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)) +(|0|(x),j(y)) -> j(+(x,y)) +(j(x),|0|(y)) -> j(+(x,y)) +(|1|(x),|1|(y)) -> j(+(+(x,y),|1|(|#|()))) +(j(x),j(y)) -> |1|(+(+(x,y),j(|#|()))) +(|1|(x),j(y)) -> |0|(+(x,y)) +(j(x),|1|(y)) -> |0|(+(x,y)) +(+(x,y),z) -> +(x,+(y,z)) opp(|#|()) -> |#|() opp(|0|(x)) -> |0|(opp(x)) opp(|1|(x)) -> j(opp(x)) opp(j(x)) -> |1|(opp(x)) -(x,y) -> +(x,opp(y)) *(|#|(),x) -> |#|() *(|0|(x),y) -> |0|(*(x,y)) *(|1|(x),y) -> +(|0|(*(x,y)),y) *(j(x),y) -> -(|0|(*(x,y)),y) *(*(x,y),z) -> *(x,*(y,z)) *(+(x,y),z) -> +(*(x,z),*(y,z)) *(x,+(y,z)) -> +(*(x,y),*(x,z)) -- 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: +#(|0|(x),j(y)) -> +#(x,y) p6: +#(j(x),|0|(y)) -> +#(x,y) p7: +#(|1|(x),|1|(y)) -> +#(+(x,y),|1|(|#|())) p8: +#(|1|(x),|1|(y)) -> +#(x,y) p9: +#(j(x),j(y)) -> +#(+(x,y),j(|#|())) p10: +#(j(x),j(y)) -> +#(x,y) p11: +#(|1|(x),j(y)) -> |0|#(+(x,y)) p12: +#(|1|(x),j(y)) -> +#(x,y) p13: +#(j(x),|1|(y)) -> |0|#(+(x,y)) p14: +#(j(x),|1|(y)) -> +#(x,y) p15: +#(+(x,y),z) -> +#(x,+(y,z)) p16: +#(+(x,y),z) -> +#(y,z) p17: opp#(|0|(x)) -> |0|#(opp(x)) p18: opp#(|0|(x)) -> opp#(x) p19: opp#(|1|(x)) -> opp#(x) p20: opp#(j(x)) -> opp#(x) p21: -#(x,y) -> +#(x,opp(y)) p22: -#(x,y) -> opp#(y) p23: *#(|0|(x),y) -> |0|#(*(x,y)) p24: *#(|0|(x),y) -> *#(x,y) p25: *#(|1|(x),y) -> +#(|0|(*(x,y)),y) p26: *#(|1|(x),y) -> |0|#(*(x,y)) p27: *#(|1|(x),y) -> *#(x,y) p28: *#(j(x),y) -> -#(|0|(*(x,y)),y) p29: *#(j(x),y) -> |0|#(*(x,y)) p30: *#(j(x),y) -> *#(x,y) p31: *#(*(x,y),z) -> *#(x,*(y,z)) p32: *#(*(x,y),z) -> *#(y,z) p33: *#(+(x,y),z) -> +#(*(x,z),*(y,z)) p34: *#(+(x,y),z) -> *#(x,z) p35: *#(+(x,y),z) -> *#(y,z) p36: *#(x,+(y,z)) -> +#(*(x,y),*(x,z)) p37: *#(x,+(y,z)) -> *#(x,y) p38: *#(x,+(y,z)) -> *#(x,z) 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: +(|0|(x),j(y)) -> j(+(x,y)) r8: +(j(x),|0|(y)) -> j(+(x,y)) r9: +(|1|(x),|1|(y)) -> j(+(+(x,y),|1|(|#|()))) r10: +(j(x),j(y)) -> |1|(+(+(x,y),j(|#|()))) r11: +(|1|(x),j(y)) -> |0|(+(x,y)) r12: +(j(x),|1|(y)) -> |0|(+(x,y)) r13: +(+(x,y),z) -> +(x,+(y,z)) r14: opp(|#|()) -> |#|() r15: opp(|0|(x)) -> |0|(opp(x)) r16: opp(|1|(x)) -> j(opp(x)) r17: opp(j(x)) -> |1|(opp(x)) r18: -(x,y) -> +(x,opp(y)) r19: *(|#|(),x) -> |#|() r20: *(|0|(x),y) -> |0|(*(x,y)) r21: *(|1|(x),y) -> +(|0|(*(x,y)),y) r22: *(j(x),y) -> -(|0|(*(x,y)),y) r23: *(*(x,y),z) -> *(x,*(y,z)) r24: *(+(x,y),z) -> +(*(x,z),*(y,z)) r25: *(x,+(y,z)) -> +(*(x,y),*(x,z)) The estimated dependency graph contains the following