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


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


YES
proof of /export/starexec/sandbox/benchmark/theBenchmark.xml
# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty


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

(0) QTRS
(1) DependencyPairsProof [EQUIVALENT, 186 ms]
(2) QDP
(3) DependencyGraphProof [EQUIVALENT, 0 ms]
(4) AND
    (5) QDP
        (6) UsableRulesProof [EQUIVALENT, 0 ms]
        (7) QDP
        (8) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (9) YES
    (10) QDP
        (11) UsableRulesProof [EQUIVALENT, 0 ms]
        (12) QDP
        (13) QDPOrderProof [EQUIVALENT, 49 ms]
        (14) QDP
        (15) PisEmptyProof [EQUIVALENT, 0 ms]
        (16) YES


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

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

   0(1(2(x1))) -> 0(0(2(1(x1))))
   0(1(2(x1))) -> 0(2(1(3(x1))))
   0(1(2(x1))) -> 0(0(2(1(4(4(x1))))))
   0(3(1(x1))) -> 0(1(3(4(0(x1)))))
   0(3(1(x1))) -> 0(1(3(4(4(x1)))))
   0(3(1(x1))) -> 1(3(4(4(4(0(x1))))))
   0(3(2(x1))) -> 0(2(1(3(x1))))
   0(3(2(x1))) -> 0(2(3(4(x1))))
   0(3(2(x1))) -> 0(0(2(4(3(x1)))))
   0(3(2(x1))) -> 0(2(1(4(3(x1)))))
   0(3(2(x1))) -> 0(2(4(3(3(x1)))))
   0(3(2(x1))) -> 0(2(1(3(3(4(x1))))))
   0(3(2(x1))) -> 0(2(3(4(5(5(x1))))))
   0(3(2(x1))) -> 2(4(4(3(4(0(x1))))))
   0(4(1(x1))) -> 0(1(4(4(x1))))
   0(4(1(x1))) -> 0(2(1(4(x1))))
   0(4(2(x1))) -> 0(2(1(4(x1))))
   0(4(2(x1))) -> 0(2(3(4(x1))))
   0(4(2(x1))) -> 0(2(4(3(x1))))
   2(0(1(x1))) -> 5(0(2(1(x1))))
   2(3(1(x1))) -> 1(3(5(2(x1))))
   2(3(1(x1))) -> 0(2(1(3(5(x1)))))
   2(3(1(x1))) -> 1(4(3(5(2(x1)))))
   0(2(0(1(x1)))) -> 5(0(0(2(1(x1)))))
   0(3(1(1(x1)))) -> 0(1(4(1(3(4(x1))))))
   0(3(2(1(x1)))) -> 0(0(3(4(2(1(x1))))))
   0(3(2(2(x1)))) -> 1(3(4(0(2(2(x1))))))
   0(4(1(2(x1)))) -> 1(4(0(2(5(x1)))))
   0(4(3(2(x1)))) -> 2(3(4(4(0(0(x1))))))
   0(5(3(1(x1)))) -> 0(1(4(3(5(4(x1))))))
   0(5(3(1(x1)))) -> 0(1(5(3(4(0(x1))))))
   0(5(3(2(x1)))) -> 0(2(4(5(3(x1)))))
   0(5(3(2(x1)))) -> 0(2(5(3(3(x1)))))
   2(0(3(1(x1)))) -> 2(0(1(3(5(2(x1))))))
   2(0(4(1(x1)))) -> 2(0(1(4(5(x1)))))
   2(5(3(2(x1)))) -> 2(5(2(3(3(x1)))))
   2(5(4(2(x1)))) -> 0(2(5(2(4(x1)))))
   0(0(3(2(1(x1))))) -> 0(0(1(3(5(2(x1))))))
   0(1(0(3(2(x1))))) -> 0(1(4(3(2(0(x1))))))
   0(1(0(3(2(x1))))) -> 2(3(1(0(0(5(x1))))))
   0(3(2(5(1(x1))))) -> 0(2(5(1(3(3(x1))))))
   0(5(1(1(2(x1))))) -> 0(2(4(1(1(5(x1))))))
   0(5(1(2(2(x1))))) -> 0(2(5(2(1(2(x1))))))
   0(5(3(2(1(x1))))) -> 0(1(3(4(2(5(x1))))))
   0(5(5(3(2(x1))))) -> 0(2(5(1(3(5(x1))))))
   2(0(3(1(1(x1))))) -> 2(1(0(1(3(4(x1))))))
   2(2(0(3(1(x1))))) -> 1(3(0(2(5(2(x1))))))
   2(2(0(5(1(x1))))) -> 2(0(2(1(5(1(x1))))))
   2(5(5(4(1(x1))))) -> 5(5(2(1(3(4(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:

