -> |0|(|#|()) r17: sum(cons(x,l)) -> +(x,sum(l)) r18: sum(app(l1,l2)) -> +(sum(l1),sum(l2)) r19: prod(nil()) -> |1|(|#|()) r20: prod(cons(x,l)) -> *(x,prod(l)) r21: prod(app(l1,l2)) -> *(prod(l1),prod(l2)) 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: prod#_A(x1) = x1 app_A(x1,x2) = x1 + x2 + 9 cons_A(x1,x2) = x1 + x2 + 8 base order 2: marked-algebra reduction triple (>=A, >=A!, >A!): max/plus interpretation on natural numbers: prod#_A(x1) = max{0, x1 - 4} app_A(x1,x2) = max{6, x1, x2 + 3} cons_A(x1,x2) = max{7, x1 - 1, x2 + 3} prod#!_A(x1) = max{1, x1 - 165} app!_A(x1,x2) = max{8, x1 - 164, x2 - 83} cons!_A(x1,x2) = max{7, x2 - 87} partial status: pi(prod#) = [1] pi(app) = [2] pi(cons) = [2] 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: *#(x,+(y,z)) -> *#(x,z) p2: *#(x,+(y,z)) -> *#(x,y) p3: *#(*(x,y),z) -> *#(y,z) p4: *#(*(x,y),z) -> *#(x,*(y,z)) p5: *#(|1|(x),y) -> *#(x,y) p6: *#(|0|(x),y) -> *#(x,y) and R consists of: r1: |0|(|#|()) -> |#|() r2: +(x,|#|()) -> x r3: +(|#|(),x) -> x r4: +(|0|(x),|0|(y)) -> |0|(+(x,y)) r5: +(|0|(x),|1|(y)) -> |1|(+(x,y)) r6: +(|1|(x),|0|(y)) -> |1|(+(x,y)) r7: +(|1|(x),|1|(y)) -> |0|(+(+(x,y),|1|(|#|()))) r8: +(+(x,y),z) -> +(x,+(y,z)) r9: *(|#|(),x) -> |#|() r10: *(|0|(x),y) -> |0|(*(x,y)) r11: *(|1|(x),y) -> +(|0|(*(x,y)),y) r12: *(*(x,y),z) -> *(x,*(y,z)) r13: *(x,+(y,z)) -> +(*(x,y),*(x,z)) r14: app(nil(),l) -> l r15: app(cons(x,l1),l2) -> cons(x,app(l1,l2)) r16: sum(nil()) -> |0|(|#|()) r17: sum(cons(x,l)) -> +(x,sum(l)) r18: sum(app(l1,l2)) -> +(sum(l1),sum(l2)) r19: prod(nil()) -> |1|(|#|()) r20: prod(cons(x,l)) -> *(x,prod(l)) r21: prod(app(l1,l2)) -> *(prod(l1),prod(l2)) The set of usable rules consists of r2, r3, r9 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: *#_A(x1,x2) = x1 + 1 +_A(x1,x2) = x1 + x2 + 3 *_A(x1,x2) = x1 + x2 + 8 |1|_A(x1) = x1 + 7 |0|_A(x1) = x1 + 2 |#|_A() = 1 base order 2: marked-algebra reduction triple (>=A, >=A!, >A!): max/plus interpretation on natural numbers: *#_A(x1,x2) = max{0, x1 - 98} +_A(x1,x2) = max{65, x1 + 1, x2 + 1} *_A(x1,x2) = max{66, x1 + 1, x2 + 1} |1|_A(x1) = max{92, x1 + 90} |0|_A(x1) = x1 + 97 |#|_A = 95 *#!_A(x1,x2) = max{1, x1 - 225} +!_A(x1,x2) = max{5, x1 - 170, x2 - 225} *!_A(x1,x2) = max{6, x1 - 96, x2 - 222} |1|!_A(x1) = max{2, x1 - 137} |0|!_A(x1) = max{4, x1 - 224} |#|!