YES Termination w.r.t. Q proof of /home/cern_httpd/provide/research/cycsrs/tpdb/TPDB-d9b80194f163/SRS_Standard/ICFP_2010/214011.srs

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

0 QTRS
↳1 DependencyPairsProof (⇔, 209 ms)
↳2 QDP
↳3 DependencyGraphProof (⇔, 0 ms)
↳4 AND
    ↳5 QDP
        ↳6 UsableRulesProof (⇔, 0 ms)
        ↳7 QDP
        ↳8 QDPOrderProof (⇔, 70 ms)
        ↳9 QDP
        ↳10 DependencyGraphProof (⇔, 0 ms)
        ↳11 QDP
        ↳12 QDPOrderProof (⇔, 25 ms)
        ↳13 QDP
        ↳14 UsableRulesProof (⇔, 0 ms)
        ↳15 QDP
        ↳16 QDPSizeChangeProof (⇔, 0 ms)
        ↳17 YES
    ↳18 QDP
        ↳19 UsableRulesProof (⇔, 0 ms)
        ↳20 QDP
        ↳21 DependencyGraphProof (⇔, 0 ms)
        ↳22 TRUE

(0) Obligation:

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

0(1(1(x))) → 0(2(1(1(x))))
0(1(1(x))) → 0(0(2(1(1(x)))))
0(1(1(x))) → 0(2(1(2(1(x)))))
0(1(1(x))) → 2(1(0(2(1(x)))))
3(0(1(x))) → 1(3(0(0(2(4(x))))))
3(5(1(x))) → 1(3(4(5(x))))
3(5(1(x))) → 2(4(5(3(1(x)))))
5(1(3(x))) → 5(3(1(2(x))))
0(1(0(1(x)))) → 1(1(0(0(2(4(x))))))
0(1(2(3(x)))) → 3(1(5(0(2(x)))))
0(1(2(3(x)))) → 0(3(1(2(1(1(x))))))
0(1(4(1(x)))) → 0(2(1(2(4(1(x))))))
0(1(4(1(x)))) → 4(0(0(2(1(1(x))))))
0(1(5(1(x)))) → 0(0(2(1(1(5(x))))))
0(4(5(1(x)))) → 0(1(3(4(5(x)))))
3(2(0(1(x)))) → 1(3(0(2(4(5(x))))))
3(5(1(1(x)))) → 1(3(4(5(1(x)))))
3(5(1(1(x)))) → 1(5(3(1(2(x)))))
3(5(1(3(x)))) → 3(5(3(1(2(x)))))
3(5(4(1(x)))) → 4(1(3(4(5(x)))))
5(1(2(3(x)))) → 5(5(3(1(2(x)))))
5(2(0(1(x)))) → 5(3(1(0(2(4(x))))))
5(4(3(3(x)))) → 3(1(3(4(5(x)))))
5(5(0(1(x)))) → 1(0(2(4(5(5(x))))))
0(1(2(0(1(x))))) → 0(3(0(2(1(1(x))))))
0(1(2(2(1(x))))) → 0(2(1(2(1(3(x))))))
0(3(0(5(1(x))))) → 0(3(4(5(0(1(x))))))
0(3(4(2(3(x))))) → 0(5(3(4(3(2(x))))))
0(3(5(4(1(x))))) → 0(5(2(4(3(1(x))))))
0(4(1(2(3(x))))) → 0(3(2(4(5(1(x))))))
0(4(1(2(3(x))))) → 4(3(1(0(0(2(x))))))
0(4(5(5(1(x))))) → 2(4(5(5(0(1(x))))))
0(5(1(0(1(x))))) → 0(1(5(5(0(1(x))))))
0(5(3(2(1(x))))) → 0(0(2(5(3(1(x))))))
3(0(1(2(3(x))))) → 0(2(3(4(3(1(x))))))
