YES Termination w.r.t. Q proof of /home/cern_httpd/provide/research/cycsrs/tpdb/TPDB-d9b80194f163/SRS_Standard/ICFP_2010/3336.srs

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

0 QTRS
↳1 DependencyPairsProof (⇔, 74 ms)
↳2 QDP
↳3 DependencyGraphProof (⇔, 0 ms)
↳4 QDP
↳5 QDPOrderProof (⇔, 207 ms)
↳6 QDP
↳7 DependencyGraphProof (⇔, 0 ms)
↳8 TRUE

(0) Obligation:

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

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

21(5(x)) → 11(3(3(0(1(0(x))))))
21(5(x)) → 31(3(0(1(0(x)))))
21(5(x)) → 31(0(1(0(x))))
21(5(x)) → 11(0(x))
21(5(x)) → 21(2(0(5(0(1(x))))))
21(5(x)) → 21(0(5(0(1(x)))))
21(5(x)) → 51(0(1(x)))
21(5(x)) → 11(x)
31(5(x)) → 11(3(2(0(0(1(x))))))
31(5(x)) → 31(2(0(0(1(x)))))
31(5(x)) → 21(0(0(1(x))))
31(5(x)) → 11(x)
31(5(x)) → 31(2(0(5(3(0(x))))))
31(5(x)) → 21(0(5(3(0(x)))))
31(5(x)) → 51(3(0(x)))
31(5(x)) → 31(0(x))
41(5(x)) → 21(2(1(3(2(1(x))))))
41(5(x)) → 21(1(3(2(1(x)))))
41(5(x)) → 11(3(2(1(x))))
41(5(x)) → 31(2(1(x)))
41(5(x)) → 21(1(x))
41(5(x)) → 11(x)
41(5(x)) → 31(2(0(5(0(0(x))))))
41(5(x)) → 21(0(5(0(0(x)))))
41(5(x)) → 51(0(0(x)))
11(2(5(x))) → 11(0(5(0(5(4(x))))))
11(2(5(x))) → 51(0(5(4(x))))
11(2(5(x))) → 51(4(x))
11(2(5(x))) → 41(x)
11(2(5(x))) → 11(2(2(1(0(1(x))))))
11(2(5(x))) → 21(2(1(0(1(x)))))
11(2(5(x))) → 21(1(0(1(x))))
11(2(5(x))) → 11(0(1(x)))
11(2(5(x))) → 11(x)
11(2(5(x))) → 21(0(1(3(1(0(x))))))
11(2(5(x))) → 11(3(1(0(x))))
11(2(5(x))) → 31(1(0(x)))
11(2(5(x))) → 11(0(x))
11(4(5(x))) → 11(2(4(0(2(1(x))))))
11(4(5(x))) → 21(4(0(2(1(x)))))
11(4(5(x))) → 41(0(2(1(x))))
11(4(5(x))) → 21(1(x))
11(4(5(x))) → 11(x)
21(5(1(x))) → 21(2(2(1(2(3(x))))))
21(5(1(x))) → 21(2(1(2(3(x)))))
21(5(1(x))) → 21(1(2(3(x))))
21(5(1(x))) → 11(2(3(x)))
21(5(1(x))) → 21(3(x))
21(5(1(x))) → 31(x)
21(5(2(x))) → 41(0(2(2(3(3(x))))))
21(5(2(x))) → 21(2(3(3(x))))
21(5(2(x))) → 21(3(3(x)))
21(5(2(x))) → 31(3(x))
21(5(2(x))) → 31(x)
21(5(3(x))) → 21(0(4(1(3(3(x))))))
21(5(3(x))) → 41(1(3(3(x))))
21(5(3(x))) → 11(3(3(x)))
21(5(3(x))) → 31(3(x))
21(5(4(x))) → 21(0(5(1(0(1(x))))))
21(5(4(x))) → 51(1(0(1(x))))
21(5(4(x))) → 11(0(1(x)))
21(5(4(x))) → 11(x)
31(2(5(x))) → 31(2(0(1(0(5(x))))))
