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 DependencyPairsProof (⇔, 167 ms)

↳2 QDP

↳3 DependencyGraphProof (⇔, 0 ms)

↳4 AND

    ↳5 QDP

        ↳6 UsableRulesProof (⇔, 0 ms)

        ↳7 QDP

        ↳8 QDPOrderProof (⇔, 29 ms)

        ↳9 QDP

        ↳10 PisEmptyProof (⇔, 0 ms)

        ↳11 YES

    ↳12 QDP

        ↳13 QDPOrderProof (⇔, 172 ms)

        ↳14 QDP

        ↳15 QDPOrderProof (⇔, 78 ms)

        ↳16 QDP

        ↳17 QDPOrderProof (⇔, 51 ms)

        ↳18 QDP

        ↳19 PisEmptyProof (⇔, 0 ms)

        ↳20 YES

(0) Obligation:

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

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

Q is empty.

(1) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(2) Obligation:

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

0¹(1(2(x))) → 1¹(3(0(2(x))))
0¹(1(2(x))) → 0¹(2(x))
0¹(1(2(x))) → 1¹(2(0(x)))
0¹(1(2(x))) → 0¹(x)
0¹(1(2(x))) → 0¹(4(1(3(x))))
0¹(1(2(x))) → 1¹(3(x))
0¹(1(2(x))) → 0¹(2(1(3(x))))
0¹(1(2(x))) → 0¹(2(1(x)))
0¹(1(2(x))) → 1¹(x)
0¹(1(2(x))) → 1¹(3(3(0(2(3(x))))))
0¹(1(2(x))) → 0¹(2(3(x)))
0¹(1(2(x))) → 0¹(4(3(1(3(x)))))
0¹(1(2(x))) → 0¹(1(3(1(2(x)))))
0¹(1(2(x))) → 1¹(3(1(2(x))))
0¹(1(2(x))) → 0¹(5(3(1(2(x)))))
0¹(1(2(x))) → 5¹(3(1(2(x))))
0¹(5(2(x))) → 1¹(5(0(2(x))))
0¹(5(2(x))) → 5¹(0(2(x)))
0¹(5(2(x))) → 0¹(2(x))
0¹(5(2(x))) → 0¹(5(1(2(x))))
0¹(5(2(x))) → 5¹(1(2(x)))
0¹(5(2(x))) → 1¹(2(x))
0¹(5(2(x))) → 5¹(3(0(2(x))))
0¹(5(2(x))) → 1¹(5(5(3(0(2(x))))))
0¹(5(2(x))) → 5¹(5(3(0(2(x)))))
0¹(5(2(x))) → 5¹(0(4(3(1(2(x))))))
0¹(5(2(x))) → 0¹(4(3(1(2(x)))))
1¹(5(2(x))) → 5¹(3(1(2(x))))
1¹(5(2(x))) → 1¹(2(x))
1¹(5(2(x))) → 1¹(5(3(1(2(x)))))
0¹(0(5(4(x)))) → 5¹(3(0(4(0(0(x))))))
0¹(0(5(4(x)))) → 0¹(4(0(0(x))))
0¹(0(5(4(x)))) → 0¹(0(x))
0¹(0(5(4(x)))) → 0¹(x)
0¹(1(0(1(x)))) → 1¹(1(3(0(3(0(x))))))
0¹(1(0(1(x)))) → 1¹(3(0(3(0(x)))))
0¹(1(0(1(x)))) → 0¹(3(0(x)))
0¹(1(0(1(x)))) → 0¹(x)
0¹(1(1(2(x)))) → 1¹(1(3(0(2(x)))))
0¹(1(1(2(x)))) → 1¹(3(0(2(x))))
0¹(1(1(2(x)))) → 0¹(2(x))
0¹(1(2(5(x)))) → 0¹(5(1(2(x))))
0¹(1(2(5(x)))) → 5¹(1(2(x)))
0¹(1(2(5(x)))) → 1¹(2(x))
0¹(1(2(5(x)))) → 5¹(3(1(2(0(x)))))
0¹(1(2(5(x)))) → 1¹(2(0(x)))
0¹(1(2(5(x)))) → 0¹(x)
0¹(1(3(2(x)))) → 0¹(2(1(3(5(x)))))
0¹(1(3(2(x)))) → 1¹(3(5(x)))
0¹(1(3(2(x)))) → 5¹(x)
0¹(1(4(2(x)))) → 0¹(0(4(1(2(x)))))
0¹(1(4(2(x)))) → 0¹(4(1(2(x))))
0¹(1(4(2(x)))) → 1¹(2(x))
0¹(1(4(2(x)))) → 0¹(1(3(1(2(x)))))
