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 (⇔, 129 ms)

↳4 QDP

↳5 DependencyGraphProof (⇔, 0 ms)

↳6 AND

    ↳7 QDP

        ↳8 QDPSizeChangeProof (⇔, 0 ms)

        ↳9 YES

    ↳10 QDP

        ↳11 QDPOrderProof (⇔, 245 ms)

        ↳12 QDP

        ↳13 QDPOrderProof (⇔, 151 ms)

        ↳14 QDP

        ↳15 DependencyGraphProof (⇔, 0 ms)

        ↳16 QDP

        ↳17 QDPOrderProof (⇔, 100 ms)

        ↳18 QDP

        ↳19 QDPOrderProof (⇔, 85 ms)

        ↳20 QDP

        ↳21 QDPOrderProof (⇔, 65 ms)

        ↳22 QDP

        ↳23 QDPOrderProof (⇔, 57 ms)

        ↳24 QDP

        ↳25 PisEmptyProof (⇔, 0 ms)

        ↳26 YES

    ↳27 QDP

        ↳28 UsableRulesProof (⇔, 0 ms)

        ↳29 QDP

        ↳30 QDPSizeChangeProof (⇔, 0 ms)

        ↳31 YES

(0) Obligation:

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

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

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

2¹(1(0(0(x)))) → 0¹(3(3(1(2(0(x))))))
2¹(1(0(0(x)))) → 2¹(0(x))
4¹(0(1(0(x)))) → 4¹(0(3(1(0(x)))))
4¹(0(1(0(x)))) → 0¹(3(1(0(x))))
5¹(0(1(0(x)))) → 0¹(5(3(1(0(x)))))
5¹(0(1(0(x)))) → 5¹(3(1(0(x))))
0¹(5(1(0(x)))) → 0¹(5(3(1(0(x)))))
0¹(5(1(0(x)))) → 5¹(3(1(0(x))))
0¹(1(3(0(x)))) → 0¹(5(3(1(0(x)))))
0¹(1(3(0(x)))) → 5¹(3(1(0(x))))
2¹(0(1(3(x)))) → 2¹(0(5(3(1(x)))))
2¹(0(1(3(x)))) → 0¹(5(3(1(x))))
2¹(0(1(3(x)))) → 5¹(3(1(x)))
2¹(0(1(3(x)))) → 0¹(3(1(2(3(x)))))
2¹(0(1(3(x)))) → 2¹(3(x))
5¹(0(1(3(x)))) → 0¹(3(5(3(1(x)))))
5¹(0(1(3(x)))) → 5¹(3(1(x)))
0¹(2(1(3(x)))) → 0¹(3(3(1(2(x)))))
0¹(2(1(3(x)))) → 2¹(x)
0¹(2(1(3(x)))) → 4¹(1(2(0(3(x)))))
0¹(2(1(3(x)))) → 2¹(0(3(x)))
0¹(2(1(3(x)))) → 0¹(3(x))
0¹(2(1(3(x)))) → 2¹(3(0(3(x))))
0¹(2(1(3(x)))) → 0¹(1(2(3(3(x)))))
0¹(2(1(3(x)))) → 2¹(3(3(x)))
0¹(2(1(3(x)))) → 0¹(1(2(5(3(x)))))
0¹(2(1(3(x)))) → 2¹(5(3(x)))
0¹(2(1(3(x)))) → 5¹(3(x))
2¹(2(1(3(x)))) → 2¹(2(5(3(x))))
2¹(2(1(3(x)))) → 2¹(5(3(x)))
