NO
proof of HirokawaMiddeldorp_04_n008.trs
# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 jera 20211004 unpublished


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, 15 ms]
(4) QTRS
(5) AAECC Innermost [EQUIVALENT, 0 ms]
(6) QTRS
(7) DependencyPairsProof [EQUIVALENT, 0 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:

   f(a) -> f(a)
   a -> b

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:

   a'(f(x)) -> a'(f(x))
   a'(x) -> b'(x)

Q is empty.

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

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

   a'(x) -> b'(x)




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

   a'(f(x)) -> a'(f(x))

Q is empty.

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

The TRS R 2 is 
   a'(f(x)) -> a'(f(x))

The signature Sigma is {a'_1}
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(6)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   a'(f(x)) -> a'(f(x))

The set Q consists of the following terms:

   a'(f(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:

   A'(f(x)) -> A'(f(x))

The TRS R consists of the following rules:

   a'(f(x)) -> a'(f(x))

The set Q consists of the following terms:

   a'(f(x0))

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

   A'(f(x)) -> A'(f(x))

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

   a'(f(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].

   a'(f(x0))


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

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

   A'(f(x)) -> A'(f(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 semiunifying a rule from P directly.

s = A'(f(x)) evaluates to  t =A'(f(x))

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

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Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from A'(f(x)) to A'(f(x)).




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(14)
NO