/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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, 842 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:

   f(s(x1), x2, x3, x4, x5, x6, x7, x8, x9, x10) -> f(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10)
   f(0, s(x2), x3, x4, x5, x6, x7, x8, x9, x10) -> f(x2, x2, x3, x4, x5, x6, x7, x8, x9, x10)
   f(0, 0, s(x3), x4, x5, x6, x7, x8, x9, x10) -> f(x3, x3, x3, x4, x5, x6, x7, x8, x9, x10)
   f(0, 0, 0, s(x4), x5, x6, x7, x8, x9, x10) -> f(x4, x4, x4, x4, x5, x6, x7, x8, x9, x10)
   f(0, 0, 0, 0, s(x5), x6, x7, x8, x9, x10) -> f(x5, x5, x5, x5, x5, x6, x7, x8, x9, x10)
   f(0, 0, 0, 0, 0, s(x6), x7, x8, x9, x10) -> f(x6, x6, x6, x6, x6, x6, x7, x8, x9, x10)
   f(0, 0, 0, 0, 0, 0, s(x7), x8, x9, x10) -> f(x7, x7, x7, x7, x7, x7, x7, x8, x9, x10)
   f(0, 0, 0, 0, 0, 0, 0, s(x8), x9, x10) -> f(x8, x8, x8, x8, x8, x8, x8, x8, x9, x10)
   f(0, 0, 0, 0, 0, 0, 0, 0, s(x9), x10) -> f(x9, x9, x9, x9, x9, x9, x9, x9, x9, x10)
   f(0, 0, 0, 0, 0, 0, 0, 0, 0, s(x10)) -> f(x10, x10, x10, x10, x10, x10, x10, x10, x10, x10)
   f(0, 0, 0, 0, 0, 0, 0, 0, 0, 0) -> 0

Q is empty.

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(1) QTRSRRRProof (EQUIVALENT)
Used ordering:
Quasi precedence:
s_1 > [f_10, 0]


Status:
f_10: [9,10,8,7,6,5,4,3,2,1]
s_1: multiset status
0: multiset status

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

   f(s(x1), x2, x3, x4, x5, x6, x7, x8, x9, x10) -> f(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10)
   f(0, s(x2), x3, x4, x5, x6, x7, x8, x9, x10) -> f(x2, x2, x3, x4, x5, x6, x7, x8, x9, x10)
   f(0, 0, s(x3), x4, x5, x6, x7, x8, x9, x10) -> f(x3, x3, x3, x4, x5, x6, x7, x8, x9, x10)
   f(0, 0, 0, s(x4), x5, x6, x7, x8, x9, x10) -> f(x4, x4, x4, x4, x5, x6, x7, x8, x9, x10)
   f(0, 0, 0, 0, s(x5), x6, x7, x8, x9, x10) -> f(x5, x5, x5, x5, x5, x6, x7, x8, x9, x10)
   f(0, 0, 0, 0, 0, s(x6), x7, x8, x9, x10) -> f(x6, x6, x6, x6, x6, x6, x7, x8, x9, x10)
   f(0, 0, 0, 0, 0, 0, s(x7), x8, x9, x10) -> f(x7, x7, x7, x7, x7, x7, x7, x8, x9, x10)
   f(0, 0, 0, 0, 0, 0, 0, s(x8), x9, x10) -> f(x8, x8, x8, x8, x8, x8, x8, x8, x9, x10)
   f(0, 0, 0, 0, 0, 0, 0, 0, s(x9), x10) -> f(x9, x9, x9, x9, x9, x9, x9, x9, x9, x10)
   f(0, 0, 0, 0, 0, 0, 0, 0, 0, s(x10)) -> f(x10, x10, x10, x10, x10, x10, x10, x10, x10, x10)
   f(0, 0, 0, 0, 0, 0, 0, 0, 0, 0) -> 0




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(2)
Obligation:
Q restricted rewrite system:
R is empty.
Q is empty.

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