Termination w.r.t. Q proof of /home/cern_httpd/provide/research/cycsrs/tpdb/TPDB-d9b80194f163/SRS_Standard/ICFP

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

0 QTRS

↳1 DependencyPairsProof (⇔, 149 ms)

↳2 QDP

↳3 DependencyGraphProof (⇔, 0 ms)

↳4 AND

    ↳5 QDP

        ↳6 UsableRulesProof (⇔, 0 ms)

        ↳7 QDP

        ↳8 QDPSizeChangeProof (⇔, 0 ms)

        ↳9 YES

    ↳10 QDP

        ↳11 UsableRulesProof (⇔, 0 ms)

        ↳12 QDP

        ↳13 QDPSizeChangeProof (⇔, 0 ms)

        ↳14 YES

(0) Obligation:

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

0(1(2(x))) → 0(2(1(0(x))))
0(1(2(x))) → 1(0(2(3(x))))
0(1(2(x))) → 0(2(4(1(5(x)))))
0(1(2(x))) → 0(3(2(1(0(x)))))
0(1(2(x))) → 1(0(3(2(3(x)))))
0(1(2(x))) → 0(1(3(4(2(3(x))))))
0(5(2(x))) → 0(2(4(5(3(x)))))
0(5(2(x))) → 5(4(2(3(0(4(x))))))
2(0(1(x))) → 3(0(2(1(x))))
2(0(1(x))) → 0(2(1(1(4(x)))))
2(0(1(x))) → 0(3(2(4(1(x)))))
2(0(1(x))) → 3(0(2(1(4(x)))))
2(0(1(x))) → 0(2(2(3(4(1(x))))))
2(0(1(x))) → 0(3(2(3(1(1(x))))))
2(0(1(x))) → 4(0(4(2(1(4(x))))))
2(5(1(x))) → 0(2(1(5(1(x)))))
2(5(1(x))) → 1(4(5(4(2(x)))))
2(5(1(x))) → 5(0(2(1(4(x)))))
2(5(1(x))) → 5(2(1(4(1(x)))))
2(5(1(x))) → 1(5(0(2(4(1(x))))))
2(5(1(x))) → 5(2(1(1(1(1(x))))))
0(1(2(1(x)))) → 3(1(4(0(2(1(x))))))
0(1(3(1(x)))) → 5(0(3(1(1(x)))))
0(1(3(1(x)))) → 1(0(3(4(2(1(x))))))
0(1(5(1(x)))) → 5(0(3(1(1(x)))))
0(2(1(2(x)))) → 0(2(2(1(5(x)))))
0(2(5(1(x)))) → 1(1(5(0(2(1(x))))))
0(5(3(1(x)))) → 0(1(4(4(3(5(x))))))
0(5(5(2(x)))) → 5(4(2(3(5(0(x))))))
2(0(1(2(x)))) → 0(2(3(2(1(1(x))))))
2(0(1(2(x)))) → 4(0(2(1(1(2(x))))))
2(0(4(1(x)))) → 3(0(2(4(1(x)))))
2(0(5(1(x)))) → 5(4(2(1(0(x)))))
2(2(5(1(x)))) → 3(2(2(4(5(1(x))))))
2(4(0(1(x)))) → 1(0(2(4(4(x)))))
2(4(0(1(x)))) → 3(0(0(2(4(1(x))))))
