YES Termination w.r.t. Q proof of /home/cern_httpd/provide/research/cycsrs/tpdb/TPDB-d9b80194f163/SRS_Standard/ICFP_2010/211857.srs

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

0 QTRS
↳1 DependencyPairsProof (⇔, 116 ms)
↳2 QDP
↳3 DependencyGraphProof (⇔, 0 ms)
↳4 QDP
↳5 QDPOrderProof (⇔, 227 ms)
↳6 QDP
↳7 DependencyGraphProof (⇔, 0 ms)
↳8 TRUE

(0) Obligation:

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

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

The TRS R consists of the following rules:

0(0(1(x))) → 0(1(2(0(x))))
0(0(1(x))) → 0(3(1(0(x))))
0(0(1(x))) → 1(0(4(0(x))))
0(0(1(x))) → 0(1(3(0(2(x)))))
0(0(1(x))) → 0(1(3(0(4(x)))))
0(0(1(x))) → 0(2(0(1(2(x)))))
0(0(1(x))) → 0(3(0(1(2(x)))))
0(0(1(x))) → 0(3(0(3(1(x)))))
0(0(1(x))) → 0(4(0(4(1(x)))))
0(0(1(x))) → 1(2(0(2(0(x)))))
0(0(1(x))) → 1(2(2(0(0(x)))))
0(0(1(x))) → 0(0(2(2(1(2(x))))))
0(0(1(x))) → 0(1(2(4(2(0(x))))))
0(0(1(x))) → 1(2(0(3(0(4(x))))))
0(1(1(x))) → 0(2(1(1(x))))
0(1(1(x))) → 0(3(1(1(x))))
0(1(1(x))) → 1(1(3(0(4(x)))))
0(1(1(x))) → 1(2(0(2(1(x)))))
0(1(1(x))) → 1(0(3(1(2(4(x))))))
0(1(1(x))) → 1(0(4(2(1(2(x))))))
0(1(1(x))) → 1(1(2(4(3(0(x))))))
0(1(1(x))) → 1(2(1(0(4(4(x))))))
0(1(1(x))) → 1(2(2(1(3(0(x))))))
0(5(1(x))) → 0(3(1(5(x))))
0(5(1(x))) → 0(4(5(1(x))))
0(5(1(x))) → 0(2(3(1(5(x)))))
0(5(1(x))) → 0(3(1(5(2(x)))))
0(5(1(x))) → 0(3(1(2(5(2(x))))))
5(0(1(x))) → 5(1(2(4(0(x)))))
5(0(1(x))) → 5(0(2(1(2(4(x))))))
5(0(1(x))) → 5(1(2(3(0(4(x))))))
0(0(1(5(x)))) → 0(4(1(0(5(x)))))
0(0(2(1(x)))) → 2(0(3(0(2(1(x))))))
0(0(2(1(x)))) → 2(3(0(2(0(1(x))))))
0(1(0(1(x)))) → 1(0(2(0(1(x)))))
0(1(1(1(x)))) → 1(1(3(1(0(x)))))
5(0(1(1(x)))) → 1(5(1(2(0(x)))))
5(3(0(1(x)))) → 5(1(2(3(0(x)))))
5(3(1(5(x)))) → 5(3(1(2(5(x)))))
5(3(2(1(x)))) → 1(2(3(5(2(x)))))
5(4(0(1(x)))) → 1(2(5(0(4(x)))))
0(0(5(1(5(x))))) → 1(2(5(5(0(0(x))))))
0(5(3(0(1(x))))) → 1(0(5(3(0(4(x))))))
0(5(3(4(1(x))))) → 1(0(3(5(4(5(x))))))
0(5(4(0(1(x))))) → 0(1(3(0(4(5(x))))))
5(4(2(1(1(x))))) → 5(4(1(2(1(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 55 less nodes.