   0^1(1(2(x1))) -> 0^1(0(2(1(x1))))
   0^1(1(2(x1))) -> 0^1(2(1(x1)))
   0^1(1(2(x1))) -> 2^1(1(x1))
   0^1(1(2(x1))) -> 0^1(2(1(3(x1))))
   0^1(1(2(x1))) -> 2^1(1(3(x1)))
   0^1(1(2(x1))) -> 0^1(0(2(1(4(4(x1))))))
   0^1(1(2(x1))) -> 0^1(2(1(4(4(x1)))))
   0^1(1(2(x1))) -> 2^1(1(4(4(x1))))
   0^1(3(1(x1))) -> 0^1(1(3(4(0(x1)))))
   0^1(3(1(x1))) -> 0^1(x1)
   0^1(3(1(x1))) -> 0^1(1(3(4(4(x1)))))
   0^1(3(2(x1))) -> 0^1(2(1(3(x1))))
   0^1(3(2(x1))) -> 2^1(1(3(x1)))
   0^1(3(2(x1))) -> 0^1(2(3(4(x1))))
   0^1(3(2(x1))) -> 2^1(3(4(x1)))
   0^1(3(2(x1))) -> 0^1(0(2(4(3(x1)))))
   0^1(3(2(x1))) -> 0^1(2(4(3(x1))))
   0^1(3(2(x1))) -> 2^1(4(3(x1)))
   0^1(3(2(x1))) -> 0^1(2(1(4(3(x1)))))
   0^1(3(2(x1))) -> 2^1(1(4(3(x1))))
   0^1(3(2(x1))) -> 0^1(2(4(3(3(x1)))))
   0^1(3(2(x1))) -> 2^1(4(3(3(x1))))
   0^1(3(2(x1))) -> 0^1(2(1(3(3(4(x1))))))
   0^1(3(2(x1))) -> 2^1(1(3(3(4(x1)))))
   0^1(3(2(x1))) -> 0^1(2(3(4(5(5(x1))))))
   0^1(3(2(x1))) -> 2^1(3(4(5(5(x1)))))
   0^1(3(2(x1))) -> 2^1(4(4(3(4(0(x1))))))
   0^1(3(2(x1))) -> 0^1(x1)
   0^1(4(1(x1))) -> 0^1(1(4(4(x1))))
   0^1(4(1(x1))) -> 0^1(2(1(4(x1))))
   0^1(4(1(x1))) -> 2^1(1(4(x1)))
   0^1(4(2(x1))) -> 0^1(2(1(4(x1))))
   0^1(4(2(x1))) -> 2^1(1(4(x1)))
   0^1(4(2(x1))) -> 0^1(2(3(4(x1))))
   0^1(4(2(x1))) -> 2^1(3(4(x1)))
   0^1(4(2(x1))) -> 0^1(2(4(3(x1))))
   0^1(4(2(x1))) -> 2^1(4(3(x1)))
   2^1(0(1(x1))) -> 0^1(2(1(x1)))
   2^1(0(1(x1))) -> 2^1(1(x1))
   2^1(3(1(x1))) -> 2^1(x1)
   2^1(3(1(x1))) -> 0^1(2(1(3(5(x1)))))
   2^1(3(1(x1))) -> 2^1(1(3(5(x1))))
   0^1(2(0(1(x1)))) -> 0^1(0(2(1(x1))))
   0^1(2(0(1(x1)))) -> 0^1(2(1(x1)))
   0^1(2(0(1(x1)))) -> 2^1(1(x1))
   0^1(3(1(1(x1)))) -> 0^1(1(4(1(3(4(x1))))))
   0^1(3(2(1(x1)))) -> 0^1(0(3(4(2(1(x1))))))
   0^1(3(2(1(x1)))) -> 0^1(3(4(2(1(x1)))))
   0^1(3(2(2(x1)))) -> 0^1(2(2(x1)))
   0^1(4(1(2(x1)))) -> 0^1(2(5(x1)))
   0^1(4(1(2(x1)))) -> 2^1(5(x1))
   0^1(4(3(2(x1)))) -> 2^1(3(4(4(0(0(x1))))))
   0^1(4(3(2(x1)))) -> 0^1(0(x1))
