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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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) QTRSRRRProof [EQUIVALENT, 38 ms]
(2) QTRS
(3) DependencyPairsProof [EQUIVALENT, 0 ms]
(4) QDP
(5) DependencyGraphProof [EQUIVALENT, 1 ms]
(6) QDP
(7) MRRProof [EQUIVALENT, 36 ms]
(8) QDP
(9) DependencyGraphProof [EQUIVALENT, 0 ms]
(10) TRUE


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

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

Q is empty.

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

   POL(a(x_1)) = 25 + x_1
   POL(b(x_1)) = 17 + x_1
   POL(c(x_1)) = 11 + x_1
   POL(d(x_1)) = 7 + 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)))




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

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

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

Q is empty.

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

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

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

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

The TRS R consists of the following rules:

   a(a(x1)) -> b(c(c(c(x1))))
   b(c(x1)) -> d(d(d(d(x1))))
   a(x1) -> d(c(d(x1)))
   c(d(d(x1))) -> 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 1 less node.
----------------------------------------

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

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

The TRS R consists of the following rules:

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

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

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


Used ordering: Polynomial interpretation [POLO]:

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


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

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

   A(x1) -> C(d(x1))

The TRS R consists of the following rules:

   a(a(x1)) -> b(c(c(c(x1))))
   b(c(x1)) -> d(d(d(d(x1))))
   a(x1) -> d(c(d(x1)))
   c(d(d(x1))) -> 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 0 SCCs with 1 less node.
----------------------------------------

(10)
TRUE