(4) Obligation:

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

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

The TRS R consists of the following rules:

0(0(1(x))) → 0(1(2(0(x))))
0(0(1(x))) → 0(3(1(0(x))))
0(0(1(x))) → 1(0(4(0(x))))
0(0(1(x))) → 0(1(3(0(2(x)))))
0(0(1(x))) → 0(1(3(0(4(x)))))
0(0(1(x))) → 0(2(0(1(2(x)))))
0(0(1(x))) → 0(3(0(1(2(x)))))
0(0(1(x))) → 0(3(0(3(1(x)))))
0(0(1(x))) → 0(4(0(4(1(x)))))
0(0(1(x))) → 1(2(0(2(0(x)))))
0(0(1(x))) → 1(2(2(0(0(x)))))
0(0(1(x))) → 0(0(2(2(1(2(x))))))
0(0(1(x))) → 0(1(2(4(2(0(x))))))
0(0(1(x))) → 1(2(0(3(0(4(x))))))
0(1(1(x))) → 0(2(1(1(x))))
0(1(1(x))) → 0(3(1(1(x))))
0(1(1(x))) → 1(1(3(0(4(x)))))
0(1(1(x))) → 1(2(0(2(1(x)))))
0(1(1(x))) → 1(0(3(1(2(4(x))))))
0(1(1(x))) → 1(0(4(2(1(2(x))))))
0(1(1(x))) → 1(1(2(4(3(0(x))))))
0(1(1(x))) → 1(2(1(0(4(4(x))))))
0(1(1(x))) → 1(2(2(1(3(0(x))))))
0(5(1(x))) → 0(3(1(5(x))))
0(5(1(x))) → 0(4(5(1(x))))
0(5(1(x))) → 0(2(3(1(5(x)))))
0(5(1(x))) → 0(3(1(5(2(x)))))
0(5(1(x))) → 0(3(1(2(5(2(x))))))
5(0(1(x))) → 5(1(2(4(0(x)))))
5(0(1(x))) → 5(0(2(1(2(4(x))))))
5(0(1(x))) → 5(1(2(3(0(4(x))))))
0(0(1(5(x)))) → 0(4(1(0(5(x)))))
0(0(2(1(x)))) → 2(0(3(0(2(1(x))))))
0(0(2(1(x)))) → 2(3(0(2(0(1(x))))))
0(1(0(1(x)))) → 1(0(2(0(1(x)))))
0(1(1(1(x)))) → 1(1(3(1(0(x)))))
5(0(1(1(x)))) → 1(5(1(2(0(x)))))
5(3(0(1(x)))) → 5(1(2(3(0(x)))))
5(3(1(5(x)))) → 5(3(1(2(5(x)))))
5(3(2(1(x)))) → 1(2(3(5(2(x)))))
5(4(0(1(x)))) → 1(2(5(0(4(x)))))
0(0(5(1(5(x))))) → 1(2(5(5(0(0(x))))))
0(5(3(0(1(x))))) → 1(0(5(3(0(4(x))))))
0(5(3(4(1(x))))) → 1(0(3(5(4(5(x))))))
0(5(4(0(1(x))))) → 0(1(3(0(4(5(x))))))
5(4(2(1(1(x))))) → 5(4(1(2(1(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(0(1(x))) → 01(0(x))
01(0(1(x))) → 01(x)
01(1(1(x))) → 01(x)
01(5(1(x))) → 51(x)
51(0(1(x))) → 01(x)
01(0(1(5(x)))) → 01(5(x))
01(1(1(1(x)))) → 01(x)
01(0(5(1(5(x))))) → 51(5(0(0(x))))
51(0(1(1(x)))) → 01(x)
01(0(5(1(5(x))))) → 51(0(0(x)))
01(0(5(1(5(x))))) → 01(0(x))
01(0(5(1(5(x))))) → 01(x)
01(5(3(4(1(x))))) → 51(x)
51(3(0(1(x)))) → 01(x)
01(5(4(0(1(x))))) → 51(x)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(0(x1)) = x1   
POL(01(x1)) = x1   
POL(1(x1)) = 1 + x1   
POL(2(x1)) = x1   
POL(3(x1)) = x1   
POL(4(x1)) = x1   
POL(5(x1)) = x1   
POL(51(x1)) = x1   

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

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

(6) Obligation:

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

01(0(2(1(x)))) → 01(1(x))

The TRS R consists of the following rules:

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

(8) TRUE