YES Termination w.r.t. Q proof of /home/cern_httpd/provide/research/cycsrs/tpdb/TPDB-d9b80194f163/SRS_Standard/ICFP_2010/3746.srs

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

0 QTRS
↳1 QTRS Reverse (⇔, 0 ms)
↳2 QTRS
↳3 QTRSRRRProof (⇔, 78 ms)
↳4 QTRS
↳5 DependencyPairsProof (⇔, 17 ms)
↳6 QDP
↳7 DependencyGraphProof (⇔, 0 ms)
↳8 AND
    ↳9 QDP
        ↳10 QDPOrderProof (⇔, 66 ms)
        ↳11 QDP
        ↳12 PisEmptyProof (⇔, 0 ms)
        ↳13 YES
    ↳14 QDP
        ↳15 UsableRulesProof (⇔, 0 ms)
        ↳16 QDP
        ↳17 QDPSizeChangeProof (⇔, 0 ms)
        ↳18 YES
    ↳19 QDP
        ↳20 QDPSizeChangeProof (⇔, 0 ms)
        ↳21 YES

(0) Obligation:

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

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

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

Q is empty.

(3) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

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

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


(4) Obligation:

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

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

Q is empty.

(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:

01(5(2(0(x)))) → 31(1(4(1(2(5(1(1(2(0(x))))))))))
01(5(2(0(x)))) → 11(4(1(2(5(1(1(2(0(x)))))))))
01(5(2(0(x)))) → 41(1(2(5(1(1(2(0(x))))))))
01(5(2(0(x)))) → 11(2(5(1(1(2(0(x)))))))
01(5(2(0(x)))) → 21(5(1(1(2(0(x))))))
01(5(2(0(x)))) → 51(1(1(2(0(x)))))
01(5(2(0(x)))) → 11(1(2(0(x))))
01(5(2(0(x)))) → 11(2(0(x)))
31(5(0(0(3(x))))) → 31(3(1(4(2(5(3(4(3(1(x))))))))))
31(5(0(0(3(x))))) → 31(1(4(2(5(3(4(3(1(x)))))))))
31(5(0(0(3(x))))) → 11(4(2(5(3(4(3(1(x))))))))
31(5(0(0(3(x))))) → 41(2(5(3(4(3(1(x)))))))
31(5(0(0(3(x))))) → 21(5(3(4(3(1(x))))))
31(5(0(0(3(x))))) → 51(3(4(3(1(x)))))
31(5(0(0(3(x))))) → 31(4(3(1(x))))
31(5(0(0(3(x))))) → 41(3(1(x)))
31(5(0(0(3(x))))) → 31(1(x))
31(5(0(0(3(x))))) → 11(x)
51(2(1(3(5(0(x)))))) → 51(5(4(4(2(1(1(3(2(0(x))))))))))
51(2(1(3(5(0(x)))))) → 51(4(4(2(1(1(3(2(0(x)))))))))
51(2(1(3(5(0(x)))))) → 41(4(2(1(1(3(2(0(x))))))))
51(2(1(3(5(0(x)))))) → 41(2(1(1(3(2(0(x)))))))
51(2(1(3(5(0(x)))))) → 21(1(1(3(2(0(x))))))
51(2(1(3(5(0(x)))))) → 11(1(3(2(0(x)))))
51(2(1(3(5(0(x)))))) → 11(3(2(0(x))))
51(2(1(3(5(0(x)))))) → 31(2(0(x)))
51(2(1(3(5(0(x)))))) → 21(0(x))
51(5(2(0(2(5(x)))))) → 41(1(4(4(5(4(4(4(5(5(x))))))))))
51(5(2(0(2(5(x)))))) → 11(4(4(5(4(4(4(5(5(x)))))))))
51(5(2(0(2(5(x)))))) → 41(4(5(4(4(4(5(5(x))))))))
51(5(2(0(2(5(x)))))) → 41(5(4(4(4(5(5(x)))))))
51(5(2(0(2(5(x)))))) → 51(4(4(4(5(5(x))))))
51(5(2(0(2(5(x)))))) → 41(4(4(5(5(x)))))
51(5(2(0(2(5(x)))))) → 41(4(5(5(x))))
51(5(2(0(2(5(x)))))) → 41(5(5(x)))
51(5(2(0(2(5(x)))))) → 51(5(x))
11(5(0(0(4(4(0(x))))))) → 21(1(1(5(1(4(0(2(3(1(x))))))))))
11(5(0(0(4(4(0(x))))))) → 11(1(5(1(4(0(2(3(1(x)))))))))
11(5(0(0(4(4(0(x))))))) → 11(5(1(4(0(2(3(1(x))))))))
11(5(0(0(4(4(0(x))))))) → 51(1(4(0(2(3(1(x)))))))
11(5(0(0(4(4(0(x))))))) → 11(4(0(2(3(1(x))))))
11(5(0(0(4(4(0(x))))))) → 41(0(2(3(1(x)))))
11(5(0(0(4(4(0(x))))))) → 01(2(3(1(x))))
11(5(0(0(4(4(0(x))))))) → 21(3(1(x)))
11(5(0(0(4(4(0(x))))))) → 31(1(x))
11(5(0(0(4(4(0(x))))))) → 11(x)
41(3(3(3(5(5(1(x))))))) → 51(3(2(2(2(0(0(1(5(1(x))))))))))
41(3(3(3(5(5(1(x))))))) → 31(2(2(2(0(0(1(5(1(x)))))))))
41(3(3(3(5(5(1(x))))))) → 21(2(2(0(0(1(5(1(x))))))))
41(3(3(3(5(5(1(x))))))) → 21(2(0(0(1(5(1(x)))))))
41(3(3(3(5(5(1(x))))))) → 21(0(0(1(5(1(x))))))
41(3(3(3(5(5(1(x))))))) → 01(0(1(5(1(x)))))
41(3(3(3(5(5(1(x))))))) → 01(1(5(1(x))))
41(3(3(3(5(5(1(x))))))) → 11(5(1(x)))
21(5(1(5(3(2(3(x))))))) → 11(5(5(1(2(3(2(3(0(2(x))))))))))
21(5(1(5(3(2(3(x))))))) → 51(5(1(2(3(2(3(0(2(x)))))))))
21(5(1(5(3(2(3(x))))))) → 51(1(2(3(2(3(0(2(x))))))))
21(5(1(5(3(2(3(x))))))) → 11(2(3(2(3(0(2(x)))))))
21(5(1(5(3(2(3(x))))))) → 21(3(2(3(0(2(x))))))
21(5(1(5(3(2(3(x))))))) → 31(2(3(0(2(x)))))
21(5(1(5(3(2(3(x))))))) → 21(3(0(2(x))))
21(5(1(5(3(2(3(x))))))) → 31(0(2(x)))
21(5(1(5(3(2(3(x))))))) → 01(2(x))
21(5(1(5(3(2(3(x))))))) → 21(x)
51(0(1(3(5(0(5(x))))))) → 51(1(5(3(1(2(4(4(2(5(x))))))))))
51(0(1(3(5(0(5(x))))))) → 11(5(3(1(2(4(4(2(5(x)))))))))
51(0(1(3(5(0(5(x))))))) → 51(3(1(2(4(4(2(5(x))))))))
51(0(1(3(5(0(5(x))))))) → 31(1(2(4(4(2(5(x)))))))
51(0(1(3(5(0(5(x))))))) → 11(2(4(4(2(5(x))))))
51(0(1(3(5(0(5(x))))))) → 21(4(4(2(5(x)))))
51(0(1(3(5(0(5(x))))))) → 41(4(2(5(x))))
51(0(1(3(5(0(5(x))))))) → 41(2(5(x)))
51(0(1(3(5(0(5(x))))))) → 21(5(x))
31(3(5(2(0(2(5(x))))))) → 31(0(2(3(1(1(5(0(2(5(x))))))))))
31(3(5(2(0(2(5(x))))))) → 01(2(3(1(1(5(0(2(5(x)))))))))
31(3(5(2(0(2(5(x))))))) → 21(3(1(1(5(0(2(5(x))))))))
31(3(5(2(0(2(5(x))))))) → 31(1(1(5(0(2(5(x)))))))
31(3(5(2(0(2(5(x))))))) → 11(1(5(0(2(5(x))))))
31(3(5(2(0(2(5(x))))))) → 11(5(0(2(5(x)))))
31(3(5(2(0(2(5(x))))))) → 51(0(2(5(x))))

