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


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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty


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

(0) QTRS
(1) QTRSRRRProof [EQUIVALENT, 59 ms]
(2) QTRS
(3) DependencyPairsProof [EQUIVALENT, 1 ms]
(4) QDP
(5) MRRProof [EQUIVALENT, 51 ms]
(6) QDP
(7) DependencyGraphProof [EQUIVALENT, 0 ms]
(8) AND
    (9) QDP
        (10) QDPOrderProof [EQUIVALENT, 21 ms]
        (11) QDP
        (12) PisEmptyProof [EQUIVALENT, 0 ms]
        (13) YES
    (14) QDP
        (15) QDPOrderProof [EQUIVALENT, 11 ms]
        (16) QDP
        (17) DependencyGraphProof [EQUIVALENT, 0 ms]
        (18) TRUE


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

   a(a(x1)) -> b(b(b(x1)))
   a(x1) -> d(c(d(x1)))
   b(b(b(x1))) -> a(f(x1))
   b(b(x1)) -> c(c(c(x1)))
   c(c(x1)) -> d(d(d(x1)))
   c(d(d(x1))) -> f(x1)
   f(f(x1)) -> f(a(x1))

Q is empty.

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

   POL(a(x_1)) = 39 + x_1
   POL(b(x_1)) = 26 + x_1
   POL(c(x_1)) = 17 + x_1
   POL(d(x_1)) = 11 + x_1
   POL(f(x_1)) = 39 + x_1
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

   b(b(x1)) -> c(c(c(x1)))
   c(c(x1)) -> d(d(d(x1)))




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

   a(a(x1)) -> b(b(b(x1)))
   a(x1) -> d(c(d(x1)))
   b(b(b(x1))) -> a(f(x1))
   c(d(d(x1))) -> f(x1)
   f(f(x1)) -> f(a(x1))

Q is empty.

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(3) DependencyPairsProof (EQUIVALENT)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
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(4)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   A(a(x1)) -> B(b(b(x1)))
   A(a(x1)) -> B(b(x1))
   A(a(x1)) -> B(x1)
   A(x1) -> C(d(x1))
   B(b(b(x1))) -> A(f(x1))
   B(b(b(x1))) -> F(x1)
   C(d(d(x1))) -> F(x1)
   F(f(x1)) -> F(a(x1))
   F(f(x1)) -> A(x1)

The TRS R consists of the following rules:

   a(a(x1)) -> b(b(b(x1)))
   a(x1) -> d(c(d(x1)))
   b(b(b(x1))) -> a(f(x1))
   c(d(d(x1))) -> f(x1)
   f(f(x1)) -> f(a(x1))

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

(5) 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:

   A(a(x1)) -> B(b(x1))
   A(a(x1)) -> B(x1)
   A(x1) -> C(d(x1))
   B(b(b(x1))) -> F(x1)
   C(d(d(x1))) -> F(x1)
   F(f(x1)) -> A(x1)


Used ordering: Polynomial interpretation [POLO]:

   POL(A(x_1)) = 2 + 2*x_1
   POL(B(x_1)) = 2*x_1
   POL(C(x_1)) = 1 + 2*x_1
   POL(F(x_1)) = 2*x_1
   POL(a(x_1)) = 3 + x_1
   POL(b(x_1)) = 2 + x_1
   POL(c(x_1)) = 3 + x_1
   POL(d(x_1)) = x_1
   POL(f(x_1)) = 3 + x_1


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

   A(a(x1)) -> B(b(b(x1)))
   B(b(b(x1))) -> A(f(x1))
   F(f(x1)) -> F(a(x1))

The TRS R consists of the following rules:

   a(a(x1)) -> b(b(b(x1)))
   a(x1) -> d(c(d(x1)))
   b(b(b(x1))) -> a(f(x1))
   c(d(d(x1))) -> f(x1)
   f(f(x1)) -> f(a(x1))

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

(7) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs.
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(8)
Complex Obligation (AND)

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

   F(f(x1)) -> F(a(x1))

The TRS R consists of the following rules:

   a(a(x1)) -> b(b(b(x1)))
   a(x1) -> d(c(d(x1)))
   b(b(b(x1))) -> a(f(x1))
   c(d(d(x1))) -> f(x1)
   f(f(x1)) -> f(a(x1))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(10) QDPOrderProof (EQUIVALENT)
We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.

   F(f(x1)) -> F(a(x1))
The remaining pairs can at least be oriented weakly.
Used ordering:  Polynomial Order [NEGPOLO,POLO] with Interpretation:

POL( F_1(x_1) ) = max{0, 2x_1 - 2}
POL( b_1(x_1) ) = 0
POL( a_1(x_1) ) = max{0, -2}
POL( f_1(x_1) ) = x_1 + 2
POL( d_1(x_1) ) = max{0, -2}
POL( c_1(x_1) ) = 0

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

   b(b(b(x1))) -> a(f(x1))
   a(a(x1)) -> b(b(b(x1)))
   a(x1) -> d(c(d(x1)))


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

   a(a(x1)) -> b(b(b(x1)))
   a(x1) -> d(c(d(x1)))
   b(b(b(x1))) -> a(f(x1))
   c(d(d(x1))) -> f(x1)
   f(f(x1)) -> f(a(x1))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(12) PisEmptyProof (EQUIVALENT)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
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(13)
YES

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

   B(b(b(x1))) -> A(f(x1))
   A(a(x1)) -> B(b(b(x1)))

The TRS R consists of the following rules:

   a(a(x1)) -> b(b(b(x1)))
   a(x1) -> d(c(d(x1)))
   b(b(b(x1))) -> a(f(x1))
   c(d(d(x1))) -> f(x1)
   f(f(x1)) -> f(a(x1))

Q is empty.
We have to consider all minimal (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.

   A(a(x1)) -> B(b(b(x1)))
The remaining pairs can at least be oriented weakly.
Used ordering:  Polynomial interpretation [POLO]:

   POL(A(x_1)) = 1 + x_1
   POL(B(x_1)) = 1
   POL(a(x_1)) = 1
   POL(b(x_1)) = 0
   POL(c(x_1)) = 1 + x_1
   POL(d(x_1)) = 1 + x_1
   POL(f(x_1)) = 0

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

   f(f(x1)) -> f(a(x1))


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

   B(b(b(x1))) -> A(f(x1))

The TRS R consists of the following rules:

   a(a(x1)) -> b(b(b(x1)))
   a(x1) -> d(c(d(x1)))
   b(b(b(x1))) -> a(f(x1))
   c(d(d(x1))) -> f(x1)
   f(f(x1)) -> f(a(x1))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(17) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.
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(18)
TRUE