31(2(5(x))) → 21(0(1(0(5(x)))))
31(2(5(x))) → 11(0(5(x)))
31(4(2(x))) → 31(4(0(2(2(2(x))))))
31(4(2(x))) → 41(0(2(2(2(x)))))
31(4(2(x))) → 21(2(2(x)))
31(4(2(x))) → 21(2(x))
31(5(1(x))) → 41(2(0(0(5(x)))))
31(5(1(x))) → 21(0(0(5(x))))
31(5(1(x))) → 51(x)
31(5(1(x))) → 41(2(2(3(4(x)))))
31(5(1(x))) → 21(2(3(4(x))))
31(5(1(x))) → 21(3(4(x)))
31(5(1(x))) → 31(4(x))
31(5(1(x))) → 41(x)
31(5(1(x))) → 21(1(4(1(0(1(x))))))
31(5(1(x))) → 11(4(1(0(1(x)))))
31(5(1(x))) → 41(1(0(1(x))))
31(5(1(x))) → 11(0(1(x)))
31(5(2(x))) → 41(3(2(2(2(x)))))
31(5(2(x))) → 31(2(2(2(x))))
31(5(2(x))) → 21(2(2(x)))
31(5(2(x))) → 21(2(x))
31(5(2(x))) → 21(0(2(2(3(0(x))))))
31(5(2(x))) → 21(2(3(0(x))))
31(5(2(x))) → 21(3(0(x)))
31(5(2(x))) → 31(0(x))
31(5(2(x))) → 21(3(3(2(1(2(x))))))
31(5(2(x))) → 31(3(2(1(2(x)))))
31(5(2(x))) → 31(2(1(2(x))))
31(5(2(x))) → 21(1(2(x)))
31(5(2(x))) → 11(2(x))
31(5(3(x))) → 21(4(3(3(0(x)))))
31(5(3(x))) → 41(3(3(0(x))))
31(5(3(x))) → 31(3(0(x)))
31(5(3(x))) → 31(0(x))
31(5(3(x))) → 51(4(3(3(0(x)))))
31(5(3(x))) → 21(3(4(0(4(2(x))))))
31(5(3(x))) → 31(4(0(4(2(x)))))
31(5(3(x))) → 41(0(4(2(x))))
31(5(3(x))) → 41(2(x))
31(5(3(x))) → 21(x)
31(5(4(x))) → 21(0(5(0(0(x)))))
31(5(4(x))) → 51(0(0(x)))
31(5(4(x))) → 51(0(0(1(2(x)))))
31(5(4(x))) → 11(2(x))
31(5(4(x))) → 21(x)
31(5(5(x))) → 51(4(1(0(5(x)))))
31(5(5(x))) → 41(1(0(5(x))))
31(5(5(x))) → 11(0(5(x)))
41(5(1(x))) → 21(1(0(5(3(3(x))))))
41(5(1(x))) → 11(0(5(3(3(x)))))
41(5(1(x))) → 51(3(3(x)))
41(5(1(x))) → 31(3(x))
41(5(1(x))) → 31(x)
41(5(2(x))) → 51(1(0(0(4(x)))))
41(5(2(x))) → 11(0(0(4(x))))
41(5(2(x))) → 41(x)
41(5(4(x))) → 21(2(1(0(4(2(x))))))
41(5(4(x))) → 21(1(0(4(2(x)))))
41(5(4(x))) → 11(0(4(2(x))))
41(5(4(x))) → 41(2(x))
41(5(4(x))) → 21(x)
41(5(4(x))) → 31(2(0(3(2(0(x))))))
41(5(4(x))) → 21(0(3(2(0(x)))))
41(5(4(x))) → 31(2(0(x)))
41(5(4(x))) → 21(0(x))
51(5(3(x))) → 51(1(0(1(2(2(x))))))
51(5(3(x))) → 11(0(1(2(2(x)))))
51(5(3(x))) → 11(2(2(x)))
51(5(3(x))) → 21(2(x))
51(5(3(x))) → 21(x)
51(5(4(x))) → 51(1(0(4(2(2(x))))))
51(5(4(x))) → 11(0(4(2(2(x)))))
51(5(4(x))) → 41(2(2(x)))
51(5(4(x))) → 21(2(x))
51(5(4(x))) → 21(x)

The TRS R consists of the following rules:

