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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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) DependencyPairsProof [EQUIVALENT, 24 ms]
(2) QDP
(3) DependencyGraphProof [EQUIVALENT, 0 ms]
(4) QDP
(5) QDPOrderProof [EQUIVALENT, 97 ms]
(6) QDP
(7) QDPOrderProof [EQUIVALENT, 52 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(d(x1)) -> d(b(x1))
   a(x1) -> b(b(b(x1)))
   d(x1) -> x1
   a(x1) -> x1
   b(d(b(x1))) -> a(d(x1))
   b(c(x1)) -> c(d(d(x1)))
   a(c(x1)) -> b(b(c(d(x1))))

Q is empty.

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(1) DependencyPairsProof (EQUIVALENT)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
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(2)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

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

The TRS R consists of the following rules:

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

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

(3) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 6 less nodes.
----------------------------------------

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

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

The TRS R consists of the following rules:

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

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

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

POL( A_1(x_1) ) = 2x_1 + 2
POL( B_1(x_1) ) = 2x_1 + 2
POL( d_1(x_1) ) = 2x_1 + 1
POL( a_1(x_1) ) = x_1
POL( b_1(x_1) ) = x_1
POL( c_1(x_1) ) = 1

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

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


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

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

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

The TRS R consists of the following rules:

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

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

   POL(A(x_1)) = 4 + 2*x_1
   POL(B(x_1)) = 2*x_1
   POL(a(x_1)) = 4 + x_1
   POL(b(x_1)) = 1 + x_1
   POL(c(x_1)) = 1
   POL(d(x_1)) = 4 + 4*x_1

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

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


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

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(9) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.
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(10)
TRUE