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


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


YES
proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml
# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty


Termination of the given RelTRS could be proven:

(0) RelTRS
(1) RelTRStoQDPProof [SOUND, 0 ms]
(2) QDP
(3) MRRProof [EQUIVALENT, 54 ms]
(4) QDP
(5) PisEmptyProof [EQUIVALENT, 0 ms]
(6) YES


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

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

   t(f(x), g(y), f(z)) -> t(z, g(x), g(y))
   t(g(x), g(y), f(z)) -> t(f(y), f(z), x)

The relative TRS consists of the following S rules:

   f(g(x)) -> g(f(x))
   g(f(x)) -> f(g(x))
   f(f(x)) -> g(g(x))
   g(g(x)) -> f(f(x))


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

(1) RelTRStoQDPProof (SOUND)
The relative termination problem is root-restricted. We can therefore treat it as a dependency pair problem.
----------------------------------------

(2)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   t(f(x), g(y), f(z)) -> t(z, g(x), g(y))
   t(g(x), g(y), f(z)) -> t(f(y), f(z), x)

The TRS R consists of the following rules:

   f(g(x)) -> g(f(x))
   g(f(x)) -> f(g(x))
   f(f(x)) -> g(g(x))
   g(g(x)) -> f(f(x))

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

(3) MRRProof (EQUIVALENT)
By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.

Strictly oriented dependency pairs:

   t(f(x), g(y), f(z)) -> t(z, g(x), g(y))
   t(g(x), g(y), f(z)) -> t(f(y), f(z), x)


Used ordering: Polynomial interpretation [POLO]:

   POL(f(x_1)) = 2 + 2*x_1
   POL(g(x_1)) = 2 + 2*x_1
   POL(t(x_1, x_2, x_3)) = 2*x_1 + 2*x_2 + 2*x_3


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

(4)
Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:

   f(g(x)) -> g(f(x))
   g(f(x)) -> f(g(x))
   f(f(x)) -> g(g(x))
   g(g(x)) -> f(f(x))

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

(5) PisEmptyProof (EQUIVALENT)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
----------------------------------------

(6)
YES