(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
0(1(2(2(3(0(4(4(x)))))))) → 0(4(5(0(3(4(0(4(x))))))))
0(2(2(2(0(5(2(5(4(x))))))))) → 0(0(0(3(4(3(0(1(0(x)))))))))
2(1(1(3(4(3(1(1(5(x))))))))) → 1(0(2(3(1(0(5(1(5(x)))))))))
5(4(0(4(3(3(1(2(5(3(0(x))))))))))) → 5(5(2(5(3(1(5(0(3(5(2(x)))))))))))
2(4(5(0(1(1(3(3(5(3(0(0(x)))))))))))) → 4(4(2(2(1(0(4(0(1(3(2(0(x))))))))))))
4(4(3(0(3(1(5(3(5(1(3(1(5(3(x)))))))))))))) → 0(2(4(5(0(0(0(5(1(0(5(4(4(x)))))))))))))
2(2(0(0(2(1(0(5(3(2(2(1(4(0(5(x))))))))))))))) → 0(0(3(5(3(0(4(3(1(3(0(2(5(5(x))))))))))))))
3(0(0(4(2(5(5(1(3(0(2(3(3(5(1(4(5(x))))))))))))))))) → 3(4(0(1(0(5(5(3(1(4(0(3(5(3(2(2(5(x)))))))))))))))))
5(1(1(4(1(5(3(0(4(3(2(5(4(1(3(3(5(x))))))))))))))))) → 5(0(1(0(4(0(2(4(5(1(5(4(1(5(3(3(5(x)))))))))))))))))
0(1(4(4(3(2(0(4(1(4(3(4(4(1(5(3(4(4(x)))))))))))))))))) → 0(2(4(2(3(1(0(1(1(1(3(0(2(4(4(1(1(2(x))))))))))))))))))
1(1(4(1(0(1(0(3(3(4(4(1(5(4(0(4(4(5(5(3(x)))))))))))))))))))) → 1(3(0(3(2(2(4(4(2(0(3(3(4(0(3(0(4(3(4(0(x))))))))))))))))))))
2(0(3(3(3(4(1(1(0(4(4(0(3(3(3(0(0(1(5(3(x)))))))))))))))))))) → 5(3(3(1(4(0(4(5(4(4(4(2(4(3(1(1(1(5(4(x)))))))))))))))))))
2(3(3(2(1(5(0(5(0(1(3(3(2(5(1(5(0(3(0(5(x)))))))))))))))))))) → 3(4(0(2(5(5(2(4(2(4(3(1(1(4(4(5(5(3(5(x)))))))))))))))))))
3(3(2(2(3(3(4(0(0(0(2(5(0(5(3(0(0(1(1(4(x)))))))))))))))))))) → 3(1(1(0(3(3(5(4(2(2(1(0(1(1(0(5(0(3(4(0(x))))))))))))))))))))
4(2(4(1(0(5(0(4(1(0(3(0(2(5(4(3(5(3(5(3(x)))))))))))))))))))) → 4(2(5(4(4(2(4(5(0(0(3(1(5(0(2(0(2(2(1(x)))))))))))))))))))
Q is empty.