2¹(2(1(3(x)))) → 5¹(3(x))
0¹(2(4(3(x)))) → 0¹(4(2(3(3(x)))))
0¹(2(4(3(x)))) → 4¹(2(3(3(x))))
0¹(2(4(3(x)))) → 2¹(3(3(x)))
0¹(1(0(4(x)))) → 0¹(4(0(3(1(x)))))
0¹(1(0(4(x)))) → 4¹(0(3(1(x))))
0¹(1(0(4(x)))) → 0¹(3(1(x)))
5¹(0(1(5(x)))) → 0¹(5(5(3(3(1(x))))))
5¹(0(1(5(x)))) → 5¹(5(3(3(1(x)))))
5¹(0(1(5(x)))) → 5¹(3(3(1(x))))
0¹(5(2(5(x)))) → 0¹(5(3(2(5(x)))))
0¹(5(2(5(x)))) → 5¹(3(2(5(x))))
0¹(1(3(5(x)))) → 5¹(5(3(1(0(x)))))
0¹(1(3(5(x)))) → 5¹(3(1(0(x))))
0¹(1(3(5(x)))) → 0¹(x)
0¹(1(4(0(0(x))))) → 0¹(0(3(1(4(0(x))))))
0¹(1(4(0(0(x))))) → 0¹(3(1(4(0(x)))))
0¹(1(4(0(0(x))))) → 4¹(0(x))
5¹(0(2(1(0(x))))) → 2¹(1(5(0(3(0(x))))))
5¹(0(2(1(0(x))))) → 5¹(0(3(0(x))))
5¹(0(2(1(0(x))))) → 0¹(3(0(x)))
2¹(1(3(1(0(x))))) → 2¹(2(0(3(x))))
2¹(1(3(1(0(x))))) → 2¹(0(3(x)))
2¹(1(3(1(0(x))))) → 0¹(3(x))
4¹(0(5(1(0(x))))) → 0¹(5(3(1(4(0(x))))))
4¹(0(5(1(0(x))))) → 5¹(3(1(4(0(x)))))
4¹(0(5(1(0(x))))) → 4¹(0(x))
5¹(3(5(1(0(x))))) → 5¹(3(5(3(1(0(x))))))
5¹(3(5(1(0(x))))) → 5¹(3(1(0(x))))
4¹(3(0(2(0(x))))) → 0¹(3(2(5(4(0(x))))))
4¹(3(0(2(0(x))))) → 2¹(5(4(0(x))))
4¹(3(0(2(0(x))))) → 5¹(4(0(x)))
4¹(3(0(2(0(x))))) → 4¹(0(x))
5¹(1(3(2(0(x))))) → 2¹(5(5(3(1(0(x))))))
5¹(1(3(2(0(x))))) → 5¹(5(3(1(0(x)))))
5¹(1(3(2(0(x))))) → 5¹(3(1(0(x))))
2¹(1(4(2(0(x))))) → 2¹(4(1(1(2(0(x))))))
2¹(1(4(2(0(x))))) → 4¹(1(1(2(0(x)))))
2¹(0(5(2(0(x))))) → 2¹(2(5(3(0(0(x))))))
2¹(0(5(2(0(x))))) → 2¹(5(3(0(0(x)))))
2¹(0(5(2(0(x))))) → 5¹(3(0(0(x))))
2¹(0(5(2(0(x))))) → 0¹(0(x))
0¹(5(5(2(0(x))))) → 5¹(1(5(2(0(0(x))))))
0¹(5(5(2(0(x))))) → 5¹(2(0(0(x))))
0¹(5(5(2(0(x))))) → 2¹(0(0(x)))
0¹(5(5(2(0(x))))) → 0¹(0(x))
0¹(5(1(3(0(x))))) → 0¹(3(5(0(3(1(x))))))
0¹(5(1(3(0(x))))) → 5¹(0(3(1(x))))
0¹(5(1(3(0(x))))) → 0¹(3(1(x)))
0¹(2(3(5(0(x))))) → 5¹(3(1(2(0(0(x))))))
0¹(2(3(5(0(x))))) → 2¹(0(0(x)))
0¹(2(3(5(0(x))))) → 0¹(0(x))
2¹(5(0(1(3(x))))) → 0¹(5(3(1(4(2(x))))))
2¹(5(0(1(3(x))))) → 5¹(3(1(4(2(x)))))
