/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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, 56 ms]
(4) QDP
(5) QDPOrderProof [EQUIVALENT, 41 ms]
(6) QDP
(7) DependencyGraphProof [EQUIVALENT, 0 ms]
(8) AND
    (9) QDP
        (10) QDPOrderProof [EQUIVALENT, 0 ms]
        (11) QDP
        (12) PisEmptyProof [EQUIVALENT, 0 ms]
        (13) YES
    (14) QDP
        (15) QDPOrderProof [EQUIVALENT, 0 ms]
        (16) QDP
        (17) PisEmptyProof [EQUIVALENT, 0 ms]
        (18) YES


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

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

   f(c(s(x), y)) -> f(c(x, s(y)))
   f(c(s(x), s(y))) -> g(c(x, y))
   g(c(x, s(y))) -> g(c(s(x), y))
   g(c(s(x), s(y))) -> f(c(x, y))

The relative TRS consists of the following S rules:

   rand(x) -> x
   rand(x) -> rand(s(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:

   f(c(s(x), y)) -> f(c(x, s(y)))
   f(c(s(x), s(y))) -> g(c(x, y))
   g(c(x, s(y))) -> g(c(s(x), y))
   g(c(s(x), s(y))) -> f(c(x, y))

The TRS R consists of the following rules:

   rand(x) -> x
   rand(x) -> rand(s(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 rules of the TRS R:

   rand(x) -> x

Used ordering: Polynomial interpretation [POLO]:

   POL(c(x_1, x_2)) = x_1 + 2*x_2
   POL(f(x_1)) = 2*x_1
   POL(g(x_1)) = 2*x_1
   POL(rand(x_1)) = 1 + 2*x_1
   POL(s(x_1)) = x_1


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

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

   f(c(s(x), y)) -> f(c(x, s(y)))
   f(c(s(x), s(y))) -> g(c(x, y))
   g(c(x, s(y))) -> g(c(s(x), y))
   g(c(s(x), s(y))) -> f(c(x, y))

The TRS R consists of the following rules:

   rand(x) -> rand(s(x))

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

(5) QDPOrderProof (EQUIVALENT)
We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.

   f(c(s(x), s(y))) -> g(c(x, y))
   g(c(s(x), s(y))) -> f(c(x, y))
The remaining pairs can at least be oriented weakly.
Used ordering:  Polynomial Order [NEGPOLO,POLO] with Interpretation:

POL( rand_1(x_1) ) = 2
POL( s_1(x_1) ) = x_1 + 2
POL( f_1(x_1) ) = x_1 + 2
POL( c_2(x_1, x_2) ) = 2x_1 + 2x_2 + 2
POL( g_1(x_1) ) = max{0, x_1 - 2}

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

   rand(x) -> rand(s(x))


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

(6)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   f(c(s(x), y)) -> f(c(x, s(y)))
   g(c(x, s(y))) -> g(c(s(x), y))

The TRS R consists of the following rules:

   rand(x) -> rand(s(x))

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

(7) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs.
----------------------------------------

(8)
Complex Obligation (AND)

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

(9)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   g(c(x, s(y))) -> g(c(s(x), y))

The TRS R consists of the following rules:

   rand(x) -> rand(s(x))

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

(10) QDPOrderProof (EQUIVALENT)
We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.

   g(c(x, s(y))) -> g(c(s(x), y))
The remaining pairs can at least be oriented weakly.
Used ordering:  Combined order from the following AFS and order.
g(x1)  =  x1

c(x1, x2)  =  x2

s(x1)  =  s(x1)

rand(x1)  =  rand


Knuth-Bendix order [KBO] with precedence:trivial

and weight map:

   s_1=1
   rand=1

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

   rand(x) -> rand(s(x))


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

(11)
Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:

   rand(x) -> rand(s(x))

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

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

(13)
YES

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

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

   f(c(s(x), y)) -> f(c(x, s(y)))

The TRS R consists of the following rules:

   rand(x) -> rand(s(x))

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

(15) QDPOrderProof (EQUIVALENT)
We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.

   f(c(s(x), y)) -> f(c(x, s(y)))
The remaining pairs can at least be oriented weakly.
Used ordering:  Combined order from the following AFS and order.
f(x1)  =  x1

c(x1, x2)  =  x1

s(x1)  =  s(x1)

rand(x1)  =  rand


Knuth-Bendix order [KBO] with precedence:trivial

and weight map:

   s_1=1
   rand=1

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

   rand(x) -> rand(s(x))


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

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

   rand(x) -> rand(s(x))

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

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

(18)
YES