3(0(1(2(3(x))))) → 1(1(3(3(0(2(x))))))
3(0(1(2(3(x))))) → 1(2(3(3(0(2(x))))))
3(0(4(1(1(x))))) → 0(0(1(3(4(1(x))))))
3(2(4(1(3(x))))) → 4(3(4(3(1(2(x))))))
3(3(4(1(1(x))))) → 1(3(4(5(3(1(x))))))
3(3(5(1(1(x))))) → 3(1(4(5(3(1(x))))))
3(5(4(1(3(x))))) → 1(4(5(3(1(3(x))))))
3(5(4(4(1(x))))) → 4(1(4(3(4(5(x))))))
5(2(4(2(3(x))))) → 3(2(4(5(3(2(x))))))
5(4(2(0(1(x))))) → 5(1(2(0(2(4(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))) → 01(2(1(1(x))))
01(1(1(x))) → 01(0(2(1(1(x)))))
01(1(1(x))) → 01(2(1(2(1(x)))))
01(1(1(x))) → 01(2(1(x)))
31(0(1(x))) → 31(0(0(2(4(x)))))
31(0(1(x))) → 01(0(2(4(x))))
31(0(1(x))) → 01(2(4(x)))
31(5(1(x))) → 31(4(5(x)))
31(5(1(x))) → 51(x)
31(5(1(x))) → 51(3(1(x)))
31(5(1(x))) → 31(1(x))
51(1(3(x))) → 51(3(1(2(x))))
51(1(3(x))) → 31(1(2(x)))
01(1(0(1(x)))) → 01(0(2(4(x))))
01(1(0(1(x)))) → 01(2(4(x)))
01(1(2(3(x)))) → 31(1(5(0(2(x)))))
01(1(2(3(x)))) → 51(0(2(x)))
01(1(2(3(x)))) → 01(2(x))
01(1(2(3(x)))) → 01(3(1(2(1(1(x))))))
01(1(2(3(x)))) → 31(1(2(1(1(x)))))
01(1(4(1(x)))) → 01(2(1(2(4(1(x))))))
01(1(4(1(x)))) → 01(0(2(1(1(x)))))
01(1(4(1(x)))) → 01(2(1(1(x))))
01(1(5(1(x)))) → 01(0(2(1(1(5(x))))))
01(1(5(1(x)))) → 01(2(1(1(5(x)))))
01(1(5(1(x)))) → 51(x)
01(4(5(1(x)))) → 01(1(3(4(5(x)))))
01(4(5(1(x)))) → 31(4(5(x)))
01(4(5(1(x)))) → 51(x)
31(2(0(1(x)))) → 31(0(2(4(5(x)))))
31(2(0(1(x)))) → 01(2(4(5(x))))
31(2(0(1(x)))) → 51(x)
31(5(1(1(x)))) → 31(4(5(1(x))))
31(5(1(1(x)))) → 51(1(x))
31(5(1(1(x)))) → 51(3(1(2(x))))
31(5(1(1(x)))) → 31(1(2(x)))
31(5(1(3(x)))) → 31(5(3(1(2(x)))))
31(5(1(3(x)))) → 51(3(1(2(x))))
31(5(1(3(x)))) → 31(1(2(x)))
31(5(4(1(x)))) → 31(4(5(x)))
31(5(4(1(x)))) → 51(x)
51(1(2(3(x)))) → 51(5(3(1(2(x)))))
51(1(2(3(x)))) → 51(3(1(2(x))))
51(1(2(3(x)))) → 31(1(2(x)))
51(2(0(1(x)))) → 51(3(1(0(2(4(x))))))
51(2(0(1(x)))) → 31(1(0(2(4(x)))))
51(2(0(1(x)))) → 01(2(4(x)))
51(4(3(3(x)))) → 31(1(3(4(5(x)))))
51(4(3(3(x)))) → 31(4(5(x)))
51(4(3(3(x)))) → 51(x)
51(5(0(1(x)))) → 01(2(4(5(5(x)))))
51(5(0(1(x)))) → 51(5(x))
