YES
proof of /export/starexec/sandbox2/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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) DependencyPairsProof [EQUIVALENT, 12 ms]
(2) QDP
(3) MRRProof [EQUIVALENT, 63 ms]
(4) QDP
(5) DependencyGraphProof [EQUIVALENT, 0 ms]
(6) QDP
(7) QDPOrderProof [EQUIVALENT, 13 ms]
(8) QDP
(9) DependencyGraphProof [EQUIVALENT, 0 ms]
(10) QDP
(11) QDPOrderProof [EQUIVALENT, 33 ms]
(12) QDP
(13) PisEmptyProof [EQUIVALENT, 0 ms]
(14) YES


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

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

Q is empty.

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(1) DependencyPairsProof (EQUIVALENT)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
----------------------------------------

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

   B(a(a(x1))) -> A(b(c(x1)))
   B(a(a(x1))) -> B(c(x1))
   B(a(a(x1))) -> C(x1)
   C(a(x1)) -> A(c(x1))
   C(a(x1)) -> C(x1)
   B(c(a(x1))) -> A(b(c(x1)))
   B(c(a(x1))) -> B(c(x1))
   B(c(a(x1))) -> C(x1)
   C(b(x1)) -> D(x1)
   D(x1) -> B(a(x1))
   D(x1) -> A(x1)
   A(d(x1)) -> D(a(x1))
   A(d(x1)) -> A(x1)
   A(a(x1)) -> A(b(a(x1)))
   A(a(x1)) -> B(a(x1))

The TRS R consists of the following rules:

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

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

   B(a(a(x1))) -> B(c(x1))
   B(a(a(x1))) -> C(x1)
   C(a(x1)) -> C(x1)
   B(c(a(x1))) -> B(c(x1))
   B(c(a(x1))) -> C(x1)
   A(d(x1)) -> A(x1)
   A(a(x1)) -> B(a(x1))


Used ordering: Polynomial interpretation [POLO]:

   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)) = 3 + x_1
   POL(a(x_1)) = 1 + x_1
   POL(b(x_1)) = x_1
   POL(c(x_1)) = 1 + x_1
   POL(d(x_1)) = 1 + x_1


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

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

   B(a(a(x1))) -> A(b(c(x1)))
   C(a(x1)) -> A(c(x1))
   B(c(a(x1))) -> A(b(c(x1)))
   C(b(x1)) -> D(x1)
   D(x1) -> B(a(x1))
   D(x1) -> A(x1)
   A(d(x1)) -> D(a(x1))
   A(a(x1)) -> A(b(a(x1)))

The TRS R consists of the following rules:

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

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

(5) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 3 less nodes.
----------------------------------------

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

   A(d(x1)) -> D(a(x1))
   D(x1) -> B(a(x1))
   B(a(a(x1))) -> A(b(c(x1)))
   A(a(x1)) -> A(b(a(x1)))
   D(x1) -> A(x1)

The TRS R consists of the following rules:

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

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

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


The following pairs can be oriented strictly and are deleted.

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

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

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

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


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

(8)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   D(x1) -> B(a(x1))
   B(a(a(x1))) -> A(b(c(x1)))
   A(a(x1)) -> A(b(a(x1)))
   D(x1) -> A(x1)

The TRS R consists of the following rules:

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

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

(9) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 3 less nodes.
----------------------------------------

(10)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

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

The TRS R consists of the following rules:

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

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

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


The following pairs can be oriented strictly and are deleted.

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

POL( A_1(x_1) ) = max{0, 2x_1 - 2}
POL( a_1(x_1) ) = x_1 + 2
POL( d_1(x_1) ) = x_1
POL( b_1(x_1) ) = max{0, x_1 - 2}
POL( c_1(x_1) ) = x_1 + 2

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

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


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

(12)
Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:

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

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

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

(14)
YES