YES
proof of CiME_04_list-sum-prod-assoc-append.trs
# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 jera 20211004 unpublished


Termination w.r.t. Q of the given QTRS could be proven:

(0) QTRS
(1) QTRSRRRProof [EQUIVALENT, 64 ms]
(2) QTRS
(3) QTRSRRRProof [EQUIVALENT, 0 ms]
(4) QTRS
(5) RisEmptyProof [EQUIVALENT, 0 ms]
(6) YES


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(0)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   +(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))
   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) -> s(0)
   prod(cons(x, l)) -> *(x, prod(l))
   prod(app(l1, l2)) -> *(prod(l1), prod(l2))

Q is empty.

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(1) QTRSRRRProof (EQUIVALENT)
Used ordering:
+/2(YES,YES)
0/0)
s/1(YES)
*/2(YES,YES)
app/2(YES,YES)
nil/0)
cons/2(YES,YES)
sum/1(YES)
prod/1)YES(

Quasi precedence:
[*_2, app_2, cons_2, sum_1] > +_2 > s_1 > 0
nil > s_1 > 0


Status:
+_2: [1,2]
0: multiset status
s_1: multiset status
*_2: [1,2]
app_2: [1,2]
nil: multiset status
cons_2: [1,2]
sum_1: multiset status

With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

   +(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))
   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) -> s(0)




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(2)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   prod(cons(x, l)) -> *(x, prod(l))
   prod(app(l1, l2)) -> *(prod(l1), prod(l2))

Q is empty.

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(3) QTRSRRRProof (EQUIVALENT)
Used ordering:
Knuth-Bendix order [KBO] with precedence:app_2 > *_2 > cons_2 > prod_1

and weight map:

   prod_1=1
   cons_2=1
   *_2=0
   app_2=2

The variable weight is 1With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

   prod(cons(x, l)) -> *(x, prod(l))
   prod(app(l1, l2)) -> *(prod(l1), prod(l2))




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(4)
Obligation:
Q restricted rewrite system:
R is empty.
Q is empty.

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(5) RisEmptyProof (EQUIVALENT)
The TRS R is empty. Hence, termination is trivially proven.
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(6)
YES