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 QTRS Reverse (⇔, 0 ms)

↳2 QTRS

↳3 DependencyPairsProof (⇔, 76 ms)

↳4 QDP

↳5 DependencyGraphProof (⇔, 0 ms)

↳6 QDP

↳7 UsableRulesProof (⇔, 0 ms)

↳8 QDP

↳9 QDPSizeChangeProof (⇔, 0 ms)

↳10 YES

(0) Obligation:

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

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

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

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

The TRS R consists of the following rules:

4(2(4(x))) → 4(0(2(5(3(3(5(0(0(2(x))))))))))
2(4(2(4(4(x))))) → 1(0(2(0(4(1(2(5(0(2(x))))))))))
3(4(2(4(5(0(x)))))) → 0(0(0(3(5(3(5(5(1(5(x))))))))))
3(3(5(4(1(1(x)))))) → 1(2(2(1(0(3(1(1(3(1(x))))))))))
2(1(3(4(1(3(x)))))) → 3(0(0(3(2(4(1(1(0(0(x))))))))))
1(4(2(4(2(3(x)))))) → 3(3(1(3(5(1(1(1(2(0(x))))))))))
2(1(4(0(3(3(x)))))) → 1(3(5(0(2(0(2(1(5(3(x))))))))))
4(5(0(5(4(1(4(x))))))) → 0(1(3(5(0(1(3(5(1(4(x))))))))))
2(2(4(5(0(4(4(x))))))) → 1(4(5(4(4(4(3(4(0(4(x))))))))))
3(4(2(3(5(4(5(x))))))) → 4(1(0(5(4(0(5(5(5(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 1 SCC with 55 less nodes.

(6) Obligation:

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

2¹(4(2(4(4(x))))) → 2¹(x)

The TRS R consists of the following rules:

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

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

(7) 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.

(8) Obligation:

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

2¹(4(2(4(4(x))))) → 2¹(x)

R is empty.
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:

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

(0) Obligation:

(1) QTRS Reverse (EQUIVALENT transformation)

(2) Obligation:

(3) DependencyPairsProof (EQUIVALENT transformation)

(4) Obligation:

(5) DependencyGraphProof (EQUIVALENT transformation)

(6) Obligation:

(7) UsableRulesProof (EQUIVALENT transformation)

(8) Obligation:

(9) QDPSizeChangeProof (EQUIVALENT transformation)

(10) YES