   0^1(4(3(2(x1)))) -> 0^1(x1)
   0^1(5(3(1(x1)))) -> 0^1(1(4(3(5(4(x1))))))
   0^1(5(3(1(x1)))) -> 0^1(1(5(3(4(0(x1))))))
   0^1(5(3(1(x1)))) -> 0^1(x1)
   0^1(5(3(2(x1)))) -> 0^1(2(4(5(3(x1)))))
   0^1(5(3(2(x1)))) -> 2^1(4(5(3(x1))))
   0^1(5(3(2(x1)))) -> 0^1(2(5(3(3(x1)))))
   0^1(5(3(2(x1)))) -> 2^1(5(3(3(x1))))
   2^1(0(3(1(x1)))) -> 2^1(0(1(3(5(2(x1))))))
   2^1(0(3(1(x1)))) -> 0^1(1(3(5(2(x1)))))
   2^1(0(3(1(x1)))) -> 2^1(x1)
   2^1(0(4(1(x1)))) -> 2^1(0(1(4(5(x1)))))
   2^1(0(4(1(x1)))) -> 0^1(1(4(5(x1))))
   2^1(5(3(2(x1)))) -> 2^1(5(2(3(3(x1)))))
   2^1(5(3(2(x1)))) -> 2^1(3(3(x1)))
   2^1(5(4(2(x1)))) -> 0^1(2(5(2(4(x1)))))
   2^1(5(4(2(x1)))) -> 2^1(5(2(4(x1))))
   2^1(5(4(2(x1)))) -> 2^1(4(x1))
   0^1(0(3(2(1(x1))))) -> 0^1(0(1(3(5(2(x1))))))
   0^1(0(3(2(1(x1))))) -> 0^1(1(3(5(2(x1)))))
   0^1(0(3(2(1(x1))))) -> 2^1(x1)
   0^1(1(0(3(2(x1))))) -> 0^1(1(4(3(2(0(x1))))))
   0^1(1(0(3(2(x1))))) -> 2^1(0(x1))
   0^1(1(0(3(2(x1))))) -> 0^1(x1)
   0^1(1(0(3(2(x1))))) -> 2^1(3(1(0(0(5(x1))))))
   0^1(1(0(3(2(x1))))) -> 0^1(0(5(x1)))
   0^1(1(0(3(2(x1))))) -> 0^1(5(x1))
   0^1(3(2(5(1(x1))))) -> 0^1(2(5(1(3(3(x1))))))
   0^1(3(2(5(1(x1))))) -> 2^1(5(1(3(3(x1)))))
   0^1(5(1(1(2(x1))))) -> 0^1(2(4(1(1(5(x1))))))
   0^1(5(1(1(2(x1))))) -> 2^1(4(1(1(5(x1)))))
   0^1(5(1(2(2(x1))))) -> 0^1(2(5(2(1(2(x1))))))
   0^1(5(1(2(2(x1))))) -> 2^1(5(2(1(2(x1)))))
   0^1(5(1(2(2(x1))))) -> 2^1(1(2(x1)))
   0^1(5(3(2(1(x1))))) -> 0^1(1(3(4(2(5(x1))))))
   0^1(5(3(2(1(x1))))) -> 2^1(5(x1))
   0^1(5(5(3(2(x1))))) -> 0^1(2(5(1(3(5(x1))))))
   0^1(5(5(3(2(x1))))) -> 2^1(5(1(3(5(x1)))))
   2^1(0(3(1(1(x1))))) -> 2^1(1(0(1(3(4(x1))))))
   2^1(0(3(1(1(x1))))) -> 0^1(1(3(4(x1))))
   2^1(2(0(3(1(x1))))) -> 0^1(2(5(2(x1))))
   2^1(2(0(3(1(x1))))) -> 2^1(5(2(x1)))
   2^1(2(0(3(1(x1))))) -> 2^1(x1)
   2^1(2(0(5(1(x1))))) -> 2^1(0(2(1(5(1(x1))))))
   2^1(2(0(5(1(x1))))) -> 0^1(2(1(5(1(x1)))))
   2^1(2(0(5(1(x1))))) -> 2^1(1(5(1(x1))))
   2^1(5(5(4(1(x1))))) -> 2^1(1(3(4(x1))))

