YES Termination w.r.t. Q proof of /home/cern_httpd/provide/research/cycsrs/tpdb/TPDB-d9b80194f163/SRS_Standard/ICFP_2010/212062.srs

(0) Obligation:

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

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

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

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

The TRS R consists of the following rules:

0(0(x)) → 2(0(1(0(x))))
0(0(x)) → 0(2(0(1(x))))
0(0(x)) → 1(0(1(0(1(x)))))
0(0(x)) → 0(2(1(0(1(x)))))
0(0(x)) → 3(0(2(0(1(x)))))
0(0(x)) → 0(2(2(0(1(x)))))
0(0(x)) → 0(2(0(1(2(x)))))
0(0(x)) → 2(1(2(0(1(0(x))))))
0(0(x)) → 2(2(0(1(0(1(x))))))
0(0(x)) → 1(0(3(1(0(1(x))))))
0(0(x)) → 2(0(1(4(0(1(x))))))
0(0(x)) → 0(2(0(1(1(1(x))))))
0(0(x)) → 2(2(0(4(0(3(x))))))
0(0(x)) → 4(0(1(0(1(3(x))))))
0(0(0(x))) → 4(0(4(0(1(0(x))))))
0(0(0(x))) → 2(0(1(0(0(3(x))))))
0(0(3(x))) → 3(0(2(0(3(x)))))
0(0(3(x))) → 2(0(4(2(0(3(x))))))
0(2(5(x))) → 5(3(2(0(x))))
0(2(5(x))) → 2(0(5(3(x))))
0(2(5(x))) → 5(3(3(2(0(x)))))
0(2(5(x))) → 5(3(2(0(1(x)))))
0(2(5(x))) → 4(2(0(1(5(x)))))
0(2(5(x))) → 2(2(2(1(0(5(x))))))
0(2(5(x))) → 2(0(1(5(3(5(x))))))
0(2(5(0(x)))) → 2(2(0(5(0(x)))))
0(0(4(3(x)))) → 5(4(0(3(3(0(x))))))
0(0(4(3(x)))) → 0(3(5(4(0(3(x))))))
0(0(1(5(x)))) → 5(1(0(1(3(0(x))))))
0(4(1(5(x)))) → 4(2(5(1(0(x)))))
0(5(1(5(x)))) → 5(3(0(1(5(x)))))
0(2(2(5(x)))) → 5(4(2(1(2(0(x))))))
0(2(3(5(x)))) → 2(1(0(3(5(x)))))
0(2(3(5(x)))) → 2(0(3(5(3(3(x))))))
0(0(4(5(x)))) → 2(0(5(5(4(0(x))))))
0(2(4(5(x)))) → 0(5(3(4(2(x)))))
0(2(4(5(x)))) → 4(2(2(0(5(x)))))
0(2(4(5(x)))) → 2(0(5(4(5(x)))))
0(2(5(3(0(x))))) → 0(3(5(2(0(3(x))))))
0(2(5(3(3(x))))) → 2(0(3(2(5(3(x))))))
0(2(5(3(3(x))))) → 2(0(5(3(3(4(x))))))
0(2(5(0(5(x))))) → 2(0(5(5(3(0(x))))))
0(0(4(1(5(x))))) → 4(1(0(5(2(0(x))))))
0(2(3(3(5(x))))) → 2(0(3(3(2(5(x))))))
0(0(3(4(5(x))))) → 5(3(0(4(0(1(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 49 less nodes.

(6) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(7) 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.

(8) Obligation:

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

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

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

(9) 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:

  • 01(0(4(3(x)))) → 01(x)
    The graph contains the following edges 1 > 1

  • 01(2(5(x))) → 01(x)
    The graph contains the following edges 1 > 1

  • 01(0(1(5(x)))) → 01(x)
    The graph contains the following edges 1 > 1

  • 01(4(1(5(x)))) → 01(x)
    The graph contains the following edges 1 > 1

  • 01(2(2(5(x)))) → 01(x)
    The graph contains the following edges 1 > 1

  • 01(0(4(5(x)))) → 01(x)
    The graph contains the following edges 1 > 1

  • 01(2(5(0(5(x))))) → 01(x)
    The graph contains the following edges 1 > 1

  • 01(0(4(1(5(x))))) → 01(x)
    The graph contains the following edges 1 > 1

(10) YES