NO
proof of Transformed_CSR_04_Ex24_Luc06_L.trs
# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 jera 20211004 unpublished


Termination w.r.t. Q of the given QTRS could be disproven:

(0) QTRS
(1) AAECC Innermost [EQUIVALENT, 0 ms]
(2) QTRS
(3) DependencyPairsProof [EQUIVALENT, 0 ms]
(4) QDP
(5) DependencyGraphProof [EQUIVALENT, 0 ms]
(6) QDP
(7) UsableRulesProof [EQUIVALENT, 0 ms]
(8) QDP
(9) QReductionProof [EQUIVALENT, 0 ms]
(10) QDP
(11) MNOCProof [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:

   f(X) -> f(c)
   c -> b

Q is empty.

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(1) AAECC Innermost (EQUIVALENT)
We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is 
   c -> b

The TRS R 2 is 
   f(X) -> f(c)

The signature Sigma is {f_1}
----------------------------------------

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

   f(X) -> f(c)
   c -> b

The set Q consists of the following terms:

   f(x0)
   c


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(3) DependencyPairsProof (EQUIVALENT)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
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(4)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   F(X) -> F(c)
   F(X) -> C

The TRS R consists of the following rules:

   f(X) -> f(c)
   c -> b

The set Q consists of the following terms:

   f(x0)
   c

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

(5) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.
----------------------------------------

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

   F(X) -> F(c)

The TRS R consists of the following rules:

   f(X) -> f(c)
   c -> b

The set Q consists of the following terms:

   f(x0)
   c

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

(7) 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.
----------------------------------------

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

   F(X) -> F(c)

The TRS R consists of the following rules:

   c -> b

The set Q consists of the following terms:

   f(x0)
   c

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

(9) 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(x0)


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

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

   F(X) -> F(c)

The TRS R consists of the following rules:

   c -> b

The set Q consists of the following terms:

   c

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

(11) MNOCProof (EQUIVALENT)
We use the modular non-overlap check [FROCOS05] to decrease Q to the empty set.
----------------------------------------

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

   F(X) -> F(c)

The TRS R consists of the following rules:

   c -> b

Q is empty.
We have to consider all (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 semiunifying a rule from P directly.

s = F(X) evaluates to  t =F(c)

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

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Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from F(X) to F(c).




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(14)
NO