The TRS R consists of the following rules:

   0(1(2(x1))) -> 0(0(2(1(x1))))
   0(1(2(x1))) -> 0(2(1(3(x1))))
   0(1(2(x1))) -> 0(0(2(1(4(4(x1))))))
   0(3(1(x1))) -> 0(1(3(4(0(x1)))))
   0(3(1(x1))) -> 0(1(3(4(4(x1)))))
   0(3(1(x1))) -> 1(3(4(4(4(0(x1))))))
   0(3(2(x1))) -> 0(2(1(3(x1))))
   0(3(2(x1))) -> 0(2(3(4(x1))))
   0(3(2(x1))) -> 0(0(2(4(3(x1)))))
   0(3(2(x1))) -> 0(2(1(4(3(x1)))))
   0(3(2(x1))) -> 0(2(4(3(3(x1)))))
   0(3(2(x1))) -> 0(2(1(3(3(4(x1))))))
   0(3(2(x1))) -> 0(2(3(4(5(5(x1))))))
   0(3(2(x1))) -> 2(4(4(3(4(0(x1))))))
   0(4(1(x1))) -> 0(1(4(4(x1))))
   0(4(1(x1))) -> 0(2(1(4(x1))))
   0(4(2(x1))) -> 0(2(1(4(x1))))
   0(4(2(x1))) -> 0(2(3(4(x1))))
   0(4(2(x1))) -> 0(2(4(3(x1))))
   2(0(1(x1))) -> 5(0(2(1(x1))))
   2(3(1(x1))) -> 1(3(5(2(x1))))
   2(3(1(x1))) -> 0(2(1(3(5(x1)))))
   2(3(1(x1))) -> 1(4(3(5(2(x1)))))
   0(2(0(1(x1)))) -> 5(0(0(2(1(x1)))))
   0(3(1(1(x1)))) -> 0(1(4(1(3(4(x1))))))
   0(3(2(1(x1)))) -> 0(0(3(4(2(1(x1))))))
   0(3(2(2(x1)))) -> 1(3(4(0(2(2(x1))))))
   0(4(1(2(x1)))) -> 1(4(0(2(5(x1)))))
   0(4(3(2(x1)))) -> 2(3(4(4(0(0(x1))))))
   0(5(3(1(x1)))) -> 0(1(4(3(5(4(x1))))))
   0(5(3(1(x1)))) -> 0(1(5(3(4(0(x1))))))
   0(5(3(2(x1)))) -> 0(2(4(5(3(x1)))))
   0(5(3(2(x1)))) -> 0(2(5(3(3(x1)))))
   2(0(3(1(x1)))) -> 2(0(1(3(5(2(x1))))))
   2(0(4(1(x1)))) -> 2(0(1(4(5(x1)))))
   2(5(3(2(x1)))) -> 2(5(2(3(3(x1)))))
   2(5(4(2(x1)))) -> 0(2(5(2(4(x1)))))
   0(0(3(2(1(x1))))) -> 0(0(1(3(5(2(x1))))))
   0(1(0(3(2(x1))))) -> 0(1(4(3(2(0(x1))))))
   0(1(0(3(2(x1))))) -> 2(3(1(0(0(5(x1))))))
   0(3(2(5(1(x1))))) -> 0(2(5(1(3(3(x1))))))
   0(5(1(1(2(x1))))) -> 0(2(4(1(1(5(x1))))))
   0(5(1(2(2(x1))))) -> 0(2(5(2(1(2(x1))))))
   0(5(3(2(1(x1))))) -> 0(1(3(4(2(5(x1))))))
   0(5(5(3(2(x1))))) -> 0(2(5(1(3(5(x1))))))
   2(0(3(1(1(x1))))) -> 2(1(0(1(3(4(x1))))))
   2(2(0(3(1(x1))))) -> 1(3(0(2(5(2(x1))))))
   2(2(0(5(1(x1))))) -> 2(0(2(1(5(1(x1))))))
   2(5(5(4(1(x1))))) -> 5(5(2(1(3(4(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 91 less nodes.
----------------------------------------

