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 (⇔, 241 ms)

↳2 QDP

↳3 DependencyGraphProof (⇔, 0 ms)

↳4 AND

    ↳5 QDP

        ↳6 QDPSizeChangeProof (⇔, 0 ms)

        ↳7 YES

    ↳8 QDP

        ↳9 UsableRulesProof (⇔, 0 ms)

        ↳10 QDP

        ↳11 QDPSizeChangeProof (⇔, 0 ms)

        ↳12 YES

    ↳13 QDP

        ↳14 QDPOrderProof (⇔, 137 ms)

        ↳15 QDP

        ↳16 PisEmptyProof (⇔, 0 ms)

        ↳17 YES

(0) Obligation:

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

0(1(0(x))) → 0(2(1(0(x))))
0(1(0(x))) → 0(0(2(1(0(x)))))
0(1(0(x))) → 0(0(2(1(2(x)))))
0(1(0(x))) → 0(2(1(0(2(x)))))
0(1(0(x))) → 0(3(2(1(0(x)))))
0(1(0(x))) → 1(0(0(0(2(x)))))
0(1(0(x))) → 1(0(0(2(0(x)))))
0(1(0(x))) → 1(0(4(2(0(x)))))
0(1(0(x))) → 1(4(0(4(0(x)))))
0(1(0(x))) → 4(0(0(2(1(x)))))
0(1(0(x))) → 5(0(0(4(1(x)))))
0(1(0(x))) → 5(1(0(4(0(x)))))
0(1(0(x))) → 0(2(1(0(3(2(x))))))
0(1(0(x))) → 0(4(0(4(1(3(x))))))
0(1(0(x))) → 0(4(2(2(1(0(x))))))
0(1(0(x))) → 0(5(2(1(2(0(x))))))
0(1(0(x))) → 0(5(2(5(1(0(x))))))
0(1(0(x))) → 1(0(0(5(4(4(x))))))
0(1(0(x))) → 1(0(4(4(4(0(x))))))
0(1(0(x))) → 1(5(0(0(4(2(x))))))
0(1(0(x))) → 3(0(0(4(1(4(x))))))
0(1(0(x))) → 4(5(1(0(2(0(x))))))
0(1(0(x))) → 5(5(1(0(0(2(x))))))
0(0(1(0(x)))) → 1(0(0(2(0(x)))))
0(0(1(0(x)))) → 0(1(5(0(0(2(x))))))
0(1(0(3(x)))) → 1(0(3(3(0(2(x))))))
0(1(0(3(x)))) → 1(0(5(3(2(0(x))))))
0(1(1(0(x)))) → 0(4(4(1(1(0(x))))))
0(1(1(3(x)))) → 3(4(5(1(1(0(x))))))
0(1(2(0(x)))) → 1(1(0(2(0(x)))))
0(1(2(0(x)))) → 3(0(2(1(0(x)))))
0(1(2(0(x)))) → 4(1(0(0(2(x)))))
0(1(2(0(x)))) → 0(0(4(2(5(1(x))))))
0(1(2(0(x)))) → 1(1(2(0(4(0(x))))))
0(1(2(0(x)))) → 3(0(2(1(0(4(x))))))
0(1(3(0(x)))) → 1(0(3(0(2(x)))))
0(1(4(0(x)))) → 0(3(4(2(1(0(x))))))
0(1(5(0(x)))) → 0(5(1(4(0(x)))))
0(1(5(0(x)))) → 1(5(3(0(2(0(x))))))
0(3(1(0(x)))) → 1(2(3(0(5(0(x))))))
5(0(1(0(x)))) → 1(4(0(0(5(1(x))))))
5(0(1(0(x)))) → 2(1(0(0(4(5(x))))))
0(1(0(0(0(x))))) → 0(0(5(1(0(0(x))))))
0(1(2(4(0(x))))) → 0(0(5(4(2(1(x))))))
0(1(2(5(0(x))))) → 1(0(2(0(5(4(x))))))
