NO
proof of Transformed_CSR_04_Ex14_AEGL02_L.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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 jera 20211004 unpublished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, 12 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:

   from(X) -> cons(X)
   length -> 0
   length -> s(length1)
   length1 -> length

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:

   from(X) -> cons(X)
   length'(x) -> 0'(x)
   length'(x) -> length1'(s(x))
   length1'(x) -> length'(x)

Q is empty.

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

   POL(0'(x_1)) = x_1
   POL(cons(x_1)) = x_1
   POL(from(x_1)) = 1 + x_1
   POL(length'(x_1)) = 1 + x_1
   POL(length1'(x_1)) = 1 + x_1
   POL(s(x_1)) = x_1
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

   from(X) -> cons(X)
   length'(x) -> 0'(x)




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

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

   length'(x) -> length1'(s(x))
   length1'(x) -> length'(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 
   length'(x) -> length1'(s(x))
   length1'(x) -> length'(x)

The signature Sigma is {length'_1, length1'_1}
----------------------------------------

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

   length'(x) -> length1'(s(x))
   length1'(x) -> length'(x)

The set Q consists of the following terms:

   length'(x0)
   length1'(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:

   LENGTH'(x) -> LENGTH1'(s(x))
   LENGTH1'(x) -> LENGTH'(x)

The TRS R consists of the following rules:

   length'(x) -> length1'(s(x))
   length1'(x) -> length'(x)

The set Q consists of the following terms:

   length'(x0)
   length1'(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:

   LENGTH'(x) -> LENGTH1'(s(x))
   LENGTH1'(x) -> LENGTH'(x)

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

   length'(x0)
   length1'(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].

   length'(x0)
   length1'(x0)


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

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

   LENGTH'(x) -> LENGTH1'(s(x))
   LENGTH1'(x) -> LENGTH'(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 = LENGTH1'(x') evaluates to  t =LENGTH1'(s(x'))

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
* Matcher: [x' / s(x')]
* Semiunifier: [ ]

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

LENGTH1'(x') -> LENGTH'(x')
with rule LENGTH1'(x'') -> LENGTH'(x'') at position [] and matcher [x'' / x']

LENGTH'(x') -> LENGTH1'(s(x'))
with rule LENGTH'(x) -> LENGTH1'(s(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