YES Termination w.r.t. Q proof of /home/cern_httpd/provide/research/cycsrs/tpdb/TPDB-d9b80194f163/SRS_Standard/ICFP_2010/212693.srs

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

0 QTRS
↳1 QTRS Reverse (⇔, 0 ms)
↳2 QTRS
↳3 DependencyPairsProof (⇔, 37 ms)
↳4 QDP
↳5 DependencyGraphProof (⇔, 0 ms)
↳6 AND
    ↳7 QDP
        ↳8 QDPSizeChangeProof (⇔, 0 ms)
        ↳9 YES
    ↳10 QDP
        ↳11 UsableRulesProof (⇔, 0 ms)
        ↳12 QDP
        ↳13 DependencyGraphProof (⇔, 0 ms)
        ↳14 QDP
        ↳15 UsableRulesProof (⇔, 0 ms)
        ↳16 QDP
        ↳17 QDPSizeChangeProof (⇔, 0 ms)
        ↳18 YES

(0) Obligation:

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

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

The TRS R consists of the following rules:

2(1(0(x))) → 1(2(0(0(x))))
2(1(0(x))) → 3(1(2(0(x))))
2(1(0(x))) → 4(4(1(2(0(0(x))))))
1(3(0(x))) → 0(4(3(1(0(x)))))
1(3(0(x))) → 4(4(3(1(0(x)))))
1(3(0(x))) → 0(4(4(4(3(1(x))))))
2(3(0(x))) → 3(1(2(0(x))))
2(3(0(x))) → 4(3(2(0(x))))
2(3(0(x))) → 3(4(2(0(0(x)))))
2(3(0(x))) → 3(4(1(2(0(x)))))
2(3(0(x))) → 3(3(4(2(0(x)))))
2(3(0(x))) → 4(3(3(1(2(0(x))))))
2(3(0(x))) → 5(5(4(3(2(0(x))))))
2(3(0(x))) → 0(4(3(4(4(2(x))))))
1(4(0(x))) → 4(4(1(0(x))))
1(4(0(x))) → 4(1(2(0(x))))
2(4(0(x))) → 4(1(2(0(x))))
2(4(0(x))) → 4(3(2(0(x))))
2(4(0(x))) → 3(4(2(0(x))))
1(0(2(x))) → 1(2(0(5(x))))
1(3(2(x))) → 2(5(3(1(x))))
1(3(2(x))) → 5(3(1(2(0(x)))))
1(3(2(x))) → 2(5(3(4(1(x)))))
1(0(2(0(x)))) → 1(2(0(0(5(x)))))
1(1(3(0(x)))) → 4(3(1(4(1(0(x))))))
1(2(3(0(x)))) → 1(2(4(3(0(0(x))))))
2(2(3(0(x)))) → 2(2(0(4(3(1(x))))))
2(1(4(0(x)))) → 5(2(0(4(1(x)))))
2(3(4(0(x)))) → 0(0(4(4(3(2(x))))))
1(3(5(0(x)))) → 4(5(3(4(1(0(x))))))
1(3(5(0(x)))) → 0(4(3(5(1(0(x))))))
2(3(5(0(x)))) → 3(5(4(2(0(x)))))
2(3(5(0(x)))) → 3(3(5(2(0(x)))))
1(3(0(2(x)))) → 2(5(3(1(0(2(x))))))
1(4(0(2(x)))) → 5(4(1(0(2(x)))))
2(3(5(2(x)))) → 3(3(2(5(2(x)))))
2(4(5(2(x)))) → 4(2(5(2(0(x)))))
1(2(3(0(0(x))))) → 2(5(3(1(0(0(x))))))
2(3(0(1(0(x))))) → 0(2(3(4(1(0(x))))))
2(3(0(1(0(x))))) → 5(0(0(1(3(2(x))))))
1(5(2(3(0(x))))) → 3(3(1(5(2(0(x))))))
2(1(1(5(0(x))))) → 5(1(1(4(2(0(x))))))
2(2(1(5(0(x))))) → 2(1(2(5(2(0(x))))))
1(2(3(5(0(x))))) → 5(2(4(3(1(0(x))))))
2(3(5(5(0(x))))) → 5(3(1(5(2(0(x))))))
1(1(3(0(2(x))))) → 4(3(1(0(1(2(x))))))
1(3(0(2(2(x))))) → 2(5(2(0(3(1(x))))))
1(5(0(2(2(x))))) → 1(5(1(2(0(2(x))))))
1(4(5(5(2(x))))) → 4(3(1(2(5(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 2 SCCs with 62 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

11(3(2(x))) → 11(x)
11(3(0(x))) → 11(x)
11(1(3(0(2(x))))) → 11(2(x))
11(3(0(2(2(x))))) → 11(x)

The TRS R consists of the following rules:

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

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

(8) 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:

  • 11(3(2(x))) → 11(x)
    The graph contains the following edges 1 > 1

  • 11(3(0(x))) → 11(x)
    The graph contains the following edges 1 > 1

  • 11(1(3(0(2(x))))) → 11(2(x))
    The graph contains the following edges 1 > 1

  • 11(3(0(2(2(x))))) → 11(x)
    The graph contains the following edges 1 > 1

(9) YES

(10) Obligation:

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

21(3(4(0(x)))) → 21(x)
21(3(0(x))) → 21(x)
21(3(0(1(0(x))))) → 21(3(4(1(0(x)))))
21(3(0(1(0(x))))) → 21(x)

The TRS R consists of the following rules:

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

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

(11) 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.

(12) Obligation:

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

21(3(4(0(x)))) → 21(x)
21(3(0(x))) → 21(x)
21(3(0(1(0(x))))) → 21(3(4(1(0(x)))))
21(3(0(1(0(x))))) → 21(x)

The TRS R consists of the following rules:

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

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

(13) DependencyGraphProof (EQUIVALENT transformation)

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

(14) Obligation:

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

21(3(0(x))) → 21(x)
21(3(4(0(x)))) → 21(x)
21(3(0(1(0(x))))) → 21(x)

The TRS R consists of the following rules:

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

21(3(0(x))) → 21(x)
21(3(4(0(x)))) → 21(x)
21(3(0(1(0(x))))) → 21(x)

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

(17) 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:

  • 21(3(0(x))) → 21(x)
    The graph contains the following edges 1 > 1

  • 21(3(4(0(x)))) → 21(x)
    The graph contains the following edges 1 > 1

  • 21(3(0(1(0(x))))) → 21(x)
    The graph contains the following edges 1 > 1

(18) YES