(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(x))))))))))))) → 0(0(0(1(0(1(1(1(0(1(1(1(1(0(0(0(0(1(0(0(1(1(1(1(0(0(0(0(0(0(0(0(1(0(0(0(1(1(1(1(1(1(1(1(1(0(0(0(0(1(1(1(1(1(1(1(0(0(0(1(0(1(1(1(0(0(0(1(1(1(1(1(1(1(1(0(1(0(0(1(0(0(1(1(0(0(0(1(0(1(0(1(0(0(1(1(1(0(1(0(0(0(0(1(0(1(1(1(1(0(0(0(0(1(0(0(0(0(0(0(0(0(0(1(1(1(1(0(1(0(1(0(0(0(0(1(1(0(0(1(1(1(1(1(x))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))
1(1(0(0(1(1(0(1(0(0(1(0(1(1(1(1(1(0(0(1(0(1(1(1(0(1(0(0(0(0(1(0(1(1(1(1(1(0(0(1(0(0(1(1(0(0(0(0(1(1(0(0(1(0(1(0(0(0(0(0(0(1(0(0(0(1(0(1(1(0(0(1(1(0(1(1(0(1(1(0(1(0(0(1(1(1(0(1(0(0(1(0(0(0(0(0(0(1(1(1(0(1(1(0(0(1(1(0(1(0(0(1(1(0(0(1(0(1(1(0(1(0(1(1(1(0(1(0(1(1(0(0(0(1(0(1(1(0(1(1(0(1(1(0(0(0(1(1(x)))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))) → 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)) = 2 + x1
POL(1(x1)) = 1 + x1
POL(2(x1)) = 17 + 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(x))))))))))))) → 0(0(0(1(0(1(1(1(0(1(1(1(1(0(0(0(0(1(0(0(1(1(1(1(0(0(0(0(0(0(0(0(1(0(0(0(1(1(1(1(1(1(1(1(1(0(0(0(0(1(1(1(1(1(1(1(0(0(0(1(0(1(1(1(0(0(0(1(1(1(1(1(1(1(1(0(1(0(0(1(0(0(1(1(0(0(0(1(0(1(0(1(0(0(1(1(1(0(1(0(0(0(0(1(0(1(1(1(1(0(0(0(0(1(0(0(0(0(0(0(0(0(0(1(1(1(1(0(1(0(1(0(0(0(0(1(1(0(0(1(1(1(1(1(x))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))
1(1(0(0(1(1(0(1(0(0(1(0(1(1(1(1(1(0(0(1(0(1(1(1(0(1(0(0(0(0(1(0(1(1(1(1(1(0(0(1(0(0(1(1(0(0(0(0(1(1(0(0(1(0(1(0(0(0(0(0(0(1(0(0(0(1(0(1(1(0(0(1(1(0(1(1(0(1(1(0(1(0(0(1(1(1(0(1(0(0(1(0(0(0(0(0(0(1(1(1(0(1(1(0(0(1(1(0(1(0(0(1(1(0(0(1(0(1(1(0(1(0(1(1(1(0(1(0(1(1(0(0(0(1(0(1(1(0(1(1(0(1(1(0(0(0(1(1(x)))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))) → 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.
(8) TRUE