/export/starexec/sandbox2/solver/bin/starexec_run_standard /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files


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


YES
proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml
# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished


Termination w.r.t. Q of the given QTRS could be proven:

(0) QTRS
(1) DependencyPairsProof [EQUIVALENT, 18 ms]
(2) QDP
(3) DependencyGraphProof [EQUIVALENT, 2 ms]
(4) AND
    (5) QDP
        (6) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (7) YES
    (8) QDP
        (9) MRRProof [EQUIVALENT, 36 ms]
        (10) QDP
        (11) PisEmptyProof [EQUIVALENT, 0 ms]
        (12) YES


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

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

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

Q is empty.

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

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

The TRS R consists of the following rules:

   b(a(a(a(x1)))) -> b(b(a(a(x1))))
   a(b(b(b(x1)))) -> b(b(b(a(x1))))
   a(b(b(b(x1)))) -> b(b(a(a(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 2 SCCs with 6 less nodes.
----------------------------------------

(4)
Complex Obligation (AND)

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

(5)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   B(a(a(a(x1)))) -> B(a(a(x1)))

The TRS R consists of the following rules:

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

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

(6) QDPSizeChangeProof (EQUIVALENT)
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem. 

From the DPs we obtained the following set of size-change graphs:
*B(a(a(a(x1)))) -> B(a(a(x1)))
The graph contains the following edges 1 > 1


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

(7)
YES

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

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

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

The TRS R consists of the following rules:

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

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

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


Used ordering: Polynomial interpretation [POLO]:

   POL(A(x_1)) = 2*x_1
   POL(a(x_1)) = 2 + x_1
   POL(b(x_1)) = 2 + x_1


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

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

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

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

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

(12)
YES