YES Termination w.r.t. Q proof of /home/cern_httpd/provide/research/cycsrs/tpdb/TPDB-d9b80194f163/SRS_Standard/ICFP_2010/213407.srs

(0) Obligation:

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

0(1(1(x))) → 1(0(2(1(x))))
0(1(1(x))) → 1(1(0(3(2(x)))))
0(1(1(x))) → 1(4(0(0(2(1(x))))))
0(1(4(x))) → 1(4(0(3(x))))
0(1(4(x))) → 4(0(2(1(x))))
0(1(4(x))) → 4(0(2(1(3(x)))))
0(4(1(x))) → 0(2(4(1(x))))
0(4(1(x))) → 4(0(2(1(x))))
0(4(1(x))) → 4(0(3(1(x))))
0(4(1(x))) → 1(4(4(0(2(x)))))
0(4(1(x))) → 2(1(4(0(2(x)))))
0(5(4(x))) → 1(4(0(0(2(5(x))))))
0(5(4(x))) → 4(0(2(5(2(5(x))))))
0(5(4(x))) → 5(0(4(0(3(3(x))))))
4(2(1(x))) → 4(0(2(1(x))))
4(2(1(x))) → 1(2(4(0(2(x)))))
0(1(0(4(x)))) → 0(0(2(2(1(4(x))))))
0(1(1(2(x)))) → 5(1(0(2(1(x)))))
0(1(1(3(x)))) → 1(1(5(0(3(x)))))
0(1(4(1(x)))) → 1(0(2(4(1(x)))))
0(1(4(3(x)))) → 1(0(2(4(3(x)))))
0(1(4(3(x)))) → 4(0(2(1(3(x)))))
0(4(1(2(x)))) → 2(0(3(1(4(x)))))
0(4(2(1(x)))) → 0(4(3(0(2(1(x))))))
0(5(0(4(x)))) → 1(5(4(0(0(2(x))))))
0(5(1(3(x)))) → 1(1(5(0(3(x)))))
0(5(1(3(x)))) → 5(3(1(0(3(x)))))
0(5(4(1(x)))) → 4(1(5(4(0(2(x))))))
0(5(4(3(x)))) → 0(2(5(0(3(4(x))))))
0(5(4(3(x)))) → 1(5(3(4(0(3(x))))))
1(0(5(4(x)))) → 1(4(5(0(3(3(x))))))
1(0(5(4(x)))) → 5(5(1(0(2(4(x))))))
1(4(2(1(x)))) → 4(0(2(1(2(1(x))))))
4(1(2(1(x)))) → 4(1(0(2(1(x)))))
4(1(2(1(x)))) → 3(4(0(2(1(1(x))))))
4(3(2(1(x)))) → 0(3(4(0(2(1(x))))))
0(0(1(2(3(x))))) → 1(0(3(0(0(2(x))))))
0(0(5(1(2(x))))) → 0(0(2(5(0(1(x))))))
0(0(5(1(3(x))))) → 4(5(0(0(3(1(x))))))
0(1(3(4(2(x))))) → 1(3(4(3(0(2(x))))))
0(4(5(3(4(x))))) → 0(3(2(5(4(4(x))))))
0(5(0(2(2(x))))) → 0(0(2(5(2(4(x))))))
0(5(1(1(3(x))))) → 0(3(2(1(1(5(x))))))
0(5(1(4(3(x))))) → 3(0(2(1(5(4(x))))))
0(5(5(1(2(x))))) → 1(5(3(0(2(5(x))))))
1(0(1(2(4(x))))) → 1(1(0(2(3(4(x))))))
1(4(2(1(2(x))))) → 0(2(2(1(1(4(x))))))
4(0(0(5(4(x))))) → 5(0(0(4(4(5(x))))))
4(2(5(4(1(x))))) → 4(4(1(5(3(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(1(x))) → 11(0(2(1(x))))
01(1(1(x))) → 01(2(1(x)))
01(1(1(x))) → 11(1(0(3(2(x)))))
01(1(1(x))) → 11(0(3(2(x))))
01(1(1(x))) → 01(3(2(x)))
01(1(1(x))) → 11(4(0(0(2(1(x))))))
01(1(1(x))) → 41(0(0(2(1(x)))))
