Termination w.r.t. Q proof of /home/cern_httpd/provide/research/cycsrs/tpdb/TPDB-d9b80194f163/SRS_Standard/ICFP

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

0 QTRS

↳1 QTRS Reverse (⇔, 0 ms)

↳2 QTRS

↳3 DependencyPairsProof (⇔, 125 ms)

↳4 QDP

↳5 DependencyGraphProof (⇔, 0 ms)

↳6 QDP

↳7 QDPOrderProof (⇔, 205 ms)

↳8 QDP

↳9 DependencyGraphProof (⇔, 0 ms)

↳10 AND

    ↳11 QDP

        ↳12 QDPOrderProof (⇔, 66 ms)

        ↳13 QDP

        ↳14 DependencyGraphProof (⇔, 0 ms)

        ↳15 QDP

        ↳16 QDPOrderProof (⇔, 0 ms)

        ↳17 QDP

        ↳18 PisEmptyProof (⇔, 0 ms)

        ↳19 YES

    ↳20 QDP

        ↳21 QDPOrderProof (⇔, 21 ms)

        ↳22 QDP

        ↳23 PisEmptyProof (⇔, 0 ms)

        ↳24 YES

(0) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

0(0(1(x))) → 0(0(2(1(2(x)))))
0(0(1(x))) → 0(0(3(1(4(x)))))
0(1(0(x))) → 0(0(1(2(4(x)))))
0(1(0(x))) → 2(1(2(0(0(x)))))
0(1(0(x))) → 0(0(1(2(2(4(x))))))
0(1(0(x))) → 0(0(2(4(1(4(x))))))
0(4(0(x))) → 3(4(3(2(0(0(x))))))
0(0(1(0(x)))) → 0(0(0(1(4(x)))))
0(0(4(1(x)))) → 0(0(2(1(4(x)))))
0(0(4(1(x)))) → 0(0(2(1(4(3(x))))))
0(1(0(4(x)))) → 0(0(3(4(1(2(x))))))
0(1(3(0(x)))) → 0(0(3(1(4(x)))))
0(1(3(0(x)))) → 0(3(0(1(2(x)))))
0(1(3(0(x)))) → 0(0(1(3(2(5(x))))))
0(1(4(0(x)))) → 0(2(0(3(1(4(x))))))
0(1(5(1(x)))) → 2(1(1(4(5(0(x))))))
0(1(5(4(x)))) → 1(2(4(2(5(0(x))))))
0(1(5(4(x)))) → 4(1(0(3(2(5(x))))))
0(1(5(4(x)))) → 5(0(2(4(1(4(x))))))
0(3(0(1(x)))) → 0(0(3(1(2(2(x))))))
0(3(1(0(x)))) → 0(0(3(1(4(x)))))
0(4(0(1(x)))) → 0(0(2(1(4(x)))))
0(4(5(1(x)))) → 1(2(4(2(5(0(x))))))
0(4(5(1(x)))) → 3(1(4(5(0(2(x))))))
0(4(5(1(x)))) → 3(2(5(1(4(0(x))))))
0(4(5(1(x)))) → 5(3(0(5(1(4(x))))))
0(4(5(1(x)))) → 5(5(0(5(1(4(x))))))
0(4(5(4(x)))) → 5(2(4(4(0(4(x))))))
0(5(1(0(x)))) → 0(0(5(1(2(x)))))
3(5(0(1(x)))) → 3(0(2(1(2(5(x))))))
3(5(1(0(x)))) → 0(5(1(3(2(x)))))
3(5(1(0(x)))) → 2(1(2(0(5(3(x))))))
3(5(1(0(x)))) → 3(1(2(2(0(5(x))))))
3(5(1(0(x)))) → 5(1(3(2(0(2(x))))))
0(1(3(3(0(x))))) → 3(3(2(0(0(1(x))))))
0(1(3(5(1(x))))) → 1(1(3(4(5(0(x))))))
0(1(3(5(1(x))))) → 1(1(5(0(3(3(x))))))
0(1(5(2(0(x))))) → 5(0(3(1(0(2(x))))))
0(1(5(4(1(x))))) → 5(3(4(1(0(1(x))))))
0(1(5(4(4(x))))) → 4(5(2(1(4(0(x))))))
0(1(5(4(4(x))))) → 5(0(4(3(4(1(x))))))
0(3(1(0(4(x))))) → 0(0(3(1(2(4(x))))))
0(4(3(3(0(x))))) → 0(2(3(4(0(3(x))))))
0(4(5(2(0(x))))) → 0(2(2(5(0(4(x))))))
0(5(1(5(1(x))))) → 5(5(3(0(1(1(x))))))
3(0(1(5(4(x))))) → 0(5(3(4(1(2(x))))))
3(5(0(1(0(x))))) → 5(1(2(0(0(3(x))))))
3(5(4(0(0(x))))) → 5(0(3(0(4(4(x))))))
3(5(5(0(1(x))))) → 5(5(0(3(1(2(x))))))

Q is empty.

(1) QTRS Reverse (EQUIVALENT transformation)

We applied the QTRS Reverse Processor [REVERSE].

(2) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

1(0(0(x))) → 2(1(2(0(0(x)))))
1(0(0(x))) → 4(1(3(0(0(x)))))
0(1(0(x))) → 4(2(1(0(0(x)))))
0(1(0(x))) → 0(0(2(1(2(x)))))
0(1(0(x))) → 4(2(2(1(0(0(x))))))
0(1(0(x))) → 4(1(4(2(0(0(x))))))
0(4(0(x))) → 0(0(2(3(4(3(x))))))
0(1(0(0(x)))) → 4(1(0(0(0(x)))))
1(4(0(0(x)))) → 4(1(2(0(0(x)))))
1(4(0(0(x)))) → 3(4(1(2(0(0(x))))))
4(0(1(0(x)))) → 2(1(4(3(0(0(x))))))
0(3(1(0(x)))) → 4(1(3(0(0(x)))))
0(3(1(0(x)))) → 2(1(0(3(0(x)))))
0(3(1(0(x)))) → 5(2(3(1(0(0(x))))))
0(4(1(0(x)))) → 4(1(3(0(2(0(x))))))
1(5(1(0(x)))) → 0(5(4(1(1(2(x))))))
4(5(1(0(x)))) → 0(5(2(4(2(1(x))))))
4(5(1(0(x)))) → 5(2(3(0(1(4(x))))))
4(5(1(0(x)))) → 4(1(4(2(0(5(x))))))
1(0(3(0(x)))) → 2(2(1(3(0(0(x))))))
0(1(3(0(x)))) → 4(1(3(0(0(x)))))
1(0(4(0(x)))) → 4(1(2(0(0(x)))))
1(5(4(0(x)))) → 0(5(2(4(2(1(x))))))
1(5(4(0(x)))) → 2(0(5(4(1(3(x))))))
1(5(4(0(x)))) → 0(4(1(5(2(3(x))))))
1(5(4(0(x)))) → 4(1(5(0(3(5(x))))))
1(5(4(0(x)))) → 4(1(5(0(5(5(x))))))
4(5(4(0(x)))) → 4(0(4(4(2(5(x))))))
0(1(5(0(x)))) → 2(1(5(0(0(x)))))
1(0(5(3(x)))) → 5(2(1(2(0(3(x))))))
0(1(5(3(x)))) → 2(3(1(5(0(x)))))
0(1(5(3(x)))) → 3(5(0(2(1(2(x))))))
0(1(5(3(x)))) → 5(0(2(2(1(3(x))))))
0(1(5(3(x)))) → 2(0(2(3(1(5(x))))))
0(3(3(1(0(x))))) → 1(0(0(2(3(3(x))))))
1(5(3(1(0(x))))) → 0(5(4(3(1(1(x))))))
1(5(3(1(0(x))))) → 3(3(0(5(1(1(x))))))
0(2(5(1(0(x))))) → 2(0(1(3(0(5(x))))))
1(4(5(1(0(x))))) → 1(0(1(4(3(5(x))))))
4(4(5(1(0(x))))) → 0(4(1(2(5(4(x))))))
4(4(5(1(0(x))))) → 1(4(3(4(0(5(x))))))
4(0(1(3(0(x))))) → 4(2(1(3(0(0(x))))))
0(3(3(4(0(x))))) → 3(0(4(3(2(0(x))))))
0(2(5(4(0(x))))) → 4(0(5(2(2(0(x))))))
1(5(1(5(0(x))))) → 1(1(0(3(5(5(x))))))
4(5(1(0(3(x))))) → 2(1(4(3(5(0(x))))))
0(1(0(5(3(x))))) → 3(0(0(2(1(5(x))))))
0(0(4(5(3(x))))) → 4(4(0(3(0(5(x))))))
1(0(5(5(3(x))))) → 2(1(3(0(5(5(x))))))

Q is empty.

