(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(x))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))) → 1(0(1(1(1(0(1(0(1(1(1(0(1(1(0(0(0(1(0(1(1(1(1(1(0(1(1(0(0(1(0(1(1(1(1(1(1(1(1(0(0(0(1(1(0(1(1(1(0(0(0(0(0(0(0(0(1(0(1(1(1(0(0(0(1(1(1(1(0(0(1(0(0(1(0(1(0(0(1(0(1(0(1(1(0(1(0(1(0(1(1(0(0(0(0(1(1(0(1(1(0(0(1(0(0(0(0(1(0(1(0(0(1(0(0(0(1(0(1(1(0(0(1(0(0(0(0(0(1(1(0(1(0(0(1(0(1(1(1(0(0(0(1(1(1(0(0(1(0(1(0(0(1(0(1(0(1(0(0(1(1(0(0(0(1(1(1(0(0(1(1(0(0(0(1(0(1(0(0(0(1(1(1(0(0(0(1(1(1(0(1(0(1(0(1(0(1(1(1(1(1(0(1(0(x))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))
0(0(0(0(0(0(0(0(0(1(1(0(1(1(1(1(0(0(1(0(0(1(1(1(1(1(0(1(1(0(0(0(1(0(0(1(1(1(0(1(0(1(0(0(1(0(1(0(1(1(0(0(1(0(0(0(0(1(0(0(0(0(1(0(1(1(1(0(0(0(1(1(1(1(1(1(1(1(1(0(1(0(0(0(1(0(1(1(1(1(0(0(0(1(0(1(1(0(0(1(1(1(0(0(0(1(0(1(0(0(0(1(1(0(0(0(0(1(0(1(1(1(0(1(0(0(1(0(0(0(1(1(1(0(0(0(0(1(1(1(1(0(1(1(0(0(1(0(0(1(0(1(1(0(1(0(0(0(0(0(0(1(1(0(1(1(0(0(1(1(0(0(1(0(1(1(1(0(1(0(0(0(1(1(0(1(0(1(1(1(1(0(1(1(0(0(1(1(1(0(0(1(0(1(1(1(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(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:
0(x) → 1(x)
0(0(x)) → 0(x)
5(4(3(x))) → 5(3(4(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(x))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))) → 0(1(0(1(1(1(1(1(0(1(0(1(0(1(0(1(1(1(0(0(0(1(1(1(0(0(0(1(0(1(0(0(0(1(1(0(0(1(1(1(0(0(0(1(1(0(0(1(0(1(0(1(0(0(1(0(1(0(0(1(1(1(0(0(0(1(1(1(0(1(0(0(1(0(1(1(0(0(0(0(0(1(0(0(1(1(0(1(0(0(0(1(0(0(1(0(1(0(0(0(0(1(0(0(1(1(0(1(1(0(0(0(0(1(1(0(1(0(1(0(1(1(0(1(0(1(0(0(1(0(1(0(0(1(0(0(1(1(1(1(0(0(0(1(1(1(0(1(0(0(0(0(0(0(0(0(1(1(1(0(1(1(0(0(0(1(1(1(1(1(1(1(1(0(1(0(0(1(1(0(1(1(1(1(1(0(1(0(0(0(1(1(0(1(1(1(0(1(0(1(1(1(0(1(x))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))
1(1(1(1(1(0(1(0(0(1(1(1(0(0(1(1(0(1(1(1(1(0(1(0(1(1(0(0(0(1(0(1(1(1(0(1(0(0(1(1(0(0(1(1(0(1(1(0(0(0(0(0(0(1(0(1(1(0(1(0(0(1(0(0(1(1(0(1(1(1(1(0(0(0(0(1(1(1(0(0(0(1(0(0(1(0(1(1(1(0(1(0(0(0(0(1(1(0(0(0(1(0(1(0(0(0(1(1(1(0(0(1(1(0(1(0(0(0(1(1(1(1(0(1(0(0(0(1(0(1(1(1(1(1(1(1(1(1(0(0(0(1(1(1(0(1(0(0(0(0(1(0(0(0(0(1(0(0(1(1(0(1(0(1(0(0(1(0(1(0(1(1(1(0(0(1(0(0(0(1(1(0(1(1(1(1(1(0(0(1(0(0(1(1(1(1(0(1(1(0(0(0(0(0(0(0(0(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(x)))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))
Q is empty.
(3) QTRSRRRProof (EQUIVALENT transformation)
Used ordering:
Polynomial interpretation [POLO]:
POL(0(x1)) = 6 + x1
POL(1(x1)) = 5 + x1
POL(2(x1)) = 9 + 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(x))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))) → 0(1(0(1(1(1(1(1(0(1(0(1(0(1(0(1(1(1(0(0(0(1(1(1(0(0(0(1(0(1(0(0(0(1(1(0(0(1(1(1(0(0(0(1(1(0(0(1(0(1(0(1(0(0(1(0(1(0(0(1(1(1(0(0(0(1(1(1(0(1(0(0(1(0(1(1(0(0(0(0(0(1(0(0(1(1(0(1(0(0(0(1(0(0(1(0(1(0(0(0(0(1(0(0(1(1(0(1(1(0(0(0(0(1(1(0(1(0(1(0(1(1(0(1(0(1(0(0(1(0(1(0(0(1(0(0(1(1(1(1(0(0(0(1(1(1(0(1(0(0(0(0(0(0(0(0(1(1(1(0(1(1(0(0(0(1(1(1(1(1(1(1(1(0(1(0(0(1(1(0(1(1(1(1(1(0(1(0(0(0(1(1(0(1(1(1(0(1(0(1(1(1(0(1(x))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))
1(1(1(1(1(0(1(0(0(1(1(1(0(0(1(1(0(1(1(1(1(0(1(0(1(1(0(0(0(1(0(1(1(1(0(1(0(0(1(1(0(0(1(1(0(1(1(0(0(0(0(0(0(1(0(1(1(0(1(0(0(1(0(0(1(1(0(1(1(1(1(0(0(0(0(1(1(1(0(0(0(1(0(0(1(0(1(1(1(0(1(0(0(0(0(1(1(0(0(0(1(0(1(0(0(0(1(1(1(0(0(1(1(0(1(0(0(0(1(1(1(1(0(1(0(0(0(1(0(1(1(1(1(1(1(1(1(1(0(0(0(1(1(1(0(1(0(0(0(0(1(0(0(0(0(1(0(0(1(1(0(1(0(1(0(0(1(0(1(0(1(1(1(0(0(1(0(0(0(1(1(0(1(1(1(1(1(0(0(1(0(0(1(1(1(1(0(1(1(0(0(0(0(0(0(0(0(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(x)))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))
(4) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
5(4(3(x))) → 5(3(4(x)))
Q is empty.
(5) AAECC Innermost (EQUIVALENT transformation)
We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is none
The TRS R 2 is
5(4(3(x))) → 5(3(4(x)))
The signature Sigma is {
5}
(6) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
5(4(3(x))) → 5(3(4(x)))
The set Q consists of the following terms:
5(4(3(x0)))
(7) DependencyPairsProof (EQUIVALENT transformation)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
(8) Obligation:
Q DP problem:
The TRS P consists of the following rules:
51(4(3(x))) → 51(3(4(x)))
The TRS R consists of the following rules:
5(4(3(x))) → 5(3(4(x)))
The set Q consists of the following terms:
5(4(3(x0)))
We have to consider all minimal (P,Q,R)-chains.
(9) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.
(10) TRUE