/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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, 157 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(x, 0, 0) -> s(x)
   f(0, y, 0) -> s(y)
   f(0, 0, z) -> s(z)
   f(s(0), y, z) -> f(0, s(y), s(z))
   f(s(x), s(y), 0) -> f(x, y, s(0))
   f(s(x), 0, s(z)) -> f(x, s(0), z)
   f(0, s(0), s(0)) -> s(s(0))
   f(s(x), s(y), s(z)) -> f(x, y, f(s(x), s(y), z))
   f(0, s(s(y)), s(0)) -> f(0, y, s(0))
   f(0, s(0), s(s(z))) -> f(0, s(0), z)
   f(0, s(s(y)), s(s(z))) -> f(0, y, f(0, s(s(y)), s(z)))

Q is empty.

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(1) QTRSRRRProof (EQUIVALENT)
Used ordering:
Quasi precedence:
f_3 > [0, s_1]


Status:
f_3: [1,2,3]
0: multiset status
s_1: [1]

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

   f(x, 0, 0) -> s(x)
   f(0, y, 0) -> s(y)
   f(0, 0, z) -> s(z)
   f(s(0), y, z) -> f(0, s(y), s(z))
   f(s(x), s(y), 0) -> f(x, y, s(0))
   f(s(x), 0, s(z)) -> f(x, s(0), z)
   f(0, s(0), s(0)) -> s(s(0))
   f(s(x), s(y), s(z)) -> f(x, y, f(s(x), s(y), z))
   f(0, s(s(y)), s(0)) -> f(0, y, s(0))
   f(0, s(0), s(s(z))) -> f(0, s(0), z)
   f(0, s(s(y)), s(s(z))) -> f(0, y, f(0, s(s(y)), s(z)))




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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