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 QTRSRRRProof (⇔, 98 ms)

↳2 QTRS

↳3 Overlay + Local Confluence (⇔, 0 ms)

↳4 QTRS

↳5 DependencyPairsProof (⇔, 0 ms)

↳6 QDP

↳7 DependencyGraphProof (⇔, 0 ms)

↳8 TRUE

(0) Obligation:

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

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

Q is empty.

(1) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(0(x₁)) = 31 + x₁
POL(1(x₁)) = 29 + x₁
POL(2(x₁)) = 31 + x₁
POL(3(x₁)) = x₁
POL(4(x₁)) = x₁
POL(5(x₁)) = x₁

With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

0(x) → 1(x)
0(0(x)) → 0(x)
2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(x))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))) → 1(1(0(0(1(0(0(1(1(1(1(1(0(1(0(0(1(1(1(0(0(1(0(0(0(1(1(0(0(1(0(1(1(0(0(0(1(1(1(1(1(1(0(1(0(1(1(1(0(0(1(1(0(1(0(1(0(0(1(0(1(0(1(0(1(0(0(1(1(1(1(1(0(0(1(0(1(0(1(0(0(0(1(0(1(0(1(1(0(1(1(1(1(1(0(1(0(1(1(1(1(1(0(0(1(1(1(0(0(1(1(0(1(0(1(1(0(0(0(0(1(0(0(0(0(1(1(1(0(0(1(0(0(0(0(0(1(1(0(0(0(1(1(0(0(0(1(0(0(0(1(0(0(0(1(1(1(1(0(0(0(1(0(0(0(1(1(0(1(0(1(1(0(0(1(0(0(1(0(1(0(0(1(1(0(0(1(0(1(0(1(1(0(0(0(1(0(1(1(1(1(0(0(0(x))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))
0(0(0(1(1(0(1(0(0(0(0(0(0(1(0(0(1(0(0(1(1(0(1(1(0(1(1(0(1(1(0(0(0(1(1(1(1(0(1(1(1(0(0(0(0(0(1(1(0(1(1(1(1(1(1(1(0(1(1(1(1(0(0(0(0(0(0(0(1(0(0(0(0(1(0(0(0(1(0(1(0(0(1(0(0(1(1(0(1(1(0(1(1(0(0(0(0(1(0(1(1(0(1(1(0(1(0(0(0(0(0(1(1(0(1(1(1(0(1(1(1(1(0(1(0(1(1(1(1(0(0(0(1(1(0(0(1(0(1(1(1(1(0(0(0(0(1(1(1(0(0(0(0(0(0(1(1(0(1(1(1(0(0(0(1(1(0(1(0(0(1(1(1(1(0(1(1(1(1(0(1(1(1(0(0(1(1(1(1(1(0(0(0(0(0(0(1(0(0(1(0(0(1(0(1(0(1(1(x)))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))) → 2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(x)))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))

(2) Obligation:

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

3(4(5(x))) → 4(3(5(x)))

Q is empty.

(3) Overlay + Local Confluence (EQUIVALENT transformation)

The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.

(4) Obligation:

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

3(4(5(x))) → 4(3(5(x)))

The set Q consists of the following terms:

3(4(5(x0)))

(5) DependencyPairsProof (EQUIVALENT transformation)

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

(6) Obligation:

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

3¹(4(5(x))) → 3¹(5(x))

The TRS R consists of the following rules:

3(4(5(x))) → 4(3(5(x)))

The set Q consists of the following terms:

3(4(5(x0)))

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

(7) DependencyGraphProof (EQUIVALENT transformation)

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

(0) Obligation:

(1) QTRSRRRProof (EQUIVALENT transformation)

(2) Obligation:

(3) Overlay + Local Confluence (EQUIVALENT transformation)

(4) Obligation:

(5) DependencyPairsProof (EQUIVALENT transformation)

(6) Obligation:

(7) DependencyGraphProof (EQUIVALENT transformation)

(8) TRUE