(1) QTRSRRRProof (EQUIVALENT transformation)
Used ordering:
Polynomial interpretation [POLO]:
POL(0(x1)) = 69 + x1
POL(1(x1)) = 88 + x1
POL(2(x1)) = 85 + x1
POL(3(x1)) = 111 + x1
POL(4(x1)) = 94 + x1
POL(5(x1)) = 94 + x1
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
0(1(2(2(3(0(4(4(x)))))))) → 0(4(5(0(3(4(0(4(x))))))))
0(2(2(2(0(5(2(5(4(x))))))))) → 0(0(0(3(4(3(0(1(0(x)))))))))
2(1(1(3(4(3(1(1(5(x))))))))) → 1(0(2(3(1(0(5(1(5(x)))))))))
5(4(0(4(3(3(1(2(5(3(0(x))))))))))) → 5(5(2(5(3(1(5(0(3(5(2(x)))))))))))
2(4(5(0(1(1(3(3(5(3(0(0(x)))))))))))) → 4(4(2(2(1(0(4(0(1(3(2(0(x))))))))))))
4(4(3(0(3(1(5(3(5(1(3(1(5(3(x)))))))))))))) → 0(2(4(5(0(0(0(5(1(0(5(4(4(x)))))))))))))
2(2(0(0(2(1(0(5(3(2(2(1(4(0(5(x))))))))))))))) → 0(0(3(5(3(0(4(3(1(3(0(2(5(5(x))))))))))))))
5(1(1(4(1(5(3(0(4(3(2(5(4(1(3(3(5(x))))))))))))))))) → 5(0(1(0(4(0(2(4(5(1(5(4(1(5(3(3(5(x)))))))))))))))))
0(1(4(4(3(2(0(4(1(4(3(4(4(1(5(3(4(4(x)))))))))))))))))) → 0(2(4(2(3(1(0(1(1(1(3(0(2(4(4(1(1(2(x))))))))))))))))))
1(1(4(1(0(1(0(3(3(4(4(1(5(4(0(4(4(5(5(3(x)))))))))))))))))))) → 1(3(0(3(2(2(4(4(2(0(3(3(4(0(3(0(4(3(4(0(x))))))))))))))))))))
2(0(3(3(3(4(1(1(0(4(4(0(3(3(3(0(0(1(5(3(x)))))))))))))))))))) → 5(3(3(1(4(0(4(5(4(4(4(2(4(3(1(1(1(5(4(x)))))))))))))))))))
2(3(3(2(1(5(0(5(0(1(3(3(2(5(1(5(0(3(0(5(x)))))))))))))))))))) → 3(4(0(2(5(5(2(4(2(4(3(1(1(4(4(5(5(3(5(x)))))))))))))))))))
3(3(2(2(3(3(4(0(0(0(2(5(0(5(3(0(0(1(1(4(x)))))))))))))))))))) → 3(1(1(0(3(3(5(4(2(2(1(0(1(1(0(5(0(3(4(0(x))))))))))))))))))))
4(2(4(1(0(5(0(4(1(0(3(0(2(5(4(3(5(3(5(3(x)))))))))))))))))))) → 4(2(5(4(4(2(4(5(0(0(3(1(5(0(2(0(2(2(1(x)))))))))))))))))))
(2) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
3(0(0(4(2(5(5(1(3(0(2(3(3(5(1(4(5(x))))))))))))))))) → 3(4(0(1(0(5(5(3(1(4(0(3(5(3(2(2(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(0(0(4(2(5(5(1(3(0(2(3(3(5(1(4(5(x))))))))))))))))) → 3(4(0(1(0(5(5(3(1(4(0(3(5(3(2(2(5(x)))))))))))))))))
The set Q consists of the following terms:
3(0(0(4(2(5(5(1(3(0(2(3(3(5(1(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(0(0(4(2(5(5(1(3(0(2(3(3(5(1(4(5(x))))))))))))))))) → 31(4(0(1(0(5(5(3(1(4(0(3(5(3(2(2(5(x)))))))))))))))))
31(0(0(4(2(5(5(1(3(0(2(3(3(5(1(4(5(x))))))))))))))))) → 31(1(4(0(3(5(3(2(2(5(x))))))))))
31(0(0(4(2(5(5(1(3(0(2(3(3(5(1(4(5(x))))))))))))))))) → 31(5(3(2(2(5(x))))))
31(0(0(4(2(5(5(1(3(0(2(3(3(5(1(4(5(x))))))))))))))))) → 31(2(2(5(x))))
The TRS R consists of the following rules:
3(0(0(4(2(5(5(1(3(0(2(3(3(5(1(4(5(x))))))))))))))))) → 3(4(0(1(0(5(5(3(1(4(0(3(5(3(2(2(5(x)))))))))))))))))
The set Q consists of the following terms:
3(0(0(4(2(5(5(1(3(0(2(3(3(5(1(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 4 less nodes.
(8) TRUE