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) -> |#|() *(|0|(x),y) -> |0|(*(x,y)) *(|1|(x),y) -> +(|0|(*(x,y)),y) sum(nil()) -> |0|(|#|()) sum(cons(x,l)) -> +(x,sum(l)) prod(nil()) -> |1|(|#|()) prod(cons(x,l)) -> *(x,prod(l)) -- 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: *#(|0|(x),y) -> |0|#(*(x,y)) p9: *#(|0|(x),y) -> *#(x,y) p10: *#(|1|(x),y) -> +#(|0|(*(x,y)),y) p11: *#(|1|(x),y) -> |0|#(*(x,y)) p12: *#(|1|(x),y) -> *#(x,y) p13: sum#(nil()) -> |0|#(|#|()) p14: sum#(cons(x,l)) -> +#(x,sum(l)) p15: sum#(cons(x,l)) -> sum#(l) p16: prod#(cons(x,l)) -> *#(x,prod(l)) p17: 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) -> |#|() r9: *(|0|(x),y) -> |0|(*(x,y)) r10: *(|1|(x),y) -> +(|0|(*(x,y)),y) r11: sum(nil()) -> |0|(|#|()) r12: sum(cons(x,l)) -> +(x,sum(l)) r13: prod(nil()) -> |1|(|#|()) r14: prod(cons(x,l)) -> *(x,prod(l)) The estimated dependency graph contains the following SCCs: {p15} {p17} {p9, p12} {p2, p3, p4, p6, p7} -- 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: |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) -> |#|() r9: *(|0|(x),y) -> |0|(*(x,y)) r10: *(|1|(x),y) -> +(|0|(*(x,y)),y) r11: sum(nil()) -> |0|(|#|()) r12: sum(cons(x,l)) -> +(x,sum(l)) r13: prod(nil()) -> |1|(|#|()) r14: prod(cons(x,l)) -> *(x,prod(l)) The set of usable rules consists of r1, r4, r7, r8, r9 Take the reduction pair: generalized weighted path order (>=MG, >MG) 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 + 7 |0|_A(x1) = 2 |#|_A() = 1 +_A(x1,x2) = 3 |1|_A(x1) = x1 + 4 *_A(x1,x2) = x1 + 1 base order 2: precedence: |1| > |0| > * > cons > sum# > + > |#| partial status: pi(sum#) = [] pi(cons) = [2] pi(|0|) = [] pi(|#|) = [] pi(+) = [] pi(|1|) = [1] pi(*) = [1] 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: |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) -> |#|() r9: *(|0|(x),y) -> |0|(*(x,y)) r10: *(|1|(x),y) -> +(|0|(*(x,y)),y) r11: sum(nil()) -> |0|(|#|()) r12: sum(cons(x,l)) -> +(x,sum(l)) r13: prod(nil()) -> |1|(|#|()) r14: prod(cons(x,l)) -> *(x,prod(l)) The set of usable rules consists of r1, r4, r7, r8, r9 Take the reduction pair: generalized weighted path order (>=MG, >MG) 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 + 7 |0|_A(x1) = 2 |#|_A() = 1 +_A(x1,x2) = 3 |1|_A(x1) = x1 + 4 *_A(x1,x2) = x1 + 1 base order 2: precedence: |1| > |0| > * > cons > prod# > + > |#| partial status: pi(prod#) = [] pi(cons) = [2] pi(|0|) = [] pi(|#|) = [] pi(+) = [] pi(|1|) = [1] pi(*) = [1] 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: *#(|1|(x),y) -> *#(x,y) p2: *#(|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) -> |#|() r9: *(|0|(x),y) -> |0|(*(x,y)) r10: *(|1|(x),y) -> +(|0|(*(x,y)),y) r11: sum(nil()) -> |0|(|#|()) r12: sum(cons(x,l)) -> +(x,sum(l)) r13: prod(nil()) -> |1|(|#|()) r14: prod(cons(x,l)) -> *(x,prod(l)) The set of usable rules consists of r3, r8 Take the reduction pair: generalized weighted path order (>=MG, >MG) base order 1: matrix interpretation with standard point-wise order: carrier: N^1 order: standard order interpretations: *#_A(x1,x2) = x1 |1|_A(x1) = x1 + 4 |0|_A(x1) = x1 + 8 +_A(x1,x2) = x2 + 3 |#|_A() = 1 *_A(x1,x2) = 2 base order 2: precedence: + > |1| > |#| = * > |0| > *# partial status: pi(*#) = [] pi(|1|) = [1] pi(|0|) = [1] pi(+) = [2] pi(|#|) = [] pi(*) = [] The following rules are removed: p2 -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: *#(|1|(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) -> |#|() r9: *(|0|(x),y) -> |0|(*(x,y)) r10: *(|1|(x),y) -> +(|0|(*(x,y)),y) r11: sum(nil()) -> |0|(|#|()) r12: sum(cons(x,l)) -> +(x,sum(l)) r13: prod(nil()) -> |1|(|#|()) r14: 