01(1(1(x))) → 01(0(2(1(x))))
01(1(4(x))) → 11(4(0(3(x))))
01(1(4(x))) → 41(0(3(x)))
01(1(4(x))) → 01(3(x))
01(1(4(x))) → 41(0(2(1(x))))
01(1(4(x))) → 01(2(1(x)))
01(1(4(x))) → 11(x)
01(1(4(x))) → 41(0(2(1(3(x)))))
01(1(4(x))) → 01(2(1(3(x))))
01(1(4(x))) → 11(3(x))
01(4(1(x))) → 01(2(4(1(x))))
01(4(1(x))) → 41(0(2(1(x))))
01(4(1(x))) → 01(2(1(x)))
01(4(1(x))) → 41(0(3(1(x))))
01(4(1(x))) → 01(3(1(x)))
01(4(1(x))) → 11(4(4(0(2(x)))))
01(4(1(x))) → 41(4(0(2(x))))
01(4(1(x))) → 41(0(2(x)))
01(4(1(x))) → 01(2(x))
01(4(1(x))) → 11(4(0(2(x))))
01(5(4(x))) → 11(4(0(0(2(5(x))))))
01(5(4(x))) → 41(0(0(2(5(x)))))
01(5(4(x))) → 01(0(2(5(x))))
01(5(4(x))) → 01(2(5(x)))
01(5(4(x))) → 41(0(2(5(2(5(x))))))
01(5(4(x))) → 01(2(5(2(5(x)))))
01(5(4(x))) → 01(4(0(3(3(x)))))
01(5(4(x))) → 41(0(3(3(x))))
01(5(4(x))) → 01(3(3(x)))
41(2(1(x))) → 41(0(2(1(x))))
41(2(1(x))) → 01(2(1(x)))
41(2(1(x))) → 11(2(4(0(2(x)))))
41(2(1(x))) → 41(0(2(x)))
41(2(1(x))) → 01(2(x))
01(1(0(4(x)))) → 01(0(2(2(1(4(x))))))
01(1(0(4(x)))) → 01(2(2(1(4(x)))))
01(1(0(4(x)))) → 11(4(x))
01(1(1(2(x)))) → 11(0(2(1(x))))
01(1(1(2(x)))) → 01(2(1(x)))
01(1(1(2(x)))) → 11(x)
01(1(1(3(x)))) → 11(1(5(0(3(x)))))
01(1(1(3(x)))) → 11(5(0(3(x))))
01(1(1(3(x)))) → 01(3(x))
01(1(4(1(x)))) → 11(0(2(4(1(x)))))
01(1(4(1(x)))) → 01(2(4(1(x))))
01(1(4(3(x)))) → 11(0(2(4(3(x)))))
01(1(4(3(x)))) → 01(2(4(3(x))))
01(1(4(3(x)))) → 41(0(2(1(3(x)))))
01(1(4(3(x)))) → 01(2(1(3(x))))
01(1(4(3(x)))) → 11(3(x))
01(4(1(2(x)))) → 01(3(1(4(x))))
01(4(1(2(x)))) → 11(4(x))
01(4(1(2(x)))) → 41(x)
01(4(2(1(x)))) → 01(4(3(0(2(1(x))))))
01(4(2(1(x)))) → 41(3(0(2(1(x)))))
01(4(2(1(x)))) → 01(2(1(x)))
01(5(0(4(x)))) → 11(5(4(0(0(2(x))))))
01(5(0(4(x)))) → 41(0(0(2(x))))
01(5(0(4(x)))) → 01(0(2(x)))
01(5(0(4(x)))) → 01(2(x))
01(5(1(3(x)))) → 11(1(5(0(3(x)))))
01(5(1(3(x)))) → 11(5(0(3(x))))
01(5(1(3(x)))) → 01(3(x))
01(5(1(3(x)))) → 11(0(3(x)))
01(5(4(1(x)))) → 41(1(5(4(0(2(x))))))
01(5(4(1(x)))) → 11(5(4(0(2(x)))))
01(5(4(1(x)))) → 41(0(2(x)))
01(5(4(1(x)))) → 01(2(x))
01(5(4(3(x)))) → 01(2(5(0(3(4(x))))))
01(5(4(3(x)))) → 01(3(4(x)))
