NO
proof of Transformed_CSR_04_Ex14_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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 jera 20211004 unpublished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) QTRS Reverse [EQUIVALENT, 0 ms]
(2) QTRS
(3) QTRSRRRProof [EQUIVALENT, 6 ms]
(4) QTRS
(5) Overlay + Local Confluence [EQUIVALENT, 0 ms]
(6) QTRS
(7) DependencyPairsProof [EQUIVALENT, 0 ms]
(8) QDP
(9) DependencyGraphProof [EQUIVALENT, 0 ms]
(10) QDP
(11) UsableRulesProof [EQUIVALENT, 0 ms]
(12) QDP
(13) QReductionProof [EQUIVALENT, 0 ms]
(14) QDP
(15) MNOCProof [EQUIVALENT, 0 ms]
(16) QDP
(17) NonTerminationLoopProof [COMPLETE, 0 ms]
(18) NO


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

   h(X) -> g(X)
   g(a) -> f(b)
   f(X) -> h(a)
   a -> b

Q is empty.

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(1) QTRS Reverse (EQUIVALENT)
We applied the QTRS Reverse Processor [REVERSE].
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(2)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   h(X) -> g(X)
   a'(g(x)) -> b'(f(x))
   f(X) -> a'(h(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)) = 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:

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




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

   h(X) -> g(X)
   a'(g(x)) -> b'(f(x))
   f(X) -> a'(h(X))

Q is empty.

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(5) Overlay + Local Confluence (EQUIVALENT)
The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.
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(6)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   h(X) -> g(X)
   a'(g(x)) -> b'(f(x))
   f(X) -> a'(h(X))

The set Q consists of the following terms:

   h(x0)
   a'(g(x0))
   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'(g(x)) -> F(x)
   F(X) -> A'(h(X))
   F(X) -> H(X)

The TRS R consists of the following rules:

   h(X) -> g(X)
   a'(g(x)) -> b'(f(x))
   f(X) -> a'(h(X))

The set Q consists of the following terms:

   h(x0)
   a'(g(x0))
   f(x0)

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

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

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

   F(X) -> A'(h(X))
   A'(g(x)) -> F(x)

The TRS R consists of the following rules:

   h(X) -> g(X)
   a'(g(x)) -> b'(f(x))
   f(X) -> a'(h(X))

The set Q consists of the following terms:

   h(x0)
   a'(g(x0))
   f(x0)

We have to consider all minimal (P,Q,R)-chains.
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(11) 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.
----------------------------------------

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

   F(X) -> A'(h(X))
   A'(g(x)) -> F(x)

The TRS R consists of the following rules:

   h(X) -> g(X)

The set Q consists of the following terms:

   h(x0)
   a'(g(x0))
   f(x0)

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

(13) 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'(g(x0))
   f(x0)


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(14)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   F(X) -> A'(h(X))
   A'(g(x)) -> F(x)

The TRS R consists of the following rules:

   h(X) -> g(X)

The set Q consists of the following terms:

   h(x0)

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

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

(16)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   F(X) -> A'(h(X))
   A'(g(x)) -> F(x)

The TRS R consists of the following rules:

   h(X) -> g(X)

Q is empty.
We have to consider all (P,Q,R)-chains.
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(17) 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 = A'(h(X)) evaluates to  t =A'(h(X))

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

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Rewriting sequence

A'(h(X)) -> A'(g(X))
with rule h(X') -> g(X') at position [0] and matcher [X' / X]

A'(g(X)) -> F(X)
with rule A'(g(x)) -> F(x) at position [] and matcher [x / X]

F(X) -> A'(h(X))
with rule F(X) -> A'(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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(18)
NO