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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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) QTRS Reverse [EQUIVALENT, 0 ms]
(2) QTRS
(3) DependencyPairsProof [EQUIVALENT, 201 ms]
(4) QDP
(5) DependencyGraphProof [EQUIVALENT, 0 ms]
(6) AND
    (7) QDP
        (8) UsableRulesProof [EQUIVALENT, 0 ms]
        (9) QDP
        (10) MRRProof [EQUIVALENT, 13 ms]
        (11) QDP
        (12) DependencyGraphProof [EQUIVALENT, 0 ms]
        (13) TRUE
    (14) QDP
        (15) QDPOrderProof [EQUIVALENT, 227 ms]
        (16) QDP
        (17) DependencyGraphProof [EQUIVALENT, 0 ms]
        (18) AND
            (19) QDP
                (20) QDPOrderProof [EQUIVALENT, 95 ms]
                (21) QDP
                (22) PisEmptyProof [EQUIVALENT, 0 ms]
                (23) YES
            (24) QDP
                (25) QDPOrderProof [EQUIVALENT, 0 ms]
                (26) QDP
                (27) PisEmptyProof [EQUIVALENT, 0 ms]
                (28) YES


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

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

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

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

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

The TRS R consists of the following rules:

   0(0(x1)) -> 2(0(2(0(1(x1)))))
   0(0(x1)) -> 2(0(4(3(2(0(x1))))))
   3(0(x1)) -> 2(3(2(0(x1))))
   3(0(x1)) -> 3(1(2(0(2(x1)))))
   3(0(x1)) -> 2(2(0(3(2(x1)))))
   3(0(x1)) -> 2(3(2(2(2(0(x1))))))
   0(0(0(x1))) -> 0(0(2(0(x1))))
   4(0(0(x1))) -> 2(3(0(0(4(2(x1))))))
   5(0(0(x1))) -> 0(2(5(0(x1))))
   3(1(0(x1))) -> 0(1(2(2(3(x1)))))
   0(3(0(x1))) -> 0(4(3(2(0(2(x1))))))
   3(4(0(x1))) -> 4(2(0(3(x1))))
   3(4(0(x1))) -> 4(3(1(2(0(1(x1))))))
   3(4(0(x1))) -> 0(4(3(4(2(2(x1))))))
   5(4(0(x1))) -> 5(2(5(0(4(2(x1))))))
   5(4(0(x1))) -> 0(4(4(2(3(5(x1))))))
   0(5(0(x1))) -> 0(4(2(5(0(x1)))))
   0(5(0(x1))) -> 0(2(3(2(0(5(x1))))))
   3(5(0(x1))) -> 0(2(3(5(2(3(x1))))))
   0(0(1(x1))) -> 2(0(5(2(0(1(x1))))))
   3(0(1(x1))) -> 2(5(0(3(1(2(x1))))))
   4(0(1(x1))) -> 4(1(2(0(3(x1)))))
   4(0(1(x1))) -> 0(4(1(2(1(2(x1))))))
   0(0(3(x1))) -> 0(2(0(3(x1))))
   3(0(3(x1))) -> 3(2(0(3(x1))))
   3(0(3(x1))) -> 0(2(2(3(3(x1)))))
   4(0(3(x1))) -> 0(4(3(2(2(x1)))))
   4(0(3(x1))) -> 3(0(2(4(3(2(x1))))))
   4(0(3(x1))) -> 3(0(4(2(5(5(x1))))))
   4(3(3(x1))) -> 4(2(3(1(2(3(x1))))))
   0(5(3(x1))) -> 1(2(0(1(5(3(x1))))))
   4(0(4(x1))) -> 0(4(4(2(0(x1)))))
   4(0(4(x1))) -> 0(4(2(0(4(1(x1))))))
   3(4(0(0(x1)))) -> 4(1(2(0(0(3(x1))))))
   3(5(0(0(x1)))) -> 3(0(5(2(5(0(x1))))))
   3(0(1(0(x1)))) -> 0(3(0(1(2(x1)))))
   0(1(1(0(x1)))) -> 2(1(0(0(1(2(x1))))))
   0(0(4(0(x1)))) -> 0(1(2(0(4(0(x1))))))
   3(1(4(0(x1)))) -> 3(0(2(4(1(2(x1))))))
   3(5(4(0(x1)))) -> 0(5(2(3(4(x1)))))
   3(5(4(0(x1)))) -> 4(2(0(5(3(4(x1))))))
   4(0(5(0(x1)))) -> 4(2(0(0(5(x1)))))
   3(5(5(0(x1)))) -> 0(5(5(3(2(x1)))))
   4(0(3(1(x1)))) -> 3(2(1(0(4(x1)))))
   5(3(0(4(x1)))) -> 0(2(4(3(2(5(x1))))))
   4(0(1(0(0(x1))))) -> 2(4(1(0(0(0(x1))))))
   3(5(5(0(0(x1))))) -> 5(2(3(0(5(0(x1))))))
   4(3(4(0(3(x1))))) -> 4(2(4(3(0(3(x1))))))
   5(4(0(3(3(x1))))) -> 3(0(3(5(4(1(x1))))))
   3(0(2(4(3(x1))))) -> 4(3(2(0(3(2(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 2 SCCs with 108 less nodes.
----------------------------------------

(6)
Complex Obligation (AND)

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

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

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

The TRS R consists of the following rules:

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

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

(8) 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.
----------------------------------------

(9)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

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

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

(10) 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:

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


Used ordering: Polynomial interpretation [POLO]:

   POL(0(x_1)) = 3 + 3*x_1
   POL(0^1(x_1)) = 2*x_1
   POL(1(x_1)) = 3 + 3*x_1
   POL(2(x_1)) = x_1


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

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

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

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

(12) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.
----------------------------------------

(13)
TRUE

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

(14)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

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

The TRS R consists of the following rules:

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

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

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


The following pairs can be oriented strictly and are deleted.