0¹(1(4(2(x)))) → 1¹(3(1(2(x))))
0¹(5(2(4(x)))) → 5¹(5(3(0(2(4(x))))))
0¹(5(2(4(x)))) → 5¹(3(0(2(4(x)))))
0¹(5(2(4(x)))) → 0¹(2(4(x)))
0¹(5(2(5(x)))) → 0¹(2(3(5(5(x)))))
0¹(5(2(5(x)))) → 5¹(5(x))
0¹(5(3(2(x)))) → 0¹(2(1(5(x))))
0¹(5(3(2(x)))) → 1¹(5(x))
0¹(5(3(2(x)))) → 5¹(x)
0¹(5(3(2(x)))) → 5¹(3(1(4(0(2(x))))))
0¹(5(3(2(x)))) → 1¹(4(0(2(x))))
0¹(5(3(2(x)))) → 0¹(2(x))
1¹(0(3(4(x)))) → 1¹(0(4(1(3(x)))))
1¹(0(3(4(x)))) → 0¹(4(1(3(x))))
1¹(0(3(4(x)))) → 1¹(3(x))
1¹(5(3(2(x)))) → 1¹(3(5(2(x))))
1¹(5(3(2(x)))) → 5¹(2(x))
5¹(0(1(2(x)))) → 1¹(5(0(4(1(2(x))))))
5¹(0(1(2(x)))) → 5¹(0(4(1(2(x)))))
5¹(0(1(2(x)))) → 0¹(4(1(2(x))))
5¹(1(0(2(x)))) → 1¹(5(0(2(x))))
5¹(1(0(2(x)))) → 5¹(0(2(x)))
5¹(1(1(2(x)))) → 1¹(3(5(2(1(3(x))))))
5¹(1(1(2(x)))) → 5¹(2(1(3(x))))
5¹(1(1(2(x)))) → 1¹(3(x))
5¹(1(5(2(x)))) → 1¹(3(5(5(2(x)))))
5¹(1(5(2(x)))) → 5¹(5(2(x)))
0¹(0(1(3(2(x))))) → 0¹(3(1(2(0(3(x))))))
0¹(0(1(3(2(x))))) → 1¹(2(0(3(x))))
0¹(0(1(3(2(x))))) → 0¹(3(x))
0¹(0(5(3(2(x))))) → 0¹(3(0(3(2(5(x))))))
0¹(0(5(3(2(x))))) → 0¹(3(2(5(x))))
0¹(0(5(3(2(x))))) → 5¹(x)
0¹(1(0(3(4(x))))) → 0¹(1(3(0(3(x)))))
0¹(1(0(3(4(x))))) → 1¹(3(0(3(x))))
0¹(1(0(3(4(x))))) → 0¹(3(x))
0¹(1(0(5(4(x))))) → 0¹(4(0(1(5(x)))))
0¹(1(0(5(4(x))))) → 0¹(1(5(x)))
0¹(1(0(5(4(x))))) → 1¹(5(x))
0¹(1(0(5(4(x))))) → 5¹(x)
0¹(1(2(4(4(x))))) → 0¹(4(1(2(1(4(x))))))
0¹(1(2(4(4(x))))) → 1¹(2(1(4(x))))
0¹(1(2(4(4(x))))) → 1¹(4(x))
0¹(1(5(3(2(x))))) → 1¹(2(3(0(4(5(x))))))
0¹(1(5(3(2(x))))) → 0¹(4(5(x)))
0¹(1(5(3(2(x))))) → 5¹(x)
0¹(5(4(2(4(x))))) → 5¹(3(0(4(2(x)))))
0¹(5(4(2(4(x))))) → 0¹(4(2(x)))
0¹(5(4(4(2(x))))) → 0¹(4(4(3(5(2(x))))))
0¹(5(4(4(2(x))))) → 5¹(2(x))
1¹(0(1(3(4(x))))) → 1¹(1(3(0(3(x)))))
1¹(0(1(3(4(x))))) → 1¹(3(0(3(x))))
1¹(0(1(3(4(x))))) → 0¹(3(x))
1¹(0(3(4(5(x))))) → 1¹(3(0(4(5(x)))))
1¹(0(3(4(5(x))))) → 0¹(4(5(x)))
1¹(1(3(4(2(x))))) → 5¹(1(3(1(4(x)))))
1¹(1(3(4(2(x))))) → 1¹(3(1(4(x))))
1¹(1(3(4(2(x))))) → 1¹(4(x))
1¹(5(3(2(2(x))))) → 5¹(3(1(2(x))))
1¹(5(3(2(2(x))))) → 1¹(2(x))
5¹(0(5(2(4(x))))) → 5¹(1(5(0(4(2(x))))))
5¹(0(5(2(4(x))))) → 1¹(5(0(4(2(x)))))
5¹(0(5(2(4(x))))) → 5¹(0(4(2(x))))
5¹(0(5(2(4(x))))) → 0¹(4(2(x)))
5¹(1(0(5(2(x))))) → 1¹(5(0(2(3(5(x))))))
5¹(1(0(5(2(x))))) → 5¹(0(2(3(5(x)))))
5¹(1(0(5(2(x))))) → 0¹(2(3(5(x))))
5¹(1(0(5(2(x))))) → 5¹(x)
5¹(1(1(2(1(x))))) → 1¹(2(5(1(1(x)))))
5¹(1(1(2(1(x))))) → 5¹(1(1(x)))
5¹(1(1(2(1(x))))) → 1¹(1(x))

