YES Termination w.r.t. Q proof of /home/cern_httpd/provide/research/cycsrs/tpdb/TPDB-d9b80194f163/SRS_Standard/ICFP_2010/212117.srs

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

0 QTRS
↳1 QTRS Reverse (⇔, 0 ms)
↳2 QTRS
↳3 DependencyPairsProof (⇔, 164 ms)
↳4 QDP
↳5 DependencyGraphProof (⇔, 0 ms)
↳6 AND
    ↳7 QDP
        ↳8 UsableRulesProof (⇔, 0 ms)
        ↳9 QDP
        ↳10 DependencyGraphProof (⇔, 0 ms)
        ↳11 QDP
        ↳12 QDPOrderProof (⇔, 258 ms)
        ↳13 QDP
        ↳14 QDPOrderProof (⇔, 32 ms)
        ↳15 QDP
        ↳16 PisEmptyProof (⇔, 0 ms)
        ↳17 YES
    ↳18 QDP
        ↳19 UsableRulesProof (⇔, 0 ms)
        ↳20 QDP
        ↳21 QDPSizeChangeProof (⇔, 0 ms)
        ↳22 YES

(0) Obligation:

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

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

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

Q is empty.

(3) DependencyPairsProof (EQUIVALENT transformation)

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

(4) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 143 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

11(1(0(x))) → 21(1(0(x)))
21(4(2(3(5(x))))) → 11(2(x))
11(1(0(x))) → 11(x)
11(1(0(x))) → 11(5(x))
11(2(5(x))) → 21(2(1(x)))
21(4(2(3(5(x))))) → 21(x)
11(2(5(x))) → 21(1(x))
11(2(5(x))) → 11(x)
11(2(4(0(x)))) → 21(1(4(0(x))))
11(2(4(0(x)))) → 11(4(0(x)))
11(2(4(0(1(x))))) → 21(1(x))
11(0(5(2(1(x))))) → 21(x)

The TRS R consists of the following rules:

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

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

(8) 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.

(9) Obligation:

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

11(1(0(x))) → 21(1(0(x)))
21(4(2(3(5(x))))) → 11(2(x))
11(1(0(x))) → 11(x)
11(1(0(x))) → 11(5(x))
11(2(5(x))) → 21(2(1(x)))
21(4(2(3(5(x))))) → 21(x)
11(2(5(x))) → 21(1(x))
11(2(5(x))) → 11(x)
11(2(4(0(x)))) → 21(1(4(0(x))))
11(2(4(0(x)))) → 11(4(0(x)))
11(2(4(0(1(x))))) → 21(1(x))
11(0(5(2(1(x))))) → 21(x)

The TRS R consists of the following rules:

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

(11) Obligation:

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

11(1(0(x))) → 11(x)
11(1(0(x))) → 11(5(x))
11(2(5(x))) → 11(x)
11(2(4(0(x)))) → 11(4(0(x)))
11(0(5(2(1(x))))) → 21(x)
21(4(2(3(5(x))))) → 11(2(x))
21(4(2(3(5(x))))) → 21(x)

The TRS R consists of the following rules:

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


11(1(0(x))) → 11(x)
11(2(5(x))) → 11(x)
11(2(4(0(x)))) → 11(4(0(x)))
11(0(5(2(1(x))))) → 21(x)
21(4(2(3(5(x))))) → 11(2(x))
21(4(2(3(5(x))))) → 21(x)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation:
POL( 11(x1) ) = max{0, 2x1 - 2}

POL( 5(x1) ) = x1 + 2

POL( 1(x1) ) = x1

POL( 0(x1) ) = x1 + 2

POL( 2(x1) ) = x1 + 2

POL( 3(x1) ) = max{0, x1 - 2}

POL( 4(x1) ) = x1

POL( 21(x1) ) = 2x1


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

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

(13) Obligation:

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

11(1(0(x))) → 11(5(x))

The TRS R consists of the following rules:

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

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

(14) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


11(1(0(x))) → 11(5(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(11(x1)) = x1   
POL(2(x1)) = 0   
POL(3(x1)) = 0   
POL(4(x1)) = x1   
POL(5(x1)) = x1   

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

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

(15) Obligation:

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

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

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

(16) PisEmptyProof (EQUIVALENT transformation)

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

(17) YES

(18) Obligation:

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

41(2(5(0(5(x))))) → 41(x)

The TRS R consists of the following rules:

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

41(2(5(0(5(x))))) → 41(x)

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

(21) 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:

  • 41(2(5(0(5(x))))) → 41(x)
    The graph contains the following edges 1 > 1

(22) YES