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 (⇔, 95 ms)

↳4 QDP

↳5 DependencyGraphProof (⇔, 0 ms)

↳6 QDP

↳7 UsableRulesProof (⇔, 0 ms)

↳8 QDP

↳9 MRRProof (⇔, 47 ms)

↳10 QDP

↳11 DependencyGraphProof (⇔, 0 ms)

↳12 TRUE

(0) Obligation:

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

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

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

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

The TRS R consists of the following rules:

2(1(0(x))) → 1(1(2(0(x))))
2(1(0(x))) → 3(1(2(0(x))))
2(1(0(x))) → 3(1(1(2(0(x)))))
2(1(0(x))) → 1(1(4(2(0(x)))))
2(1(0(x))) → 3(1(4(2(0(x)))))
2(3(0(x))) → 3(1(2(0(x))))
2(3(0(x))) → 3(1(1(2(0(x)))))
2(3(0(x))) → 3(1(1(2(0(0(x))))))
2(1(1(0(x)))) → 3(1(1(2(0(x)))))
2(2(1(0(x)))) → 1(1(2(2(0(x)))))
2(2(1(0(x)))) → 1(2(3(2(0(x)))))
2(3(1(0(x)))) → 3(4(1(2(0(x)))))
2(3(1(0(x)))) → 1(2(0(3(3(x)))))
2(3(1(0(x)))) → 1(1(4(2(0(3(x))))))
2(5(1(0(x)))) → 1(1(2(0(5(x)))))
2(2(3(0(x)))) → 2(3(2(0(0(x)))))
2(2(3(0(x)))) → 2(3(2(4(0(0(x))))))
2(3(0(2(x)))) → 2(2(0(3(1(x)))))
2(3(0(3(x)))) → 3(2(0(0(0(3(x))))))
2(5(1(3(x)))) → 3(5(1(2(0(x)))))
2(5(3(3(x)))) → 3(5(1(2(0(3(x))))))
2(2(5(4(x)))) → 1(4(2(0(5(2(x))))))
2(1(0(5(x)))) → 5(4(3(1(2(0(x))))))
2(2(0(5(x)))) → 5(1(4(2(0(2(x))))))
2(3(0(5(x)))) → 5(4(1(2(0(3(x))))))
2(2(1(5(x)))) → 1(2(5(1(2(0(x))))))
2(2(1(5(x)))) → 1(1(2(5(1(2(x))))))
2(4(2(1(0(x))))) → 1(2(0(1(2(4(x))))))
2(5(2(1(0(x))))) → 4(5(1(2(0(2(x))))))
2(1(5(1(0(x))))) → 5(1(1(2(0(1(x))))))
2(5(5(1(0(x))))) → 1(5(5(1(2(0(x))))))
2(1(0(5(0(x))))) → 1(2(0(5(4(0(x))))))
2(3(0(5(0(x))))) → 3(2(0(0(5(0(x))))))
2(2(5(5(0(x))))) → 2(5(5(1(2(0(x))))))
2(1(3(0(3(x))))) → 3(2(0(3(3(1(x))))))
2(4(3(0(3(x))))) → 3(2(0(4(3(1(x))))))
2(4(1(5(4(x))))) → 5(1(4(4(4(2(x))))))
2(3(1(0(5(x))))) → 3(1(2(0(5(3(x))))))
2(1(3(0(5(x))))) → 5(3(4(1(2(0(x))))))
2(2(4(0(5(x))))) → 5(1(4(2(0(2(x))))))
2(3(0(1(5(x))))) → 2(1(5(3(4(0(x))))))
2(3(0(1(5(x))))) → 2(0(1(1(3(5(x))))))
2(5(0(1(5(x))))) → 1(2(0(5(5(3(x))))))
2(5(0(1(5(x))))) → 5(5(1(2(0(5(x))))))
2(3(0(2(5(x))))) → 1(3(2(5(2(0(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 45 less nodes.

(6) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(9) MRRProof (EQUIVALENT transformation)

By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

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

Used ordering: Polynomial interpretation [POLO]:

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

(10) Obligation:

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

2¹(2(1(0(x)))) → 2¹(2(0(x)))

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

(11) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.

(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) MRRProof (EQUIVALENT transformation)

(10) Obligation:

(11) DependencyGraphProof (EQUIVALENT transformation)

(12) TRUE