(3) DependencyPairsProof (EQUIVALENT transformation)

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:

1¹(0(0(x))) → 1¹(2(0(0(x))))
1¹(0(0(x))) → 4¹(1(3(0(0(x)))))
1¹(0(0(x))) → 1¹(3(0(0(x))))
0¹(1(0(x))) → 4¹(2(1(0(0(x)))))
0¹(1(0(x))) → 1¹(0(0(x)))
0¹(1(0(x))) → 0¹(0(x))
0¹(1(0(x))) → 0¹(0(2(1(2(x)))))
0¹(1(0(x))) → 0¹(2(1(2(x))))
0¹(1(0(x))) → 1¹(2(x))
0¹(1(0(x))) → 4¹(2(2(1(0(0(x))))))
0¹(1(0(x))) → 4¹(1(4(2(0(0(x))))))
0¹(1(0(x))) → 1¹(4(2(0(0(x)))))
0¹(1(0(x))) → 4¹(2(0(0(x))))
0¹(4(0(x))) → 0¹(0(2(3(4(3(x))))))
0¹(4(0(x))) → 0¹(2(3(4(3(x)))))
0¹(4(0(x))) → 4¹(3(x))
0¹(1(0(0(x)))) → 4¹(1(0(0(0(x)))))
0¹(1(0(0(x)))) → 1¹(0(0(0(x))))
0¹(1(0(0(x)))) → 0¹(0(0(x)))
1¹(4(0(0(x)))) → 4¹(1(2(0(0(x)))))
1¹(4(0(0(x)))) → 1¹(2(0(0(x))))
4¹(0(1(0(x)))) → 1¹(4(3(0(0(x)))))
4¹(0(1(0(x)))) → 4¹(3(0(0(x))))
4¹(0(1(0(x)))) → 0¹(0(x))
0¹(3(1(0(x)))) → 4¹(1(3(0(0(x)))))
0¹(3(1(0(x)))) → 1¹(3(0(0(x))))
0¹(3(1(0(x)))) → 0¹(0(x))
0¹(3(1(0(x)))) → 1¹(0(3(0(x))))
0¹(3(1(0(x)))) → 0¹(3(0(x)))
0¹(3(1(0(x)))) → 1¹(0(0(x)))
0¹(4(1(0(x)))) → 4¹(1(3(0(2(0(x))))))
0¹(4(1(0(x)))) → 1¹(3(0(2(0(x)))))
0¹(4(1(0(x)))) → 0¹(2(0(x)))
1¹(5(1(0(x)))) → 0¹(5(4(1(1(2(x))))))
1¹(5(1(0(x)))) → 4¹(1(1(2(x))))
1¹(5(1(0(x)))) → 1¹(1(2(x)))
1¹(5(1(0(x)))) → 1¹(2(x))
4¹(5(1(0(x)))) → 0¹(5(2(4(2(1(x))))))
4¹(5(1(0(x)))) → 4¹(2(1(x)))
4¹(5(1(0(x)))) → 1¹(x)
4¹(5(1(0(x)))) → 0¹(1(4(x)))
4¹(5(1(0(x)))) → 1¹(4(x))
4¹(5(1(0(x)))) → 4¹(x)
4¹(5(1(0(x)))) → 4¹(1(4(2(0(5(x))))))
4¹(5(1(0(x)))) → 1¹(4(2(0(5(x)))))
4¹(5(1(0(x)))) → 4¹(2(0(5(x))))
4¹(5(1(0(x)))) → 0¹(5(x))
1¹(0(3(0(x)))) → 1¹(3(0(0(x))))
1¹(0(3(0(x)))) → 0¹(0(x))
0¹(1(3(0(x)))) → 4¹(1(3(0(0(x)))))
0¹(1(3(0(x)))) → 1¹(3(0(0(x))))
0¹(1(3(0(x)))) → 0¹(0(x))
1¹(0(4(0(x)))) → 4¹(1(2(0(0(x)))))
1¹(0(4(0(x)))) → 1¹(2(0(0(x))))
1¹(0(4(0(x)))) → 0¹(0(x))
1¹(5(4(0(x)))) → 0¹(5(2(4(2(1(x))))))
1¹(5(4(0(x)))) → 4¹(2(1(x)))
1¹(5(4(0(x)))) → 1¹(x)
1¹(5(4(0(x)))) → 0¹(5(4(1(3(x)))))
1¹(5(4(0(x)))) → 4¹(1(3(x)))
1¹(5(4(0(x)))) → 1¹(3(x))
1¹(5(4(0(x)))) → 0¹(4(1(5(2(3(x))))))
1¹(5(4(0(x)))) → 4¹(1(5(2(3(x)))))
1¹(5(4(0(x)))) → 1¹(5(2(3(x))))
1¹(5(4(0(x)))) → 4¹(1(5(0(3(5(x))))))
1¹(5(4(0(x)))) → 1¹(5(0(3(5(x)))))
1¹(5(4(0(x)))) → 0¹(3(5(x)))
1¹(5(4(0(x)))) → 4¹(1(5(0(5(5(x))))))
1¹(5(4(0(x)))) → 1¹(5(0(5(5(x)))))
1¹(5(4(0(x)))) → 0¹(5(5(x)))
4¹(5(4(0(x)))) → 4¹(0(4(4(2(5(x))))))
4¹(5(4(0(x)))) → 0¹(4(4(2(5(x)))))
4¹(5(4(0(x)))) → 4¹(4(2(5(x))))
4¹(5(4(0(x)))) → 4¹(2(5(x)))
0¹(1(5(0(x)))) → 1¹(5(0(0(x))))
0¹(1(5(0(x)))) → 0¹(0(x))
1¹(0(5(3(x)))) → 1¹(2(0(3(x))))
1¹(0(5(3(x)))) → 0¹(3(x))
0¹(1(5(3(x)))) → 1¹(5(0(x)))
0¹(1(5(3(x)))) → 0¹(x)
0¹(1(5(3(x)))) → 0¹(2(1(2(x))))
0¹(1(5(3(x)))) → 1¹(2(x))
0¹(1(5(3(x)))) → 0¹(2(2(1(3(x)))))
0¹(1(5(3(x)))) → 1¹(3(x))
0¹(1(5(3(x)))) → 0¹(2(3(1(5(x)))))
0¹(1(5(3(x)))) → 1¹(5(x))
0¹(3(3(1(0(x))))) → 1¹(0(0(2(3(3(x))))))
0¹(3(3(1(0(x))))) → 0¹(0(2(3(3(x)))))
0¹(3(3(1(0(x))))) → 0¹(2(3(3(x))))
1¹(5(3(1(0(x))))) → 0¹(5(4(3(1(1(x))))))
1¹(5(3(1(0(x))))) → 4¹(3(1(1(x))))
1¹(5(3(1(0(x))))) → 1¹(1(x))
1¹(5(3(1(0(x))))) → 1¹(x)
1¹(5(3(1(0(x))))) → 0¹(5(1(1(x))))
0¹(2(5(1(0(x))))) → 0¹(1(3(0(5(x)))))
0¹(2(5(1(0(x))))) → 1¹(3(0(5(x))))
0¹(2(5(1(0(x))))) → 0¹(5(x))
1¹(4(5(1(0(x))))) → 1¹(0(1(4(3(5(x))))))
1¹(4(5(1(0(x))))) → 0¹(1(4(3(5(x)))))
1¹(4(5(1(0(x))))) → 1¹(4(3(5(x))))
1¹(4(5(1(0(x))))) → 4¹(3(5(x)))
4¹(4(5(1(0(x))))) → 0¹(4(1(2(5(4(x))))))
4¹(4(5(1(0(x))))) → 4¹(1(2(5(4(x)))))
4¹(4(5(1(0(x))))) → 1¹(2(5(4(x))))
4¹(4(5(1(0(x))))) → 4¹(x)
4¹(4(5(1(0(x))))) → 1¹(4(3(4(0(5(x))))))
4¹(4(5(1(0(x))))) → 4¹(3(4(0(5(x)))))
4¹(4(5(1(0(x))))) → 4¹(0(5(x)))
4¹(4(5(1(0(x))))) → 0¹(5(x))
4¹(0(1(3(0(x))))) → 4¹(2(1(3(0(0(x))))))
4¹(0(1(3(0(x))))) → 1¹(3(0(0(x))))
4¹(0(1(3(0(x))))) → 0¹(0(x))
0¹(3(3(4(0(x))))) → 0¹(4(3(2(0(x)))))
0¹(3(3(4(0(x))))) → 4¹(3(2(0(x))))
0¹(2(5(4(0(x))))) → 4¹(0(5(2(2(0(x))))))
0¹(2(5(4(0(x))))) → 0¹(5(2(2(0(x)))))
1¹(5(1(5(0(x))))) → 1¹(1(0(3(5(5(x))))))
1¹(5(1(5(0(x))))) → 1¹(0(3(5(5(x)))))
1¹(5(1(5(0(x))))) → 0¹(3(5(5(x))))
4¹(5(1(0(3(x))))) → 1¹(4(3(5(0(x)))))
4¹(5(1(0(3(x))))) → 4¹(3(5(0(x))))
4¹(5(1(0(3(x))))) → 0¹(x)
0¹(1(0(5(3(x))))) → 0¹(0(2(1(5(x)))))
0¹(1(0(5(3(x))))) → 0¹(2(1(5(x))))
0¹(1(0(5(3(x))))) → 1¹(5(x))
0¹(0(4(5(3(x))))) → 4¹(4(0(3(0(5(x))))))
0¹(0(4(5(3(x))))) → 4¹(0(3(0(5(x)))))
0¹(0(4(5(3(x))))) → 0¹(3(0(5(x))))
0¹(0(4(5(3(x))))) → 0¹(5(x))
1¹(0(5(5(3(x))))) → 1¹(3(0(5(5(x)))))
1¹(0(5(5(3(x))))) → 0¹(5(5(x)))

