/export/starexec/sandbox/solver/bin/starexec_run_standard /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files


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YES
proof of /export/starexec/sandbox/benchmark/theBenchmark.xml
# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty


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

(0) QTRS
(1) QTRSRRRProof [EQUIVALENT, 153 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:

   terms(N) -> cons(recip(sqr(N)), n__terms(s(N)))
   sqr(0) -> 0
   sqr(s(X)) -> s(add(sqr(X), dbl(X)))
   dbl(0) -> 0
   dbl(s(X)) -> s(s(dbl(X)))
   add(0, X) -> X
   add(s(X), Y) -> s(add(X, Y))
   first(0, X) -> nil
   first(s(X), cons(Y, Z)) -> cons(Y, n__first(X, activate(Z)))
   half(0) -> 0
   half(s(0)) -> 0
   half(s(s(X))) -> s(half(X))
   half(dbl(X)) -> X
   terms(X) -> n__terms(X)
   first(X1, X2) -> n__first(X1, X2)
   activate(n__terms(X)) -> terms(X)
   activate(n__first(X1, X2)) -> first(X1, X2)
   activate(X) -> X

Q is empty.

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(1) QTRSRRRProof (EQUIVALENT)
Used ordering:
terms/1(YES)
cons/2(YES,YES)
recip/1)YES(
sqr/1(YES)
n__terms/1)YES(
s/1(YES)
0/0)
add/2(YES,YES)
dbl/1(YES)
first/2(YES,YES)
nil/0)
n__first/2(YES,YES)
activate/1(YES)
half/1)YES(

Quasi precedence:
[terms_1, first_2, activate_1] > [cons_2, n__first_2] > nil
[terms_1, first_2, activate_1] > [sqr_1, dbl_1] > add_2 > s_1 > 0 > nil


Status:
terms_1: multiset status
cons_2: multiset status
sqr_1: multiset status
s_1: multiset status
0: multiset status
add_2: [2,1]
dbl_1: multiset status
first_2: multiset status
nil: multiset status
n__first_2: multiset status
activate_1: multiset status

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

   terms(N) -> cons(recip(sqr(N)), n__terms(s(N)))
   sqr(0) -> 0
   sqr(s(X)) -> s(add(sqr(X), dbl(X)))
   dbl(0) -> 0
   dbl(s(X)) -> s(s(dbl(X)))
   add(0, X) -> X
   add(s(X), Y) -> s(add(X, Y))
   first(0, X) -> nil
   first(s(X), cons(Y, Z)) -> cons(Y, n__first(X, activate(Z)))
   half(s(0)) -> 0
   half(s(s(X))) -> s(half(X))
   half(dbl(X)) -> X
   terms(X) -> n__terms(X)
   first(X1, X2) -> n__first(X1, X2)
   activate(n__first(X1, X2)) -> first(X1, X2)
   activate(X) -> X




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

   half(0) -> 0
   activate(n__terms(X)) -> terms(X)

Q is empty.

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(3) QTRSRRRProof (EQUIVALENT)
Used ordering:
Knuth-Bendix order [KBO] with precedence:n__terms_1 > activate_1 > half_1 > 0 > terms_1

and weight map:

   0=1
   half_1=1
   activate_1=1
   n__terms_1=1
   terms_1=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:

   half(0) -> 0
   activate(n__terms(X)) -> terms(X)




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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