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

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

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

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


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

(16)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

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

The TRS R consists of the following rules:

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

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

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

(18)
Complex Obligation (AND)

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

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

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

The TRS R consists of the following rules:

   0(0(x1)) -> 2(0(2(0(1(x1)))))
   0(0(x1)) -> 2(0(4(3(2(0(x1))))))
   3(0(x1)) -> 2(3(2(0(x1))))
   3(0(x1)) -> 3(1(2(0(2(x1)))))
   3(0(x1)) -> 2(2(0(3(2(x1)))))
   3(0(x1)) -> 2(3(2(2(2(0(x1))))))
   0(0(0(x1))) -> 0(0(2(0(x1))))
   4(0(0(x1))) -> 2(3(0(0(4(2(x1))))))
   5(0(0(x1))) -> 0(2(5(0(x1))))
   3(1(0(x1))) -> 0(1(2(2(3(x1)))))
   0(3(0(x1))) -> 0(4(3(2(0(2(x1))))))
   3(4(0(x1))) -> 4(2(0(3(x1))))
   3(4(0(x1))) -> 4(3(1(2(0(1(x1))))))
   3(4(0(x1))) -> 0(4(3(4(2(2(x1))))))
   5(4(0(x1))) -> 5(2(5(0(4(2(x1))))))
   5(4(0(x1))) -> 0(4(4(2(3(5(x1))))))
   0(5(0(x1))) -> 0(4(2(5(0(x1)))))
   0(5(0(x1))) -> 0(2(3(2(0(5(x1))))))
   3(5(0(x1))) -> 0(2(3(5(2(3(x1))))))
   0(0(1(x1))) -> 2(0(5(2(0(1(x1))))))
   3(0(1(x1))) -> 2(5(0(3(1(2(x1))))))
   4(0(1(x1))) -> 4(1(2(0(3(x1)))))
   4(0(1(x1))) -> 0(4(1(2(1(2(x1))))))
   0(0(3(x1))) -> 0(2(0(3(x1))))
   3(0(3(x1))) -> 3(2(0(3(x1))))
   3(0(3(x1))) -> 0(2(2(3(3(x1)))))
   4(0(3(x1))) -> 0(4(3(2(2(x1)))))
   4(0(3(x1))) -> 3(0(2(4(3(2(x1))))))
   4(0(3(x1))) -> 3(0(4(2(5(5(x1))))))
   4(3(3(x1))) -> 4(2(3(1(2(3(x1))))))
   0(5(3(x1))) -> 1(2(0(1(5(3(x1))))))
   4(0(4(x1))) -> 0(4(4(2(0(x1)))))
   4(0(4(x1))) -> 0(4(2(0(4(1(x1))))))
   3(4(0(0(x1)))) -> 4(1(2(0(0(3(x1))))))
   3(5(0(0(x1)))) -> 3(0(5(2(5(0(x1))))))
   3(0(1(0(x1)))) -> 0(3(0(1(2(x1)))))
   0(1(1(0(x1)))) -> 2(1(0(0(1(2(x1))))))
   0(0(4(0(x1)))) -> 0(1(2(0(4(0(x1))))))
   3(1(4(0(x1)))) -> 3(0(2(4(1(2(x1))))))
   3(5(4(0(x1)))) -> 0(5(2(3(4(x1)))))
   3(5(4(0(x1)))) -> 4(2(0(5(3(4(x1))))))
   4(0(5(0(x1)))) -> 4(2(0(0(5(x1)))))
   3(5(5(0(x1)))) -> 0(5(5(3(2(x1)))))
   4(0(3(1(x1)))) -> 3(2(1(0(4(x1)))))
   5(3(0(4(x1)))) -> 0(2(4(3(2(5(x1))))))
   4(0(1(0(0(x1))))) -> 2(4(1(0(0(0(x1))))))
   3(5(5(0(0(x1))))) -> 5(2(3(0(5(0(x1))))))
   4(3(4(0(3(x1))))) -> 4(2(4(3(0(3(x1))))))
   5(4(0(3(3(x1))))) -> 3(0(3(5(4(1(x1))))))
   3(0(2(4(3(x1))))) -> 4(3(2(0(3(2(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(3(4(0(3(x1))))) -> 4^1(3(0(3(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(2(x_1)) = x_1
   POL(3(x_1)) = 1 + x_1
   POL(4(x_1)) = 1
   POL(4^1(x_1)) = x_1
   POL(5(x_1)) = 1 + x_1

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

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


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

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

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

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

The TRS R consists of the following rules:

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

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

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


The following pairs can be oriented strictly and are deleted.

   3^1(5(5(0(0(x1))))) -> 3^1(0(5(0(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(2(x_1)) = 0
   POL(3(x_1)) = 1 + x_1
   POL(3^1(x_1)) = x_1
   POL(4(x_1)) = 0
   POL(5(x_1)) = 1

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

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


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

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

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(27) PisEmptyProof (EQUIVALENT)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
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(28)
YES