The TRS R consists of the following rules:

1(0(0(x))) → 2(1(2(0(0(x)))))
1(0(0(x))) → 4(1(3(0(0(x)))))
0(1(0(x))) → 4(2(1(0(0(x)))))
0(1(0(x))) → 0(0(2(1(2(x)))))
0(1(0(x))) → 4(2(2(1(0(0(x))))))
0(1(0(x))) → 4(1(4(2(0(0(x))))))
0(4(0(x))) → 0(0(2(3(4(3(x))))))
0(1(0(0(x)))) → 4(1(0(0(0(x)))))
1(4(0(0(x)))) → 4(1(2(0(0(x)))))
1(4(0(0(x)))) → 3(4(1(2(0(0(x))))))
4(0(1(0(x)))) → 2(1(4(3(0(0(x))))))
0(3(1(0(x)))) → 4(1(3(0(0(x)))))
0(3(1(0(x)))) → 2(1(0(3(0(x)))))
0(3(1(0(x)))) → 5(2(3(1(0(0(x))))))
0(4(1(0(x)))) → 4(1(3(0(2(0(x))))))
1(5(1(0(x)))) → 0(5(4(1(1(2(x))))))
4(5(1(0(x)))) → 0(5(2(4(2(1(x))))))
4(5(1(0(x)))) → 5(2(3(0(1(4(x))))))
4(5(1(0(x)))) → 4(1(4(2(0(5(x))))))
1(0(3(0(x)))) → 2(2(1(3(0(0(x))))))
0(1(3(0(x)))) → 4(1(3(0(0(x)))))
1(0(4(0(x)))) → 4(1(2(0(0(x)))))
1(5(4(0(x)))) → 0(5(2(4(2(1(x))))))
1(5(4(0(x)))) → 2(0(5(4(1(3(x))))))
1(5(4(0(x)))) → 0(4(1(5(2(3(x))))))
1(5(4(0(x)))) → 4(1(5(0(3(5(x))))))
1(5(4(0(x)))) → 4(1(5(0(5(5(x))))))
4(5(4(0(x)))) → 4(0(4(4(2(5(x))))))
0(1(5(0(x)))) → 2(1(5(0(0(x)))))
1(0(5(3(x)))) → 5(2(1(2(0(3(x))))))
0(1(5(3(x)))) → 2(3(1(5(0(x)))))
0(1(5(3(x)))) → 3(5(0(2(1(2(x))))))
0(1(5(3(x)))) → 5(0(2(2(1(3(x))))))
0(1(5(3(x)))) → 2(0(2(3(1(5(x))))))
0(3(3(1(0(x))))) → 1(0(0(2(3(3(x))))))
1(5(3(1(0(x))))) → 0(5(4(3(1(1(x))))))
1(5(3(1(0(x))))) → 3(3(0(5(1(1(x))))))
0(2(5(1(0(x))))) → 2(0(1(3(0(5(x))))))
1(4(5(1(0(x))))) → 1(0(1(4(3(5(x))))))
4(4(5(1(0(x))))) → 0(4(1(2(5(4(x))))))
4(4(5(1(0(x))))) → 1(4(3(4(0(5(x))))))
4(0(1(3(0(x))))) → 4(2(1(3(0(0(x))))))
0(3(3(4(0(x))))) → 3(0(4(3(2(0(x))))))
0(2(5(4(0(x))))) → 4(0(5(2(2(0(x))))))
1(5(1(5(0(x))))) → 1(1(0(3(5(5(x))))))
4(5(1(0(3(x))))) → 2(1(4(3(5(0(x))))))
0(1(0(5(3(x))))) → 3(0(0(2(1(5(x))))))
0(0(4(5(3(x))))) → 4(4(0(3(0(5(x))))))
1(0(5(5(3(x))))) → 2(1(3(0(5(5(x))))))

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 97 less nodes.

(6) Obligation:

Q DP problem:
The TRS P consists of the following rules:

1¹(0(3(0(x)))) → 0¹(0(x))
0¹(1(0(x))) → 1¹(0(0(x)))
1¹(0(4(0(x)))) → 0¹(0(x))
0¹(1(0(x))) → 0¹(0(x))
0¹(1(0(0(x)))) → 4¹(1(0(0(0(x)))))
4¹(0(1(0(x)))) → 0¹(0(x))
0¹(1(0(0(x)))) → 1¹(0(0(0(x))))
1¹(5(4(0(x)))) → 1¹(x)
1¹(0(5(3(x)))) → 0¹(3(x))
0¹(3(1(0(x)))) → 0¹(0(x))
0¹(1(0(0(x)))) → 0¹(0(0(x)))
0¹(3(1(0(x)))) → 1¹(0(3(0(x))))
1¹(5(3(1(0(x))))) → 1¹(1(x))
1¹(5(3(1(0(x))))) → 1¹(x)
0¹(3(1(0(x)))) → 0¹(3(0(x)))
0¹(3(1(0(x)))) → 1¹(0(0(x)))
0¹(4(1(0(x)))) → 0¹(2(0(x)))
0¹(2(5(1(0(x))))) → 0¹(1(3(0(5(x)))))
0¹(1(3(0(x)))) → 0¹(0(x))
0¹(1(5(0(x)))) → 1¹(5(0(0(x))))
0¹(1(5(0(x)))) → 0¹(0(x))
0¹(1(5(3(x)))) → 1¹(5(0(x)))
0¹(1(5(3(x)))) → 0¹(x)
0¹(1(5(3(x)))) → 1¹(5(x))
0¹(1(0(5(3(x))))) → 0¹(0(2(1(5(x)))))
0¹(1(0(5(3(x))))) → 0¹(2(1(5(x))))
0¹(1(0(5(3(x))))) → 1¹(5(x))
4¹(5(1(0(x)))) → 1¹(x)
4¹(5(1(0(x)))) → 0¹(1(4(x)))
4¹(5(1(0(x)))) → 1¹(4(x))
4¹(5(1(0(x)))) → 4¹(x)
4¹(4(5(1(0(x))))) → 4¹(x)
4¹(0(1(3(0(x))))) → 0¹(0(x))
4¹(5(1(0(3(x))))) → 0¹(x)

The TRS R consists of the following rules:

1(0(0(x))) → 2(1(2(0(0(x)))))
1(0(0(x))) → 4(1(3(0(0(x)))))
0(1(0(x))) → 4(2(1(0(0(x)))))
0(1(0(x))) → 0(0(2(1(2(x)))))
0(1(0(x))) → 4(2(2(1(0(0(x))))))
0(1(0(x))) → 4(1(4(2(0(0(x))))))
0(4(0(x))) → 0(0(2(3(4(3(x))))))
0(1(0(0(x)))) → 4(1(0(0(0(x)))))
1(4(0(0(x)))) → 4(1(2(0(0(x)))))
1(4(0(0(x)))) → 3(4(1(2(0(0(x))))))
4(0(1(0(x)))) → 2(1(4(3(0(0(x))))))
0(3(1(0(x)))) → 4(1(3(0(0(x)))))
0(3(1(0(x)))) → 2(1(0(3(0(x)))))
0(3(1(0(x)))) → 5(2(3(1(0(0(x))))))
0(4(1(0(x)))) → 4(1(3(0(2(0(x))))))
1(5(1(0(x)))) → 0(5(4(1(1(2(x))))))
4(5(1(0(x)))) → 0(5(2(4(2(1(x))))))
4(5(1(0(x)))) → 5(2(3(0(1(4(x))))))
4(5(1(0(x)))) → 4(1(4(2(0(5(x))))))
1(0(3(0(x)))) → 2(2(1(3(0(0(x))))))
0(1(3(0(x)))) → 4(1(3(0(0(x)))))
1(0(4(0(x)))) → 4(1(2(0(0(x)))))
1(5(4(0(x)))) → 0(5(2(4(2(1(x))))))
1(5(4(0(x)))) → 2(0(5(4(1(3(x))))))
1(5(4(0(x)))) → 0(4(1(5(2(3(x))))))
1(5(4(0(x)))) → 4(1(5(0(3(5(x))))))
1(5(4(0(x)))) → 4(1(5(0(5(5(x))))))
4(5(4(0(x)))) → 4(0(4(4(2(5(x))))))
0(1(5(0(x)))) → 2(1(5(0(0(x)))))
1(0(5(3(x)))) → 5(2(1(2(0(3(x))))))
0(1(5(3(x)))) → 2(3(1(5(0(x)))))
0(1(5(3(x)))) → 3(5(0(2(1(2(x))))))
0(1(5(3(x)))) → 5(0(2(2(1(3(x))))))
0(1(5(3(x)))) → 2(0(2(3(1(5(x))))))
0(3(3(1(0(x))))) → 1(0(0(2(3(3(x))))))
1(5(3(1(0(x))))) → 0(5(4(3(1(1(x))))))
1(5(3(1(0(x))))) → 3(3(0(5(1(1(x))))))
0(2(5(1(0(x))))) → 2(0(1(3(0(5(x))))))
1(4(5(1(0(x))))) → 1(0(1(4(3(5(x))))))
4(4(5(1(0(x))))) → 0(4(1(2(5(4(x))))))
4(4(5(1(0(x))))) → 1(4(3(4(0(5(x))))))
4(0(1(3(0(x))))) → 4(2(1(3(0(0(x))))))
0(3(3(4(0(x))))) → 3(0(4(3(2(0(x))))))
0(2(5(4(0(x))))) → 4(0(5(2(2(0(x))))))
1(5(1(5(0(x))))) → 1(1(0(3(5(5(x))))))
4(5(1(0(3(x))))) → 2(1(4(3(5(0(x))))))
0(1(0(5(3(x))))) → 3(0(0(2(1(5(x))))))
0(0(4(5(3(x))))) → 4(4(0(3(0(5(x))))))
1(0(5(5(3(x))))) → 2(1(3(0(5(5(x))))))

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

(7) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].

The following pairs can be oriented strictly and are deleted.

0¹(1(0(x))) → 1¹(0(0(x)))
0¹(1(0(x))) → 0¹(0(x))
4¹(0(1(0(x)))) → 0¹(0(x))
0¹(1(0(0(x)))) → 1¹(0(0(0(x))))
0¹(3(1(0(x)))) → 0¹(0(x))
0¹(1(0(0(x)))) → 0¹(0(0(x)))
0¹(3(1(0(x)))) → 1¹(0(3(0(x))))
1¹(5(3(1(0(x))))) → 1¹(x)
0¹(3(1(0(x)))) → 0¹(3(0(x)))
0¹(3(1(0(x)))) → 1¹(0(0(x)))
0¹(4(1(0(x)))) → 0¹(2(0(x)))
0¹(1(3(0(x)))) → 0¹(0(x))
0¹(1(5(0(x)))) → 1¹(5(0(0(x))))
0¹(1(5(0(x)))) → 0¹(0(x))
0¹(1(5(3(x)))) → 1¹(5(0(x)))
0¹(1(5(3(x)))) → 0¹(x)
0¹(1(5(3(x)))) → 1¹(5(x))
0¹(1(0(5(3(x))))) → 1¹(5(x))
4¹(5(1(0(x)))) → 1¹(x)
4¹(5(1(0(x)))) → 1¹(4(x))
4¹(5(1(0(x)))) → 4¹(x)
4¹(4(5(1(0(x))))) → 4¹(x)
4¹(0(1(3(0(x))))) → 0¹(0(x))
4¹(5(1(0(3(x))))) → 0¹(x)

The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(0(x₁)) = x₁
POL(0¹(x₁)) = x₁
POL(1(x₁)) = 1 + x₁
POL(1¹(x₁)) = x₁
POL(2(x₁)) = x₁
POL(3(x₁)) = x₁
POL(4(x₁)) = x₁
POL(4¹(x₁)) = x₁
POL(5(x₁)) = x₁

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

0(1(0(x))) → 4(2(1(0(0(x)))))
0(1(0(x))) → 0(0(2(1(2(x)))))
0(1(0(x))) → 4(2(2(1(0(0(x))))))
0(1(0(x))) → 4(1(4(2(0(0(x))))))
0(4(0(x))) → 0(0(2(3(4(3(x))))))
0(1(0(0(x)))) → 4(1(0(0(0(x)))))
0(3(1(0(x)))) → 4(1(3(0(0(x)))))
0(3(1(0(x)))) → 2(1(0(3(0(x)))))
0(3(1(0(x)))) → 5(2(3(1(0(0(x))))))
0(4(1(0(x)))) → 4(1(3(0(2(0(x))))))
0(1(3(0(x)))) → 4(1(3(0(0(x)))))
0(1(5(0(x)))) → 2(1(5(0(0(x)))))
0(1(5(3(x)))) → 2(3(1(5(0(x)))))
0(1(5(3(x)))) → 3(5(0(2(1(2(x))))))
0(1(5(3(x)))) → 5(0(2(2(1(3(x))))))
0(1(5(3(x)))) → 2(0(2(3(1(5(x))))))
0(3(3(1(0(x))))) → 1(0(0(2(3(3(x))))))
0(2(5(1(0(x))))) → 2(0(1(3(0(5(x))))))
0(3(3(4(0(x))))) → 3(0(4(3(2(0(x))))))
0(2(5(4(0(x))))) → 4(0(5(2(2(0(x))))))
0(1(0(5(3(x))))) → 3(0(0(2(1(5(x))))))
0(0(4(5(3(x))))) → 4(4(0(3(0(5(x))))))
1(0(0(x))) → 2(1(2(0(0(x)))))
1(0(0(x))) → 4(1(3(0(0(x)))))
1(4(0(0(x)))) → 4(1(2(0(0(x)))))
1(4(0(0(x)))) → 3(4(1(2(0(0(x))))))
1(5(1(0(x)))) → 0(5(4(1(1(2(x))))))
1(0(3(0(x)))) → 2(2(1(3(0(0(x))))))
1(0(4(0(x)))) → 4(1(2(0(0(x)))))
1(5(4(0(x)))) → 0(5(2(4(2(1(x))))))
1(5(4(0(x)))) → 2(0(5(4(1(3(x))))))
1(5(4(0(x)))) → 0(4(1(5(2(3(x))))))
1(5(4(0(x)))) → 4(1(5(0(3(5(x))))))
1(5(4(0(x)))) → 4(1(5(0(5(5(x))))))
1(0(5(3(x)))) → 5(2(1(2(0(3(x))))))
1(5(3(1(0(x))))) → 0(5(4(3(1(1(x))))))
1(5(3(1(0(x))))) → 3(3(0(5(1(1(x))))))
1(4(5(1(0(x))))) → 1(0(1(4(3(5(x))))))
1(5(1(5(0(x))))) → 1(1(0(3(5(5(x))))))
1(0(5(5(3(x))))) → 2(1(3(0(5(5(x))))))
4(0(1(0(x)))) → 2(1(4(3(0(0(x))))))
4(5(1(0(x)))) → 0(5(2(4(2(1(x))))))
4(5(1(0(x)))) → 5(2(3(0(1(4(x))))))
4(5(1(0(x)))) → 4(1(4(2(0(5(x))))))
4(5(4(0(x)))) → 4(0(4(4(2(5(x))))))
4(4(5(1(0(x))))) → 0(4(1(2(5(4(x))))))
4(4(5(1(0(x))))) → 1(4(3(4(0(5(x))))))
4(0(1(3(0(x))))) → 4(2(1(3(0(0(x))))))
4(5(1(0(3(x))))) → 2(1(4(3(5(0(x))))))

(8) Obligation:

Q DP problem:
The TRS P consists of the following rules:

1¹(0(3(0(x)))) → 0¹(0(x))
1¹(0(4(0(x)))) → 0¹(0(x))
0¹(1(0(0(x)))) → 4¹(1(0(0(0(x)))))
1¹(5(4(0(x)))) → 1¹(x)
1¹(0(5(3(x)))) → 0¹(3(x))
1¹(5(3(1(0(x))))) → 1¹(1(x))
0¹(2(5(1(0(x))))) → 0¹(1(3(0(5(x)))))
0¹(1(0(5(3(x))))) → 0¹(0(2(1(5(x)))))
0¹(1(0(5(3(x))))) → 0¹(2(1(5(x))))
4¹(5(1(0(x)))) → 0¹(1(4(x)))