2¹(5(0(1(3(x))))) → 4¹(2(x))
2¹(5(0(1(3(x))))) → 2¹(x)
2¹(0(3(1(3(x))))) → 0¹(3(3(3(1(2(x))))))
2¹(0(3(1(3(x))))) → 2¹(x)
0¹(2(3(1(3(x))))) → 2¹(5(0(3(1(3(x))))))
0¹(2(3(1(3(x))))) → 5¹(0(3(1(3(x)))))
0¹(2(3(1(3(x))))) → 0¹(3(1(3(x))))
2¹(1(4(1(3(x))))) → 2¹(2(3(4(1(x)))))
2¹(1(4(1(3(x))))) → 2¹(3(4(1(x))))
2¹(1(4(1(3(x))))) → 4¹(1(x))
0¹(2(4(1(3(x))))) → 4¹(0(3(3(1(2(x))))))
0¹(2(4(1(3(x))))) → 0¹(3(3(1(2(x)))))
0¹(2(4(1(3(x))))) → 2¹(x)
2¹(0(5(1(3(x))))) → 5¹(0(3(1(5(2(x))))))
2¹(0(5(1(3(x))))) → 0¹(3(1(5(2(x)))))
2¹(0(5(1(3(x))))) → 5¹(2(x))
2¹(0(5(1(3(x))))) → 2¹(x)
0¹(0(1(3(3(x))))) → 0¹(0(3(3(1(5(x))))))
0¹(0(1(3(3(x))))) → 0¹(3(3(1(5(x)))))
0¹(0(1(3(3(x))))) → 5¹(x)
0¹(2(0(4(3(x))))) → 4¹(1(2(0(0(3(x))))))
0¹(2(0(4(3(x))))) → 2¹(0(0(3(x))))
0¹(2(0(4(3(x))))) → 0¹(0(3(x)))
0¹(2(0(4(3(x))))) → 0¹(3(x))
2¹(3(2(4(3(x))))) → 2¹(4(5(2(3(3(x))))))
2¹(3(2(4(3(x))))) → 4¹(5(2(3(3(x)))))
2¹(3(2(4(3(x))))) → 5¹(2(3(3(x))))
2¹(3(2(4(3(x))))) → 2¹(3(3(x)))
2¹(0(1(3(4(x))))) → 0¹(3(1(2(4(x)))))
2¹(0(1(3(4(x))))) → 2¹(4(x))
0¹(2(1(3(4(x))))) → 2¹(4(1(0(3(1(x))))))
0¹(2(1(3(4(x))))) → 4¹(1(0(3(1(x)))))
0¹(2(1(3(4(x))))) → 0¹(3(1(x)))
2¹(1(5(5(4(x))))) → 2¹(5(4(5(x))))
2¹(1(5(5(4(x))))) → 5¹(4(5(x)))
2¹(1(5(5(4(x))))) → 4¹(5(x))
2¹(1(5(5(4(x))))) → 5¹(x)
0¹(5(5(5(4(x))))) → 4¹(0(5(1(5(5(x))))))
0¹(5(5(5(4(x))))) → 0¹(5(1(5(5(x)))))
0¹(5(5(5(4(x))))) → 5¹(1(5(5(x))))
0¹(5(5(5(4(x))))) → 5¹(5(x))
0¹(5(5(5(4(x))))) → 5¹(x)
2¹(1(0(1(5(x))))) → 0¹(5(3(1(1(2(x))))))
2¹(1(0(1(5(x))))) → 5¹(3(1(1(2(x)))))
2¹(1(0(1(5(x))))) → 2¹(x)
4¹(0(1(1(5(x))))) → 0¹(3(4(1(1(5(x))))))
4¹(0(1(1(5(x))))) → 4¹(1(1(5(x))))
0¹(2(5(1(5(x))))) → 5¹(0(5(3(1(2(x))))))
0¹(2(5(1(5(x))))) → 0¹(5(3(1(2(x)))))
0¹(2(5(1(5(x))))) → 5¹(3(1(2(x))))
0¹(2(5(1(5(x))))) → 2¹(x)
5¹(0(1(4(5(x))))) → 0¹(5(5(3(1(4(x))))))
5¹(0(1(4(5(x))))) → 5¹(5(3(1(4(x)))))
5¹(0(1(4(5(x))))) → 5¹(3(1(4(x))))
5¹(0(1(4(5(x))))) → 4¹(x)