51(5(0(1(x)))) → 51(x)
01(1(2(0(1(x))))) → 01(3(0(2(1(1(x))))))
01(1(2(0(1(x))))) → 31(0(2(1(1(x)))))
01(1(2(0(1(x))))) → 01(2(1(1(x))))
01(1(2(2(1(x))))) → 01(2(1(2(1(3(x))))))
01(1(2(2(1(x))))) → 31(x)
01(3(0(5(1(x))))) → 01(3(4(5(0(1(x))))))
01(3(0(5(1(x))))) → 31(4(5(0(1(x)))))
01(3(0(5(1(x))))) → 51(0(1(x)))
01(3(0(5(1(x))))) → 01(1(x))
01(3(4(2(3(x))))) → 01(5(3(4(3(2(x))))))
01(3(4(2(3(x))))) → 51(3(4(3(2(x)))))
01(3(4(2(3(x))))) → 31(4(3(2(x))))
01(3(4(2(3(x))))) → 31(2(x))
01(3(5(4(1(x))))) → 01(5(2(4(3(1(x))))))
01(3(5(4(1(x))))) → 51(2(4(3(1(x)))))
01(3(5(4(1(x))))) → 31(1(x))
01(4(1(2(3(x))))) → 01(3(2(4(5(1(x))))))
01(4(1(2(3(x))))) → 31(2(4(5(1(x)))))
01(4(1(2(3(x))))) → 51(1(x))
01(4(1(2(3(x))))) → 31(1(0(0(2(x)))))
01(4(1(2(3(x))))) → 01(0(2(x)))
01(4(1(2(3(x))))) → 01(2(x))
01(4(5(5(1(x))))) → 51(5(0(1(x))))
01(4(5(5(1(x))))) → 51(0(1(x)))
01(4(5(5(1(x))))) → 01(1(x))
01(5(1(0(1(x))))) → 01(1(5(5(0(1(x))))))
01(5(1(0(1(x))))) → 51(5(0(1(x))))
01(5(1(0(1(x))))) → 51(0(1(x)))
01(5(3(2(1(x))))) → 01(0(2(5(3(1(x))))))
01(5(3(2(1(x))))) → 01(2(5(3(1(x)))))
01(5(3(2(1(x))))) → 51(3(1(x)))
01(5(3(2(1(x))))) → 31(1(x))
31(0(1(2(3(x))))) → 01(2(3(4(3(1(x))))))
31(0(1(2(3(x))))) → 31(4(3(1(x))))
31(0(1(2(3(x))))) → 31(1(x))
31(0(1(2(3(x))))) → 31(3(0(2(x))))
31(0(1(2(3(x))))) → 31(0(2(x)))
31(0(1(2(3(x))))) → 01(2(x))
31(0(4(1(1(x))))) → 01(0(1(3(4(1(x))))))
31(0(4(1(1(x))))) → 01(1(3(4(1(x)))))
31(0(4(1(1(x))))) → 31(4(1(x)))
31(2(4(1(3(x))))) → 31(4(3(1(2(x)))))
31(2(4(1(3(x))))) → 31(1(2(x)))
31(3(4(1(1(x))))) → 31(4(5(3(1(x)))))
31(3(4(1(1(x))))) → 51(3(1(x)))
31(3(4(1(1(x))))) → 31(1(x))
31(3(5(1(1(x))))) → 31(1(4(5(3(1(x))))))
31(3(5(1(1(x))))) → 51(3(1(x)))
31(3(5(1(1(x))))) → 31(1(x))
31(5(4(1(3(x))))) → 51(3(1(3(x))))
31(5(4(1(3(x))))) → 31(1(3(x)))
31(5(4(4(1(x))))) → 31(4(5(x)))
31(5(4(4(1(x))))) → 51(x)
51(2(4(2(3(x))))) → 31(2(4(5(3(2(x))))))
51(2(4(2(3(x))))) → 51(3(2(x)))
51(2(4(2(3(x))))) → 31(2(x))
51(4(2(0(1(x))))) → 51(1(2(0(2(4(x))))))
51(4(2(0(1(x))))) → 01(2(4(x)))