0(1(4(0(0(x))))) → 0(0(0(4(1(0(x))))))
0(1(4(5(0(x))))) → 1(5(0(0(4(2(x))))))
0(3(0(1(0(x))))) → 0(3(0(0(2(1(x))))))
3(0(3(1(0(x))))) → 0(1(3(2(3(0(x))))))
5(0(1(2(0(x))))) → 0(0(5(2(1(0(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(0(x))) → 0¹(2(1(0(x))))
0¹(1(0(x))) → 0¹(0(2(1(0(x)))))
0¹(1(0(x))) → 0¹(0(2(1(2(x)))))
0¹(1(0(x))) → 0¹(2(1(2(x))))
0¹(1(0(x))) → 0¹(2(1(0(2(x)))))
0¹(1(0(x))) → 0¹(2(x))
0¹(1(0(x))) → 0¹(3(2(1(0(x)))))
0¹(1(0(x))) → 3¹(2(1(0(x))))
0¹(1(0(x))) → 0¹(0(0(2(x))))
0¹(1(0(x))) → 0¹(0(2(x)))
0¹(1(0(x))) → 0¹(0(2(0(x))))
0¹(1(0(x))) → 0¹(2(0(x)))
0¹(1(0(x))) → 0¹(4(2(0(x))))
0¹(1(0(x))) → 0¹(4(0(x)))
0¹(1(0(x))) → 0¹(0(2(1(x))))
0¹(1(0(x))) → 0¹(2(1(x)))
0¹(1(0(x))) → 5¹(0(0(4(1(x)))))
0¹(1(0(x))) → 0¹(0(4(1(x))))
0¹(1(0(x))) → 0¹(4(1(x)))
0¹(1(0(x))) → 5¹(1(0(4(0(x)))))
0¹(1(0(x))) → 0¹(2(1(0(3(2(x))))))
0¹(1(0(x))) → 0¹(3(2(x)))
0¹(1(0(x))) → 3¹(2(x))
0¹(1(0(x))) → 0¹(4(0(4(1(3(x))))))
0¹(1(0(x))) → 0¹(4(1(3(x))))
0¹(1(0(x))) → 3¹(x)
0¹(1(0(x))) → 0¹(4(2(2(1(0(x))))))
0¹(1(0(x))) → 0¹(5(2(1(2(0(x))))))
0¹(1(0(x))) → 5¹(2(1(2(0(x)))))
0¹(1(0(x))) → 0¹(5(2(5(1(0(x))))))
0¹(1(0(x))) → 5¹(2(5(1(0(x)))))
0¹(1(0(x))) → 5¹(1(0(x)))
0¹(1(0(x))) → 0¹(0(5(4(4(x)))))
0¹(1(0(x))) → 0¹(5(4(4(x))))
0¹(1(0(x))) → 5¹(4(4(x)))
0¹(1(0(x))) → 0¹(4(4(4(0(x)))))
0¹(1(0(x))) → 5¹(0(0(4(2(x)))))
0¹(1(0(x))) → 0¹(0(4(2(x))))
0¹(1(0(x))) → 0¹(4(2(x)))
0¹(1(0(x))) → 3¹(0(0(4(1(4(x))))))
0¹(1(0(x))) → 0¹(0(4(1(4(x)))))
0¹(1(0(x))) → 0¹(4(1(4(x))))
0¹(1(0(x))) → 5¹(1(0(2(0(x)))))
0¹(1(0(x))) → 5¹(5(1(0(0(2(x))))))
0¹(1(0(x))) → 5¹(1(0(0(2(x)))))
0¹(0(1(0(x)))) → 0¹(0(2(0(x))))
0¹(0(1(0(x)))) → 0¹(2(0(x)))
0¹(0(1(0(x)))) → 0¹(1(5(0(0(2(x))))))
0¹(0(1(0(x)))) → 5¹(0(0(2(x))))
0¹(0(1(0(x)))) → 0¹(0(2(x)))
0¹(0(1(0(x)))) → 0¹(2(x))
0¹(1(0(3(x)))) → 0¹(3(3(0(2(x)))))
0¹(1(0(3(x)))) → 3¹(3(0(2(x))))
0¹(1(0(3(x)))) → 3¹(0(2(x)))
0¹(1(0(3(x)))) → 0¹(2(x))
0¹(1(0(3(x)))) → 0¹(5(3(2(0(x)))))
0¹(1(0(3(x)))) → 5¹(3(2(0(x))))
0¹(1(0(3(x)))) → 3¹(2(0(x)))
0¹(1(0(3(x)))) → 0¹(x)
