YES Termination w.r.t. Q proof of /home/cern_httpd/provide/research/cycsrs/tpdb/TPDB-d9b80194f163/SRS_Standard/ICFP_2010/213281.srs

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

0 QTRS
↳1 DependencyPairsProof (⇔, 135 ms)
↳2 QDP
↳3 DependencyGraphProof (⇔, 1 ms)
↳4 QDP
↳5 QDPOrderProof (⇔, 181 ms)
↳6 QDP
↳7 DependencyGraphProof (⇔, 0 ms)
↳8 QDP
↳9 UsableRulesProof (⇔, 0 ms)
↳10 QDP
↳11 QDPSizeChangeProof (⇔, 1 ms)
↳12 YES

(0) Obligation:

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

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

The TRS R consists of the following rules:

0(1(0(2(x)))) → 0(0(3(1(2(x)))))
0(1(3(4(x)))) → 0(4(1(0(3(x)))))
0(1(3(4(x)))) → 0(4(1(1(3(x)))))
0(1(3(4(x)))) → 0(4(1(3(1(x)))))
0(2(1(4(x)))) → 0(4(1(2(3(x)))))
0(2(1(4(x)))) → 0(4(1(3(2(x)))))
0(2(1(4(x)))) → 2(0(4(1(4(x)))))
0(2(1(4(x)))) → 5(5(0(4(1(2(x))))))
0(2(1(5(x)))) → 5(0(4(1(2(x)))))
0(2(2(4(x)))) → 0(4(2(2(5(x)))))
0(2(2(4(x)))) → 0(4(2(5(2(x)))))
3(4(0(2(x)))) → 3(0(4(5(2(x)))))
3(4(0(2(x)))) → 3(5(0(4(2(x)))))
0(0(1(4(5(x))))) → 0(4(1(0(3(5(x))))))
0(1(0(2(4(x))))) → 2(0(0(4(1(1(x))))))
0(1(2(3(4(x))))) → 2(0(4(1(0(3(x))))))
0(1(3(3(4(x))))) → 0(0(3(1(3(4(x))))))
0(1(4(0(2(x))))) → 0(4(1(5(0(2(x))))))
0(1(4(1(5(x))))) → 2(5(0(4(1(1(x))))))
0(1(4(3(4(x))))) → 0(4(0(3(1(4(x))))))
0(1(4(3(4(x))))) → 3(0(4(1(5(4(x))))))
0(1(4(3(5(x))))) → 5(4(5(0(3(1(x))))))
0(1(5(0(2(x))))) → 0(0(4(1(2(5(x))))))
0(1(5(1(4(x))))) → 4(5(0(3(1(1(x))))))
0(2(1(4(4(x))))) → 0(4(1(2(4(3(x))))))
0(2(1(4(5(x))))) → 0(4(1(2(5(2(x))))))
0(2(1(5(4(x))))) → 5(0(2(0(4(1(x))))))
0(2(4(1(5(x))))) → 5(0(4(1(5(2(x))))))
0(2(4(3(5(x))))) → 0(4(5(2(5(3(x))))))
0(2(5(1(4(x))))) → 0(0(5(4(1(2(x))))))
3(0(1(3(2(x))))) → 0(3(1(0(3(2(x))))))
3(0(2(1(4(x))))) → 4(0(4(1(3(2(x))))))
3(0(2(1(5(x))))) → 5(3(2(0(4(1(x))))))
3(0(4(0(2(x))))) → 0(3(4(0(4(2(x))))))
3(0(4(0(2(x))))) → 0(4(1(2(0(3(x))))))
3(0(5(1(4(x))))) → 3(0(4(1(1(5(x))))))
3(0(5(1(5(x))))) → 0(4(1(3(5(5(x))))))
3(2(4(1(2(x))))) → 3(1(2(2(5(4(x))))))
3(2(4(1(5(x))))) → 3(1(4(5(2(5(x))))))
3(4(0(1(2(x))))) → 0(4(2(0(3(1(x))))))
3(4(0(1(4(x))))) → 0(4(1(5(3(4(x))))))
3(4(0(1(5(x))))) → 0(4(1(5(5(3(x))))))
3(4(0(2(4(x))))) → 0(3(4(0(4(2(x))))))
3(4(1(2(4(x))))) → 0(4(1(2(4(3(x))))))
3(4(1(3(5(x))))) → 4(3(0(3(1(5(x))))))
3(4(3(0(2(x))))) → 3(3(0(4(1(2(x))))))
3(4(5(0(2(x))))) → 0(3(0(4(2(5(x))))))
3(5(0(2(2(x))))) → 0(3(2(5(2(5(x))))))
3(5(2(1(4(x))))) → 3(5(1(0(4(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 1 SCC with 93 less nodes.

(4) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(5) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


01(2(4(3(5(x))))) → 31(x)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(0(x1)) = 0   
POL(01(x1)) = x1   
POL(1(x1)) = 0   
POL(2(x1)) = 1   
POL(3(x1)) = 0   
POL(31(x1)) = 0   
POL(4(x1)) = 0   
POL(5(x1)) = 0   

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

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

(6) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(7) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes.

(8) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(9) 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.

(10) Obligation:

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

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

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

(11) 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:

  • 31(0(4(0(2(x))))) → 31(x)
    The graph contains the following edges 1 > 1

  • 31(4(0(1(4(x))))) → 31(4(x))
    The graph contains the following edges 1 > 1

  • 31(4(0(1(5(x))))) → 31(x)
    The graph contains the following edges 1 > 1

  • 31(4(1(2(4(x))))) → 31(x)
    The graph contains the following edges 1 > 1

(12) YES