2(5(x)) → 1(3(3(0(1(0(x))))))
2(5(x)) → 2(2(0(5(0(1(x))))))
3(5(x)) → 1(3(2(0(0(1(x))))))
3(5(x)) → 3(2(0(5(3(0(x))))))
4(5(x)) → 2(2(1(3(2(1(x))))))
4(5(x)) → 3(2(0(5(0(0(x))))))
1(2(5(x))) → 1(0(5(0(5(4(x))))))
1(2(5(x))) → 1(2(2(1(0(1(x))))))
1(2(5(x))) → 2(0(1(3(1(0(x))))))
1(4(5(x))) → 1(2(4(0(2(1(x))))))
2(5(1(x))) → 2(2(2(1(2(3(x))))))
2(5(2(x))) → 4(0(2(2(3(3(x))))))
2(5(3(x))) → 2(0(4(1(3(3(x))))))
2(5(4(x))) → 2(0(5(1(0(1(x))))))
3(2(5(x))) → 3(2(0(1(0(5(x))))))
3(4(2(x))) → 3(4(0(2(2(2(x))))))
3(5(1(x))) → 0(4(2(0(0(5(x))))))
3(5(1(x))) → 0(4(2(2(3(4(x))))))
3(5(1(x))) → 2(1(4(1(0(1(x))))))
3(5(2(x))) → 0(4(3(2(2(2(x))))))
3(5(2(x))) → 2(0(2(2(3(0(x))))))
3(5(2(x))) → 2(3(3(2(1(2(x))))))
3(5(3(x))) → 0(2(4(3(3(0(x))))))
3(5(3(x))) → 0(5(4(3(3(0(x))))))
3(5(3(x))) → 2(3(4(0(4(2(x))))))
3(5(4(x))) → 0(2(0(5(0(0(x))))))
3(5(4(x))) → 0(5(0(0(1(2(x))))))
3(5(5(x))) → 0(5(4(1(0(5(x))))))
4(5(1(x))) → 2(1(0(5(3(3(x))))))
4(5(2(x))) → 0(5(1(0(0(4(x))))))
4(5(4(x))) → 2(2(1(0(4(2(x))))))
4(5(4(x))) → 3(2(0(3(2(0(x))))))
5(5(3(x))) → 5(1(0(1(2(2(x))))))
5(5(4(x))) → 5(1(0(4(2(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 79 less nodes.

(4) Obligation:

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

21(5(x)) → 11(x)
11(2(5(x))) → 51(4(x))
51(5(3(x))) → 11(2(2(x)))
11(2(5(x))) → 41(x)
41(5(x)) → 21(2(1(3(2(1(x))))))
21(5(1(x))) → 21(2(2(1(2(3(x))))))
21(5(1(x))) → 21(2(1(2(3(x)))))
21(5(1(x))) → 21(1(2(3(x))))
21(5(1(x))) → 11(2(3(x)))
11(2(5(x))) → 11(x)
11(4(5(x))) → 21(1(x))
21(5(1(x))) → 21(3(x))
21(5(1(x))) → 31(x)
31(5(x)) → 11(x)
11(4(5(x))) → 11(x)
31(4(2(x))) → 21(2(2(x)))
21(5(2(x))) → 21(2(3(3(x))))
21(5(2(x))) → 21(3(3(x)))
21(5(2(x))) → 31(3(x))
31(4(2(x))) → 21(2(x))
21(5(2(x))) → 31(x)
31(5(1(x))) → 51(x)
51(5(3(x))) → 21(2(x))
21(5(3(x))) → 41(1(3(3(x))))
41(5(x)) → 21(1(3(2(1(x)))))
21(5(3(x))) → 11(3(3(x)))
21(5(3(x))) → 31(3(x))
31(5(1(x))) → 41(2(2(3(4(x)))))
41(5(x)) → 11(3(2(1(x))))
41(5(x)) → 31(2(1(x)))
31(5(1(x))) → 21(2(3(4(x))))
21(5(4(x))) → 11(x)
31(5(1(x))) → 21(3(4(x)))
31(5(1(x))) → 31(4(x))
31(5(1(x))) → 41(x)
41(5(x)) → 21(1(x))
41(5(x)) → 11(x)
41(5(1(x))) → 51(3(3(x)))
51(5(3(x))) → 21(x)
51(5(4(x))) → 41(2(2(x)))
41(5(1(x))) → 31(3(x))
31(5(2(x))) → 41(3(2(2(2(x)))))
41(5(1(x))) → 31(x)
31(5(2(x))) → 31(2(2(2(x))))
31(5(2(x))) → 21(2(2(x)))
31(5(2(x))) → 21(2(x))
31(5(2(x))) → 21(3(3(2(1(2(x))))))
31(5(2(x))) → 31(3(2(1(2(x)))))
31(5(2(x))) → 31(2(1(2(x))))
31(5(2(x))) → 21(1(2(x)))
31(5(2(x))) → 11(2(x))
31(5(3(x))) → 41(2(x))
41(5(2(x))) → 41(x)
41(5(4(x))) → 41(2(x))
41(5(4(x))) → 21(x)
31(5(3(x))) → 21(x)
31(5(4(x))) → 11(2(x))
31(5(4(x))) → 21(x)
51(5(4(x))) → 21(2(x))
51(5(4(x))) → 21(x)