The TRS R consists of the following rules:

1(0(0(x))) → 2(1(2(0(0(x)))))
1(0(0(x))) → 4(1(3(0(0(x)))))
0(1(0(x))) → 4(2(1(0(0(x)))))
0(1(0(x))) → 0(0(2(1(2(x)))))
0(1(0(x))) → 4(2(2(1(0(0(x))))))
0(1(0(x))) → 4(1(4(2(0(0(x))))))
0(4(0(x))) → 0(0(2(3(4(3(x))))))
0(1(0(0(x)))) → 4(1(0(0(0(x)))))
1(4(0(0(x)))) → 4(1(2(0(0(x)))))
1(4(0(0(x)))) → 3(4(1(2(0(0(x))))))
4(0(1(0(x)))) → 2(1(4(3(0(0(x))))))
0(3(1(0(x)))) → 4(1(3(0(0(x)))))
0(3(1(0(x)))) → 2(1(0(3(0(x)))))
0(3(1(0(x)))) → 5(2(3(1(0(0(x))))))
0(4(1(0(x)))) → 4(1(3(0(2(0(x))))))
1(5(1(0(x)))) → 0(5(4(1(1(2(x))))))
4(5(1(0(x)))) → 0(5(2(4(2(1(x))))))
4(5(1(0(x)))) → 5(2(3(0(1(4(x))))))
4(5(1(0(x)))) → 4(1(4(2(0(5(x))))))
1(0(3(0(x)))) → 2(2(1(3(0(0(x))))))
0(1(3(0(x)))) → 4(1(3(0(0(x)))))
1(0(4(0(x)))) → 4(1(2(0(0(x)))))
1(5(4(0(x)))) → 0(5(2(4(2(1(x))))))
1(5(4(0(x)))) → 2(0(5(4(1(3(x))))))
1(5(4(0(x)))) → 0(4(1(5(2(3(x))))))
1(5(4(0(x)))) → 4(1(5(0(3(5(x))))))
1(5(4(0(x)))) → 4(1(5(0(5(5(x))))))
4(5(4(0(x)))) → 4(0(4(4(2(5(x))))))
0(1(5(0(x)))) → 2(1(5(0(0(x)))))
1(0(5(3(x)))) → 5(2(1(2(0(3(x))))))
0(1(5(3(x)))) → 2(3(1(5(0(x)))))
0(1(5(3(x)))) → 3(5(0(2(1(2(x))))))
0(1(5(3(x)))) → 5(0(2(2(1(3(x))))))
0(1(5(3(x)))) → 2(0(2(3(1(5(x))))))
0(3(3(1(0(x))))) → 1(0(0(2(3(3(x))))))
1(5(3(1(0(x))))) → 0(5(4(3(1(1(x))))))
1(5(3(1(0(x))))) → 3(3(0(5(1(1(x))))))
0(2(5(1(0(x))))) → 2(0(1(3(0(5(x))))))
1(4(5(1(0(x))))) → 1(0(1(4(3(5(x))))))
4(4(5(1(0(x))))) → 0(4(1(2(5(4(x))))))
4(4(5(1(0(x))))) → 1(4(3(4(0(5(x))))))
4(0(1(3(0(x))))) → 4(2(1(3(0(0(x))))))
0(3(3(4(0(x))))) → 3(0(4(3(2(0(x))))))
0(2(5(4(0(x))))) → 4(0(5(2(2(0(x))))))
1(5(1(5(0(x))))) → 1(1(0(3(5(5(x))))))
4(5(1(0(3(x))))) → 2(1(4(3(5(0(x))))))
0(1(0(5(3(x))))) → 3(0(0(2(1(5(x))))))
0(0(4(5(3(x))))) → 4(4(0(3(0(5(x))))))
1(0(5(5(3(x))))) → 2(1(3(0(5(5(x))))))

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

(9) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 5 less nodes.

(10) Complex Obligation (AND)

(11) Obligation:

Q DP problem:
The TRS P consists of the following rules:

4¹(5(1(0(x)))) → 0¹(1(4(x)))
0¹(1(0(0(x)))) → 4¹(1(0(0(0(x)))))
0¹(1(0(5(3(x))))) → 0¹(0(2(1(5(x)))))

The TRS R consists of the following rules:

1(0(0(x))) → 2(1(2(0(0(x)))))
1(0(0(x))) → 4(1(3(0(0(x)))))
0(1(0(x))) → 4(2(1(0(0(x)))))
0(1(0(x))) → 0(0(2(1(2(x)))))
0(1(0(x))) → 4(2(2(1(0(0(x))))))
0(1(0(x))) → 4(1(4(2(0(0(x))))))
0(4(0(x))) → 0(0(2(3(4(3(x))))))
0(1(0(0(x)))) → 4(1(0(0(0(x)))))
1(4(0(0(x)))) → 4(1(2(0(0(x)))))
1(4(0(0(x)))) → 3(4(1(2(0(0(x))))))
4(0(1(0(x)))) → 2(1(4(3(0(0(x))))))
0(3(1(0(x)))) → 4(1(3(0(0(x)))))
0(3(1(0(x)))) → 2(1(0(3(0(x)))))
0(3(1(0(x)))) → 5(2(3(1(0(0(x))))))
0(4(1(0(x)))) → 4(1(3(0(2(0(x))))))
1(5(1(0(x)))) → 0(5(4(1(1(2(x))))))
4(5(1(0(x)))) → 0(5(2(4(2(1(x))))))
4(5(1(0(x)))) → 5(2(3(0(1(4(x))))))
4(5(1(0(x)))) → 4(1(4(2(0(5(x))))))
1(0(3(0(x)))) → 2(2(1(3(0(0(x))))))
0(1(3(0(x)))) → 4(1(3(0(0(x)))))
1(0(4(0(x)))) → 4(1(2(0(0(x)))))
1(5(4(0(x)))) → 0(5(2(4(2(1(x))))))
1(5(4(0(x)))) → 2(0(5(4(1(3(x))))))
1(5(4(0(x)))) → 0(4(1(5(2(3(x))))))
1(5(4(0(x)))) → 4(1(5(0(3(5(x))))))
1(5(4(0(x)))) → 4(1(5(0(5(5(x))))))
4(5(4(0(x)))) → 4(0(4(4(2(5(x))))))
0(1(5(0(x)))) → 2(1(5(0(0(x)))))
1(0(5(3(x)))) → 5(2(1(2(0(3(x))))))
0(1(5(3(x)))) → 2(3(1(5(0(x)))))
0(1(5(3(x)))) → 3(5(0(2(1(2(x))))))
0(1(5(3(x)))) → 5(0(2(2(1(3(x))))))
0(1(5(3(x)))) → 2(0(2(3(1(5(x))))))
0(3(3(1(0(x))))) → 1(0(0(2(3(3(x))))))
1(5(3(1(0(x))))) → 0(5(4(3(1(1(x))))))
1(5(3(1(0(x))))) → 3(3(0(5(1(1(x))))))
0(2(5(1(0(x))))) → 2(0(1(3(0(5(x))))))
1(4(5(1(0(x))))) → 1(0(1(4(3(5(x))))))
4(4(5(1(0(x))))) → 0(4(1(2(5(4(x))))))
4(4(5(1(0(x))))) → 1(4(3(4(0(5(x))))))
4(0(1(3(0(x))))) → 4(2(1(3(0(0(x))))))
0(3(3(4(0(x))))) → 3(0(4(3(2(0(x))))))
0(2(5(4(0(x))))) → 4(0(5(2(2(0(x))))))
1(5(1(5(0(x))))) → 1(1(0(3(5(5(x))))))
4(5(1(0(3(x))))) → 2(1(4(3(5(0(x))))))
0(1(0(5(3(x))))) → 3(0(0(2(1(5(x))))))
0(0(4(5(3(x))))) → 4(4(0(3(0(5(x))))))
1(0(5(5(3(x))))) → 2(1(3(0(5(5(x))))))

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

(12) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].

The following pairs can be oriented strictly and are deleted.

