YES Termination w.r.t. Q proof of /home/cern_httpd/provide/research/cycsrs/tpdb/TPDB-d9b80194f163/SRS_Standard/ICFP_2010/96485.srs

(0) Obligation:

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

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

Q is empty.

(1) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(0(x1)) = 235 + x1   
POL(1(x1)) = 191 + x1   
POL(2(x1)) = 157 + x1   
POL(3(x1)) = 381 + x1   
POL(4(x1)) = 237 + x1   
POL(5(x1)) = 136 + x1   
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

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


(2) Obligation:

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

4(5(1(4(x)))) → 1(5(4(4(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:

41(5(1(4(x)))) → 41(4(x))

The TRS R consists of the following rules:

4(5(1(4(x)))) → 1(5(4(4(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 0 SCCs with 1 less node.

(6) TRUE