The TRS R consists of the following rules:

0(1(1(x))) → 0(2(1(1(x))))
0(1(1(x))) → 0(0(2(1(1(x)))))
0(1(1(x))) → 0(2(1(2(1(x)))))
0(1(1(x))) → 2(1(0(2(1(x)))))
3(0(1(x))) → 1(3(0(0(2(4(x))))))
3(5(1(x))) → 1(3(4(5(x))))
3(5(1(x))) → 2(4(5(3(1(x)))))
5(1(3(x))) → 5(3(1(2(x))))
0(1(0(1(x)))) → 1(1(0(0(2(4(x))))))
0(1(2(3(x)))) → 3(1(5(0(2(x)))))
0(1(2(3(x)))) → 0(3(1(2(1(1(x))))))
0(1(4(1(x)))) → 0(2(1(2(4(1(x))))))
0(1(4(1(x)))) → 4(0(0(2(1(1(x))))))
0(1(5(1(x)))) → 0(0(2(1(1(5(x))))))
0(4(5(1(x)))) → 0(1(3(4(5(x)))))
3(2(0(1(x)))) → 1(3(0(2(4(5(x))))))
3(5(1(1(x)))) → 1(3(4(5(1(x)))))
3(5(1(1(x)))) → 1(5(3(1(2(x)))))
3(5(1(3(x)))) → 3(5(3(1(2(x)))))
3(5(4(1(x)))) → 4(1(3(4(5(x)))))
5(1(2(3(x)))) → 5(5(3(1(2(x)))))
5(2(0(1(x)))) → 5(3(1(0(2(4(x))))))
5(4(3(3(x)))) → 3(1(3(4(5(x)))))
5(5(0(1(x)))) → 1(0(2(4(5(5(x))))))
0(1(2(0(1(x))))) → 0(3(0(2(1(1(x))))))
0(1(2(2(1(x))))) → 0(2(1(2(1(3(x))))))
0(3(0(5(1(x))))) → 0(3(4(5(0(1(x))))))
0(3(4(2(3(x))))) → 0(5(3(4(3(2(x))))))
0(3(5(4(1(x))))) → 0(5(2(4(3(1(x))))))
0(4(1(2(3(x))))) → 0(3(2(4(5(1(x))))))
0(4(1(2(3(x))))) → 4(3(1(0(0(2(x))))))
0(4(5(5(1(x))))) → 2(4(5(5(0(1(x))))))
0(5(1(0(1(x))))) → 0(1(5(5(0(1(x))))))
0(5(3(2(1(x))))) → 0(0(2(5(3(1(x))))))
3(0(1(2(3(x))))) → 0(2(3(4(3(1(x))))))
3(0(1(2(3(x))))) → 1(1(3(3(0(2(x))))))
3(0(1(2(3(x))))) → 1(2(3(3(0(2(x))))))
3(0(4(1(1(x))))) → 0(0(1(3(4(1(x))))))
3(2(4(1(3(x))))) → 4(3(4(3(1(2(x))))))
3(3(4(1(1(x))))) → 1(3(4(5(3(1(x))))))
3(3(5(1(1(x))))) → 3(1(4(5(3(1(x))))))
3(5(4(1(3(x))))) → 1(4(5(3(1(3(x))))))
3(5(4(4(1(x))))) → 4(1(4(3(4(5(x))))))
5(2(4(2(3(x))))) → 3(2(4(5(3(2(x))))))
5(4(2(0(1(x))))) → 5(1(2(0(2(4(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 104 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

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

The TRS R consists of the following rules:

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

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

The TRS R consists of the following rules:

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

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

(8) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


51(5(0(1(x)))) → 51(5(x))
51(5(0(1(x)))) → 51(x)
31(2(0(1(x)))) → 51(x)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(0(x1)) = 1 + x1   
POL(1(x1)) = x1   
POL(2(x1)) = x1   
POL(3(x1)) = x1   
POL(31(x1)) = 1 + x1   
POL(4(x1)) = x1   
POL(5(x1)) = x1   
POL(51(x1)) = 1 + x1   

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

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

(9) Obligation:

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

51(4(3(3(x)))) → 51(x)
51(2(4(2(3(x))))) → 51(3(2(x)))
51(2(4(2(3(x))))) → 31(2(x))

The TRS R consists of the following rules:

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

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

(10) DependencyGraphProof (EQUIVALENT transformation)

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

(11) Obligation:

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

51(2(4(2(3(x))))) → 51(3(2(x)))
51(4(3(3(x)))) → 51(x)

The TRS R consists of the following rules:

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

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

(12) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


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

POL(0(x1)) = 0   
POL(1(x1)) = 1   
POL(2(x1)) = 1 + x1   
POL(3(x1)) = x1   
POL(4(x1)) = x1   
POL(5(x1)) = 0   
POL(51(x1)) = x1   

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

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

(13) Obligation:

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

51(4(3(3(x)))) → 51(x)

The TRS R consists of the following rules:

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

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

(14) 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.

(15) Obligation:

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

51(4(3(3(x)))) → 51(x)

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

(16) 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:

  • 51(4(3(3(x)))) → 51(x)
    The graph contains the following edges 1 > 1

(17) YES

(18) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(19) 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.

(20) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(21) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.

(22) TRUE