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


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YES
proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml
# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty


Quasi decreasingness of the given CTRS could be proven:

(0) CTRS
(1) CTRSToQTRSProof [SOUND, 0 ms]
(2) QTRS
(3) QTRSRRRProof [EQUIVALENT, 47 ms]
(4) QTRS
(5) AAECC Innermost [EQUIVALENT, 0 ms]
(6) QTRS
(7) DependencyPairsProof [EQUIVALENT, 15 ms]
(8) QDP
(9) DependencyGraphProof [EQUIVALENT, 0 ms]
(10) TRUE


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(0)
Obligation:
Conditional term rewrite system:
The TRS R consists of the following rules:

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

The conditional TRS C consists of the following conditional rules:

   h(x) -> h(g(x)) <= f(x) -> g(x)


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(1) CTRSToQTRSProof (SOUND)
The conditional rules have been transormed into unconditional rules according to [CTRS,AAECCNOC].
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(2)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   h(x) -> U1(f(x), x)
   U1(g(x), x) -> h(g(x))
   f(g(a)) -> g(b)
   g(a) -> b

Q is empty.

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

   POL(U1(x_1, x_2)) = x_1 + x_2
   POL(a) = 1
   POL(b) = 0
   POL(f(x_1)) = x_1
   POL(g(x_1)) = x_1
   POL(h(x_1)) = 2*x_1
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

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




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

   h(x) -> U1(f(x), x)
   U1(g(x), x) -> h(g(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 
   h(x) -> U1(f(x), x)
   U1(g(x), x) -> h(g(x))

The signature Sigma is {h_1, U1_2}
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(6)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   h(x) -> U1(f(x), x)
   U1(g(x), x) -> h(g(x))

The set Q consists of the following terms:

   h(x0)
   U1(g(x0), 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:

   H(x) -> U1^1(f(x), x)
   U1^1(g(x), x) -> H(g(x))

The TRS R consists of the following rules:

   h(x) -> U1(f(x), x)
   U1(g(x), x) -> h(g(x))

The set Q consists of the following terms:

   h(x0)
   U1(g(x0), x0)

We have to consider all minimal (P,Q,R)-chains.
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(9) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 2 less nodes.
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(10)
TRUE