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 (⇔, 140 ms)

↳2 QDP

↳3 DependencyGraphProof (⇔, 0 ms)

↳4 QDP

↳5 QDPOrderProof (⇔, 219 ms)

↳6 QDP

↳7 DependencyGraphProof (⇔, 0 ms)

↳8 AND

    ↳9 QDP

        ↳10 QDPOrderProof (⇔, 44 ms)

        ↳11 QDP

        ↳12 PisEmptyProof (⇔, 0 ms)

        ↳13 YES

    ↳14 QDP

        ↳15 QDPOrderProof (⇔, 49 ms)

        ↳16 QDP

        ↳17 PisEmptyProof (⇔, 0 ms)

        ↳18 YES

    ↳19 QDP

        ↳20 QDPOrderProof (⇔, 124 ms)

        ↳21 QDP

        ↳22 DependencyGraphProof (⇔, 0 ms)

        ↳23 TRUE

(0) Obligation:

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

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

The TRS R consists of the following rules:

0(1(2(x))) → 2(1(0(3(x))))
0(1(2(x))) → 2(1(3(0(x))))
0(1(2(x))) → 2(1(3(0(4(x)))))
0(1(2(x))) → 2(1(3(4(0(x)))))
0(1(2(x))) → 2(1(3(4(0(4(x))))))
0(0(1(2(x)))) → 0(2(1(0(3(x)))))
0(1(1(2(x)))) → 1(0(4(1(2(x)))))
0(1(1(2(x)))) → 1(2(1(0(3(x)))))
0(1(2(0(x)))) → 2(1(0(3(0(3(x))))))
0(1(2(5(x)))) → 2(1(0(3(5(x)))))
0(1(2(5(x)))) → 5(1(0(3(2(x)))))
0(1(4(2(x)))) → 2(1(3(4(0(x)))))
0(1(4(2(x)))) → 2(4(1(0(3(x)))))
0(1(4(2(x)))) → 0(2(1(0(3(4(x))))))
2(1(1(5(x)))) → 2(5(1(4(1(x)))))
2(1(1(5(x)))) → 5(1(3(2(1(x)))))
2(1(2(0(x)))) → 2(2(1(0(3(x)))))
2(3(5(5(x)))) → 5(1(0(3(2(5(x))))))
2(5(3(2(x)))) → 5(1(3(2(2(x)))))
5(1(2(0(x)))) → 2(5(1(0(3(x)))))
0(1(0(1(2(x))))) → 0(2(1(1(3(0(x))))))
0(1(1(2(5(x))))) → 0(1(3(5(1(2(x))))))
0(1(1(4(2(x))))) → 4(1(0(3(2(1(x))))))
0(1(1(5(0(x))))) → 0(1(0(5(4(1(x))))))
0(1(1(5(4(x))))) → 1(4(1(0(3(5(x))))))
0(1(2(3(5(x))))) → 0(1(3(4(5(2(x))))))
0(1(2(3(5(x))))) → 0(2(5(1(0(3(x))))))
0(1(2(3(5(x))))) → 1(0(3(0(2(5(x))))))
0(1(2(3(5(x))))) → 5(1(3(0(2(0(x))))))
0(1(4(0(2(x))))) → 4(0(1(3(0(2(x))))))
0(1(4(0(2(x))))) → 4(0(2(1(3(0(x))))))
0(4(5(3(5(x))))) → 5(1(0(3(4(5(x))))))
2(1(0(1(5(x))))) → 1(0(3(2(5(1(x))))))
2(1(4(3(5(x))))) → 1(3(5(4(4(2(x))))))
2(1(4(3(5(x))))) → 5(4(4(1(3(2(x))))))
2(1(4(5(0(x))))) → 2(4(1(0(3(5(x))))))
2(2(3(5(0(x))))) → 5(2(4(2(0(3(x))))))
2(2(4(3(5(x))))) → 5(1(3(4(2(2(x))))))
2(3(1(1(2(x))))) → 1(3(2(1(2(0(x))))))
2(3(1(1(2(x))))) → 4(1(2(2(1(3(x))))))
2(3(2(0(5(x))))) → 2(2(1(0(3(5(x))))))
2(3(3(1(5(x))))) → 1(3(5(1(3(2(x))))))
2(5(0(3(5(x))))) → 5(2(1(3(0(5(x))))))
4(2(0(1(2(x))))) → 4(2(2(1(3(0(x))))))
5(0(1(2(2(x))))) → 4(1(5(2(0(2(x))))))
5(1(4(2(2(x))))) → 5(1(3(2(4(2(x))))))
5(1(4(3(2(x))))) → 4(5(1(3(4(2(x))))))
5(5(4(3(2(x))))) → 5(1(3(4(5(2(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 1 SCC with 95 less nodes.

