NO
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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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 disproven:

(0) QTRS
(1) QTRS Reverse [EQUIVALENT, 0 ms]
(2) QTRS
(3) QTRSRRRProof [EQUIVALENT, 18 ms]
(4) QTRS
(5) AAECC Innermost [EQUIVALENT, 0 ms]
(6) QTRS
(7) DependencyPairsProof [EQUIVALENT, 2 ms]
(8) QDP
(9) UsableRulesProof [EQUIVALENT, 0 ms]
(10) QDP
(11) QReductionProof [EQUIVALENT, 0 ms]
(12) QDP
(13) NonTerminationLoopProof [COMPLETE, 0 ms]
(14) NO


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

   g -> h
   c -> d
   h -> g

Q is empty.

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(1) QTRS Reverse (EQUIVALENT)
We applied the QTRS Reverse Processor [REVERSE].
----------------------------------------

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

   g'(x) -> h'(x)
   c'(x) -> d'(x)
   h'(x) -> g'(x)

Q is empty.

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

   POL(c'(x_1)) = 1 + x_1
   POL(d'(x_1)) = x_1
   POL(g'(x_1)) = x_1
   POL(h'(x_1)) = x_1
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

   c'(x) -> d'(x)




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

(4)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   g'(x) -> h'(x)
   h'(x) -> g'(x)

Q is empty.

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

(5) AAECC Innermost (EQUIVALENT)
We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is none

The TRS R 2 is 
   g'(x) -> h'(x)
   h'(x) -> g'(x)

The signature Sigma is {g'_1, h'_1}
----------------------------------------

(6)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   g'(x) -> h'(x)
   h'(x) -> g'(x)

The set Q consists of the following terms:

   g'(x0)
   h'(x0)


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(7) DependencyPairsProof (EQUIVALENT)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
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(8)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   G'(x) -> H'(x)
   H'(x) -> G'(x)

The TRS R consists of the following rules:

   g'(x) -> h'(x)
   h'(x) -> g'(x)

The set Q consists of the following terms:

   g'(x0)
   h'(x0)

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

(9) 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.
----------------------------------------

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

   G'(x) -> H'(x)
   H'(x) -> G'(x)

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

   g'(x0)
   h'(x0)

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

(11) 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].

   g'(x0)
   h'(x0)


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

(12)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   G'(x) -> H'(x)
   H'(x) -> G'(x)

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

(13) NonTerminationLoopProof (COMPLETE)
We used the non-termination processor [FROCOS05] to show that the DP problem is infinite.
Found a loop by narrowing to the left:

s = H'(x') evaluates to  t =H'(x')

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
* Matcher: [ ]
* Semiunifier: [ ]

--------------------------------------------------------------------------------
Rewriting sequence

H'(x') -> G'(x')
with rule H'(x'') -> G'(x'') at position [] and matcher [x'' / x']

G'(x') -> H'(x')
with rule G'(x) -> H'(x)

Now applying the matcher to the start term leads to a term which is equal to the last term in the rewriting sequence


All these steps are and every following step will be a correct step w.r.t to Q.




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(14)
NO