(4)
Complex Obligation (AND)

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

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

   2^1(0(3(1(x1)))) -> 2^1(x1)
   2^1(3(1(x1))) -> 2^1(x1)
   2^1(2(0(3(1(x1))))) -> 2^1(x1)

The TRS R consists of the following rules:

   0(1(2(x1))) -> 0(0(2(1(x1))))
   0(1(2(x1))) -> 0(2(1(3(x1))))
   0(1(2(x1))) -> 0(0(2(1(4(4(x1))))))
   0(3(1(x1))) -> 0(1(3(4(0(x1)))))
   0(3(1(x1))) -> 0(1(3(4(4(x1)))))
   0(3(1(x1))) -> 1(3(4(4(4(0(x1))))))
   0(3(2(x1))) -> 0(2(1(3(x1))))
   0(3(2(x1))) -> 0(2(3(4(x1))))
   0(3(2(x1))) -> 0(0(2(4(3(x1)))))
   0(3(2(x1))) -> 0(2(1(4(3(x1)))))
   0(3(2(x1))) -> 0(2(4(3(3(x1)))))
   0(3(2(x1))) -> 0(2(1(3(3(4(x1))))))
   0(3(2(x1))) -> 0(2(3(4(5(5(x1))))))
   0(3(2(x1))) -> 2(4(4(3(4(0(x1))))))
   0(4(1(x1))) -> 0(1(4(4(x1))))
   0(4(1(x1))) -> 0(2(1(4(x1))))
   0(4(2(x1))) -> 0(2(1(4(x1))))
   0(4(2(x1))) -> 0(2(3(4(x1))))
   0(4(2(x1))) -> 0(2(4(3(x1))))
   2(0(1(x1))) -> 5(0(2(1(x1))))
   2(3(1(x1))) -> 1(3(5(2(x1))))
   2(3(1(x1))) -> 0(2(1(3(5(x1)))))
   2(3(1(x1))) -> 1(4(3(5(2(x1)))))
   0(2(0(1(x1)))) -> 5(0(0(2(1(x1)))))
   0(3(1(1(x1)))) -> 0(1(4(1(3(4(x1))))))
   0(3(2(1(x1)))) -> 0(0(3(4(2(1(x1))))))
   0(3(2(2(x1)))) -> 1(3(4(0(2(2(x1))))))
   0(4(1(2(x1)))) -> 1(4(0(2(5(x1)))))
   0(4(3(2(x1)))) -> 2(3(4(4(0(0(x1))))))
   0(5(3(1(x1)))) -> 0(1(4(3(5(4(x1))))))
   0(5(3(1(x1)))) -> 0(1(5(3(4(0(x1))))))
   0(5(3(2(x1)))) -> 0(2(4(5(3(x1)))))
   0(5(3(2(x1)))) -> 0(2(5(3(3(x1)))))
   2(0(3(1(x1)))) -> 2(0(1(3(5(2(x1))))))
   2(0(4(1(x1)))) -> 2(0(1(4(5(x1)))))
   2(5(3(2(x1)))) -> 2(5(2(3(3(x1)))))
   2(5(4(2(x1)))) -> 0(2(5(2(4(x1)))))
   0(0(3(2(1(x1))))) -> 0(0(1(3(5(2(x1))))))
   0(1(0(3(2(x1))))) -> 0(1(4(3(2(0(x1))))))
   0(1(0(3(2(x1))))) -> 2(3(1(0(0(5(x1))))))
   0(3(2(5(1(x1))))) -> 0(2(5(1(3(3(x1))))))
   0(5(1(1(2(x1))))) -> 0(2(4(1(1(5(x1))))))
   0(5(1(2(2(x1))))) -> 0(2(5(2(1(2(x1))))))
   0(5(3(2(1(x1))))) -> 0(1(3(4(2(5(x1))))))
   0(5(5(3(2(x1))))) -> 0(2(5(1(3(5(x1))))))
   2(0(3(1(1(x1))))) -> 2(1(0(1(3(4(x1))))))
   2(2(0(3(1(x1))))) -> 1(3(0(2(5(2(x1))))))
   2(2(0(5(1(x1))))) -> 2(0(2(1(5(1(x1))))))
   2(5(5(4(1(x1))))) -> 5(5(2(1(3(4(x1))))))

