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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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) DependencyPairsProof [EQUIVALENT, 0 ms]
(2) QDP
(3) UsableRulesProof [EQUIVALENT, 0 ms]
(4) QDP
(5) QReductionProof [EQUIVALENT, 0 ms]
(6) QDP
(7) TransformationProof [EQUIVALENT, 0 ms]
(8) QDP
(9) TransformationProof [EQUIVALENT, 0 ms]
(10) QDP
(11) DependencyGraphProof [EQUIVALENT, 0 ms]
(12) TRUE


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

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

The set Q consists of the following terms:

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


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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(a(x), y) -> G(x, y)
   G(x, y) -> H(x, y)
   H(b, y) -> F(y, y)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

We have to consider all minimal (P,Q,R)-chains.
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(3) UsableRulesProof (EQUIVALENT)
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

(4)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   F(a(x), y) -> G(x, y)
   G(x, y) -> H(x, y)
   H(b, y) -> F(y, y)

R is empty.
The set Q consists of the following terms:

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

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(5) QReductionProof (EQUIVALENT)
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

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


----------------------------------------

(6)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   F(a(x), y) -> G(x, y)
   G(x, y) -> H(x, y)
   H(b, y) -> F(y, y)

R is empty.
The set Q consists of the following terms:

   a(b)

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(7) TransformationProof (EQUIVALENT)
By instantiating [LPAR04] the rule F(a(x), y) -> G(x, y) we obtained the following new rules [LPAR04]:

   (F(a(x0), a(x0)) -> G(x0, a(x0)),F(a(x0), a(x0)) -> G(x0, a(x0)))


----------------------------------------

(8)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   G(x, y) -> H(x, y)
   H(b, y) -> F(y, y)
   F(a(x0), a(x0)) -> G(x0, a(x0))

R is empty.
The set Q consists of the following terms:

   a(b)

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(9) TransformationProof (EQUIVALENT)
By instantiating [LPAR04] the rule G(x, y) -> H(x, y) we obtained the following new rules [LPAR04]:

   (G(z0, a(z0)) -> H(z0, a(z0)),G(z0, a(z0)) -> H(z0, a(z0)))


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(10)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   H(b, y) -> F(y, y)
   F(a(x0), a(x0)) -> G(x0, a(x0))
   G(z0, a(z0)) -> H(z0, a(z0))

R is empty.
The set Q consists of the following terms:

   a(b)

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(11) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 3 less nodes.
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(12)
TRUE