_A = 3 partial status: pi(*#) = [1] pi(+) = [1] pi(*) = [1, 2] pi(|1|) = [1] pi(|0|) = [1] pi(|#|) = [] The following rules are removed: p3, p4, p5, p6 -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: *#(x,+(y,z)) -> *#(x,z) p2: *#(x,+(y,z)) -> *#(x,y) and R consists of: r1: |0|(|#|()) -> |#|() r2: +(x,|#|()) -> x r3: +(|#|(),x) -> x r4: +(|0|(x),|0|(y)) -> |0|(+(x,y)) r5: +(|0|(x),|1|(y)) -> |1|(+(x,y)) r6: +(|1|(x),|0|(y)) -> |1|(+(x,y)) r7: +(|1|(x),|1|(y)) -> |0|(+(+(x,y),|1|(|#|()))) r8: +(+(x,y),z) -> +(x,+(y,z)) r9: *(|#|(),x) -> |#|() r10: *(|0|(x),y) -> |0|(*(x,y)) r11: *(|1|(x),y) -> +(|0|(*(x,y)),y) r12: *(*(x,y),z) -> *(x,*(y,z)) r13: *(x,+(y,z)) -> +(*(x,y),*(x,z)) r14: app(nil(),l) -> l r15: app(cons(x,l1),l2) -> cons(x,app(l1,l2)) r16: sum(nil()) -> |0|(|#|()) r17: sum(cons(x,l)) -> +(x,sum(l)) r18: sum(app(l1,l2)) -> +(sum(l1),sum(l2)) r19: prod(nil()) -> |1|(|#|()) r20: prod(cons(x,l)) -> *(x,prod(l)) r21: prod(app(l1,l2)) -> *(prod(l1),prod(l2)) The estimated dependency graph contains the following SCCs: {p1, p2} -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: *#(x,+(y,z)) -> *#(x,z) p2: *#(x,+(y,z)) -> *#(x,y) and R consists of: r1: |0|(|#|()) -> |#|() r2: +(x,|#|()) -> x r3: +(|#|(),x) -> x r4: +(|0|(x),|0|(y)) -> |0|(+(x,y)) r5: +(|0|(x),|1|(y)) -> |1|(+(x,y)) r6: +(|1|(x),|0|(y)) -> |1|(+(x,y)) r7: +(|1|(x),|1|(y)) -> |0|(+(+(x,y),|1|(|#|()))) r8: +(+(x,y),z) -> +(x,+(y,z)) r9: *(|#|(),x) -> |#|() r10: *(|0|(x),y) -> |0|(*(x,y)) r11: *(|1|(x),y) -> +(|0|(*(x,y)),y) r12: *(*(x,y),z) -> *(x,*(y,z)) r13: *(x,+(y,z)) -> +(*(x,y),*(x,z)) r14: app(nil(),l) -> l r15: app(cons(x,l1),l2) -> cons(x,app(l1,l2)) r16: sum(nil()) -> |0|(|#|()) r17: sum(cons(x,l)) -> +(x,sum(l)) r18: sum(app(l1,l2)) -> +(sum(l1),sum(l2)) r19: prod(nil()) -> |1|(|#|()) r20: prod(cons(x,l)) -> *(x,prod(l)) r21: prod(app(l1,l2)) -> *(prod(l1),prod(l2)) The set of usable rules consists of r2, r3 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: *#_A(x1,x2) = x1 + x2 +_A(x1,x2) = x1 + x2 + 2 |#|_A() = 2 base order 2: marked-algebra reduction triple (>=A, >=A!, >A!): max/plus interpretation on natural numbers: *#_A(x1,x2) = max{0, x2 - 18} +_A(x1,x2) = max{33, x1 + 14, x2 + 17} |#|_A = 46 *#!_A(x1,x2) = max{4, x1 - 1, x2 - 35} +!_A(x1,x2) = max{3, x1 - 35, x2 - 36} |#|!_A = 1 partial status: pi(*#) = [1] pi(+) = [1, 2] pi(|#|) = [] 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: +#(|0|(x),|0|(y)) -> +#(x,y) p2: +#(+(x,y),z) -> +#(y,z) p3: +#(+(x,y),z) -> +#(x,+(y,z)) p4: +#(|1|(x),|1|(y)) -> +#(x,y) p5: +#(|1|(x),|1|(y)) -> +#(+(x,y),|1|(|#|())) p6: +#(|0|(x),|1|(y)) -> +#(x,y) p7: +#(|1|(x),|0|(y)) -> +#(x,y) and R consists of: r1: |0|(|#|()) -> |#|() r2: +(x,|#|()) -> x r3: +(|#|(),x) -> x r4: +(|0|(x),|0|(y)) -> |0|(+(x,y)) r5: +(|0|(x),|1|(y)) -> |1|(+(x,y)) r6: +(|1|(x),|0|(y)) -> |1|(+(x,y)) r7: +(|1|(x),|1|(y)) -> |0|(+(+(x,y),|1|(|#|()))) r8: +(+(x,y),z) -> +(x,+(y,z)) r9: *(|#|(),x) -> |#|() r10: *(|0|(x),y) -> |0|(*(x,y)) r11: *(|1|(x),y) -> +(|0|(*(x,y)),y) r12: *(*(x,y),z) -> *(x,*(y,z)) r13: *(x,+(y,z)) -> +(*(x,y),*(x,z)) r14: app(nil(),l) -> l r15: app(cons(x,l1),l2) -> cons(x,app(l1,l2)) r16: sum(nil()) -> |0|(|#|()) r17: sum(cons(x,l)) -> +(x,sum(l)) r18: sum(app(l1,l2)) -> +(sum(l1),sum(l2)) r19: prod(nil()) -> |1|(|#|()) r20: prod(cons(x,l)) -> *(x,prod(l)) r21: prod(app(l1,l2)) -> *(prod(l1),prod(l2)) The set of usable rules consists of r1, r2, r3, r4, r5, r6, r7, r8, r9 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: +#_A(x1,x2) = x1 + x2 + 2 |0|_A(x1) = x1 + 1 +_A(x1,x2) = x1 + x2 + 3 |1|_A(x1) = x1 + 5 |#|_A() = 0 *_A(x1,x2) = x1 + x2 base order 2: marked-algebra reduction triple (>=A, >=A!, >A!): max/plus interpretation on natural numbers: +#_A(x1,x2) = 0 |0|_A(x1) = 40 +_A(x1,x2) = max{9, x1 + 4, x2} |1|_A(x1) = 39 |#|_A = 14 *_A(x1,x2) = max{35, x2 - 36} +#!_A(x1,x2) = 55 |0|!_A(x1) = max{22, x1 - 12} +!_A(x1,x2) = max{54, x1 + 6, x2 + 2} |1|!_A(x1) = max{55, x1 - 15} |#|!_A = 21 *!_A(x1,x2) = max{53, x1 + 6, x2 + 4} partial status: pi(+#) = [1] pi(|0|) = [] pi(+) = [] pi(|1|) = [] pi(|#|) = [] pi(*) = [1, 2] The following rules are removed: p1, p2, p3, p4, p5, p6, p7 Then no dependency pair remains. -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: app#(cons(x,l1),l2) -> app#(l1,l2) and R consists of: r1: |0|(|#|()) -> |#|() r2: +(x,|#|()) -> x r3: +(|#|(),x) -> x r4: +(|0|(x),|0|(y)) -> |0|(+(x,y)) r5: +(|0|(x),|1|(y)) -> |1|(+(x,y)) r6: +(|1|(x),|0|(y)) -> |1|(+(x,y)) r7: +(|1|(x),|1|(y)) -> |0|(+(+(x,y),|1|(|#|()))) r8: +(+(x,y),z) -> +(x,+(y,z)) r9: *(|#|(),x) -> |#|() r10: *(|0|(x),y) -> |0|(*(x,y)) r11: *(|1|(x),y) -> +(|0|(*(x,y)),y) r12: *(*(x,y),z) -> *(x,*(y,z)) r13: *(x,+(y,z)) -> +(*(x,y),*(x,z)) r14: app(nil(),l) -> l r15: app(cons(x,l1),l2) -> cons(x,app(l1,l2)) r16: sum(nil()) -> |0|(|#|()) r17: sum(cons(x,l)) -> +(x,sum(l)) r18: sum(app(l1,l2)) -> +(sum(l1),sum(l2)) r19: prod(nil()) -> |1|(|#|()) r20: prod(cons(x,l)) -> *(x,prod(l)) r21: prod(app(l1,l2)) -> *(prod(l1),prod(l2)) 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: app#_A(x1,x2) = x1 cons_A(x1,x2) = x1 + x2 + 8 base order 2: marked-algebra reduction triple (>=A, >=A!, >A!): max/plus interpretation on natural numbers: app#_A(x1,x2) = max{0, x1 - 2} cons_A(x1,x2) = max{11, x1 - 1, x2 + 1} app#!_A(x1,x2) = max{7, x1 - 112, x2 - 1} cons!_A(x1,x2) = max{6, x1 - 50, x2 - 113} partial status: pi(app#) = [1] pi(cons) = [1] The following rules are removed: p1 Then no dependency pair remains.