The TRS R consists of the following rules:

2(1(0(0(x)))) → 0(3(3(1(2(0(x))))))
4(0(1(0(x)))) → 4(0(3(1(0(x)))))
5(0(1(0(x)))) → 0(5(3(1(0(x)))))
0(5(1(0(x)))) → 0(5(3(1(0(x)))))
0(1(3(0(x)))) → 0(5(3(1(0(x)))))
2(0(1(3(x)))) → 2(0(5(3(1(x)))))
2(0(1(3(x)))) → 0(3(1(2(3(x)))))
5(0(1(3(x)))) → 0(3(5(3(1(x)))))
0(2(1(3(x)))) → 0(3(3(1(2(x)))))
0(2(1(3(x)))) → 4(1(2(0(3(x)))))
0(2(1(3(x)))) → 1(2(3(0(3(x)))))
0(2(1(3(x)))) → 0(1(2(3(3(x)))))
0(2(1(3(x)))) → 0(1(2(5(3(x)))))
2(2(1(3(x)))) → 1(2(2(5(3(x)))))
0(2(4(3(x)))) → 0(4(2(3(3(x)))))
0(1(0(4(x)))) → 0(4(0(3(1(x)))))
5(0(1(5(x)))) → 0(5(5(3(3(1(x))))))
0(5(2(5(x)))) → 0(5(3(2(5(x)))))
0(1(3(5(x)))) → 5(5(3(1(0(x)))))
0(1(4(0(0(x))))) → 0(0(3(1(4(0(x))))))
5(0(2(1(0(x))))) → 2(1(5(0(3(0(x))))))
2(1(3(1(0(x))))) → 1(1(2(2(0(3(x))))))
4(0(5(1(0(x))))) → 0(5(3(1(4(0(x))))))
5(3(5(1(0(x))))) → 5(3(5(3(1(0(x))))))
4(3(0(2(0(x))))) → 0(3(2(5(4(0(x))))))
5(1(3(2(0(x))))) → 2(5(5(3(1(0(x))))))
2(1(4(2(0(x))))) → 2(4(1(1(2(0(x))))))
2(0(5(2(0(x))))) → 2(2(5(3(0(0(x))))))
0(5(5(2(0(x))))) → 5(1(5(2(0(0(x))))))
0(5(1(3(0(x))))) → 0(3(5(0(3(1(x))))))
0(2(3(5(0(x))))) → 5(3(1(2(0(0(x))))))
2(5(0(1(3(x))))) → 0(5(3(1(4(2(x))))))
2(0(3(1(3(x))))) → 0(3(3(3(1(2(x))))))
0(2(3(1(3(x))))) → 2(5(0(3(1(3(x))))))
2(1(4(1(3(x))))) → 1(2(2(3(4(1(x))))))
0(2(4(1(3(x))))) → 4(0(3(3(1(2(x))))))
2(0(5(1(3(x))))) → 5(0(3(1(5(2(x))))))
0(0(1(3(3(x))))) → 0(0(3(3(1(5(x))))))
0(2(0(4(3(x))))) → 4(1(2(0(0(3(x))))))
2(3(2(4(3(x))))) → 2(4(5(2(3(3(x))))))
2(0(1(3(4(x))))) → 1(0(3(1(2(4(x))))))
0(2(1(3(4(x))))) → 2(4(1(0(3(1(x))))))
2(1(5(5(4(x))))) → 1(1(2(5(4(5(x))))))
0(5(5(5(4(x))))) → 4(0(5(1(5(5(x))))))
2(1(0(1(5(x))))) → 0(5(3(1(1(2(x))))))
4(0(1(1(5(x))))) → 0(3(4(1(1(5(x))))))
0(2(5(1(5(x))))) → 5(0(5(3(1(2(x))))))
5(0(1(4(5(x))))) → 0(5(5(3(1(4(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 3 SCCs with 108 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

4¹(0(5(1(0(x))))) → 4¹(0(x))

The TRS R consists of the following rules:

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

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

(8) QDPSizeChangeProof (EQUIVALENT transformation)

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:

4¹(0(5(1(0(x))))) → 4¹(0(x))
The graph contains the following edges 1 > 1

(9) YES

(10) Obligation:

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

2¹(0(5(2(0(x))))) → 2¹(2(5(3(0(0(x))))))
2¹(1(0(0(x)))) → 2¹(0(x))
2¹(0(5(2(0(x))))) → 2¹(5(3(0(0(x)))))
2¹(0(5(2(0(x))))) → 0¹(0(x))
0¹(5(2(5(x)))) → 0¹(5(3(2(5(x)))))
0¹(5(5(2(0(x))))) → 5¹(2(0(0(x))))
5¹(0(1(4(5(x))))) → 4¹(x)
4¹(3(0(2(0(x))))) → 2¹(5(4(0(x))))
2¹(1(5(5(4(x))))) → 2¹(5(4(5(x))))
2¹(1(5(5(4(x))))) → 5¹(4(5(x)))
2¹(1(5(5(4(x))))) → 5¹(x)
2¹(1(0(1(5(x))))) → 2¹(x)
2¹(2(1(3(x)))) → 2¹(2(5(3(x))))
2¹(2(1(3(x)))) → 2¹(5(3(x)))
2¹(1(3(1(0(x))))) → 2¹(2(0(3(x))))
2¹(1(3(1(0(x))))) → 2¹(0(3(x)))
2¹(0(3(1(3(x))))) → 2¹(x)
2¹(5(0(1(3(x))))) → 2¹(x)
2¹(0(5(1(3(x))))) → 5¹(2(x))
2¹(0(5(1(3(x))))) → 2¹(x)
2¹(0(1(3(4(x))))) → 2¹(4(x))
4¹(3(0(2(0(x))))) → 5¹(4(0(x)))
0¹(5(5(2(0(x))))) → 2¹(0(0(x)))
0¹(5(5(2(0(x))))) → 0¹(0(x))
0¹(0(1(3(3(x))))) → 5¹(x)
0¹(5(5(5(4(x))))) → 0¹(5(1(5(5(x)))))
0¹(5(5(5(4(x))))) → 5¹(5(x))
0¹(5(5(5(4(x))))) → 5¹(x)
0¹(2(5(1(5(x))))) → 2¹(x)

The TRS R consists of the following rules:

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

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

(11) QDPOrderProof (EQUIVALENT transformation)

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

The following pairs can be oriented strictly and are deleted.

0¹(5(2(5(x)))) → 0¹(5(3(2(5(x)))))

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

POL(0(x₁)) = 1
POL(0¹(x₁)) = x₁
POL(1(x₁)) = 1
POL(2(x₁)) = 1
POL(2¹(x₁)) = 1
POL(3(x₁)) = 0
POL(4(x₁)) = 1
POL(4¹(x₁)) = 1
POL(5(x₁)) = x₁
POL(5¹(x₁)) = 1

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

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

(12) Obligation:

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

2¹(0(5(2(0(x))))) → 2¹(2(5(3(0(0(x))))))
2¹(1(0(0(x)))) → 2¹(0(x))
2¹(0(5(2(0(x))))) → 2¹(5(3(0(0(x)))))
2¹(0(5(2(0(x))))) → 0¹(0(x))
0¹(5(5(2(0(x))))) → 5¹(2(0(0(x))))
5¹(0(1(4(5(x))))) → 4¹(x)
4¹(3(0(2(0(x))))) → 2¹(5(4(0(x))))
2¹(1(5(5(4(x))))) → 2¹(5(4(5(x))))
2¹(1(5(5(4(x))))) → 5¹(4(5(x)))
2¹(1(5(5(4(x))))) → 5¹(x)
2¹(1(0(1(5(x))))) → 2¹(x)
2¹(2(1(3(x)))) → 2¹(2(5(3(x))))
2¹(2(1(3(x)))) → 2¹(5(3(x)))
2¹(1(3(1(0(x))))) → 2¹(2(0(3(x))))
2¹(1(3(1(0(x))))) → 2¹(0(3(x)))
2¹(0(3(1(3(x))))) → 2¹(x)
2¹(5(0(1(3(x))))) → 2¹(x)
2¹(0(5(1(3(x))))) → 5¹(2(x))
2¹(0(5(1(3(x))))) → 2¹(x)
2¹(0(1(3(4(x))))) → 2¹(4(x))
4¹(3(0(2(0(x))))) → 5¹(4(0(x)))
0¹(5(5(2(0(x))))) → 2¹(0(0(x)))
0¹(5(5(2(0(x))))) → 0¹(0(x))
0¹(0(1(3(3(x))))) → 5¹(x)
0¹(5(5(5(4(x))))) → 0¹(5(1(5(5(x)))))
0¹(5(5(5(4(x))))) → 5¹(5(x))
0¹(5(5(5(4(x))))) → 5¹(x)
0¹(2(5(1(5(x))))) → 2¹(x)

The TRS R consists of the following rules:

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

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

(13) QDPOrderProof (EQUIVALENT transformation)

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

The following pairs can be oriented strictly and are deleted.