2(4(0(1(x)))) → 5(4(0(2(1(1(x))))))
2(5(2(1(x)))) → 1(5(2(2(3(1(x))))))
2(5(4(1(x)))) → 4(5(2(1(4(4(x))))))
2(5(5(1(x)))) → 1(5(4(2(4(5(x))))))
2(5(5(2(x)))) → 5(5(2(3(2(x)))))
0(1(3(0(1(x))))) → 0(3(1(0(1(1(x))))))
0(2(4(3(1(x))))) → 1(3(4(2(3(0(x))))))
0(2(4(3(1(x))))) → 4(0(3(2(1(0(x))))))
0(2(5(3(1(x))))) → 5(0(2(3(5(1(x))))))
2(0(5(4(1(x))))) → 0(4(5(3(2(1(x))))))
2(2(0(1(2(x))))) → 2(4(0(2(2(1(x))))))
2(4(0(5(1(x))))) → 1(4(5(0(4(2(x))))))
2(4(2(3(1(x))))) → 4(2(2(3(3(1(x))))))
2(5(2(0(1(x))))) → 0(2(4(1(5(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(1(2(x))) → 0¹(2(1(0(x))))
0¹(1(2(x))) → 2¹(1(0(x)))
0¹(1(2(x))) → 0¹(x)
0¹(1(2(x))) → 0¹(2(3(x)))
0¹(1(2(x))) → 2¹(3(x))
0¹(1(2(x))) → 0¹(2(4(1(5(x)))))
0¹(1(2(x))) → 2¹(4(1(5(x))))
0¹(1(2(x))) → 0¹(3(2(1(0(x)))))
0¹(1(2(x))) → 0¹(3(2(3(x))))
0¹(1(2(x))) → 0¹(1(3(4(2(3(x))))))
0¹(5(2(x))) → 0¹(2(4(5(3(x)))))
0¹(5(2(x))) → 2¹(4(5(3(x))))
0¹(5(2(x))) → 2¹(3(0(4(x))))
0¹(5(2(x))) → 0¹(4(x))
2¹(0(1(x))) → 0¹(2(1(x)))
2¹(0(1(x))) → 2¹(1(x))
2¹(0(1(x))) → 0¹(2(1(1(4(x)))))
2¹(0(1(x))) → 2¹(1(1(4(x))))
2¹(0(1(x))) → 0¹(3(2(4(1(x)))))
2¹(0(1(x))) → 2¹(4(1(x)))
2¹(0(1(x))) → 0¹(2(1(4(x))))
2¹(0(1(x))) → 2¹(1(4(x)))
2¹(0(1(x))) → 0¹(2(2(3(4(1(x))))))
2¹(0(1(x))) → 2¹(2(3(4(1(x)))))
2¹(0(1(x))) → 2¹(3(4(1(x))))
2¹(0(1(x))) → 0¹(3(2(3(1(1(x))))))
2¹(0(1(x))) → 2¹(3(1(1(x))))
2¹(0(1(x))) → 0¹(4(2(1(4(x)))))
2¹(5(1(x))) → 0¹(2(1(5(1(x)))))
2¹(5(1(x))) → 2¹(1(5(1(x))))
2¹(5(1(x))) → 2¹(x)
2¹(5(1(x))) → 0¹(2(1(4(x))))
2¹(5(1(x))) → 2¹(1(4(x)))
2¹(5(1(x))) → 2¹(1(4(1(x))))
2¹(5(1(x))) → 0¹(2(4(1(x))))
2¹(5(1(x))) → 2¹(4(1(x)))
2¹(5(1(x))) → 2¹(1(1(1(1(x)))))
0¹(1(2(1(x)))) → 0¹(2(1(x)))
0¹(1(3(1(x)))) → 0¹(3(1(1(x))))
0¹(1(3(1(x)))) → 0¹(3(4(2(1(x)))))