(4) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(5) QDPOrderProof (EQUIVALENT transformation)

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

The following pairs can be oriented strictly and are deleted.

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

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

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

(6) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(7) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs with 11 less nodes.

(8) Complex Obligation (AND)

(9) Obligation:

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

5¹(0(1(2(2(x))))) → 5¹(2(0(2(x))))

The TRS R consists of the following rules:

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

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

(10) QDPOrderProof (EQUIVALENT transformation)

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

The following pairs can be oriented strictly and are deleted.

5¹(0(1(2(2(x))))) → 5¹(2(0(2(x))))

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

POL(0(x₁)) = 1 + x₁
POL(1(x₁)) = 0
POL(2(x₁)) = 0
POL(3(x₁)) = 1
POL(4(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:

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

(11) Obligation:

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

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

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

(12) PisEmptyProof (EQUIVALENT transformation)

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

(13) YES

(14) Obligation:

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

2¹(1(0(1(5(x))))) → 2¹(5(1(x)))

The TRS R consists of the following rules:

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

2¹(1(0(1(5(x))))) → 2¹(5(1(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)) = 0
POL(2¹(x₁)) = x₁
POL(3(x₁)) = 0
POL(4(x₁)) = x₁
POL(5(x₁)) = 1

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

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

(16) Obligation:

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

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

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

(17) PisEmptyProof (EQUIVALENT transformation)

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

(18) YES

(19) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(20) 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(x))) → 0¹(4(x))
0¹(1(2(x))) → 4¹(x)
4¹(2(0(1(2(x))))) → 0¹(x)
0¹(1(2(x))) → 4¹(0(x))
0¹(1(2(x))) → 4¹(0(4(x)))
0¹(1(4(2(x)))) → 4¹(0(x))
0¹(1(4(2(x)))) → 0¹(x)
0¹(1(4(2(x)))) → 4¹(x)
0¹(1(0(1(2(x))))) → 0¹(x)
0¹(1(2(3(5(x))))) → 4¹(5(2(x)))
0¹(1(4(0(2(x))))) → 0¹(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)) = x₁
POL(2(x₁)) = 1 + 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:

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

(21) Obligation:

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

0¹(1(2(3(5(x))))) → 0¹(2(5(x)))

The TRS R consists of the following rules:

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

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

(22) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.

(0) Obligation:

(1) DependencyPairsProof (EQUIVALENT transformation)

(2) Obligation:

(3) DependencyGraphProof (EQUIVALENT transformation)

(4) Obligation:

(5) QDPOrderProof (EQUIVALENT transformation)

(6) Obligation:

(7) DependencyGraphProof (EQUIVALENT transformation)

(8) Complex Obligation (AND)

(9) Obligation:

(10) QDPOrderProof (EQUIVALENT transformation)

(11) Obligation:

(12) PisEmptyProof (EQUIVALENT transformation)

(13) YES

(14) Obligation:

(15) QDPOrderProof (EQUIVALENT transformation)

(16) Obligation:

(17) PisEmptyProof (EQUIVALENT transformation)

(18) YES

(19) Obligation:

(20) QDPOrderProof (EQUIVALENT transformation)

(21) Obligation:

(22) DependencyGraphProof (EQUIVALENT transformation)

(23) TRUE