(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
0(1(0(2(x)))) → 3(4(4(x)))
2(0(0(3(x)))) → 3(4(1(x)))
5(3(5(0(1(x))))) → 5(2(3(0(x))))
0(0(1(2(2(3(x)))))) → 4(0(2(2(3(x)))))
0(5(3(1(4(3(x)))))) → 1(4(0(5(2(x)))))
2(0(4(3(5(3(3(1(x)))))))) → 2(1(2(2(2(0(3(1(x))))))))
2(4(0(2(3(0(0(2(x)))))))) → 2(4(0(3(4(4(2(x)))))))
2(1(4(0(4(1(5(0(2(x))))))))) → 3(3(2(2(3(0(3(3(4(x)))))))))
5(0(4(0(0(1(3(5(0(x))))))))) → 5(0(0(0(3(5(5(3(2(x)))))))))
5(3(5(4(4(2(2(2(1(x))))))))) → 5(0(3(0(0(4(5(2(1(x)))))))))
2(0(4(2(3(3(3(5(4(2(x)))))))))) → 3(4(4(2(0(2(3(3(5(2(x))))))))))
4(4(3(2(0(1(1(4(0(2(x)))))))))) → 2(3(4(0(0(1(2(1(3(4(x))))))))))
1(5(5(2(4(2(4(0(3(3(0(x))))))))))) → 1(3(5(4(5(5(5(3(2(4(2(x)))))))))))
2(5(5(0(1(1(5(1(4(2(3(3(x)))))))))))) → 4(2(3(0(3(0(0(2(3(2(3(x)))))))))))
3(0(2(4(5(2(1(2(0(2(5(1(x)))))))))))) → 3(1(3(0(5(0(2(2(4(0(4(x)))))))))))
3(3(3(4(3(0(0(4(4(5(0(5(x)))))))))))) → 3(1(5(5(0(3(1(0(5(2(5(x)))))))))))
0(5(3(5(1(0(5(1(2(4(4(5(0(x))))))))))))) → 0(3(1(0(4(5(5(2(5(5(2(4(0(x)))))))))))))
2(5(2(4(1(5(3(3(1(0(2(5(3(x))))))))))))) → 2(3(1(3(3(5(1(2(0(3(2(3(0(x)))))))))))))
2(5(5(3(5(3(1(3(1(2(5(0(0(x))))))))))))) → 2(5(5(5(0(4(5(5(1(1(3(2(2(x)))))))))))))
3(5(3(3(3(2(3(0(2(4(2(1(3(x))))))))))))) → 5(1(3(1(4(0(0(1(5(0(4(5(x))))))))))))
2(0(3(0(2(3(1(0(0(3(0(3(5(3(x)))))))))))))) → 3(4(0(2(1(4(4(4(0(5(5(5(2(x)))))))))))))
3(5(3(3(1(2(4(0(4(3(5(4(2(1(x)))))))))))))) → 5(2(0(1(4(4(3(3(3(5(3(4(1(2(x))))))))))))))
3(5(5(0(2(2(0(0(3(5(1(3(3(5(3(5(x)))))))))))))))) → 3(1(4(2(0(3(4(0(4(1(5(4(2(5(x))))))))))))))
5(5(3(5(0(5(3(5(2(3(1(3(3(4(0(2(x)))))))))))))))) → 5(2(4(2(4(4(0(5(5(1(3(4(0(4(0(x)))))))))))))))
2(2(1(3(4(0(1(1(4(2(2(2(0(4(4(2(4(4(x)))))))))))))))))) → 4(3(4(4(5(5(0(1(0(1(3(2(5(5(1(1(4(4(x))))))))))))))))))
0(3(0(2(2(2(5(5(2(4(4(3(5(1(0(0(0(3(3(x))))))))))))))))))) → 3(3(1(3(4(5(5(2(0(4(2(3(1(1(5(1(2(3(x))))))))))))))))))
3(3(4(3(3(4(3(0(2(5(3(1(4(5(2(5(2(4(3(5(x)))))))))))))))))))) → 5(2(5(0(5(0(1(1(1(3(2(4(4(0(3(2(4(1(5(x)))))))))))))))))))
2(1(5(5(3(0(1(3(3(3(1(2(0(5(2(0(3(5(2(2(2(x))))))))))))))))))))) → 3(0(3(3(4(0(1(4(5(1(3(3(3(4(1(2(1(3(2(1(x))))))))))))))))))))
5(1(3(4(0(3(0(2(5(3(0(2(2(0(1(2(3(3(4(1(1(x))))))))))))))))))))) → 5(0(1(1(4(2(0(1(1(2(1(4(0(4(3(1(2(0(0(2(x))))))))))))))))))))
5(4(0(0(3(2(5(3(0(3(0(4(5(0(4(4(5(1(1(1(3(x))))))))))))))))))))) → 5(2(5(1(0(3(5(3(2(1(0(0(5(5(4(5(4(4(2(2(2(x)))))))))))))))))))))
Q is empty.