01(5(4(3(x)))) → 41(x)
01(5(4(3(x)))) → 11(5(3(4(0(3(x))))))
01(5(4(3(x)))) → 41(0(3(x)))
01(5(4(3(x)))) → 01(3(x))
11(0(5(4(x)))) → 11(4(5(0(3(3(x))))))
11(0(5(4(x)))) → 41(5(0(3(3(x)))))
11(0(5(4(x)))) → 01(3(3(x)))
11(0(5(4(x)))) → 11(0(2(4(x))))
11(0(5(4(x)))) → 01(2(4(x)))
11(4(2(1(x)))) → 41(0(2(1(2(1(x))))))
11(4(2(1(x)))) → 01(2(1(2(1(x)))))
11(4(2(1(x)))) → 11(2(1(x)))
41(1(2(1(x)))) → 41(1(0(2(1(x)))))
41(1(2(1(x)))) → 11(0(2(1(x))))
41(1(2(1(x)))) → 01(2(1(x)))
41(1(2(1(x)))) → 41(0(2(1(1(x)))))
41(1(2(1(x)))) → 01(2(1(1(x))))
41(1(2(1(x)))) → 11(1(x))
41(3(2(1(x)))) → 01(3(4(0(2(1(x))))))
41(3(2(1(x)))) → 41(0(2(1(x))))
41(3(2(1(x)))) → 01(2(1(x)))
01(0(1(2(3(x))))) → 11(0(3(0(0(2(x))))))
01(0(1(2(3(x))))) → 01(3(0(0(2(x)))))
01(0(1(2(3(x))))) → 01(0(2(x)))
01(0(1(2(3(x))))) → 01(2(x))
01(0(5(1(2(x))))) → 01(0(2(5(0(1(x))))))
01(0(5(1(2(x))))) → 01(2(5(0(1(x)))))
01(0(5(1(2(x))))) → 01(1(x))
01(0(5(1(2(x))))) → 11(x)
01(0(5(1(3(x))))) → 41(5(0(0(3(1(x))))))
01(0(5(1(3(x))))) → 01(0(3(1(x))))
01(0(5(1(3(x))))) → 01(3(1(x)))
01(0(5(1(3(x))))) → 11(x)
01(1(3(4(2(x))))) → 11(3(4(3(0(2(x))))))
01(1(3(4(2(x))))) → 41(3(0(2(x))))
01(1(3(4(2(x))))) → 01(2(x))
01(4(5(3(4(x))))) → 01(3(2(5(4(4(x))))))
01(4(5(3(4(x))))) → 41(4(x))
01(5(0(2(2(x))))) → 01(0(2(5(2(4(x))))))
01(5(0(2(2(x))))) → 01(2(5(2(4(x)))))
01(5(0(2(2(x))))) → 41(x)
01(5(1(1(3(x))))) → 01(3(2(1(1(5(x))))))
01(5(1(1(3(x))))) → 11(1(5(x)))
01(5(1(1(3(x))))) → 11(5(x))
01(5(1(4(3(x))))) → 01(2(1(5(4(x)))))
01(5(1(4(3(x))))) → 11(5(4(x)))
01(5(1(4(3(x))))) → 41(x)
01(5(5(1(2(x))))) → 11(5(3(0(2(5(x))))))
01(5(5(1(2(x))))) → 01(2(5(x)))
11(0(1(2(4(x))))) → 11(1(0(2(3(4(x))))))
11(0(1(2(4(x))))) → 11(0(2(3(4(x)))))
11(0(1(2(4(x))))) → 01(2(3(4(x))))
11(4(2(1(2(x))))) → 01(2(2(1(1(4(x))))))
11(4(2(1(2(x))))) → 11(1(4(x)))
11(4(2(1(2(x))))) → 11(4(x))
11(4(2(1(2(x))))) → 41(x)
41(0(0(5(4(x))))) → 01(0(4(4(5(x)))))
41(0(0(5(4(x))))) → 01(4(4(5(x))))
41(0(0(5(4(x))))) → 41(4(5(x)))
41(0(0(5(4(x))))) → 41(5(x))
41(2(5(4(1(x))))) → 41(4(1(5(3(2(x))))))
41(2(5(4(1(x))))) → 41(1(5(3(2(x)))))
41(2(5(4(1(x))))) → 11(5(3(2(x))))