4¹(5(1(0(x)))) → 0¹(1(4(x)))

The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(0(x₁)) = 0
POL(0¹(x₁)) = 1
POL(1(x₁)) = 1
POL(2(x₁)) = 0
POL(3(x₁)) = 0
POL(4(x₁)) = 0
POL(4¹(x₁)) = x₁
POL(5(x₁)) = 1 + x₁

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

1(0(0(x))) → 2(1(2(0(0(x)))))
1(0(0(x))) → 4(1(3(0(0(x)))))
1(4(0(0(x)))) → 4(1(2(0(0(x)))))
1(4(0(0(x)))) → 3(4(1(2(0(0(x))))))
1(5(1(0(x)))) → 0(5(4(1(1(2(x))))))
1(0(3(0(x)))) → 2(2(1(3(0(0(x))))))
1(0(4(0(x)))) → 4(1(2(0(0(x)))))
1(5(4(0(x)))) → 0(5(2(4(2(1(x))))))
1(5(4(0(x)))) → 2(0(5(4(1(3(x))))))
1(5(4(0(x)))) → 0(4(1(5(2(3(x))))))
1(5(4(0(x)))) → 4(1(5(0(3(5(x))))))
1(5(4(0(x)))) → 4(1(5(0(5(5(x))))))
1(0(5(3(x)))) → 5(2(1(2(0(3(x))))))
1(5(3(1(0(x))))) → 0(5(4(3(1(1(x))))))
1(5(3(1(0(x))))) → 3(3(0(5(1(1(x))))))
1(4(5(1(0(x))))) → 1(0(1(4(3(5(x))))))
1(5(1(5(0(x))))) → 1(1(0(3(5(5(x))))))
1(0(5(5(3(x))))) → 2(1(3(0(5(5(x))))))

(13) Obligation:

Q DP problem:
The TRS P consists of the following rules:

0¹(1(0(0(x)))) → 4¹(1(0(0(0(x)))))
0¹(1(0(5(3(x))))) → 0¹(0(2(1(5(x)))))

The TRS R consists of the following rules:

1(0(0(x))) → 2(1(2(0(0(x)))))
1(0(0(x))) → 4(1(3(0(0(x)))))
0(1(0(x))) → 4(2(1(0(0(x)))))
0(1(0(x))) → 0(0(2(1(2(x)))))
0(1(0(x))) → 4(2(2(1(0(0(x))))))
0(1(0(x))) → 4(1(4(2(0(0(x))))))
0(4(0(x))) → 0(0(2(3(4(3(x))))))
0(1(0(0(x)))) → 4(1(0(0(0(x)))))
1(4(0(0(x)))) → 4(1(2(0(0(x)))))
1(4(0(0(x)))) → 3(4(1(2(0(0(x))))))
4(0(1(0(x)))) → 2(1(4(3(0(0(x))))))
0(3(1(0(x)))) → 4(1(3(0(0(x)))))
0(3(1(0(x)))) → 2(1(0(3(0(x)))))
0(3(1(0(x)))) → 5(2(3(1(0(0(x))))))
0(4(1(0(x)))) → 4(1(3(0(2(0(x))))))
1(5(1(0(x)))) → 0(5(4(1(1(2(x))))))
4(5(1(0(x)))) → 0(5(2(4(2(1(x))))))
4(5(1(0(x)))) → 5(2(3(0(1(4(x))))))
4(5(1(0(x)))) → 4(1(4(2(0(5(x))))))
1(0(3(0(x)))) → 2(2(1(3(0(0(x))))))
0(1(3(0(x)))) → 4(1(3(0(0(x)))))
1(0(4(0(x)))) → 4(1(2(0(0(x)))))
1(5(4(0(x)))) → 0(5(2(4(2(1(x))))))
1(5(4(0(x)))) → 2(0(5(4(1(3(x))))))
1(5(4(0(x)))) → 0(4(1(5(2(3(x))))))
1(5(4(0(x)))) → 4(1(5(0(3(5(x))))))
1(5(4(0(x)))) → 4(1(5(0(5(5(x))))))
4(5(4(0(x)))) → 4(0(4(4(2(5(x))))))
0(1(5(0(x)))) → 2(1(5(0(0(x)))))
1(0(5(3(x)))) → 5(2(1(2(0(3(x))))))
0(1(5(3(x)))) → 2(3(1(5(0(x)))))
0(1(5(3(x)))) → 3(5(0(2(1(2(x))))))
0(1(5(3(x)))) → 5(0(2(2(1(3(x))))))
0(1(5(3(x)))) → 2(0(2(3(1(5(x))))))
0(3(3(1(0(x))))) → 1(0(0(2(3(3(x))))))
1(5(3(1(0(x))))) → 0(5(4(3(1(1(x))))))
1(5(3(1(0(x))))) → 3(3(0(5(1(1(x))))))
0(2(5(1(0(x))))) → 2(0(1(3(0(5(x))))))
1(4(5(1(0(x))))) → 1(0(1(4(3(5(x))))))
4(4(5(1(0(x))))) → 0(4(1(2(5(4(x))))))
4(4(5(1(0(x))))) → 1(4(3(4(0(5(x))))))
4(0(1(3(0(x))))) → 4(2(1(3(0(0(x))))))
0(3(3(4(0(x))))) → 3(0(4(3(2(0(x))))))
0(2(5(4(0(x))))) → 4(0(5(2(2(0(x))))))
1(5(1(5(0(x))))) → 1(1(0(3(5(5(x))))))
4(5(1(0(3(x))))) → 2(1(4(3(5(0(x))))))
0(1(0(5(3(x))))) → 3(0(0(2(1(5(x))))))
0(0(4(5(3(x))))) → 4(4(0(3(0(5(x))))))
1(0(5(5(3(x))))) → 2(1(3(0(5(5(x))))))

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

(14) DependencyGraphProof (EQUIVALENT transformation)

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:

0¹(1(0(5(3(x))))) → 0¹(0(2(1(5(x)))))

The TRS R consists of the following rules:

1(0(0(x))) → 2(1(2(0(0(x)))))
1(0(0(x))) → 4(1(3(0(0(x)))))
0(1(0(x))) → 4(2(1(0(0(x)))))
0(1(0(x))) → 0(0(2(1(2(x)))))
0(1(0(x))) → 4(2(2(1(0(0(x))))))
0(1(0(x))) → 4(1(4(2(0(0(x))))))
0(4(0(x))) → 0(0(2(3(4(3(x))))))
0(1(0(0(x)))) → 4(1(0(0(0(x)))))
1(4(0(0(x)))) → 4(1(2(0(0(x)))))
1(4(0(0(x)))) → 3(4(1(2(0(0(x))))))
4(0(1(0(x)))) → 2(1(4(3(0(0(x))))))
0(3(1(0(x)))) → 4(1(3(0(0(x)))))
0(3(1(0(x)))) → 2(1(0(3(0(x)))))
0(3(1(0(x)))) → 5(2(3(1(0(0(x))))))
0(4(1(0(x)))) → 4(1(3(0(2(0(x))))))
1(5(1(0(x)))) → 0(5(4(1(1(2(x))))))
4(5(1(0(x)))) → 0(5(2(4(2(1(x))))))
4(5(1(0(x)))) → 5(2(3(0(1(4(x))))))
4(5(1(0(x)))) → 4(1(4(2(0(5(x))))))
1(0(3(0(x)))) → 2(2(1(3(0(0(x))))))
0(1(3(0(x)))) → 4(1(3(0(0(x)))))
1(0(4(0(x)))) → 4(1(2(0(0(x)))))
1(5(4(0(x)))) → 0(5(2(4(2(1(x))))))
1(5(4(0(x)))) → 2(0(5(4(1(3(x))))))
1(5(4(0(x)))) → 0(4(1(5(2(3(x))))))
1(5(4(0(x)))) → 4(1(5(0(3(5(x))))))
1(5(4(0(x)))) → 4(1(5(0(5(5(x))))))
4(5(4(0(x)))) → 4(0(4(4(2(5(x))))))
0(1(5(0(x)))) → 2(1(5(0(0(x)))))
1(0(5(3(x)))) → 5(2(1(2(0(3(x))))))
0(1(5(3(x)))) → 2(3(1(5(0(x)))))
0(1(5(3(x)))) → 3(5(0(2(1(2(x))))))
0(1(5(3(x)))) → 5(0(2(2(1(3(x))))))
0(1(5(3(x)))) → 2(0(2(3(1(5(x))))))
0(3(3(1(0(x))))) → 1(0(0(2(3(3(x))))))
1(5(3(1(0(x))))) → 0(5(4(3(1(1(x))))))
1(5(3(1(0(x))))) → 3(3(0(5(1(1(x))))))
0(2(5(1(0(x))))) → 2(0(1(3(0(5(x))))))
1(4(5(1(0(x))))) → 1(0(1(4(3(5(x))))))
4(4(5(1(0(x))))) → 0(4(1(2(5(4(x))))))
4(4(5(1(0(x))))) → 1(4(3(4(0(5(x))))))
4(0(1(3(0(x))))) → 4(2(1(3(0(0(x))))))
0(3(3(4(0(x))))) → 3(0(4(3(2(0(x))))))
0(2(5(4(0(x))))) → 4(0(5(2(2(0(x))))))
1(5(1(5(0(x))))) → 1(1(0(3(5(5(x))))))
4(5(1(0(3(x))))) → 2(1(4(3(5(0(x))))))
0(1(0(5(3(x))))) → 3(0(0(2(1(5(x))))))
0(0(4(5(3(x))))) → 4(4(0(3(0(5(x))))))
1(0(5(5(3(x))))) → 2(1(3(0(5(5(x))))))

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

(16) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].

