YES Termination w.r.t. Q proof of /home/cern_httpd/provide/research/cycsrs/tpdb/TPDB-d9b80194f163/SRS_Standard/ICFP_2010/249459.srs

(0) Obligation:

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

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

Q is empty.

(1) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(0(x1)) = 4 + x1   
POL(1(x1)) = 3 + x1   
POL(2(x1)) = 18 + x1   
POL(3(x1)) = x1   
POL(4(x1)) = x1   
POL(5(x1)) = x1   
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

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


(2) Obligation:

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

3(4(5(x))) → 4(3(5(x)))

Q is empty.

(3) Overlay + Local Confluence (EQUIVALENT transformation)

The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.

(4) Obligation:

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

3(4(5(x))) → 4(3(5(x)))

The set Q consists of the following terms:

3(4(5(x0)))

(5) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(6) Obligation:

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

31(4(5(x))) → 31(5(x))

The TRS R consists of the following rules:

3(4(5(x))) → 4(3(5(x)))

The set Q consists of the following terms:

3(4(5(x0)))

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