0¹(1(3(1(x)))) → 2¹(1(x))
0¹(1(5(1(x)))) → 0¹(3(1(1(x))))
0¹(2(1(2(x)))) → 0¹(2(2(1(5(x)))))
0¹(2(1(2(x)))) → 2¹(2(1(5(x))))
0¹(2(1(2(x)))) → 2¹(1(5(x)))
0¹(2(5(1(x)))) → 0¹(2(1(x)))
0¹(2(5(1(x)))) → 2¹(1(x))
0¹(5(3(1(x)))) → 0¹(1(4(4(3(5(x))))))
0¹(5(5(2(x)))) → 2¹(3(5(0(x))))
0¹(5(5(2(x)))) → 0¹(x)
2¹(0(1(2(x)))) → 0¹(2(3(2(1(1(x))))))
2¹(0(1(2(x)))) → 2¹(3(2(1(1(x)))))
2¹(0(1(2(x)))) → 2¹(1(1(x)))
2¹(0(1(2(x)))) → 0¹(2(1(1(2(x)))))
2¹(0(1(2(x)))) → 2¹(1(1(2(x))))
2¹(0(4(1(x)))) → 0¹(2(4(1(x))))
2¹(0(4(1(x)))) → 2¹(4(1(x)))
2¹(0(5(1(x)))) → 2¹(1(0(x)))
2¹(0(5(1(x)))) → 0¹(x)
2¹(2(5(1(x)))) → 2¹(2(4(5(1(x)))))
2¹(2(5(1(x)))) → 2¹(4(5(1(x))))
2¹(4(0(1(x)))) → 0¹(2(4(4(x))))
2¹(4(0(1(x)))) → 2¹(4(4(x)))
2¹(4(0(1(x)))) → 0¹(0(2(4(1(x)))))
2¹(4(0(1(x)))) → 0¹(2(4(1(x))))
2¹(4(0(1(x)))) → 2¹(4(1(x)))
2¹(4(0(1(x)))) → 0¹(2(1(1(x))))
2¹(4(0(1(x)))) → 2¹(1(1(x)))
2¹(5(2(1(x)))) → 2¹(2(3(1(x))))
2¹(5(2(1(x)))) → 2¹(3(1(x)))
2¹(5(4(1(x)))) → 2¹(1(4(4(x))))
2¹(5(5(1(x)))) → 2¹(4(5(x)))
2¹(5(5(2(x)))) → 2¹(3(2(x)))
0¹(1(3(0(1(x))))) → 0¹(3(1(0(1(1(x))))))
0¹(1(3(0(1(x))))) → 0¹(1(1(x)))
0¹(2(4(3(1(x))))) → 2¹(3(0(x)))
0¹(2(4(3(1(x))))) → 0¹(x)
0¹(2(4(3(1(x))))) → 0¹(3(2(1(0(x)))))
0¹(2(4(3(1(x))))) → 2¹(1(0(x)))
0¹(2(5(3(1(x))))) → 0¹(2(3(5(1(x)))))
0¹(2(5(3(1(x))))) → 2¹(3(5(1(x))))
2¹(0(5(4(1(x))))) → 0¹(4(5(3(2(1(x))))))
2¹(0(5(4(1(x))))) → 2¹(1(x))
2¹(2(0(1(2(x))))) → 2¹(4(0(2(2(1(x))))))
2¹(2(0(1(2(x))))) → 0¹(2(2(1(x))))
2¹(2(0(1(2(x))))) → 2¹(2(1(x)))
2¹(2(0(1(2(x))))) → 2¹(1(x))
2¹(4(0(5(1(x))))) → 0¹(4(2(x)))
2¹(4(0(5(1(x))))) → 2¹(x)
2¹(4(2(3(1(x))))) → 2¹(2(3(3(1(x)))))
2¹(4(2(3(1(x))))) → 2¹(3(3(1(x))))
2¹(5(2(0(1(x))))) → 0¹(2(4(1(5(2(x))))))
2¹(5(2(0(1(x))))) → 2¹(4(1(5(2(x)))))
2¹(5(2(0(1(x))))) → 2¹(x)