The following pairs can be oriented strictly and are deleted.

0¹(1(0(5(3(x))))) → 0¹(0(2(1(5(x)))))

The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(0(x₁)) = 0
POL(0¹(x₁)) = x₁
POL(1(x₁)) = 1
POL(2(x₁)) = 0
POL(3(x₁)) = 0
POL(4(x₁)) = 0
POL(5(x₁)) = 0

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

0(2(5(1(0(x))))) → 2(0(1(3(0(5(x))))))
0(2(5(4(0(x))))) → 4(0(5(2(2(0(x))))))

(17) Obligation:

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

1(0(0(x))) → 2(1(2(0(0(x)))))
1(0(0(x))) → 4(1(3(0(0(x)))))
0(1(0(x))) → 4(2(1(0(0(x)))))
0(1(0(x))) → 0(0(2(1(2(x)))))
0(1(0(x))) → 4(2(2(1(0(0(x))))))
0(1(0(x))) → 4(1(4(2(0(0(x))))))
0(4(0(x))) → 0(0(2(3(4(3(x))))))
0(1(0(0(x)))) → 4(1(0(0(0(x)))))
1(4(0(0(x)))) → 4(1(2(0(0(x)))))
1(4(0(0(x)))) → 3(4(1(2(0(0(x))))))
4(0(1(0(x)))) → 2(1(4(3(0(0(x))))))
0(3(1(0(x)))) → 4(1(3(0(0(x)))))
0(3(1(0(x)))) → 2(1(0(3(0(x)))))
0(3(1(0(x)))) → 5(2(3(1(0(0(x))))))
0(4(1(0(x)))) → 4(1(3(0(2(0(x))))))
1(5(1(0(x)))) → 0(5(4(1(1(2(x))))))
4(5(1(0(x)))) → 0(5(2(4(2(1(x))))))
4(5(1(0(x)))) → 5(2(3(0(1(4(x))))))
4(5(1(0(x)))) → 4(1(4(2(0(5(x))))))
1(0(3(0(x)))) → 2(2(1(3(0(0(x))))))
0(1(3(0(x)))) → 4(1(3(0(0(x)))))
1(0(4(0(x)))) → 4(1(2(0(0(x)))))
1(5(4(0(x)))) → 0(5(2(4(2(1(x))))))
1(5(4(0(x)))) → 2(0(5(4(1(3(x))))))
1(5(4(0(x)))) → 0(4(1(5(2(3(x))))))
1(5(4(0(x)))) → 4(1(5(0(3(5(x))))))
1(5(4(0(x)))) → 4(1(5(0(5(5(x))))))
4(5(4(0(x)))) → 4(0(4(4(2(5(x))))))
0(1(5(0(x)))) → 2(1(5(0(0(x)))))
1(0(5(3(x)))) → 5(2(1(2(0(3(x))))))
0(1(5(3(x)))) → 2(3(1(5(0(x)))))
0(1(5(3(x)))) → 3(5(0(2(1(2(x))))))
0(1(5(3(x)))) → 5(0(2(2(1(3(x))))))
0(1(5(3(x)))) → 2(0(2(3(1(5(x))))))
0(3(3(1(0(x))))) → 1(0(0(2(3(3(x))))))
1(5(3(1(0(x))))) → 0(5(4(3(1(1(x))))))
1(5(3(1(0(x))))) → 3(3(0(5(1(1(x))))))
0(2(5(1(0(x))))) → 2(0(1(3(0(5(x))))))
1(4(5(1(0(x))))) → 1(0(1(4(3(5(x))))))
4(4(5(1(0(x))))) → 0(4(1(2(5(4(x))))))
4(4(5(1(0(x))))) → 1(4(3(4(0(5(x))))))
4(0(1(3(0(x))))) → 4(2(1(3(0(0(x))))))
0(3(3(4(0(x))))) → 3(0(4(3(2(0(x))))))
0(2(5(4(0(x))))) → 4(0(5(2(2(0(x))))))
1(5(1(5(0(x))))) → 1(1(0(3(5(5(x))))))
4(5(1(0(3(x))))) → 2(1(4(3(5(0(x))))))
0(1(0(5(3(x))))) → 3(0(0(2(1(5(x))))))
0(0(4(5(3(x))))) → 4(4(0(3(0(5(x))))))
1(0(5(5(3(x))))) → 2(1(3(0(5(5(x))))))

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

(18) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(19) YES

(20) Obligation:

Q DP problem:
The TRS P consists of the following rules:

1¹(5(3(1(0(x))))) → 1¹(1(x))
1¹(5(4(0(x)))) → 1¹(x)

The TRS R consists of the following rules:

1(0(0(x))) → 2(1(2(0(0(x)))))
1(0(0(x))) → 4(1(3(0(0(x)))))
0(1(0(x))) → 4(2(1(0(0(x)))))
0(1(0(x))) → 0(0(2(1(2(x)))))
0(1(0(x))) → 4(2(2(1(0(0(x))))))
0(1(0(x))) → 4(1(4(2(0(0(x))))))
0(4(0(x))) → 0(0(2(3(4(3(x))))))
0(1(0(0(x)))) → 4(1(0(0(0(x)))))
1(4(0(0(x)))) → 4(1(2(0(0(x)))))
1(4(0(0(x)))) → 3(4(1(2(0(0(x))))))
4(0(1(0(x)))) → 2(1(4(3(0(0(x))))))
0(3(1(0(x)))) → 4(1(3(0(0(x)))))
0(3(1(0(x)))) → 2(1(0(3(0(x)))))
0(3(1(0(x)))) → 5(2(3(1(0(0(x))))))
0(4(1(0(x)))) → 4(1(3(0(2(0(x))))))
1(5(1(0(x)))) → 0(5(4(1(1(2(x))))))
4(5(1(0(x)))) → 0(5(2(4(2(1(x))))))
4(5(1(0(x)))) → 5(2(3(0(1(4(x))))))
4(5(1(0(x)))) → 4(1(4(2(0(5(x))))))
1(0(3(0(x)))) → 2(2(1(3(0(0(x))))))
0(1(3(0(x)))) → 4(1(3(0(0(x)))))
1(0(4(0(x)))) → 4(1(2(0(0(x)))))
1(5(4(0(x)))) → 0(5(2(4(2(1(x))))))
1(5(4(0(x)))) → 2(0(5(4(1(3(x))))))
1(5(4(0(x)))) → 0(4(1(5(2(3(x))))))
1(5(4(0(x)))) → 4(1(5(0(3(5(x))))))
1(5(4(0(x)))) → 4(1(5(0(5(5(x))))))
4(5(4(0(x)))) → 4(0(4(4(2(5(x))))))
0(1(5(0(x)))) → 2(1(5(0(0(x)))))
1(0(5(3(x)))) → 5(2(1(2(0(3(x))))))
0(1(5(3(x)))) → 2(3(1(5(0(x)))))
0(1(5(3(x)))) → 3(5(0(2(1(2(x))))))
0(1(5(3(x)))) → 5(0(2(2(1(3(x))))))
0(1(5(3(x)))) → 2(0(2(3(1(5(x))))))
0(3(3(1(0(x))))) → 1(0(0(2(3(3(x))))))
1(5(3(1(0(x))))) → 0(5(4(3(1(1(x))))))
1(5(3(1(0(x))))) → 3(3(0(5(1(1(x))))))
0(2(5(1(0(x))))) → 2(0(1(3(0(5(x))))))
1(4(5(1(0(x))))) → 1(0(1(4(3(5(x))))))
4(4(5(1(0(x))))) → 0(4(1(2(5(4(x))))))
4(4(5(1(0(x))))) → 1(4(3(4(0(5(x))))))
4(0(1(3(0(x))))) → 4(2(1(3(0(0(x))))))
0(3(3(4(0(x))))) → 3(0(4(3(2(0(x))))))
0(2(5(4(0(x))))) → 4(0(5(2(2(0(x))))))
1(5(1(5(0(x))))) → 1(1(0(3(5(5(x))))))
4(5(1(0(3(x))))) → 2(1(4(3(5(0(x))))))
0(1(0(5(3(x))))) → 3(0(0(2(1(5(x))))))
0(0(4(5(3(x))))) → 4(4(0(3(0(5(x))))))
1(0(5(5(3(x))))) → 2(1(3(0(5(5(x))))))

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

(21) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].

The following pairs can be oriented strictly and are deleted.

