YES Termination w.r.t. Q proof of /home/cern_httpd/provide/research/cycsrs/tpdb/TPDB-d9b80194f163/SRS_Standard/ICFP_2010/211978.srs

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

0 QTRS
↳1 DependencyPairsProof (⇔, 153 ms)
↳2 QDP
↳3 DependencyGraphProof (⇔, 2 ms)
↳4 AND
    ↳5 QDP
        ↳6 QDPSizeChangeProof (⇔, 0 ms)
        ↳7 YES
    ↳8 QDP
        ↳9 QDPSizeChangeProof (⇔, 0 ms)
        ↳10 YES

(0) Obligation:

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

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

The TRS R consists of the following rules:

0(0(1(x))) → 0(2(3(0(1(x)))))
0(0(1(x))) → 0(4(0(5(4(1(x))))))
0(0(1(x))) → 2(1(0(0(3(4(x))))))
0(0(1(x))) → 4(0(5(4(0(1(x))))))
0(1(0(x))) → 0(0(2(1(2(x)))))
0(1(0(x))) → 1(0(0(5(4(x)))))
0(1(0(x))) → 0(0(2(5(4(1(x))))))
0(1(1(x))) → 1(0(3(4(1(x)))))
0(1(1(x))) → 5(0(3(4(1(1(x))))))
5(0(1(x))) → 0(5(4(1(x))))
5(0(1(x))) → 2(5(4(0(1(x)))))
5(0(1(x))) → 5(0(2(1(2(x)))))
5(0(1(x))) → 0(1(4(5(4(4(x))))))
5(0(1(x))) → 0(5(4(1(4(4(x))))))
5(0(1(x))) → 5(0(4(3(0(1(x))))))
5(1(0(x))) → 5(0(2(2(1(x)))))
5(1(0(x))) → 5(0(5(4(1(x)))))
5(1(0(x))) → 0(5(0(2(2(1(x))))))
5(1(0(x))) → 1(4(0(5(2(3(x))))))
5(1(0(x))) → 1(5(0(4(4(2(x))))))
5(1(0(x))) → 4(4(1(0(4(5(x))))))
5(1(1(x))) → 1(1(5(4(x))))
5(1(1(x))) → 5(4(1(1(x))))
5(1(1(x))) → 1(5(3(4(1(x)))))
5(1(1(x))) → 1(1(4(5(4(4(x))))))
5(1(1(x))) → 3(5(2(3(1(1(x))))))
5(1(1(x))) → 4(1(2(1(5(4(x))))))
0(1(3(0(x)))) → 0(2(0(2(1(3(x))))))
0(1(5(0(x)))) → 0(0(5(4(1(5(x))))))
0(1(5(0(x)))) → 0(5(4(2(1(0(x))))))
0(3(0(1(x)))) → 0(0(4(1(3(0(x))))))
0(3(1(0(x)))) → 0(0(2(3(1(x)))))
0(3(1(1(x)))) → 5(1(1(0(3(4(x))))))
5(0(1(0(x)))) → 5(0(0(4(1(3(x))))))
5(1(2(0(x)))) → 1(4(0(5(4(2(x))))))
5(1(2(0(x)))) → 5(0(4(2(2(1(x))))))
5(1(4(0(x)))) → 1(5(4(0(2(3(x))))))
5(1(4(0(x)))) → 4(5(2(1(3(0(x))))))
5(1(5(1(x)))) → 5(4(1(5(1(x)))))
5(3(0(1(x)))) → 0(1(5(2(3(x)))))
5(3(1(0(x)))) → 1(4(3(5(0(x)))))
5(3(1(0(x)))) → 1(5(0(4(3(x)))))
5(3(1(0(x)))) → 5(4(3(1(0(x)))))
5(3(1(0(x)))) → 1(3(0(4(3(5(x))))))
5(3(1(1(x)))) → 1(1(5(3(3(4(x))))))
0(1(2(5(0(x))))) → 1(5(4(0(2(0(x))))))
0(1(4(2(0(x))))) → 1(0(4(2(3(0(x))))))
1(4(5(1(0(x))))) → 5(4(2(1(1(0(x))))))
5(0(1(4(0(x))))) → 1(4(5(4(0(0(x))))))
5(5(1(0(0(x))))) → 5(5(0(4(1(0(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 2 SCCs with 125 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

11(4(5(1(0(x))))) → 11(1(0(x)))

The TRS R consists of the following rules:

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

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

(6) 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(4(5(1(0(x))))) → 11(1(0(x)))
    The graph contains the following edges 1 > 1

(7) YES

(8) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(9) 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(0(1(4(0(x))))) → 01(0(x))
    The graph contains the following edges 1 > 1

  • 01(3(0(1(x)))) → 01(x)
    The graph contains the following edges 1 > 1

  • 01(1(5(0(x)))) → 51(x)
    The graph contains the following edges 1 > 1

  • 51(1(0(x))) → 51(x)
    The graph contains the following edges 1 > 1

  • 51(3(1(0(x)))) → 51(0(x))
    The graph contains the following edges 1 > 1

  • 51(3(1(0(x)))) → 51(x)
    The graph contains the following edges 1 > 1

(10) YES