The TRS R consists of the following rules:

0(1(2(x))) → 0(2(1(0(x))))
0(1(2(x))) → 1(0(2(3(x))))
0(1(2(x))) → 0(2(4(1(5(x)))))
0(1(2(x))) → 0(3(2(1(0(x)))))
0(1(2(x))) → 1(0(3(2(3(x)))))
0(1(2(x))) → 0(1(3(4(2(3(x))))))
0(5(2(x))) → 0(2(4(5(3(x)))))
0(5(2(x))) → 5(4(2(3(0(4(x))))))
2(0(1(x))) → 3(0(2(1(x))))
2(0(1(x))) → 0(2(1(1(4(x)))))
2(0(1(x))) → 0(3(2(4(1(x)))))
2(0(1(x))) → 3(0(2(1(4(x)))))
2(0(1(x))) → 0(2(2(3(4(1(x))))))
2(0(1(x))) → 0(3(2(3(1(1(x))))))
2(0(1(x))) → 4(0(4(2(1(4(x))))))
2(5(1(x))) → 0(2(1(5(1(x)))))
2(5(1(x))) → 1(4(5(4(2(x)))))
2(5(1(x))) → 5(0(2(1(4(x)))))
2(5(1(x))) → 5(2(1(4(1(x)))))
2(5(1(x))) → 1(5(0(2(4(1(x))))))
2(5(1(x))) → 5(2(1(1(1(1(x))))))
0(1(2(1(x)))) → 3(1(4(0(2(1(x))))))
0(1(3(1(x)))) → 5(0(3(1(1(x)))))
0(1(3(1(x)))) → 1(0(3(4(2(1(x))))))
0(1(5(1(x)))) → 5(0(3(1(1(x)))))
0(2(1(2(x)))) → 0(2(2(1(5(x)))))
0(2(5(1(x)))) → 1(1(5(0(2(1(x))))))
0(5(3(1(x)))) → 0(1(4(4(3(5(x))))))
0(5(5(2(x)))) → 5(4(2(3(5(0(x))))))
2(0(1(2(x)))) → 0(2(3(2(1(1(x))))))
2(0(1(2(x)))) → 4(0(2(1(1(2(x))))))
2(0(4(1(x)))) → 3(0(2(4(1(x)))))
2(0(5(1(x)))) → 5(4(2(1(0(x)))))
2(2(5(1(x)))) → 3(2(2(4(5(1(x))))))
2(4(0(1(x)))) → 1(0(2(4(4(x)))))
2(4(0(1(x)))) → 3(0(0(2(4(1(x))))))
2(4(0(1(x)))) → 5(4(0(2(1(1(x))))))
2(5(2(1(x)))) → 1(5(2(2(3(1(x))))))
2(5(4(1(x)))) → 4(5(2(1(4(4(x))))))
2(5(5(1(x)))) → 1(5(4(2(4(5(x))))))
2(5(5(2(x)))) → 5(5(2(3(2(x)))))
0(1(3(0(1(x))))) → 0(3(1(0(1(1(x))))))
0(2(4(3(1(x))))) → 1(3(4(2(3(0(x))))))
0(2(4(3(1(x))))) → 4(0(3(2(1(0(x))))))
0(2(5(3(1(x))))) → 5(0(2(3(5(1(x))))))
2(0(5(4(1(x))))) → 0(4(5(3(2(1(x))))))
2(2(0(1(2(x))))) → 2(4(0(2(2(1(x))))))
2(4(0(5(1(x))))) → 1(4(5(0(4(2(x))))))
2(4(2(3(1(x))))) → 4(2(2(3(3(1(x))))))
2(5(2(0(1(x))))) → 0(2(4(1(5(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 2 SCCs with 88 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(6) 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.

(7) Obligation:

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

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

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

(8) 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:

0¹(5(5(2(x)))) → 0¹(x)
The graph contains the following edges 1 > 1
0¹(1(2(x))) → 0¹(x)
The graph contains the following edges 1 > 1
0¹(2(4(3(1(x))))) → 0¹(x)
The graph contains the following edges 1 > 1

(9) YES

(10) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(11) UsableRulesProof (EQUIVALENT transformation)

(12) Obligation:

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

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

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

(13) 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:

2¹(4(0(5(1(x))))) → 2¹(x)
The graph contains the following edges 1 > 1
2¹(5(1(x))) → 2¹(x)
The graph contains the following edges 1 > 1
2¹(5(2(0(1(x))))) → 2¹(x)
The graph contains the following edges 1 > 1

(0) Obligation:

(1) DependencyPairsProof (EQUIVALENT transformation)

(2) Obligation:

(3) DependencyGraphProof (EQUIVALENT transformation)

(4) Complex Obligation (AND)

(5) Obligation:

(6) UsableRulesProof (EQUIVALENT transformation)

(7) Obligation:

(8) QDPSizeChangeProof (EQUIVALENT transformation)

(9) YES

(10) Obligation:

(11) UsableRulesProof (EQUIVALENT transformation)

(12) Obligation:

(13) QDPSizeChangeProof (EQUIVALENT transformation)

(14) YES