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


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YES
proof of /export/starexec/sandbox/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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished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) QTRS Reverse [EQUIVALENT, 0 ms]
(2) QTRS
(3) DependencyPairsProof [EQUIVALENT, 194 ms]
(4) QDP
(5) DependencyGraphProof [EQUIVALENT, 0 ms]
(6) QDP
(7) MRRProof [EQUIVALENT, 1157 ms]
(8) QDP
(9) DependencyGraphProof [EQUIVALENT, 1 ms]
(10) AND
    (11) QDP
        (12) QDPOrderProof [EQUIVALENT, 73 ms]
        (13) QDP
        (14) DependencyGraphProof [EQUIVALENT, 0 ms]
        (15) QDP
        (16) QDPOrderProof [EQUIVALENT, 68 ms]
        (17) QDP
        (18) QDPOrderProof [EQUIVALENT, 530 ms]
        (19) QDP
        (20) QDPOrderProof [EQUIVALENT, 58 ms]
        (21) QDP
        (22) PisEmptyProof [EQUIVALENT, 0 ms]
        (23) YES
    (24) QDP
        (25) UsableRulesProof [EQUIVALENT, 0 ms]
        (26) QDP
        (27) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (28) YES


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

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

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

Q is empty.

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

(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:

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

Q is empty.

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

(3) DependencyPairsProof (EQUIVALENT)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
----------------------------------------

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

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

The TRS R consists of the following rules:

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

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

(5) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 109 less nodes.
----------------------------------------

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

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

The TRS R consists of the following rules:

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

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

(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:

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


Used ordering: Polynomial interpretation [POLO]:

   POL(0(x_1)) = x_1
   POL(1(x_1)) = x_1
   POL(1^1(x_1)) = 1 + x_1
   POL(2(x_1)) = 1 + x_1
   POL(2^1(x_1)) = 1 + x_1
   POL(3(x_1)) = x_1
   POL(4(x_1)) = x_1
   POL(4^1(x_1)) = 1 + x_1
   POL(5(x_1)) = x_1


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

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

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

The TRS R consists of the following rules:

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

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

(9) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 1 less node.
----------------------------------------

(10)
Complex Obligation (AND)

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

(11)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

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

The TRS R consists of the following rules:

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

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

(12) QDPOrderProof (EQUIVALENT)
We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.

   1^1(4(5(1(0(x1))))) -> 4^1(x1)
The remaining pairs can at least be oriented weakly.
Used ordering:  Polynomial interpretation [POLO]:

   POL(0(x_1)) = 0
   POL(1(x_1)) = 0
   POL(1^1(x_1)) = 1 + x_1
   POL(2(x_1)) = 0
   POL(3(x_1)) = 0
   POL(4(x_1)) = 1
   POL(4^1(x_1)) = 1
   POL(5(x_1)) = 0

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

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


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

(13)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

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

The TRS R consists of the following rules:

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

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

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

(15)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

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

The TRS R consists of the following rules:

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

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

(16) QDPOrderProof (EQUIVALENT)
We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.

   4^1(4(5(0(x1)))) -> 4^1(x1)
The remaining pairs can at least be oriented weakly.
Used ordering:  Polynomial interpretation [POLO]:

   POL(0(x_1)) = x_1
   POL(1(x_1)) = 0
   POL(2(x_1)) = 0
   POL(3(x_1)) = 0
   POL(4(x_1)) = 1 + x_1
   POL(4^1(x_1)) = x_1
   POL(5(x_1)) = x_1

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

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


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

(17)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

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

The TRS R consists of the following rules:

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

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

(18) QDPOrderProof (EQUIVALENT)
We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.

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

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

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

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


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

(19)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

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

The TRS R consists of the following rules:

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

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

(20) QDPOrderProof (EQUIVALENT)
We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.

   4^1(2(0(2(x1)))) -> 4^1(2(2(x1)))
The remaining pairs can at least be oriented weakly.
Used ordering:  Polynomial interpretation [POLO]:

   POL(0(x_1)) = 1 + x_1
   POL(1(x_1)) = 0
   POL(2(x_1)) = x_1
   POL(3(x_1)) = 0
   POL(4(x_1)) = 0
   POL(4^1(x_1)) = x_1
   POL(5(x_1)) = 0

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

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


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

(21)
Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:

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

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

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

(23)
YES

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

(24)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

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

The TRS R consists of the following rules:

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

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

(25) 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.
----------------------------------------

(26)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

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

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

(27) 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(5(0(5(x1))))) -> 2^1(x1)
The graph contains the following edges 1 > 1


*2^1(0(1(5(x1)))) -> 2^1(x1)
The graph contains the following edges 1 > 1


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

(28)
YES