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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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


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

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

The set Q consists of the following terms:

   f(0, 1, g(x0, x1), x2)
   g(0, 1)
   h(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, g(x, y), z) -> F(g(x, y), g(x, y), g(x, y), h(x))
   F(0, 1, g(x, y), z) -> H(x)
   H(g(x, y)) -> H(x)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

   f(0, 1, g(x0, x1), x2)
   g(0, 1)
   h(g(x0, x1))

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 1 SCC with 2 less nodes.
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(4)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   H(g(x, y)) -> H(x)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

   f(0, 1, g(x0, x1), x2)
   g(0, 1)
   h(g(x0, x1))

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

(5) 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.
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(6)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   H(g(x, y)) -> H(x)

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

   f(0, 1, g(x0, x1), x2)
   g(0, 1)
   h(g(x0, x1))

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

(7) 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, g(x0, x1), x2)
   h(g(x0, x1))


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

   H(g(x, y)) -> H(x)

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

   g(0, 1)

We have to consider all minimal (P,Q,R)-chains.
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(9) QDPSizeChangeProof (EQUIVALENT)
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem. 

From the DPs we obtained the following set of size-change graphs:
*H(g(x, y)) -> H(x)
The graph contains the following edges 1 > 1


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(10)
YES