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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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, 48 ms]
(2) QTRS
(3) DependencyPairsProof [EQUIVALENT, 1 ms]
(4) QDP
(5) DependencyGraphProof [EQUIVALENT, 0 ms]
(6) QDP
(7) MRRProof [EQUIVALENT, 27 ms]
(8) QDP
(9) PisEmptyProof [EQUIVALENT, 0 ms]
(10) YES


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

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

Q is empty.

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

   POL(a(x_1)) = x_1
   POL(b(x_1)) = 1 + x_1
   POL(c(x_1)) = 1 + x_1
   POL(d(x_1)) = 2 + x_1
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

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




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

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

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(5) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 3 less nodes.
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(6)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

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

The TRS R consists of the following rules:

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

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

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


Used ordering: Polynomial interpretation [POLO]:

   POL(C(x_1)) = 2*x_1
   POL(a(x_1)) = x_1
   POL(b(x_1)) = 1 + x_1
   POL(c(x_1)) = 1 + x_1
   POL(d(x_1)) = 2 + x_1


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

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(9) PisEmptyProof (EQUIVALENT)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
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(10)
YES