2¹(1(0(0(x)))) → 2¹(0(x))
5¹(0(1(4(5(x))))) → 4¹(x)
4¹(3(0(2(0(x))))) → 2¹(5(4(0(x))))
2¹(1(0(1(5(x))))) → 2¹(x)
2¹(0(3(1(3(x))))) → 2¹(x)
2¹(5(0(1(3(x))))) → 2¹(x)
2¹(0(5(1(3(x))))) → 5¹(2(x))
2¹(0(5(1(3(x))))) → 2¹(x)
2¹(0(1(3(4(x))))) → 2¹(4(x))
4¹(3(0(2(0(x))))) → 5¹(4(0(x)))
0¹(0(1(3(3(x))))) → 5¹(x)
0¹(5(5(5(4(x))))) → 5¹(5(x))
0¹(5(5(5(4(x))))) → 5¹(x)
0¹(2(5(1(5(x))))) → 2¹(x)

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

POL(0(x₁)) = 1 + x₁
POL(0¹(x₁)) = 1 + x₁
POL(1(x₁)) = x₁
POL(2(x₁)) = x₁
POL(2¹(x₁)) = x₁
POL(3(x₁)) = x₁
POL(4(x₁)) = x₁
POL(4¹(x₁)) = x₁
POL(5(x₁)) = x₁
POL(5¹(x₁)) = x₁

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

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

(14) Obligation:

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

2¹(0(5(2(0(x))))) → 2¹(2(5(3(0(0(x))))))
2¹(0(5(2(0(x))))) → 2¹(5(3(0(0(x)))))
2¹(0(5(2(0(x))))) → 0¹(0(x))
0¹(5(5(2(0(x))))) → 5¹(2(0(0(x))))
2¹(1(5(5(4(x))))) → 2¹(5(4(5(x))))
2¹(1(5(5(4(x))))) → 5¹(4(5(x)))
2¹(1(5(5(4(x))))) → 5¹(x)
2¹(2(1(3(x)))) → 2¹(2(5(3(x))))
2¹(2(1(3(x)))) → 2¹(5(3(x)))
2¹(1(3(1(0(x))))) → 2¹(2(0(3(x))))
2¹(1(3(1(0(x))))) → 2¹(0(3(x)))
0¹(5(5(2(0(x))))) → 2¹(0(0(x)))
0¹(5(5(2(0(x))))) → 0¹(0(x))
0¹(5(5(5(4(x))))) → 0¹(5(1(5(5(x)))))

The TRS R consists of the following rules:

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

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

(15) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 7 less nodes.

(16) Obligation:

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

2¹(0(5(2(0(x))))) → 2¹(5(3(0(0(x)))))
2¹(0(5(2(0(x))))) → 2¹(2(5(3(0(0(x))))))
2¹(0(5(2(0(x))))) → 0¹(0(x))
0¹(5(5(2(0(x))))) → 2¹(0(0(x)))
2¹(1(5(5(4(x))))) → 2¹(5(4(5(x))))
0¹(5(5(2(0(x))))) → 0¹(0(x))
0¹(5(5(5(4(x))))) → 0¹(5(1(5(5(x)))))

The TRS R consists of the following rules:

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

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

(17) QDPOrderProof (EQUIVALENT transformation)

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

The following pairs can be oriented strictly and are deleted.

2¹(0(5(2(0(x))))) → 2¹(5(3(0(0(x)))))

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

POL(0(x₁)) = 1
POL(0¹(x₁)) = 1 + x₁
POL(1(x₁)) = 1
POL(2(x₁)) = 1
POL(2¹(x₁)) = 1 + x₁
POL(3(x₁)) = 0
POL(4(x₁)) = 1
POL(5(x₁)) = x₁

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

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

(18) Obligation:

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

2¹(0(5(2(0(x))))) → 2¹(2(5(3(0(0(x))))))
2¹(0(5(2(0(x))))) → 0¹(0(x))
0¹(5(5(2(0(x))))) → 2¹(0(0(x)))
2¹(1(5(5(4(x))))) → 2¹(5(4(5(x))))
0¹(5(5(2(0(x))))) → 0¹(0(x))
0¹(5(5(5(4(x))))) → 0¹(5(1(5(5(x)))))

The TRS R consists of the following rules:

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

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

(19) QDPOrderProof (EQUIVALENT transformation)

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

The following pairs can be oriented strictly and are deleted.