SCCs: {p24, p27, p30, p31, p32, p34, p35, p37, p38} {p2, p3, p4, p5, p6, p7, p8, p9, p10, p12, p14, p15, p16} {p18, p19, p20} -- 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,z) p5: *#(*(x,y),z) -> *#(y,z) p6: *#(*(x,y),z) -> *#(x,*(y,z)) p7: *#(j(x),y) -> *#(x,y) p8: *#(|1|(x),y) -> *#(x,y) p9: *#(|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: +(|0|(x),j(y)) -> j(+(x,y)) r8: +(j(x),|0|(y)) -> j(+(x,y)) r9: +(|1|(x),|1|(y)) -> j(+(+(x,y),|1|(|#|()))) r10: +(j(x),j(y)) -> |1|(+(+(x,y),j(|#|()))) r11: +(|1|(x),j(y)) -> |0|(+(x,y)) r12: +(j(x),|1|(y)) -> |0|(+(x,y)) r13: +(+(x,y),z) -> +(x,+(y,z)) r14: opp(|#|()) -> |#|() r15: opp(|0|(x)) -> |0|(opp(x)) r16: opp(|1|(x)) -> j(opp(x)) r17: opp(j(x)) -> |1|(opp(x)) r18: -(x,y) -> +(x,opp(y)) r19: *(|#|(),x) -> |#|() r20: *(|0|(x),y) -> |0|(*(x,y)) r21: *(|1|(x),y) -> +(|0|(*(x,y)),y) r22: *(j(x),y) -> -(|0|(*(x,y)),y) r23: *(*(x,y),z) -> *(x,*(y,z)) r24: *(+(x,y),z) -> +(*(x,z),*(y,z)) r25: *(x,+(y,z)) -> +(*(x,y),*(x,z)) The set of usable rules consists of r1, r19 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: *#_A(x1,x2) = 0 +_A(x1,x2) = max{38, x1 - 38, x2 + 22} *_A(x1,x2) = max{23, x1 - 85} j_A(x1) = x1 + 9 |1|_A(x1) = 8 |0|_A(x1) = x1 + 85 |#|_A = 7 *#!_A(x1,x2) = 159 +!_A(x1,x2) = max{53, x1 + 49, x2 - 5} *!_A(x1,x2) = max{158, x1 - 38} j!_A(x1) = 0 |1|!_A(x1) = max{57, x1 + 20} |0|!_A(x1) = 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: *#_A(x1,x2) = x1 +_A(x1,x2) = x1 + x2 + 9 *_A(x1,x2) = x1 + x2 + 1 j_A(x1) = x1 + 7 |1|_A(x1) = x1 + 8 |0|_A(x1) = x1 + 1 |#|_A() = 1 *#!_A(x1,x2) = x1 + 3 +!_A(x1,x2) = x1 + 9 *!_A(x1,x2) = x1 + x2 + 2 j!_A(x1) = 1 |1|!_A(x1) = 4 |0|!_A(x1) = 18 |#|!_A() = 0 partial status: pi(*#) = [] pi(+) = [1] pi(*) = [1] pi(j) = [] pi(|1|) = [] pi(|0|) = [1] pi(|#|) = [] The following rules are removed: p3, p4, p5, p6, p7, p8, p9 -- 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: +(|0|(x),j(y)) -> j(+(x,y)) r8: +(j(x),|0|(y)) -> j(+(x,y)) r9: +(|1|(x),|1|(y)) -> j(+(+(x,y),|1|(|#|()))) r10: +(j(x),j(y)) -> |1|(+(+(x,y),j(|#|()))) r11: +(|1|(x),j(y)) -> |0|(+(x,y)) r12: +(j(x),|1|(y)) -> |0|(+(x,y)) r13: +(+(x,y),z) -> +(x,+(y,z)) r14: opp(|#|()) -> |#|() r15: opp(|0|(x)) -> |0|(opp(x)) r16: opp(|1|(x)) -> j(opp(x)) r17: opp(j(x)) -> |1|(opp(x)) r18: -(x,y) -> +(x,opp(y)) r19: *(|#|(),x) -> |#|() r20: *(|0|(x),y) -> |0|(*(x,y)) r21: *(|1|(x),y) -> +(|0|(*(x,y)),y) r22: *(j(x),y) -> -(|0|(*(x,y)),y) r23: *(*(x,y),z) -> *(x,*(y,z)) r24: *(+(x,y),z) -> +(*(x,z),*(y,z)) r25: *(x,+(y,z)) -> +(*(x,y),*(x,z)) 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: +(|0|(x),j(y)) -> j(+(x,y)) r8: +(j(x),|0|(y)) -> j(+(x,y)) r9: +(|1|(x),|1|(y)) -> j(+(+(x,y),|1|(|#|()))) r10: +(j(x),j(y)) -> |1|(+(+(x,y),j(|#|()))) r11: +(|1|(x),j(y)) -> |0|(+(x,y)) r12: +(j(x),|1|(y)) -> |0|(+(x,y)) r13: +(+(x,y),z) -> +(x,+(y,z)) r14: opp(|#|()) -> |#|() r15: opp(|0|(x)) -> |0|(opp(x)) r16: opp(|1|(x)) -> j(opp(x)) r17: opp(j(x)) -> |1|(opp(x)) r18: -(x,y) -> +(x,opp(y)) r19: *(|#|(),x) -> |#|() r20: *(|0|(x),y) -> |0|(*(x,y)) r21: *(|1|(x),y) -> +(|0|(*(x,y)),y) r22: *(j(x),y) -> -(|0|(*(x,y)),y) r23: *(*(x,y),z) -> *(x,*(y,z)) r24: *(+(x,y),z) -> +(*(x,z),*(y,z)) r25: *(x,+(y,z)) -> +(*(x,y),*(x,z)) 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: *#_A(x1,x2) = max{0, x2 - 6} +_A(x1,x2) = max{15, x1 + 5, x2 + 4} *#!_A(x1,x2) = 0 +!_A(x1,x2) = max{0, x2 - 26} 2. marked-algebra reduction triple (>=A, >=A!, >A!): matrix interpretation with standard point-wise order: carrier: N^1 order: standard order interpretations: *#_A(x1,x2) = x2 +_A(x1,x2) = x1 + x2 + 3 *#!_A(x1,x2) = x2 +!_A(x1,x2) = x1 + x2 partial status: pi(*#) = [] 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: +#(j(x),|1|(y)) -> +#(x,y) p5: +#(|1|(x),j(y)) -> +#(x,y) p6: +#(j(x),j(y)) -> +#(x,y) p7: +#(j(x),j(y)) -> +#(+(x,y),j(|#|())) p8: +#(|0|(x),j(y)) -> +#(x,y) p9: +#(|1|(x),|1|(y)) -> +#(x,y) p10: +#(|1|(x),|1|(y)) -> +#(+(x,y),|1|(|#|())) p11: +#(|0|(x),|1|(y)) -> +#(x,y) p12: +#(j(x),|0|(y)) -> +#(x,y) p13: +#(|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: +(|0|(x),j(y)) -> j(+(x,y)) r8: +(j(x),|0|(y)) -> j(+(x,y)) r9: +(|1|(x),|1|(y)) -> j(+(+(x,y),|1|(|#|()))) r10: +(j(x),j(y)) -> |1|(+(+(x,y),j(|#|()))) r11: +(|1|(x),j(y)) -> |0|(+(x,y)) r12: +(j(x),|1|(y)) -> |0|(+(x,y)) r13: +(+(x,y),z) -> +(x,+(y,z)) r14: opp(|#|()) -> |#|() r15: opp(|0|(x)) -> |0|(opp(x)) r16: opp(|1|(x)) -> j(opp(x)) r17: opp(j(x)) -> |1|(opp(x)) r18: -(x,y) -> +(x,opp(y)) r19: *(|#|(),x) -> |#|() r20: *(|0|(x),y) -> |0|(*(x,y)) r21: *(|1|(x),y) -> +(|0|(*(x,y)),y) r22: *(j(x),y) -> -(|0|(*(x,y)),y) r23: *(*(x,y),z) -> *(x,*(y,z)) r24: *(+(x,y),z) -> +(*(x,z),*(y,z)) r25: *(x,+(y,z)) -> +(*(x,y),*(x,z)) The set of usable rules consists of r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13 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: +#_A(x1,x2) = max{41, x1 - 28, x2 - 28} |0|_A(x1) = max{47, x1} +_A(x1,x2) = max{69, x1, x2} j_A(x1) = max{15, x1} |1|_A(x1) = max{12, x1} |#|_A = 29 +#!