/export/starexec/sandbox2/solver/bin/starexec_run_standard /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files


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NO
proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml
# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished


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

(0) QTRS
(1) QTRSRRRProof [EQUIVALENT, 78 ms]
(2) QTRS
(3) AAECC Innermost [EQUIVALENT, 0 ms]
(4) QTRS
(5) DependencyPairsProof [EQUIVALENT, 17 ms]
(6) QDP
(7) DependencyGraphProof [EQUIVALENT, 0 ms]
(8) QDP
(9) MNOCProof [EQUIVALENT, 0 ms]
(10) QDP
(11) NonLoopProof [COMPLETE, 0 ms]
(12) NO


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

   f(x, x) -> f(i(x), g(g(x)))
   f(x, y) -> x
   g(x) -> i(x)
   f(x, i(x)) -> f(x, x)
   f(i(x), i(g(x))) -> a

Q is empty.

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

   POL(a) = 0
   POL(f(x_1, x_2)) = 1 + 2*x_1 + 2*x_2
   POL(g(x_1)) = x_1
   POL(i(x_1)) = x_1
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

   f(x, y) -> x
   f(i(x), i(g(x))) -> a




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

   f(x, x) -> f(i(x), g(g(x)))
   g(x) -> i(x)
   f(x, i(x)) -> f(x, x)

Q is empty.

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

The TRS R 2 is 
   f(x, x) -> f(i(x), g(g(x)))
   f(x, i(x)) -> f(x, x)

The signature Sigma is {f_2}
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(4)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   f(x, x) -> f(i(x), g(g(x)))
   g(x) -> i(x)
   f(x, i(x)) -> f(x, x)

The set Q consists of the following terms:

   f(x0, x0)
   g(x0)
   f(x0, i(x0))


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(5) DependencyPairsProof (EQUIVALENT)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
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(6)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   F(x, x) -> F(i(x), g(g(x)))
   F(x, x) -> G(g(x))
   F(x, x) -> G(x)
   F(x, i(x)) -> F(x, x)

The TRS R consists of the following rules:

   f(x, x) -> f(i(x), g(g(x)))
   g(x) -> i(x)
   f(x, i(x)) -> f(x, x)

The set Q consists of the following terms:

   f(x0, x0)
   g(x0)
   f(x0, i(x0))

We have to consider all minimal (P,Q,R)-chains.
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(7) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes.
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(8)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   F(x, i(x)) -> F(x, x)
   F(x, x) -> F(i(x), g(g(x)))

The TRS R consists of the following rules:

   f(x, x) -> f(i(x), g(g(x)))
   g(x) -> i(x)
   f(x, i(x)) -> f(x, x)

The set Q consists of the following terms:

   f(x0, x0)
   g(x0)
   f(x0, i(x0))

We have to consider all minimal (P,Q,R)-chains.
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(9) MNOCProof (EQUIVALENT)
We use the modular non-overlap check [FROCOS05] to decrease Q to the empty set.
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(10)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   F(x, i(x)) -> F(x, x)
   F(x, x) -> F(i(x), g(g(x)))

The TRS R consists of the following rules:

   f(x, x) -> f(i(x), g(g(x)))
   g(x) -> i(x)
   f(x, i(x)) -> f(x, x)

Q is empty.
We have to consider all (P,Q,R)-chains.
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(11) NonLoopProof (COMPLETE)
By Theorem 8 [NONLOOP] we deduce infiniteness of the QDP.
We apply the theorem with m = 1, b = 0, 
σ' = [ ], and μ' = [x0 / i(x0)] on the rule
F(i(x0), i(i(x0)))[ ]^n[ ] -> F(i(x0), i(i(x0)))[ ]^n[x0 / i(x0)]
This rule is correct for the QDP as the following derivation shows:

F(i(x0), i(i(x0)))[ ]^n[ ] -> F(i(x0), i(i(x0)))[ ]^n[x0 / i(x0)]
    by Equivalency by Simplifying Mu with mu1: [x0 / i(x0)] mu2: [ ]
    intermediate steps: Instantiate mu
    F(x0, i(x0))[ ]^n[ ] -> F(i(x0), i(i(x0)))[ ]^n[ ]
        by Rewrite t with the rewrite sequence <Pos,Rule>: [([1],g(x) -> i(x)), ([1,0],g(x) -> i(x))]
        F(x0, i(x0))[ ]^n[ ] -> F(i(x0), g(g(x0)))[ ]^n[ ]
            by Narrowing at position: []
            intermediate steps: Instantiation
            F(x, i(x))[ ]^n[ ] -> F(x, x)[ ]^n[ ]
                by Rule from TRS P

            intermediate steps: Instantiation - Instantiation
            F(x, x)[ ]^n[ ] -> F(i(x), g(g(x)))[ ]^n[ ]
                by Rule from TRS P
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(12)
NO