/export/starexec/sandbox2/solver/bin/starexec_run_standard /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files


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YES
proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml
# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished


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

(0) QTRS
(1) QTRSRRRProof [EQUIVALENT, 87 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(0) -> 1
   f(s(x)) -> g(f(x))
   g(x) -> +(x, s(x))
   f(s(x)) -> +(f(x), s(f(x)))

Q is empty.

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(1) QTRSRRRProof (EQUIVALENT)
Used ordering:
Quasi precedence:
f_1 > 1 > s_1
f_1 > g_1 > +_2 > s_1
0 > 1 > s_1


Status:
f_1: multiset status
0: multiset status
1: multiset status
s_1: multiset status
g_1: multiset status
+_2: [1,2]

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

   f(0) -> 1
   f(s(x)) -> g(f(x))
   g(x) -> +(x, s(x))
   f(s(x)) -> +(f(x), s(f(x)))




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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