YES Termination w.r.t. Q proof of /home/cern_httpd/provide/research/cycsrs/tpdb/TPDB-d9b80194f163/SRS_Standard/ICFP_2010/214183.srs

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

0 QTRS
↳1 QTRS Reverse (⇔, 0 ms)
↳2 QTRS
↳3 DependencyPairsProof (⇔, 149 ms)
↳4 QDP
↳5 DependencyGraphProof (⇔, 0 ms)
↳6 QDP
↳7 QDPOrderProof (⇔, 236 ms)
↳8 QDP
↳9 DependencyGraphProof (⇔, 0 ms)
↳10 QDP
↳11 QDPOrderProof (⇔, 0 ms)
↳12 QDP
↳13 PisEmptyProof (⇔, 0 ms)
↳14 YES

(0) Obligation:

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

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

Q is empty.

(1) QTRS Reverse (EQUIVALENT transformation)

We applied the QTRS Reverse Processor [REVERSE].

(2) Obligation:

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

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

Q is empty.

(3) DependencyPairsProof (EQUIVALENT transformation)

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

(4) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 79 less nodes.

(6) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(7) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


21(4(1(0(x)))) → 21(x)
21(1(0(0(x)))) → 21(x)
21(4(1(0(x)))) → 41(x)
41(0(1(0(x)))) → 41(x)
41(0(1(0(x)))) → 41(0(5(x)))
41(0(1(0(x)))) → 51(x)
51(0(1(0(x)))) → 51(x)
51(0(5(0(x)))) → 51(x)
51(0(5(0(x)))) → 51(5(x))
51(0(2(1(0(x))))) → 21(5(0(4(x))))
51(0(2(1(0(x))))) → 51(0(4(x)))
51(0(2(1(0(x))))) → 41(x)
41(0(5(0(x)))) → 41(x)
41(0(5(0(x)))) → 51(x)
51(0(0(3(0(x))))) → 51(x)
41(2(4(1(0(x))))) → 41(x)
41(0(1(4(0(x))))) → 41(x)
41(0(1(5(0(x))))) → 41(2(5(x)))
41(0(1(5(0(x))))) → 21(5(x))
21(4(0(2(x)))) → 41(x)
41(0(1(5(0(x))))) → 51(x)
21(4(0(3(x)))) → 41(x)
21(3(1(0(0(x))))) → 51(2(x))
21(3(1(0(0(x))))) → 21(x)
21(4(3(1(0(x))))) → 21(4(x))
21(4(3(1(0(x))))) → 41(x)
21(2(3(4(0(x))))) → 21(4(x))
21(2(3(4(0(x))))) → 41(x)
21(4(1(5(0(x))))) → 41(x)
21(5(0(0(3(x))))) → 21(x)
21(4(5(0(3(x))))) → 41(x)
21(1(0(2(3(x))))) → 21(x)
21(5(0(2(3(x))))) → 21(x)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(0(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)) = 1 + x1   

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

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

(8) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(9) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 3 less nodes.

(10) Obligation:

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

21(5(0(4(2(x))))) → 21(4(0(x)))

The TRS R consists of the following rules:

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


21(5(0(4(2(x))))) → 21(4(0(x)))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(0(x1)) = x1   
POL(1(x1)) = 0   
POL(2(x1)) = 1 + x1   
POL(21(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:

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

(12) Obligation:

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

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

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

(13) PisEmptyProof (EQUIVALENT transformation)

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

(14) YES