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

(6) UsableRulesProof (EQUIVALENT)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

(7)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   2^1(0(3(1(x1)))) -> 2^1(x1)
   2^1(3(1(x1))) -> 2^1(x1)
   2^1(2(0(3(1(x1))))) -> 2^1(x1)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(8) 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:
*2^1(0(3(1(x1)))) -> 2^1(x1)
The graph contains the following edges 1 > 1


*2^1(3(1(x1))) -> 2^1(x1)
The graph contains the following edges 1 > 1


*2^1(2(0(3(1(x1))))) -> 2^1(x1)
The graph contains the following edges 1 > 1


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(9)
YES

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

   0^1(3(2(x1))) -> 0^1(x1)
   0^1(3(1(x1))) -> 0^1(x1)
   0^1(4(3(2(x1)))) -> 0^1(x1)
   0^1(5(3(1(x1)))) -> 0^1(x1)
   0^1(1(0(3(2(x1))))) -> 0^1(x1)
   0^1(1(0(3(2(x1))))) -> 0^1(5(x1))

The TRS R consists of the following rules:

   0(1(2(x1))) -> 0(0(2(1(x1))))
   0(1(2(x1))) -> 0(2(1(3(x1))))
   0(1(2(x1))) -> 0(0(2(1(4(4(x1))))))
   0(3(1(x1))) -> 0(1(3(4(0(x1)))))
   0(3(1(x1))) -> 0(1(3(4(4(x1)))))
   0(3(1(x1))) -> 1(3(4(4(4(0(x1))))))
   0(3(2(x1))) -> 0(2(1(3(x1))))
   0(3(2(x1))) -> 0(2(3(4(x1))))
   0(3(2(x1))) -> 0(0(2(4(3(x1)))))
   0(3(2(x1))) -> 0(2(1(4(3(x1)))))
   0(3(2(x1))) -> 0(2(4(3(3(x1)))))
   0(3(2(x1))) -> 0(2(1(3(3(4(x1))))))
   0(3(2(x1))) -> 0(2(3(4(5(5(x1))))))
   0(3(2(x1))) -> 2(4(4(3(4(0(x1))))))
   0(4(1(x1))) -> 0(1(4(4(x1))))
   0(4(1(x1))) -> 0(2(1(4(x1))))
   0(4(2(x1))) -> 0(2(1(4(x1))))
   0(4(2(x1))) -> 0(2(3(4(x1))))
   0(4(2(x1))) -> 0(2(4(3(x1))))
   2(0(1(x1))) -> 5(0(2(1(x1))))
   2(3(1(x1))) -> 1(3(5(2(x1))))
   2(3(1(x1))) -> 0(2(1(3(5(x1)))))
   2(3(1(x1))) -> 1(4(3(5(2(x1)))))
   0(2(0(1(x1)))) -> 5(0(0(2(1(x1)))))
   0(3(1(1(x1)))) -> 0(1(4(1(3(4(x1))))))
   0(3(2(1(x1)))) -> 0(0(3(4(2(1(x1))))))
   0(3(2(2(x1)))) -> 1(3(4(0(2(2(x1))))))
   0(4(1(2(x1)))) -> 1(4(0(2(5(x1)))))
   0(4(3(2(x1)))) -> 2(3(4(4(0(0(x1))))))
   0(5(3(1(x1)))) -> 0(1(4(3(5(4(x1))))))
   0(5(3(1(x1)))) -> 0(1(5(3(4(0(x1))))))
   0(5(3(2(x1)))) -> 0(2(4(5(3(x1)))))
