YES We show the termination of the TRS R: +(x,|0|()) -> x +(|0|(),x) -> x +(s(x),s(y)) -> s(s(+(x,y))) +(+(x,y),z) -> +(x,+(y,z)) *(x,|0|()) -> |0|() *(|0|(),x) -> |0|() *(s(x),s(y)) -> s(+(*(x,y),+(x,y))) *(*(x,y),z) -> *(x,*(y,z)) sum(nil()) -> |0|() sum(cons(x,l)) -> +(x,sum(l)) prod(nil()) -> s(|0|()) prod(cons(x,l)) -> *(x,prod(l)) -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: +#(s(x),s(y)) -> +#(x,y) p2: +#(+(x,y),z) -> +#(x,+(y,z)) p3: +#(+(x,y),z) -> +#(y,z) p4: *#(s(x),s(y)) -> +#(*(x,y),+(x,y)) p5: *#(s(x),s(y)) -> *#(x,y) p6: *#(s(x),s(y)) -> +#(x,y) p7: *#(*(x,y),z) -> *#(x,*(y,z)) p8: *#(*(x,y),z) -> *#(y,z) p9: sum#(cons(x,l)) -> +#(x,sum(l)) p10: sum#(cons(x,l)) -> sum#(l) p11: prod#(cons(x,l)) -> *#(x,prod(l)) p12: prod#(cons(x,l)) -> prod#(l) and R consists of: r1: +(x,|0|()) -> x r2: +(|0|(),x) -> x r3: +(s(x),s(y)) -> s(s(+(x,y))) r4: +(+(x,y),z) -> +(x,+(y,z)) r5: *(x,|0|()) -> |0|() r6: *(|0|(),x) -> |0|() r7: *(s(x),s(y)) -> s(+(*(x,y),+(x,y))) r8: *(*(x,y),z) -> *(x,*(y,z)) r9: sum(nil()) -> |0|() r10: sum(cons(x,l)) -> +(x,sum(l)) r11: prod(nil()) -> s(|0|()) r12: prod(cons(x,l)) -> *(x,prod(l)) The estimated dependency graph contains the following SCCs: {p10} {p12} {p5, p7, p8} {p1, p2, p3} -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: sum#(cons(x,l)) -> sum#(l) and R consists of: r1: +(x,|0|()) -> x r2: +(|0|(),x) -> x r3: +(s(x),s(y)) -> s(s(+(x,y))) r4: +(+(x,y),z) -> +(x,+(y,z)) r5: *(x,|0|()) -> |0|() r6: *(|0|(),x) -> |0|() r7: *(s(x),s(y)) -> s(+(*(x,y),+(x,y))) r8: *(*(x,y),z) -> *(x,*(y,z)) r9: sum(nil()) -> |0|() r10: sum(cons(x,l)) -> +(x,sum(l)) r11: prod(nil()) -> s(|0|()) r12: prod(cons(x,l)) -> *(x,prod(l)) 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 cons_A(x1,x2) = x2 + 6 base order 2: marked-algebra reduction triple (>=A, >=A!, >A!): max/plus interpretation on natural numbers: sum#_A(x1) = x1 cons_A(x1,x2) = max{x1 + 10, x2 + 1} sum#!_A(x1) = max{5, x1 - 4} cons!_A(x1,x2) = max{x1 - 22, x2 + 4} partial status: pi(sum#) = [] 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: prod#(cons(x,l)) -> prod#(l) and R consists of: r1: +(x,|0|()) -> x r2: +(|0|(),x) -> x r3: +(s(x),s(y)) -> s(s(+(x,y))) r4: +(+(x,y),z) -> +(x,+(y,z)) r5: *(x,|0|()) -> |0|() r6: *(|0|(),x) -> |0|() r7: *(s(x),s(y)) -> s(+(*(x,y),+(x,y))) r8: *(*(x,y),z) -> *(x,*(y,z)) r9: sum(nil()) -> |0|() r10: sum(cons(x,l)) -> +(x,sum(l)) r11: prod(nil()) -> s(|0|()) r12: prod(cons(x,l)) -> *(x,prod(l)) 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 cons_A(x1,x2) = x2 + 6 base order 2: marked-algebra reduction triple (>=A, >=A!, >A!): max/plus interpretation on natural numbers: prod#_A(x1) = x1 cons_A(x1,x2) = max{x1 + 8, x2 + 13} prod#!