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: *#(|1|(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) -> |#|() r9: *(|0|(x),y) -> |0|(*(x,y)) r10: *(|1|(x),y) -> +(|0|(*(x,y)),y) r11: sum(nil()) -> |0|(|#|()) r12: sum(cons(x,l)) -> +(x,sum(l)) r13: prod(nil()) -> |1|(|#|()) r14: prod(cons(x,l)) -> *(x,prod(l)) The set of usable rules consists of r1, r8, r9 Take the reduction pair: generalized weighted path order (>=MG, >MG) base order 1: matrix interpretation with standard point-wise order: carrier: N^1 order: standard order interpretations: *#_A(x1,x2) = x1 |1|_A(x1) = x1 + 3 |0|_A(x1) = 10 |#|_A() = 1 *_A(x1,x2) = x1 + 6 base order 2: precedence: *# = |0| > |1| = * > |#| partial status: pi(*#) = [1] pi(|1|) = [1] pi(|0|) = [] pi(|#|) = [] pi(*) = [] 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: +#(|0|(x),|0|(y)) -> +#(x,y) p2: +#(|1|(x),|1|(y)) -> +#(x,y) p3: +#(|1|(x),|1|(y)) -> +#(+(x,y),|1|(|#|())) p4: +#(|0|(x),|1|(y)) -> +#(x,y) p5: +#(|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) -> |#|() r9: *(|0|(x),y) -> |0|(*(x,y)) r10: *(|1|(x),y) -> +(|0|(*(x,y)),y) r11: sum(nil()) -> |0|(|#|()) r12: sum(cons(x,l)) -> +(x,sum(l)) r13: prod(nil()) -> |1|(|#|()) r14: prod(cons(x,l)) -> *(x,prod(l)) The set of usable rules consists of r1, r2, r3, r4, r5, r6, r7, r8 Take the reduction pair: generalized weighted path order (>=MG, >MG) base order 1: matrix interpretation with standard point-wise order: carrier: N^1 order: standard order interpretations: +#_A(x1,x2) = x1 + x2 + 6 |0|_A(x1) = x1 + 1 |1|_A(x1) = x1 + 5 +_A(x1,x2) = x1 + x2 + 2 |#|_A() = 1 *_A(x1,x2) = x1 + 9 base order 2: precedence: +# = * > |0| > + > |#| > |1| partial status: pi(+#) = [2] pi(|0|) = [] pi(|1|) = [1] pi(+) = [] pi(|#|) = [] pi(*) = [1] The following rules are removed: p1 -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: +#(|1|(x),|1|(y)) -> +#(x,y) p2: +#(|1|(x),|1|(y)) -> +#(+(x,y),|1|(|#|())) p3: +#(|0|(x),|1|(y)) -> +#(x,y) p4: +#(|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) -> |#|() r9: *(|0|(x),y) -> |0|(*(x,y)) r10: *(|1|(x),y) -> +(|0|(*(x,y)),y) r11: sum(nil()) -> |0|(|#|()) r12: sum(cons(x,l)) -> +(x,sum(l)) r13: prod(nil()) -> |1|(|#|()) r14: prod(cons(x,l)) -> *(x,prod(l)) The estimated dependency graph contains the following SCCs: {p1, p2, p3, p4} -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: +#(|1|(x),|1|(y)) -> +#(x,y) p2: +#(|1|(x),|0|(y)) -> +#(x,y) p3: +#(|0|(x),|1|(y)) -> +#(x,y) p4: +#(|1|(x),|1|(y)) -> +#(+(x,y),|1|(|#|())) 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) -> |#|() r9: *(|0|(x),y) -> |0|(*(x,y)) r10: *(|1|(x),y) -> +(|0|(*(x,y)),y) r11: sum(nil()) -> |0|(|#|()) r12: sum(cons(x,l)) -> +(x,sum(l)) r13: prod(nil()) -> |1|(|#|()) r14: prod(cons(x,l)) -> *(x,prod(l)) The set of usable rules consists of r1, r2, r3, r4, r5, r6, r7, r8 Take the reduction pair: generalized weighted path order (>=MG, >MG) base order 1: matrix interpretation with standard point-wise order: carrier: N^1 order: standard order interpretations: +#_A(x1,x2) = x1 + x2 + 6 |1|_A(x1) = x1 + 5 |0|_A(x1) = x1 + 1 +_A(x1,x2) = x1 + x2 + 2 |#|_A() = 1 *_A(x1,x2) = x1 + 3 base order 2: precedence: + > +# = |1| > * > |0| > |#| partial status: pi(+#) = [] pi(|1|) = [] pi(|0|) = [] pi(+) = [1] pi(|#|) = [] pi(*) = [1] The following rules are removed: p1 -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: +#(|1|(x),|0|(y)) -> +#(x,y) p2: +#(|0|(x),|1|(y)) -> +#(x,y) p3: +#(|1|(x),|1|(y)) -> +#(+(x,y),|1|(|#|())) 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) -> |#|() r9: *(|0|(x),y) -> |0|(*(x,y)) r10: *(|1|(x),y) -> +(|0|(*(x,y)),y) r11: sum(nil()) -> |0|(|#|()) r12: sum(cons(x,l)) -> +(x,sum(l)) r13: prod(nil()) -> |1|(|#|()) r14: prod(cons(x,l)) -> *(x,prod(l)) The estimated dependency graph contains the following SCCs: {p1, p2, p3} -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: +#(|1|(x),|0|(y)) -> +#(x,y) p2: +#(|1|(x),|1|(y)) -> +#(+(x,y),|1|(|#|())) p3: +#(|0|(x),|1|(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) -> |#|() r9: *(|0|(x),y) -> |0|(*(x,y)) r10: *(|1|(x),y) -> +(|0|(*(x,y)),y) r11: sum(nil()) -> |0|(|#|()) r12: sum(cons(x,l)) -> +(x,sum(l)) r13: prod(nil()) -> |1|(|#|()) r14: prod(cons(x,l)) -> *(x,prod(l)) The set of usable rules consists of r1, r2, r3, r4, r5, r6, r7, r8 Take the reduction pair: generalized weighted path order (>=MG, >MG) base order 1: matrix interpretation with standard point-wise order: carrier: N^1 order: standard order interpretations: +#_A(x1,x2) = x1 + x2 + 6 |1|_A(x1) = x1 + 5 |0|_A(x1) = x1 + 1 +_A(x1,x2) = x1 + x2 + 2 |#|_A() = 1 *_A(x1,x2) = x1 + x2 + 6 base order 2: precedence: +# > |1| = + > * > |0| > |#| partial status: pi(+#) = [1] pi(|1|) = [1] pi(|0|) = [] pi(+) = [] pi(|#|) = [] pi(*) = [1, 2] The following rules are removed: p1 -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: +#(|1|(x),|1|(y)) -> +#(+(x,y),|1|(|#|())) p2: +#(|0|(x),|1|(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) -> |#|() r9: *(|0|(x),y) -> |0|(*(x,y)) r10: *(|1|(x),y) -> +(|0|(*(x,y)),y) r11: sum(nil()) -> |0|(|#|()) r12: sum(cons(x,l)) -> +(x,sum(l)) r13: prod(nil()) -> |1|(|#|()) r14: 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: +#(|1|(x),|1|(y)) -> +#(+(x,y),|1|(|#|())) p2: +#(|0|(x),|1|(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) -> |#|() r9: *(|0|(x),y) -> |0|(*(x,y)) r10: *(|1|(x),y) -> +(|0|(*(x,y)),y) r11: sum(nil()) -> |0|(|#|()) r12: sum(cons(x,l)) -> +(x,sum(l)) r13: prod(nil()) -> |1|(|#|()) r14: prod(cons(x,l)) -> *(x,prod(l)) The set of usable rules consists of r1, r2, r3, r4, r5, r6, r7, r8 Take the reduction pair: generalized weighted path order (>=MG, >MG) base order 1: matrix interpretation with standard point-wise order: carrier: N^1 order: standard order interpretations: +#_A(x1,x2) = x1 + x2 + 11 |1|_A(x1) = x1 + 10 +_A(x1,x2) = x1 + x2 + 2 |#|_A() = 1 |0|_A(x1) = x1 + 6 *_A(x1,x2) = x1 + 5 base order 2: precedence: * > +# = |1| > + > |0| > |#| partial status: pi(+#) = [1] pi(|1|) = [1] pi(+) = [] pi(|#|) = [] pi(|0|) = [1] pi(*) = [1] The following rules are removed: p1 -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: +#(|0|(x),|1|(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) -> |#|() r9: *(|0|(x),y) -> |0|(*(x,y)) r10: *(|1|(x),y) -> +(|0|(*(x,y)),y) r11: sum(nil()) -> |0|(|#|()) r12: sum(cons(x,l)) -> +(x,sum(l)) r13: prod(nil()) -> |1|(|#|()) r14: 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: +#(|0|(x),|1|(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) -> |#|() r9: *(|0|(x),y) -> |0|(*(x,y)) r10: *(|1|(x),y) -> +(|0|(*(x,y)),y) r11: sum(nil()) -> |0|(|#|()) r12: sum(cons(x,l)) -> +(x,sum(l)) r13: prod(nil()) -> |1|(|#|()) r14: prod(cons(x,l)) -> *(x,prod(l)) The set of usable rules consists of r3, r8 Take the reduction pair: generalized weighted path order (>=MG, >MG) base order 1: matrix interpretation with standard point-wise order: carrier: N^1 order: standard order interpretations: +#_A(x1,x2) = x1 + x2 + 6 |0|_A(x1) = x1 + 5 |1|_A(x1) = x1 + 3 +_A(x1,x2) = x2 + 1 |#|_A() = 2 *_A(x1,x2) = x1 + 1 base order 2: precedence: |1| > * > |0| > +# = + > |#| partial status: pi(+#) = [1] pi(|0|) = [] pi(|1|) = [1] pi(+) = [] pi(|#|) = [] pi(*) = [1] The following rules are removed: p1 Then no dependency pair remains.