0¹(5(5(5(4(x))))) → 0¹(5(1(5(5(x)))))

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

POL(0(x₁)) = 1
POL(0¹(x₁)) = x₁
POL(1(x₁)) = 0
POL(2(x₁)) = x₁
POL(2¹(x₁)) = 1
POL(3(x₁)) = 0
POL(4(x₁)) = 1
POL(5(x₁)) = x₁

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

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

(20) Obligation:

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

2¹(0(5(2(0(x))))) → 2¹(2(5(3(0(0(x))))))
2¹(0(5(2(0(x))))) → 0¹(0(x))
0¹(5(5(2(0(x))))) → 2¹(0(0(x)))
2¹(1(5(5(4(x))))) → 2¹(5(4(5(x))))
0¹(5(5(2(0(x))))) → 0¹(0(x))

The TRS R consists of the following rules:

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

2¹(0(5(2(0(x))))) → 0¹(0(x))
0¹(5(5(2(0(x))))) → 2¹(0(0(x)))
2¹(1(5(5(4(x))))) → 2¹(5(4(5(x))))
0¹(5(5(2(0(x))))) → 0¹(0(x))

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

POL(0(x₁)) = x₁
POL(0¹(x₁)) = 1 + x₁
POL(1(x₁)) = 1
POL(2(x₁)) = 1 + x₁
POL(2¹(x₁)) = 1 + x₁
POL(3(x₁)) = 0
POL(4(x₁)) = 0
POL(5(x₁)) = x₁

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

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

(22) Obligation:

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

2¹(0(5(2(0(x))))) → 2¹(2(5(3(0(0(x))))))

The TRS R consists of the following rules:

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

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

(23) QDPOrderProof (EQUIVALENT transformation)

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

The following pairs can be oriented strictly and are deleted.

2¹(0(5(2(0(x))))) → 2¹(2(5(3(0(0(x))))))

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

POL(0(x₁)) = 1 + x₁
POL(1(x₁)) = 1
POL(2(x₁)) = x₁
POL(2¹(x₁)) = x₁
POL(3(x₁)) = 0
POL(4(x₁)) = 0
POL(5(x₁)) = x₁

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

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

(24) Obligation:

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

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

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

(25) PisEmptyProof (EQUIVALENT transformation)

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

(26) YES

(27) Obligation:

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

0¹(1(3(5(x)))) → 0¹(x)

The TRS R consists of the following rules:

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

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

(28) UsableRulesProof (EQUIVALENT transformation)

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.

(29) Obligation:

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

0¹(1(3(5(x)))) → 0¹(x)

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

(30) QDPSizeChangeProof (EQUIVALENT transformation)

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:

0¹(1(3(5(x)))) → 0¹(x)
The graph contains the following edges 1 > 1

(0) Obligation:

(1) QTRS Reverse (EQUIVALENT transformation)

(2) Obligation:

(3) DependencyPairsProof (EQUIVALENT transformation)

(4) Obligation:

(5) DependencyGraphProof (EQUIVALENT transformation)

(6) Complex Obligation (AND)

(7) Obligation:

(8) QDPSizeChangeProof (EQUIVALENT transformation)

(9) YES

(10) Obligation:

(11) QDPOrderProof (EQUIVALENT transformation)

(12) Obligation:

(13) QDPOrderProof (EQUIVALENT transformation)

(14) Obligation:

(15) DependencyGraphProof (EQUIVALENT transformation)

(16) Obligation:

(17) QDPOrderProof (EQUIVALENT transformation)

(18) Obligation:

(19) QDPOrderProof (EQUIVALENT transformation)

(20) Obligation:

(21) QDPOrderProof (EQUIVALENT transformation)

(22) Obligation:

(23) QDPOrderProof (EQUIVALENT transformation)

(24) Obligation:

(25) PisEmptyProof (EQUIVALENT transformation)

(26) YES

(27) Obligation:

(28) UsableRulesProof (EQUIVALENT transformation)

(29) Obligation:

(30) QDPSizeChangeProof (EQUIVALENT transformation)

(31) YES