YES
proof of /export/starexec/sandbox2/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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, 60 ms]
(2) QTRS
(3) DependencyPairsProof [EQUIVALENT, 12 ms]
(4) QDP
(5) DependencyGraphProof [EQUIVALENT, 0 ms]
(6) TRUE


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

   g(c, g(a(x), y)) -> g(f(a(b)), g(a(y), x))
   f(a(x)) -> c
   a(b) -> d

The set Q consists of the following terms:

   g(c, g(a(x0), x1))
   f(a(x0))
   a(b)


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(1) QTRSRRRProof (EQUIVALENT)
Used ordering:
Polynomial interpretation [POLO]:

   POL(a(x_1)) = 1 + x_1
   POL(b) = 0
   POL(c) = 1
   POL(d) = 0
   POL(f(x_1)) = x_1
   POL(g(x_1, x_2)) = 2*x_1 + 2*x_2
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

   a(b) -> d




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(2)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   g(c, g(a(x), y)) -> g(f(a(b)), g(a(y), x))
   f(a(x)) -> c

The set Q consists of the following terms:

   g(c, g(a(x0), x1))
   f(a(x0))
   a(b)


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(3) DependencyPairsProof (EQUIVALENT)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
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(4)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   G(c, g(a(x), y)) -> G(f(a(b)), g(a(y), x))
   G(c, g(a(x), y)) -> F(a(b))
   G(c, g(a(x), y)) -> G(a(y), x)

The TRS R consists of the following rules:

   g(c, g(a(x), y)) -> g(f(a(b)), g(a(y), x))
   f(a(x)) -> c

The set Q consists of the following terms:

   g(c, g(a(x0), x1))
   f(a(x0))
   a(b)

We have to consider all minimal (P,Q,R)-chains.
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(5) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 3 less nodes.
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(6)
TRUE