The TRS R consists of the following rules:

0(5(2(0(x)))) → 3(1(4(1(2(5(1(1(2(0(x))))))))))
3(5(0(0(3(x))))) → 3(3(1(4(2(5(3(4(3(1(x))))))))))
5(2(1(3(5(0(x)))))) → 5(5(4(4(2(1(1(3(2(0(x))))))))))
5(5(2(0(2(5(x)))))) → 4(1(4(4(5(4(4(4(5(5(x))))))))))
1(5(0(0(4(4(0(x))))))) → 2(1(1(5(1(4(0(2(3(1(x))))))))))
4(3(3(3(5(5(1(x))))))) → 5(3(2(2(2(0(0(1(5(1(x))))))))))
2(5(1(5(3(2(3(x))))))) → 1(5(5(1(2(3(2(3(0(2(x))))))))))
5(0(1(3(5(0(5(x))))))) → 5(1(5(3(1(2(4(4(2(5(x))))))))))
3(3(5(2(0(2(5(x))))))) → 3(0(2(3(1(1(5(0(2(5(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 3 SCCs with 75 less nodes.

(8) Complex Obligation (AND)

(9) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(10) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


21(5(1(5(3(2(3(x))))))) → 21(3(0(2(x))))
21(5(1(5(3(2(3(x))))))) → 21(3(2(3(0(2(x))))))
21(5(1(5(3(2(3(x))))))) → 21(x)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(0(x1)) = x1   
POL(1(x1)) = x1   
POL(2(x1)) = x1   
POL(21(x1)) = x1   
POL(3(x1)) = x1   
POL(4(x1)) = x1   
POL(5(x1)) = 1 + x1   

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

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

(11) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

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

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

(12) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(13) YES

(14) Obligation:

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

11(5(0(0(4(4(0(x))))))) → 11(x)

The TRS R consists of the following rules:

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

11(5(0(0(4(4(0(x))))))) → 11(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:

  • 11(5(0(0(4(4(0(x))))))) → 11(x)
    The graph contains the following edges 1 > 1

(18) YES

(19) Obligation:

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

51(5(2(0(2(5(x)))))) → 51(5(x))

The TRS R consists of the following rules:

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

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

(20) 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(5(2(0(2(5(x)))))) → 51(5(x))
    The graph contains the following edges 1 > 1

(21) YES