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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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) QTRS Reverse [EQUIVALENT, 0 ms]
(2) QTRS
(3) DependencyPairsProof [EQUIVALENT, 1 ms]
(4) QDP
(5) QDPOrderProof [EQUIVALENT, 51 ms]
(6) QDP
(7) QDPOrderProof [EQUIVALENT, 4254 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(b(x1))) -> b(a(a(a(x1))))
   b(a(b(a(x1)))) -> a(b(b(x1)))

Q is empty.

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(1) QTRS Reverse (EQUIVALENT)
We applied the QTRS Reverse Processor [REVERSE].
----------------------------------------

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

   b(a(a(x1))) -> a(a(a(b(x1))))
   a(b(a(b(x1)))) -> b(b(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:

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

The TRS R consists of the following rules:

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

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


The following pairs can be oriented strictly and are deleted.

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

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

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

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


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

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

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

The TRS R consists of the following rules:

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

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

   POL(A(x_1)) = [1/2] + [1/2]x_1
   POL(B(x_1)) = [1/2] + [3/4]x_1
   POL(a(x_1)) = [1/4] + x_1
   POL(b(x_1)) = [1/4] + [3/2]x_1
The value of delta used in the strict ordering is 1/32.
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

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


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

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

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

The TRS R consists of the following rules:

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