1¹(5(3(1(0(x))))) → 1¹(1(x))
1¹(5(4(0(x)))) → 1¹(x)

The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(0(x₁)) = 1 + x₁
POL(1(x₁)) = 1 + x₁
POL(1¹(x₁)) = x₁
POL(2(x₁)) = 0
POL(3(x₁)) = x₁
POL(4(x₁)) = x₁
POL(5(x₁)) = x₁

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

1(0(0(x))) → 2(1(2(0(0(x)))))
1(0(0(x))) → 4(1(3(0(0(x)))))
1(4(0(0(x)))) → 4(1(2(0(0(x)))))
1(4(0(0(x)))) → 3(4(1(2(0(0(x))))))
1(5(1(0(x)))) → 0(5(4(1(1(2(x))))))
1(0(3(0(x)))) → 2(2(1(3(0(0(x))))))
1(0(4(0(x)))) → 4(1(2(0(0(x)))))
1(5(4(0(x)))) → 0(5(2(4(2(1(x))))))
1(5(4(0(x)))) → 2(0(5(4(1(3(x))))))
1(5(4(0(x)))) → 0(4(1(5(2(3(x))))))
1(5(4(0(x)))) → 4(1(5(0(3(5(x))))))
1(5(4(0(x)))) → 4(1(5(0(5(5(x))))))
1(0(5(3(x)))) → 5(2(1(2(0(3(x))))))
1(5(3(1(0(x))))) → 0(5(4(3(1(1(x))))))
1(5(3(1(0(x))))) → 3(3(0(5(1(1(x))))))
1(4(5(1(0(x))))) → 1(0(1(4(3(5(x))))))
1(5(1(5(0(x))))) → 1(1(0(3(5(5(x))))))
1(0(5(5(3(x))))) → 2(1(3(0(5(5(x))))))
0(1(0(x))) → 4(2(1(0(0(x)))))
0(1(0(x))) → 4(2(2(1(0(0(x))))))
0(1(0(x))) → 4(1(4(2(0(0(x))))))
0(1(0(0(x)))) → 4(1(0(0(0(x)))))
0(3(1(0(x)))) → 4(1(3(0(0(x)))))
0(3(1(0(x)))) → 2(1(0(3(0(x)))))
0(3(1(0(x)))) → 5(2(3(1(0(0(x))))))
0(4(1(0(x)))) → 4(1(3(0(2(0(x))))))
0(1(3(0(x)))) → 4(1(3(0(0(x)))))
0(1(5(0(x)))) → 2(1(5(0(0(x)))))
0(1(5(3(x)))) → 2(3(1(5(0(x)))))
0(1(5(3(x)))) → 2(0(2(3(1(5(x))))))
0(3(3(1(0(x))))) → 1(0(0(2(3(3(x))))))
0(2(5(1(0(x))))) → 2(0(1(3(0(5(x))))))
0(3(3(4(0(x))))) → 3(0(4(3(2(0(x))))))
0(2(5(4(0(x))))) → 4(0(5(2(2(0(x))))))
0(1(0(5(3(x))))) → 3(0(0(2(1(5(x))))))
4(0(1(0(x)))) → 2(1(4(3(0(0(x))))))
4(5(1(0(x)))) → 0(5(2(4(2(1(x))))))
4(5(1(0(x)))) → 5(2(3(0(1(4(x))))))
4(4(5(1(0(x))))) → 0(4(1(2(5(4(x))))))
4(0(1(3(0(x))))) → 4(2(1(3(0(0(x))))))
4(5(1(0(3(x))))) → 2(1(4(3(5(0(x))))))
0(1(0(x))) → 0(0(2(1(2(x)))))
0(4(0(x))) → 0(0(2(3(4(3(x))))))
0(1(5(3(x)))) → 3(5(0(2(1(2(x))))))
0(1(5(3(x)))) → 5(0(2(2(1(3(x))))))
0(0(4(5(3(x))))) → 4(4(0(3(0(5(x))))))
4(5(1(0(x)))) → 4(1(4(2(0(5(x))))))
4(5(4(0(x)))) → 4(0(4(4(2(5(x))))))
4(4(5(1(0(x))))) → 1(4(3(4(0(5(x))))))

(22) Obligation:

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

1(0(0(x))) → 2(1(2(0(0(x)))))
1(0(0(x))) → 4(1(3(0(0(x)))))
0(1(0(x))) → 4(2(1(0(0(x)))))
0(1(0(x))) → 0(0(2(1(2(x)))))
0(1(0(x))) → 4(2(2(1(0(0(x))))))
0(1(0(x))) → 4(1(4(2(0(0(x))))))
0(4(0(x))) → 0(0(2(3(4(3(x))))))
0(1(0(0(x)))) → 4(1(0(0(0(x)))))
1(4(0(0(x)))) → 4(1(2(0(0(x)))))
1(4(0(0(x)))) → 3(4(1(2(0(0(x))))))
4(0(1(0(x)))) → 2(1(4(3(0(0(x))))))
0(3(1(0(x)))) → 4(1(3(0(0(x)))))
0(3(1(0(x)))) → 2(1(0(3(0(x)))))
0(3(1(0(x)))) → 5(2(3(1(0(0(x))))))
0(4(1(0(x)))) → 4(1(3(0(2(0(x))))))
1(5(1(0(x)))) → 0(5(4(1(1(2(x))))))
4(5(1(0(x)))) → 0(5(2(4(2(1(x))))))
4(5(1(0(x)))) → 5(2(3(0(1(4(x))))))
4(5(1(0(x)))) → 4(1(4(2(0(5(x))))))
1(0(3(0(x)))) → 2(2(1(3(0(0(x))))))
0(1(3(0(x)))) → 4(1(3(0(0(x)))))
1(0(4(0(x)))) → 4(1(2(0(0(x)))))
1(5(4(0(x)))) → 0(5(2(4(2(1(x))))))
1(5(4(0(x)))) → 2(0(5(4(1(3(x))))))
1(5(4(0(x)))) → 0(4(1(5(2(3(x))))))
1(5(4(0(x)))) → 4(1(5(0(3(5(x))))))
1(5(4(0(x)))) → 4(1(5(0(5(5(x))))))
4(5(4(0(x)))) → 4(0(4(4(2(5(x))))))
0(1(5(0(x)))) → 2(1(5(0(0(x)))))
1(0(5(3(x)))) → 5(2(1(2(0(3(x))))))
0(1(5(3(x)))) → 2(3(1(5(0(x)))))
0(1(5(3(x)))) → 3(5(0(2(1(2(x))))))
0(1(5(3(x)))) → 5(0(2(2(1(3(x))))))
0(1(5(3(x)))) → 2(0(2(3(1(5(x))))))
0(3(3(1(0(x))))) → 1(0(0(2(3(3(x))))))
1(5(3(1(0(x))))) → 0(5(4(3(1(1(x))))))
1(5(3(1(0(x))))) → 3(3(0(5(1(1(x))))))
0(2(5(1(0(x))))) → 2(0(1(3(0(5(x))))))
1(4(5(1(0(x))))) → 1(0(1(4(3(5(x))))))
4(4(5(1(0(x))))) → 0(4(1(2(5(4(x))))))
4(4(5(1(0(x))))) → 1(4(3(4(0(5(x))))))
4(0(1(3(0(x))))) → 4(2(1(3(0(0(x))))))
0(3(3(4(0(x))))) → 3(0(4(3(2(0(x))))))
0(2(5(4(0(x))))) → 4(0(5(2(2(0(x))))))
1(5(1(5(0(x))))) → 1(1(0(3(5(5(x))))))
4(5(1(0(3(x))))) → 2(1(4(3(5(0(x))))))
0(1(0(5(3(x))))) → 3(0(0(2(1(5(x))))))
0(0(4(5(3(x))))) → 4(4(0(3(0(5(x))))))
1(0(5(5(3(x))))) → 2(1(3(0(5(5(x))))))

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

(23) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(0) Obligation:

(1) QTRS Reverse (EQUIVALENT transformation)

(2) Obligation:

(3) DependencyPairsProof (EQUIVALENT transformation)

(4) Obligation:

(5) DependencyGraphProof (EQUIVALENT transformation)

(6) Obligation:

(7) QDPOrderProof (EQUIVALENT transformation)

(8) Obligation:

(9) DependencyGraphProof (EQUIVALENT transformation)

(10) Complex Obligation (AND)

(11) Obligation:

(12) QDPOrderProof (EQUIVALENT transformation)

(13) Obligation:

(14) DependencyGraphProof (EQUIVALENT transformation)

(15) Obligation:

(16) QDPOrderProof (EQUIVALENT transformation)

(17) Obligation:

(18) PisEmptyProof (EQUIVALENT transformation)

(19) YES

(20) Obligation:

(21) QDPOrderProof (EQUIVALENT transformation)

(22) Obligation:

(23) PisEmptyProof (EQUIVALENT transformation)

(24) YES