_A(x1) = max{8, x1 - 14} cons!_A(x1,x2) = max{7, x1 - 30, x2 + 1} partial status: pi(prod#) = [] 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: *#(*(x,y),z) -> *#(y,z) p2: *#(*(x,y),z) -> *#(x,*(y,z)) p3: *#(s(x),s(y)) -> *#(x,y) and R consists of: r1: +(x,|0|()) -> x r2: +(|0|(),x) -> x r3: +(s(x),s(y)) -> s(s(+(x,y))) r4: +(+(x,y),z) -> +(x,+(y,z)) r5: *(x,|0|()) -> |0|() r6: *(|0|(),x) -> |0|() r7: *(s(x),s(y)) -> s(+(*(x,y),+(x,y))) r8: *(*(x,y),z) -> *(x,*(y,z)) r9: sum(nil()) -> |0|() r10: sum(cons(x,l)) -> +(x,sum(l)) r11: prod(nil()) -> s(|0|()) r12: prod(cons(x,l)) -> *(x,prod(l)) The set of usable rules consists of r1, r2, r3, r4, r5, r6 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) = 1 *_A(x1,x2) = 2 s_A(x1) = x1 + 3 +_A(x1,x2) = x1 + x2 + 4 |0|_A() = 1 base order 2: marked-algebra reduction triple (>=A, >=A!, >A!): max/plus interpretation on natural numbers: *#_A(x1,x2) = 0 *_A(x1,x2) = max{x1, x2 + 2} s_A(x1) = x1 +_A(x1,x2) = max{x1, x2} |0|_A = 0 *#!_A(x1,x2) = x1 + 2 *!_A(x1,x2) = max{3, x1 - 3, x2 - 4} s!_A(x1) = 6 +!_A(x1,x2) = max{7, x1 + 5, x2 + 4} |0|!_A = 0 partial status: pi(*#) = [] pi(*) = [] pi(s) = [1] pi(+) = [1] pi(|0|) = [] The following rules are removed: p1 -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: *#(*(x,y),z) -> *#(x,*(y,z)) p2: *#(s(x),s(y)) -> *#(x,y) and R consists of: r1: +(x,|0|()) -> x r2: +(|0|(),x) -> x r3: +(s(x),s(y)) -> s(s(+(x,y))) r4: +(+(x,y),z) -> +(x,+(y,z)) r5: *(x,|0|()) -> |0|() r6: *(|0|(),x) -> |0|() r7: *(s(x),s(y)) -> s(+(*(x,y),+(x,y))) r8: *(*(x,y),z) -> *(x,*(y,z)) r9: sum(nil()) -> |0|() r10: sum(cons(x,l)) -> +(x,sum(l)) r11: prod(nil()) -> s(|0|()) r12: prod(cons(x,l)) -> *(x,prod(l)) 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,*(y,z)) p2: *#(s(x),s(y)) -> *#(x,y) and R consists of: r1: +(x,|0|()) -> x r2: +(|0|(),x) -> x r3: +(s(x),s(y)) -> s(s(+(x,y))) r4: +(+(x,y),z) -> +(x,+(y,z)) r5: *(x,|0|()) -> |0|() r6: *(|0|(),x) -> |0|() r7: *(s(x),s(y)) -> s(+(*(x,y),+(x,y))) r8: *(*(x,y),z) -> *(x,*(y,z)) r9: sum(nil()) -> |0|() r10: sum(cons(x,l)) -> +(x,sum(l)) r11: prod(nil()) -> s(|0|()) r12: prod(cons(x,l)) -> *(x,prod(l)) The set of usable rules consists of r1, r2, r6 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 + 7 *_A(x1,x2) = x1 + 6 s_A(x1) = x1 + 1 +_A(x1,x2) = x1 + x2 + 8 |0|_A() = 2 base order 2: marked-algebra reduction triple (>=A, >=A!, >A!): max/plus interpretation on natural numbers: *#_A(x1,x2) = max{0, x1 - 11} *_A(x1,x2) = max{10, x1 + 6, x2 - 12} s_A(x1) = max{11, x1 + 4} +_A(x1,x2) = max{x1 + 66, x2 + 40} |0|_A = 11 *#!_A(x1,x2) = max{0, x1 - 55} *!_A(x1,x2) = 0 s!_A(x1) = max{1, x1 - 55} +!_A(x1,x2) = max{1, x1 - 67, x2 - 94} |0|!