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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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) DependencyGraphProof [EQUIVALENT, 0 ms]
(10) TRUE


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

   h(x, y) -> f(x, y, x)
   f(0, 1, x) -> h(x, x)
   g(x, y) -> x
   g(x, y) -> y

The set Q consists of the following terms:

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

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

The TRS R consists of the following rules:

   h(x, y) -> f(x, y, x)
   f(0, 1, x) -> h(x, x)
   g(x, y) -> x
   g(x, y) -> y

The set Q consists of the following terms:

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

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

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

   h(x0, x1)
   f(0, 1, 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].

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


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

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

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

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

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

   (H(z0, z0) -> F(z0, z0, z0),H(z0, z0) -> F(z0, z0, z0))


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

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

   F(0, 1, x) -> H(x, x)
   H(z0, z0) -> F(z0, z0, z0)

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.
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(10)
TRUE