_A(x1,x2) = 2 |0|!_A(x1) = max{32, x1 - 5} +!_A(x1,x2) = max{86, x1 + 16, x2 + 16} j!_A(x1) = x1 + 3 |1|!_A(x1) = max{14, x1 - 7} |#|!_A = 16 2. marked-algebra reduction triple (>=A, >=A!, >A!): matrix interpretation with standard point-wise order: carrier: N^1 order: standard order interpretations: +#_A(x1,x2) = x1 + x2 |0|_A(x1) = x1 + 3 +_A(x1,x2) = x1 + x2 j_A(x1) = x1 + 5 |1|_A(x1) = x1 + 5 |#|_A() = 0 +#!_A(x1,x2) = x1 + x2 |0|!_A(x1) = 16 +!_A(x1,x2) = x1 + x2 + 9 j!_A(x1) = x1 + 1 |1|!_A(x1) = x1 + 8 |#|!_A() = 15 partial status: pi(+#) = [] pi(|0|) = [] pi(+) = [1, 2] pi(j) = [] pi(|1|) = [] pi(|#|) = [] The following rules are removed: p1, p4, p5, p6, p7, p8, p9, p10, p11, p12, p13 -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: +#(+(x,y),z) -> +#(y,z) p2: +#(+(x,y),z) -> +#(x,+(y,z)) 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: +(|0|(x),j(y)) -> j(+(x,y)) r8: +(j(x),|0|(y)) -> j(+(x,y)) r9: +(|1|(x),|1|(y)) -> j(+(+(x,y),|1|(|#|()))) r10: +(j(x),j(y)) -> |1|(+(+(x,y),j(|#|()))) r11: +(|1|(x),j(y)) -> |0|(+(x,y)) r12: +(j(x),|1|(y)) -> |0|(+(x,y)) r13: +(+(x,y),z) -> +(x,+(y,z)) r14: opp(|#|()) -> |#|() r15: opp(|0|(x)) -> |0|(opp(x)) r16: opp(|1|(x)) -> j(opp(x)) r17: opp(j(x)) -> |1|(opp(x)) r18: -(x,y) -> +(x,opp(y)) r19: *(|#|(),x) -> |#|() r20: *(|0|(x),y) -> |0|(*(x,y)) r21: *(|1|(x),y) -> +(|0|(*(x,y)),y) r22: *(j(x),y) -> -(|0|(*(x,y)),y) r23: *(*(x,y),z) -> *(x,*(y,z)) r24: *(+(x,y),z) -> +(*(x,z),*(y,z)) r25: *(x,+(y,z)) -> +(*(x,y),*(x,z)) 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) -> +#(y,z) p2: +#(+(x,y),z) -> +#(x,+(y,z)) 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: +(|0|(x),j(y)) -> j(+(x,y)) r8: +(j(x),|0|(y)) -> j(+(x,y)) r9: +(|1|(x),|1|(y)) -> j(+(+(x,y),|1|(|#|()))) r10: +(j(x),j(y)) -> |1|(+(+(x,y),j(|#|()))) r11: +(|1|(x),j(y)) -> |0|(+(x,y)) r12: +(j(x),|1|(y)) -> |0|(+(x,y)) r13: +(+(x,y),z) -> +(x,+(y,z)) r14: opp(|#|()) -> |#|() r15: opp(|0|(x)) -> |0|(opp(x)) r16: opp(|1|(x)) -> j(opp(x)) r17: opp(j(x)) -> |1|(opp(x)) r18: -(x,y) -> +(x,opp(y)) r19: *(|#|(),x) -> |#|() r20: *(|0|(x),y) -> |0|(*(x,y)) r21: *(|1|(x),y) -> +(|0|(*(x,y)),y) r22: *(j(x),y) -> -(|0|(*(x,y)),y) r23: *(*(x,y),z) -> *(x,*(y,z)) r24: *(+(x,y),z) -> +(*(x,z),*(y,z)) r25: *(x,+(y,z)) -> +(*(x,y),*(x,z)) The set of usable rules consists of r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14 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: +#_A(x1,x2) = max{0, x1 - 12, x2 - 12} +_A(x1,x2) = max{11, x1, x2} |0|_A(x1) = max{11, x1 - 12} |#|_A = 2 |1|_A(x1) = max{0, x1 - 12} j_A(x1) = max{3, x1 - 12} opp_A(x1) = 8 +#!