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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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) DependencyPairsProof [EQUIVALENT, 0 ms]
(2) QDP
(3) DependencyGraphProof [EQUIVALENT, 0 ms]
(4) TRUE


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(0)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

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

The set Q consists of the following terms:

   f(s(x0))
   0


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(1) DependencyPairsProof (EQUIVALENT)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
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(2)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   F(s(x)) -> F(g(x, x))
   F(s(x)) -> G(x, x)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

   f(s(x0))
   0

We have to consider all minimal (P,Q,R)-chains.
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(3) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 2 less nodes.
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(4)
TRUE