YES
proof of /export/starexec/sandbox/benchmark/theBenchmark.xml
# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty


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

(0) QTRS
(1) QTRSRRRProof [EQUIVALENT, 0 ms]
(2) QTRS
(3) RisEmptyProof [EQUIVALENT, 0 ms]
(4) 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)))
   sqr(0) -> 0
   sqr(s) -> s
   dbl(0) -> 0
   dbl(s) -> s
   add(0, X) -> X
   add(s, Y) -> s
   first(0, X) -> nil
   first(s, cons(Y)) -> cons(Y)

The set Q consists of the following terms:

   terms(x0)
   sqr(0)
   sqr(s)
   dbl(0)
   dbl(s)
   add(0, x0)
   add(s, x0)
   first(0, x0)
   first(s, cons(x0))


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(1) QTRSRRRProof (EQUIVALENT)
Used ordering:
Knuth-Bendix order [KBO] with precedence:dbl_1 > first_2 > nil > add_2 > s > 0 > recip_1 > sqr_1 > terms_1 > cons_1

and weight map:

   0=1
   s=1
   nil=2
   terms_1=3
   cons_1=1
   recip_1=1
   sqr_1=1
   dbl_1=0
   add_2=0
   first_2=0

The variable weight is 1With 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)))
   sqr(0) -> 0
   sqr(s) -> s
   dbl(0) -> 0
   dbl(s) -> s
   add(0, X) -> X
   add(s, Y) -> s
   first(0, X) -> nil
   first(s, cons(Y)) -> cons(Y)




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(2)
Obligation:
Q restricted rewrite system:
R is empty.
The set Q consists of the following terms:

   terms(x0)
   sqr(0)
   sqr(s)
   dbl(0)
   dbl(s)
   add(0, x0)
   add(s, x0)
   first(0, x0)
   first(s, cons(x0))


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(3) RisEmptyProof (EQUIVALENT)
The TRS R is empty. Hence, termination is trivially proven.
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(4)
YES