   0(5(3(2(x1)))) -> 0(2(5(3(3(x1)))))
   2(0(3(1(x1)))) -> 2(0(1(3(5(2(x1))))))
   2(0(4(1(x1)))) -> 2(0(1(4(5(x1)))))
   2(5(3(2(x1)))) -> 2(5(2(3(3(x1)))))
   2(5(4(2(x1)))) -> 0(2(5(2(4(x1)))))
   0(0(3(2(1(x1))))) -> 0(0(1(3(5(2(x1))))))
   0(1(0(3(2(x1))))) -> 0(1(4(3(2(0(x1))))))
   0(1(0(3(2(x1))))) -> 2(3(1(0(0(5(x1))))))
   0(3(2(5(1(x1))))) -> 0(2(5(1(3(3(x1))))))
   0(5(1(1(2(x1))))) -> 0(2(4(1(1(5(x1))))))
   0(5(1(2(2(x1))))) -> 0(2(5(2(1(2(x1))))))
   0(5(3(2(1(x1))))) -> 0(1(3(4(2(5(x1))))))
   0(5(5(3(2(x1))))) -> 0(2(5(1(3(5(x1))))))
   2(0(3(1(1(x1))))) -> 2(1(0(1(3(4(x1))))))
   2(2(0(3(1(x1))))) -> 1(3(0(2(5(2(x1))))))
   2(2(0(5(1(x1))))) -> 2(0(2(1(5(1(x1))))))
   2(5(5(4(1(x1))))) -> 5(5(2(1(3(4(x1))))))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(11) UsableRulesProof (EQUIVALENT)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
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(12)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   0^1(3(2(x1))) -> 0^1(x1)
   0^1(3(1(x1))) -> 0^1(x1)
   0^1(4(3(2(x1)))) -> 0^1(x1)
   0^1(5(3(1(x1)))) -> 0^1(x1)
   0^1(1(0(3(2(x1))))) -> 0^1(x1)
   0^1(1(0(3(2(x1))))) -> 0^1(5(x1))

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


The following pairs can be oriented strictly and are deleted.

   0^1(3(2(x1))) -> 0^1(x1)
   0^1(3(1(x1))) -> 0^1(x1)
   0^1(4(3(2(x1)))) -> 0^1(x1)
   0^1(5(3(1(x1)))) -> 0^1(x1)
   0^1(1(0(3(2(x1))))) -> 0^1(x1)
   0^1(1(0(3(2(x1))))) -> 0^1(5(x1))
The remaining pairs can at least be oriented weakly.
Used ordering:  Polynomial Order [NEGPOLO,POLO] with Interpretation:

POL( 0^1_1(x_1) ) = max{0, 2x_1 - 1}
POL( 3_1(x_1) ) = 2x_1 + 1
POL( 2_1(x_1) ) = x_1
POL( 1_1(x_1) ) = x_1
POL( 4_1(x_1) ) = 2x_1
POL( 5_1(x_1) ) = 2x_1
POL( 0_1(x_1) ) = x_1 + 1

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
none


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(14)
Obligation:
Q DP problem:
P is empty.
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(15) PisEmptyProof (EQUIVALENT)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
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(16)
YES