0¹(1(1(0(x)))) → 0¹(4(4(1(1(0(x))))))
0¹(1(1(3(x)))) → 3¹(4(5(1(1(0(x))))))
0¹(1(1(3(x)))) → 5¹(1(1(0(x))))
0¹(1(1(3(x)))) → 0¹(x)
0¹(1(2(0(x)))) → 0¹(2(0(x)))
0¹(1(2(0(x)))) → 3¹(0(2(1(0(x)))))
0¹(1(2(0(x)))) → 0¹(2(1(0(x))))
0¹(1(2(0(x)))) → 0¹(0(2(x)))
0¹(1(2(0(x)))) → 0¹(2(x))
0¹(1(2(0(x)))) → 0¹(0(4(2(5(1(x))))))
0¹(1(2(0(x)))) → 0¹(4(2(5(1(x)))))
0¹(1(2(0(x)))) → 5¹(1(x))
0¹(1(2(0(x)))) → 0¹(4(0(x)))
0¹(1(2(0(x)))) → 3¹(0(2(1(0(4(x))))))
0¹(1(2(0(x)))) → 0¹(2(1(0(4(x)))))
0¹(1(2(0(x)))) → 0¹(4(x))
0¹(1(3(0(x)))) → 0¹(3(0(2(x))))
0¹(1(3(0(x)))) → 3¹(0(2(x)))
0¹(1(3(0(x)))) → 0¹(2(x))
0¹(1(4(0(x)))) → 0¹(3(4(2(1(0(x))))))
0¹(1(4(0(x)))) → 3¹(4(2(1(0(x)))))
0¹(1(5(0(x)))) → 0¹(5(1(4(0(x)))))
0¹(1(5(0(x)))) → 5¹(1(4(0(x))))
0¹(1(5(0(x)))) → 5¹(3(0(2(0(x)))))
0¹(1(5(0(x)))) → 3¹(0(2(0(x))))
0¹(1(5(0(x)))) → 0¹(2(0(x)))
0¹(3(1(0(x)))) → 3¹(0(5(0(x))))
0¹(3(1(0(x)))) → 0¹(5(0(x)))
0¹(3(1(0(x)))) → 5¹(0(x))
5¹(0(1(0(x)))) → 0¹(0(5(1(x))))
5¹(0(1(0(x)))) → 0¹(5(1(x)))
5¹(0(1(0(x)))) → 5¹(1(x))
5¹(0(1(0(x)))) → 0¹(0(4(5(x))))
5¹(0(1(0(x)))) → 0¹(4(5(x)))
5¹(0(1(0(x)))) → 5¹(x)
0¹(1(0(0(0(x))))) → 0¹(0(5(1(0(0(x))))))
0¹(1(0(0(0(x))))) → 0¹(5(1(0(0(x)))))
0¹(1(0(0(0(x))))) → 5¹(1(0(0(x))))
0¹(1(2(4(0(x))))) → 0¹(0(5(4(2(1(x))))))
0¹(1(2(4(0(x))))) → 0¹(5(4(2(1(x)))))
0¹(1(2(4(0(x))))) → 5¹(4(2(1(x))))
0¹(1(2(5(0(x))))) → 0¹(2(0(5(4(x)))))
0¹(1(2(5(0(x))))) → 0¹(5(4(x)))
0¹(1(2(5(0(x))))) → 5¹(4(x))
0¹(1(4(0(0(x))))) → 0¹(0(0(4(1(0(x))))))
0¹(1(4(0(0(x))))) → 0¹(0(4(1(0(x)))))
0¹(1(4(0(0(x))))) → 0¹(4(1(0(x))))
0¹(1(4(5(0(x))))) → 5¹(0(0(4(2(x)))))
0¹(1(4(5(0(x))))) → 0¹(0(4(2(x))))
0¹(1(4(5(0(x))))) → 0¹(4(2(x)))
0¹(3(0(1(0(x))))) → 0¹(3(0(0(2(1(x))))))
0¹(3(0(1(0(x))))) → 3¹(0(0(2(1(x)))))
0¹(3(0(1(0(x))))) → 0¹(0(2(1(x))))
0¹(3(0(1(0(x))))) → 0¹(2(1(x)))
3¹(0(3(1(0(x))))) → 0¹(1(3(2(3(0(x))))))
3¹(0(3(1(0(x))))) → 3¹(2(3(0(x))))
3¹(0(3(1(0(x))))) → 3¹(0(x))
5¹(0(1(2(0(x))))) → 0¹(0(5(2(1(0(x))))))
5¹(0(1(2(0(x))))) → 0¹(5(2(1(0(x)))))
5¹(0(1(2(0(x))))) → 5¹(2(1(0(x))))