The TRS R consists of the following rules:

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


21(5(x)) → 11(x)
11(2(5(x))) → 51(4(x))
51(5(3(x))) → 11(2(2(x)))
11(2(5(x))) → 41(x)
41(5(x)) → 21(2(1(3(2(1(x))))))
21(5(1(x))) → 21(2(2(1(2(3(x))))))
21(5(1(x))) → 21(2(1(2(3(x)))))
21(5(1(x))) → 21(1(2(3(x))))
21(5(1(x))) → 11(2(3(x)))
11(2(5(x))) → 11(x)
11(4(5(x))) → 21(1(x))
21(5(1(x))) → 21(3(x))
21(5(1(x))) → 31(x)
31(5(x)) → 11(x)
11(4(5(x))) → 11(x)
21(5(2(x))) → 21(2(3(3(x))))
21(5(2(x))) → 21(3(3(x)))
21(5(2(x))) → 31(3(x))
21(5(2(x))) → 31(x)
31(5(1(x))) → 51(x)
51(5(3(x))) → 21(2(x))
21(5(3(x))) → 41(1(3(3(x))))
41(5(x)) → 21(1(3(2(1(x)))))
21(5(3(x))) → 11(3(3(x)))
21(5(3(x))) → 31(3(x))
31(5(1(x))) → 41(2(2(3(4(x)))))
41(5(x)) → 11(3(2(1(x))))
41(5(x)) → 31(2(1(x)))
31(5(1(x))) → 21(2(3(4(x))))
21(5(4(x))) → 11(x)
31(5(1(x))) → 21(3(4(x)))
31(5(1(x))) → 31(4(x))
31(5(1(x))) → 41(x)
41(5(x)) → 21(1(x))
41(5(x)) → 11(x)
41(5(1(x))) → 51(3(3(x)))
51(5(3(x))) → 21(x)
51(5(4(x))) → 41(2(2(x)))
41(5(1(x))) → 31(3(x))
31(5(2(x))) → 41(3(2(2(2(x)))))
41(5(1(x))) → 31(x)
31(5(2(x))) → 31(2(2(2(x))))
31(5(2(x))) → 21(2(2(x)))
31(5(2(x))) → 21(2(x))
31(5(2(x))) → 21(3(3(2(1(2(x))))))
31(5(2(x))) → 31(3(2(1(2(x)))))
31(5(2(x))) → 31(2(1(2(x))))
31(5(2(x))) → 21(1(2(x)))
31(5(2(x))) → 11(2(x))
31(5(3(x))) → 41(2(x))
41(5(2(x))) → 41(x)
41(5(4(x))) → 41(2(x))
41(5(4(x))) → 21(x)
31(5(3(x))) → 21(x)
31(5(4(x))) → 11(2(x))
31(5(4(x))) → 21(x)
51(5(4(x))) → 21(2(x))
51(5(4(x))) → 21(x)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(0(x1)) = 0   
POL(1(x1)) = x1   
POL(11(x1)) = x1   
POL(2(x1)) = x1   
POL(21(x1)) = x1   
POL(3(x1)) = x1   
POL(31(x1)) = x1   
POL(4(x1)) = x1   
POL(41(x1)) = x1   
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:

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

(6) Obligation:

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

31(4(2(x))) → 21(2(2(x)))
31(4(2(x))) → 21(2(x))

The TRS R consists of the following rules:

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

(8) TRUE