The TRS R consists of the following rules:

0(1(1(x))) → 1(0(2(1(x))))
0(1(1(x))) → 1(1(0(3(2(x)))))
0(1(1(x))) → 1(4(0(0(2(1(x))))))
0(1(4(x))) → 1(4(0(3(x))))
0(1(4(x))) → 4(0(2(1(x))))
0(1(4(x))) → 4(0(2(1(3(x)))))
0(4(1(x))) → 0(2(4(1(x))))
0(4(1(x))) → 4(0(2(1(x))))
0(4(1(x))) → 4(0(3(1(x))))
0(4(1(x))) → 1(4(4(0(2(x)))))
0(4(1(x))) → 2(1(4(0(2(x)))))
0(5(4(x))) → 1(4(0(0(2(5(x))))))
0(5(4(x))) → 4(0(2(5(2(5(x))))))
0(5(4(x))) → 5(0(4(0(3(3(x))))))
4(2(1(x))) → 4(0(2(1(x))))
4(2(1(x))) → 1(2(4(0(2(x)))))
0(1(0(4(x)))) → 0(0(2(2(1(4(x))))))
0(1(1(2(x)))) → 5(1(0(2(1(x)))))
0(1(1(3(x)))) → 1(1(5(0(3(x)))))
0(1(4(1(x)))) → 1(0(2(4(1(x)))))
0(1(4(3(x)))) → 1(0(2(4(3(x)))))
0(1(4(3(x)))) → 4(0(2(1(3(x)))))
0(4(1(2(x)))) → 2(0(3(1(4(x)))))
0(4(2(1(x)))) → 0(4(3(0(2(1(x))))))
0(5(0(4(x)))) → 1(5(4(0(0(2(x))))))
0(5(1(3(x)))) → 1(1(5(0(3(x)))))
0(5(1(3(x)))) → 5(3(1(0(3(x)))))
0(5(4(1(x)))) → 4(1(5(4(0(2(x))))))
0(5(4(3(x)))) → 0(2(5(0(3(4(x))))))
0(5(4(3(x)))) → 1(5(3(4(0(3(x))))))
1(0(5(4(x)))) → 1(4(5(0(3(3(x))))))
1(0(5(4(x)))) → 5(5(1(0(2(4(x))))))
1(4(2(1(x)))) → 4(0(2(1(2(1(x))))))
4(1(2(1(x)))) → 4(1(0(2(1(x)))))
4(1(2(1(x)))) → 3(4(0(2(1(1(x))))))
4(3(2(1(x)))) → 0(3(4(0(2(1(x))))))
0(0(1(2(3(x))))) → 1(0(3(0(0(2(x))))))
0(0(5(1(2(x))))) → 0(0(2(5(0(1(x))))))
0(0(5(1(3(x))))) → 4(5(0(0(3(1(x))))))
0(1(3(4(2(x))))) → 1(3(4(3(0(2(x))))))
0(4(5(3(4(x))))) → 0(3(2(5(4(4(x))))))
0(5(0(2(2(x))))) → 0(0(2(5(2(4(x))))))
0(5(1(1(3(x))))) → 0(3(2(1(1(5(x))))))
0(5(1(4(3(x))))) → 3(0(2(1(5(4(x))))))
0(5(5(1(2(x))))) → 1(5(3(0(2(5(x))))))
1(0(1(2(4(x))))) → 1(1(0(2(3(4(x))))))
1(4(2(1(2(x))))) → 0(2(2(1(1(4(x))))))
4(0(0(5(4(x))))) → 5(0(0(4(4(5(x))))))
4(2(5(4(1(x))))) → 4(4(1(5(3(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 2 SCCs with 135 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

41(1(2(1(x)))) → 11(1(x))
11(4(2(1(2(x))))) → 11(1(4(x)))
11(4(2(1(2(x))))) → 11(4(x))
11(4(2(1(2(x))))) → 41(x)

The TRS R consists of the following rules:

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

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

(6) 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.

(7) Obligation:

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

41(1(2(1(x)))) → 11(1(x))
11(4(2(1(2(x))))) → 11(1(4(x)))
11(4(2(1(2(x))))) → 11(4(x))
11(4(2(1(2(x))))) → 41(x)

The TRS R consists of the following rules:

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

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

(8) DependencyGraphProof (EQUIVALENT transformation)

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

(9) Obligation:

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

11(4(2(1(2(x))))) → 11(4(x))

The TRS R consists of the following rules:

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


11(4(2(1(2(x))))) → 11(4(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)) = 1 + x1   
POL(3(x1)) = 0   
POL(4(x1)) = x1   
POL(5(x1)) = 0   

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

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

(11) Obligation:

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

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

01(0(5(1(2(x))))) → 01(1(x))

The TRS R consists of the following rules:

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

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

(15) 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.

(16) Obligation:

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

01(0(5(1(2(x))))) → 01(1(x))

The TRS R consists of the following rules:

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

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

(17) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


01(0(5(1(2(x))))) → 01(1(x))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(0(x1)) = 1 + x1   
POL(01(x1)) = x1   
POL(1(x1)) = 1   
POL(2(x1)) = 0   
POL(3(x1)) = 0   
POL(4(x1)) = 0   
POL(5(x1)) = x1   

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

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

(18) Obligation:

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

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

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

(19) PisEmptyProof (EQUIVALENT transformation)

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

(20) YES