The TRS R consists of the following rules:

0(1(0(x))) → 0(2(1(0(x))))
0(1(0(x))) → 0(0(2(1(0(x)))))
0(1(0(x))) → 0(0(2(1(2(x)))))
0(1(0(x))) → 0(2(1(0(2(x)))))
0(1(0(x))) → 0(3(2(1(0(x)))))
0(1(0(x))) → 1(0(0(0(2(x)))))
0(1(0(x))) → 1(0(0(2(0(x)))))
0(1(0(x))) → 1(0(4(2(0(x)))))
0(1(0(x))) → 1(4(0(4(0(x)))))
0(1(0(x))) → 4(0(0(2(1(x)))))
0(1(0(x))) → 5(0(0(4(1(x)))))
0(1(0(x))) → 5(1(0(4(0(x)))))
0(1(0(x))) → 0(2(1(0(3(2(x))))))
0(1(0(x))) → 0(4(0(4(1(3(x))))))
0(1(0(x))) → 0(4(2(2(1(0(x))))))
0(1(0(x))) → 0(5(2(1(2(0(x))))))
0(1(0(x))) → 0(5(2(5(1(0(x))))))
0(1(0(x))) → 1(0(0(5(4(4(x))))))
0(1(0(x))) → 1(0(4(4(4(0(x))))))
0(1(0(x))) → 1(5(0(0(4(2(x))))))
0(1(0(x))) → 3(0(0(4(1(4(x))))))
0(1(0(x))) → 4(5(1(0(2(0(x))))))
0(1(0(x))) → 5(5(1(0(0(2(x))))))
0(0(1(0(x)))) → 1(0(0(2(0(x)))))
0(0(1(0(x)))) → 0(1(5(0(0(2(x))))))
0(1(0(3(x)))) → 1(0(3(3(0(2(x))))))
0(1(0(3(x)))) → 1(0(5(3(2(0(x))))))
0(1(1(0(x)))) → 0(4(4(1(1(0(x))))))
0(1(1(3(x)))) → 3(4(5(1(1(0(x))))))
0(1(2(0(x)))) → 1(1(0(2(0(x)))))
0(1(2(0(x)))) → 3(0(2(1(0(x)))))
0(1(2(0(x)))) → 4(1(0(0(2(x)))))
0(1(2(0(x)))) → 0(0(4(2(5(1(x))))))
0(1(2(0(x)))) → 1(1(2(0(4(0(x))))))
0(1(2(0(x)))) → 3(0(2(1(0(4(x))))))
0(1(3(0(x)))) → 1(0(3(0(2(x)))))
0(1(4(0(x)))) → 0(3(4(2(1(0(x))))))
0(1(5(0(x)))) → 0(5(1(4(0(x)))))
0(1(5(0(x)))) → 1(5(3(0(2(0(x))))))
0(3(1(0(x)))) → 1(2(3(0(5(0(x))))))
5(0(1(0(x)))) → 1(4(0(0(5(1(x))))))
5(0(1(0(x)))) → 2(1(0(0(4(5(x))))))
0(1(0(0(0(x))))) → 0(0(5(1(0(0(x))))))
0(1(2(4(0(x))))) → 0(0(5(4(2(1(x))))))
0(1(2(5(0(x))))) → 1(0(2(0(5(4(x))))))
0(1(4(0(0(x))))) → 0(0(0(4(1(0(x))))))
0(1(4(5(0(x))))) → 1(5(0(0(4(2(x))))))
0(3(0(1(0(x))))) → 0(3(0(0(2(1(x))))))
3(0(3(1(0(x))))) → 0(1(3(2(3(0(x))))))
5(0(1(2(0(x))))) → 0(0(5(2(1(0(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 3 SCCs with 114 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

3¹(0(3(1(0(x))))) → 3¹(0(x))

The TRS R consists of the following rules:

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

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

(6) 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:

3¹(0(3(1(0(x))))) → 3¹(0(x))
The graph contains the following edges 1 > 1

(7) YES

(8) Obligation:

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

5¹(0(1(0(x)))) → 5¹(x)

The TRS R consists of the following rules:

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

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

(9) 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.

(10) Obligation:

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

5¹(0(1(0(x)))) → 5¹(x)

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

(11) 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:

5¹(0(1(0(x)))) → 5¹(x)
The graph contains the following edges 1 > 1

(12) YES

(13) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(14) QDPOrderProof (EQUIVALENT transformation)

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

The following pairs can be oriented strictly and are deleted.

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

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

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

(15) Obligation:

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

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

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

(16) 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) QDPSizeChangeProof (EQUIVALENT transformation)

(7) YES

(8) Obligation:

(9) UsableRulesProof (EQUIVALENT transformation)

(10) Obligation:

(11) QDPSizeChangeProof (EQUIVALENT transformation)

(12) YES

(13) Obligation:

(14) QDPOrderProof (EQUIVALENT transformation)

(15) Obligation:

(16) PisEmptyProof (EQUIVALENT transformation)

(17) YES