The TRS R consists of the following rules:

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

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

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 117 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

5¹(1(1(2(1(x))))) → 5¹(1(1(x)))
5¹(1(0(5(2(x))))) → 5¹(x)

The TRS R consists of the following rules:

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

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

(6) 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.

(7) Obligation:

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

5¹(1(1(2(1(x))))) → 5¹(1(1(x)))
5¹(1(0(5(2(x))))) → 5¹(x)

The TRS R consists of the following rules:

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

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

(8) QDPOrderProof (EQUIVALENT transformation)

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

The following pairs can be oriented strictly and are deleted.

5¹(1(1(2(1(x))))) → 5¹(1(1(x)))
5¹(1(0(5(2(x))))) → 5¹(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(2(x₁)) = x₁
POL(3(x₁)) = 0
POL(4(x₁)) = 0
POL(5(x₁)) = 1 + x₁
POL(5¹(x₁)) = x₁

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

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

(9) Obligation:

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

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

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

(10) PisEmptyProof (EQUIVALENT transformation)

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

(11) YES

(12) Obligation:

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

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

The TRS R consists of the following rules:

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

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

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

POL(0(x₁)) = 1 + x₁
POL(0¹(x₁)) = x₁
POL(1(x₁)) = x₁
POL(2(x₁)) = x₁
POL(3(x₁)) = 0
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:

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

(14) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(15) QDPOrderProof (EQUIVALENT transformation)

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

The following pairs can be oriented strictly and are deleted.

0¹(1(2(x))) → 0¹(x)
0¹(1(2(5(x)))) → 0¹(x)

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

POL(0(x₁)) = 1 + x₁
POL(0¹(x₁)) = x₁
POL(1(x₁)) = x₁
POL(2(x₁)) = 1 + x₁
POL(3(x₁)) = 0
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:

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

(16) Obligation:

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

0¹(0(5(4(x)))) → 0¹(0(x))

The TRS R consists of the following rules:

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

0¹(0(5(4(x)))) → 0¹(0(x))

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

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

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

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

(18) Obligation:

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

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

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

(19) PisEmptyProof (EQUIVALENT transformation)

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

(0) Obligation:

(1) DependencyPairsProof (EQUIVALENT transformation)

(2) Obligation:

(3) DependencyGraphProof (EQUIVALENT transformation)

(4) Complex Obligation (AND)

(5) Obligation:

(6) UsableRulesProof (EQUIVALENT transformation)

(7) Obligation:

(8) QDPOrderProof (EQUIVALENT transformation)

(9) Obligation:

(10) PisEmptyProof (EQUIVALENT transformation)

(11) YES

(12) Obligation:

(13) QDPOrderProof (EQUIVALENT transformation)

(14) Obligation:

(15) QDPOrderProof (EQUIVALENT transformation)

(16) Obligation:

(17) QDPOrderProof (EQUIVALENT transformation)

(18) Obligation:

(19) PisEmptyProof (EQUIVALENT transformation)

(20) YES