_A(x1,x2) = 0 +!_A(x1,x2) = max{31, x1 + 26, x2 + 23} |0|!_A(x1) = 30 |#|!_A = 10 |1|!_A(x1) = 30 j!_A(x1) = 31 opp!_A(x1) = x1 + 44 2. marked-algebra reduction triple (>=A, >=A!, >A!): matrix interpretation with standard point-wise order: carrier: N^1 order: standard order interpretations: +#_A(x1,x2) = x1 + x2 +_A(x1,x2) = x1 + x2 + 5 |0|_A(x1) = 3 |#|_A() = 2 |1|_A(x1) = 2 j_A(x1) = 4 opp_A(x1) = x1 + 1 +#!_A(x1,x2) = x1 + 4 +!_A(x1,x2) = x1 + 5 |0|!_A(x1) = 6 |#|!_A() = 0 |1|!_A(x1) = 4 j!_A(x1) = 7 opp!_A(x1) = 1 partial status: pi(+#) = [] pi(+) = [1, 2] pi(|0|) = [] pi(|#|) = [] pi(|1|) = [] pi(j) = [] pi(opp) = [] 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: opp#(|0|(x)) -> opp#(x) p2: opp#(j(x)) -> opp#(x) p3: opp#(|1|(x)) -> opp#(x) 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: +(|0|(x),j(y)) -> j(+(x,y)) r8: +(j(x),|0|(y)) -> j(+(x,y)) r9: +(|1|(x),|1|(y)) -> j(+(+(x,y),|1|(|#|()))) r10: +(j(x),j(y)) -> |1|(+(+(x,y),j(|#|()))) r11: +(|1|(x),j(y)) -> |0|(+(x,y)) r12: +(j(x),|1|(y)) -> |0|(+(x,y)) r13: +(+(x,y),z) -> +(x,+(y,z)) r14: opp(|#|()) -> |#|() r15: opp(|0|(x)) -> |0|(opp(x)) r16: opp(|1|(x)) -> j(opp(x)) r17: opp(j(x)) -> |1|(opp(x)) r18: -(x,y) -> +(x,opp(y)) r19: *(|#|(),x) -> |#|() r20: *(|0|(x),y) -> |0|(*(x,y)) r21: *(|1|(x),y) -> +(|0|(*(x,y)),y) r22: *(j(x),y) -> -(|0|(*(x,y)),y) r23: *(*(x,y),z) -> *(x,*(y,z)) r24: *(+(x,y),z) -> +(*(x,z),*(y,z)) r25: *(x,+(y,z)) -> +(*(x,y),*(x,z)) The set of usable rules consists of r1, r19 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: opp#_A(x1) = max{0, x1 - 8} |0|_A(x1) = max{17, x1 + 7} j_A(x1) = max{20, x1 + 7} |1|_A(x1) = max{8, x1 + 7} |#|_A = 3 *_A(x1,x2) = max{x1 + 23, x2 - 24} opp#!_A(x1) = 1 |0|!_A(x1) = 3 j!_A(x1) = 5 |1|!_A(x1) = max{0, x1 - 2} |#|!_A = 0 *!_A(x1,x2) = x1 + 46 2. marked-algebra reduction triple (>=A, >=A!, >A!): matrix interpretation with standard point-wise order: carrier: N^1 order: standard order interpretations: opp#_A(x1) = x1 |0|_A(x1) = x1 + 10 j_A(x1) = x1 + 2 |1|_A(x1) = x1 + 7 |#|_A() = 4 *_A(x1,x2) = x1 + x2 + 8 opp#!_A(x1) = x1 + 2 |0|!_A(x1) = 6 j!_A(x1) = x1 + 3 |1|!_A(x1) = 6 |#|!_A() = 0 *!_A(x1,x2) = x1 + 5 partial status: pi(opp#) = [] pi(|0|) = [] pi(j) = [1] pi(|1|) = [1] pi(|#|) = [] pi(*) = [1] The following rules are removed: p1, p2, p3 Then no dependency pair remains.