_A = 0 partial status: pi(*#) = [1] pi(*) = [] pi(s) = [1] pi(+) = [] pi(|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: +#(s(x),s(y)) -> +#(x,y) p2: +#(+(x,y),z) -> +#(y,z) p3: +#(+(x,y),z) -> +#(x,+(y,z)) and R consists of: r1: +(x,|0|()) -> x r2: +(|0|(),x) -> x r3: +(s(x),s(y)) -> s(s(+(x,y))) r4: +(+(x,y),z) -> +(x,+(y,z)) r5: *(x,|0|()) -> |0|() r6: *(|0|(),x) -> |0|() r7: *(s(x),s(y)) -> s(+(*(x,y),+(x,y))) r8: *(*(x,y),z) -> *(x,*(y,z)) r9: sum(nil()) -> |0|() r10: sum(cons(x,l)) -> +(x,sum(l)) r11: prod(nil()) -> s(|0|()) r12: prod(cons(x,l)) -> *(x,prod(l)) The set of usable rules consists of r1, r2, r3, r4 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 + 4 s_A(x1) = x1 + 3 +_A(x1,x2) = x1 + x2 + 2 |0|_A() = 1 base order 2: marked-algebra reduction triple (>=A, >=A!, >A!): max/plus interpretation on natural numbers: +#_A(x1,x2) = max{0, x1 - 53, x2 - 105} s_A(x1) = max{11, x1} +_A(x1,x2) = max{x1 + 52, x2} |0|_A = 8 +#!_A(x1,x2) = max{86, x1 + 34, x2 - 18} s!_A(x1) = max{30, x1 - 28} +!_A(x1,x2) = max{x1 + 33, x2 - 19} |0|!_A = 1 partial status: pi(+#) = [1] pi(s) = [1] pi(+) = [] pi(|0|) = [] The following rules are removed: p1, p2 -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: +#(+(x,y),z) -> +#(x,+(y,z)) and R consists of: r1: +(x,|0|()) -> x r2: +(|0|(),x) -> x r3: +(s(x),s(y)) -> s(s(+(x,y))) r4: +(+(x,y),z) -> +(x,+(y,z)) r5: *(x,|0|()) -> |0|() r6: *(|0|(),x) -> |0|() r7: *(s(x),s(y)) -> s(+(*(x,y),+(x,y))) r8: *(*(x,y),z) -> *(x,*(y,z)) r9: sum(nil()) -> |0|() r10: sum(cons(x,l)) -> +(x,sum(l)) r11: prod(nil()) -> s(|0|()) r12: prod(cons(x,l)) -> *(x,prod(l)) The estimated dependency graph contains the following SCCs: {p1} -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: +#(+(x,y),z) -> +#(x,+(y,z)) and R consists of: r1: +(x,|0|()) -> x r2: +(|0|(),x) -> x r3: +(s(x),s(y)) -> s(s(+(x,y))) r4: +(+(x,y),z) -> +(x,+(y,z)) r5: *(x,|0|()) -> |0|() r6: *(|0|(),x) -> |0|() r7: *(s(x),s(y)) -> s(+(*(x,y),+(x,y))) r8: *(*(x,y),z) -> *(x,*(y,z)) r9: sum(nil()) -> |0|() r10: sum(cons(x,l)) -> +(x,sum(l)) r11: prod(nil()) -> s(|0|()) r12: prod(cons(x,l)) -> *(x,prod(l)) The set of usable rules consists of r1, r2, r3, r4 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 + 1 +_A(x1,x2) = x1 + x2 + 1 |0|_A() = 1 s_A(x1) = 2 base order 2: marked-algebra reduction triple (>=A, >=A!, >A!): max/plus interpretation on natural numbers: +#_A(x1,x2) = max{0, x1 - 13, x2 - 18} +_A(x1,x2) = max{12, x1 + 5, x2} |0|_A = 13 s_A(x1) = max{13, x1 - 14} +#!_A(x1,x2) = max{2, x1 - 1, x2 - 7} +!_A(x1,x2) = max{3, x1 - 58, x2 - 79} |0|!_A = 0 s!_A(x1) = max{1, x1 - 42} partial status: pi(+#) = [1] pi(+) = [1, 2] pi(|0|) = [] pi(s) = [] The following rules are removed: p1 Then no dependency pair remains.