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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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


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

   f(0, 1, X) -> h(X, X)
   h(0, X) -> f(0, X, X)
   g(X, Y) -> X
   g(X, Y) -> Y

The set Q consists of the following terms:

   f(0, 1, x0)
   h(0, x0)
   g(x0, x1)


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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(0, 1, X) -> H(X, X)
   H(0, X) -> F(0, X, X)

The TRS R consists of the following rules:

   f(0, 1, X) -> h(X, X)
   h(0, X) -> f(0, X, X)
   g(X, Y) -> X
   g(X, Y) -> Y

The set Q consists of the following terms:

   f(0, 1, x0)
   h(0, x0)
   g(x0, x1)

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(0, 1, X) -> H(X, X)
   H(0, X) -> F(0, X, X)

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

   f(0, 1, x0)
   h(0, x0)
   g(x0, x1)

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(0, 1, x0)
   h(0, x0)
   g(x0, x1)


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

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

   F(0, 1, X) -> H(X, X)
   H(0, X) -> F(0, X, X)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(7) TransformationProof (EQUIVALENT)
By instantiating [LPAR04] the rule F(0, 1, X) -> H(X, X) we obtained the following new rules [LPAR04]:

   (F(0, 1, 1) -> H(1, 1),F(0, 1, 1) -> H(1, 1))


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

   H(0, X) -> F(0, X, X)
   F(0, 1, 1) -> H(1, 1)

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

(9) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 2 less nodes.
----------------------------------------

(10)
TRUE