(1) QTRSRRRProof (EQUIVALENT transformation)
Used ordering:
Polynomial interpretation [POLO]:
POL(0(x1)) = 130 + x1
POL(1(x1)) = 129 + x1
POL(2(x1)) = 120 + x1
POL(3(x1)) = 132 + x1
POL(4(x1)) = 151 + x1
POL(5(x1)) = 122 + x1
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
0(1(0(2(x)))) → 3(4(4(x)))
2(0(0(3(x)))) → 3(4(1(x)))
5(3(5(0(1(x))))) → 5(2(3(0(x))))
0(0(1(2(2(3(x)))))) → 4(0(2(2(3(x)))))
0(5(3(1(4(3(x)))))) → 1(4(0(5(2(x)))))
2(0(4(3(5(3(3(1(x)))))))) → 2(1(2(2(2(0(3(1(x))))))))
2(4(0(2(3(0(0(2(x)))))))) → 2(4(0(3(4(4(2(x)))))))
2(1(4(0(4(1(5(0(2(x))))))))) → 3(3(2(2(3(0(3(3(4(x)))))))))
5(0(4(0(0(1(3(5(0(x))))))))) → 5(0(0(0(3(5(5(3(2(x)))))))))
5(3(5(4(4(2(2(2(1(x))))))))) → 5(0(3(0(0(4(5(2(1(x)))))))))
4(4(3(2(0(1(1(4(0(2(x)))))))))) → 2(3(4(0(0(1(2(1(3(4(x))))))))))
1(5(5(2(4(2(4(0(3(3(0(x))))))))))) → 1(3(5(4(5(5(5(3(2(4(2(x)))))))))))
2(5(5(0(1(1(5(1(4(2(3(3(x)))))))))))) → 4(2(3(0(3(0(0(2(3(2(3(x)))))))))))
3(0(2(4(5(2(1(2(0(2(5(1(x)))))))))))) → 3(1(3(0(5(0(2(2(4(0(4(x)))))))))))
3(3(3(4(3(0(0(4(4(5(0(5(x)))))))))))) → 3(1(5(5(0(3(1(0(5(2(5(x)))))))))))
0(5(3(5(1(0(5(1(2(4(4(5(0(x))))))))))))) → 0(3(1(0(4(5(5(2(5(5(2(4(0(x)))))))))))))
2(5(2(4(1(5(3(3(1(0(2(5(3(x))))))))))))) → 2(3(1(3(3(5(1(2(0(3(2(3(0(x)))))))))))))
2(5(5(3(5(3(1(3(1(2(5(0(0(x))))))))))))) → 2(5(5(5(0(4(5(5(1(1(3(2(2(x)))))))))))))
3(5(3(3(3(2(3(0(2(4(2(1(3(x))))))))))))) → 5(1(3(1(4(0(0(1(5(0(4(5(x))))))))))))
2(0(3(0(2(3(1(0(0(3(0(3(5(3(x)))))))))))))) → 3(4(0(2(1(4(4(4(0(5(5(5(2(x)))))))))))))
3(5(5(0(2(2(0(0(3(5(1(3(3(5(3(5(x)))))))))))))))) → 3(1(4(2(0(3(4(0(4(1(5(4(2(5(x))))))))))))))
5(5(3(5(0(5(3(5(2(3(1(3(3(4(0(2(x)))))))))))))))) → 5(2(4(2(4(4(0(5(5(1(3(4(0(4(0(x)))))))))))))))
2(2(1(3(4(0(1(1(4(2(2(2(0(4(4(2(4(4(x)))))))))))))))))) → 4(3(4(4(5(5(0(1(0(1(3(2(5(5(1(1(4(4(x))))))))))))))))))
0(3(0(2(2(2(5(5(2(4(4(3(5(1(0(0(0(3(3(x))))))))))))))))))) → 3(3(1(3(4(5(5(2(0(4(2(3(1(1(5(1(2(3(x))))))))))))))))))
3(3(4(3(3(4(3(0(2(5(3(1(4(5(2(5(2(4(3(5(x)))))))))))))))))))) → 5(2(5(0(5(0(1(1(1(3(2(4(4(0(3(2(4(1(5(x)))))))))))))))))))
2(1(5(5(3(0(1(3(3(3(1(2(0(5(2(0(3(5(2(2(2(x))))))))))))))))))))) → 3(0(3(3(4(0(1(4(5(1(3(3(3(4(1(2(1(3(2(1(x))))))))))))))))))))
5(1(3(4(0(3(0(2(5(3(0(2(2(0(1(2(3(3(4(1(1(x))))))))))))))))))))) → 5(0(1(1(4(2(0(1(1(2(1(4(0(4(3(1(2(0(0(2(x))))))))))))))))))))
5(4(0(0(3(2(5(3(0(3(0(4(5(0(4(4(5(1(1(1(3(x))))))))))))))))))))) → 5(2(5(1(0(3(5(3(2(1(0(0(5(5(4(5(4(4(2(2(2(x)))))))))))))))))))))
(2) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
2(0(4(2(3(3(3(5(4(2(x)))))))))) → 3(4(4(2(0(2(3(3(5(2(x))))))))))
3(5(3(3(1(2(4(0(4(3(5(4(2(1(x)))))))))))))) → 5(2(0(1(4(4(3(3(3(5(3(4(1(2(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(0(4(2(3(3(3(5(4(2(x)))))))))) → 31(4(4(2(0(2(3(3(5(2(x))))))))))
21(0(4(2(3(3(3(5(4(2(x)))))))))) → 21(0(2(3(3(5(2(x)))))))
21(0(4(2(3(3(3(5(4(2(x)))))))))) → 21(3(3(5(2(x)))))
21(0(4(2(3(3(3(5(4(2(x)))))))))) → 31(3(5(2(x))))
21(0(4(2(3(3(3(5(4(2(x)))))))))) → 31(5(2(x)))
31(5(3(3(1(2(4(0(4(3(5(4(2(1(x)))))))))))))) → 21(0(1(4(4(3(3(3(5(3(4(1(2(x)))))))))))))
31(5(3(3(1(2(4(0(4(3(5(4(2(1(x)))))))))))))) → 31(3(3(5(3(4(1(2(x))))))))
31(5(3(3(1(2(4(0(4(3(5(4(2(1(x)))))))))))))) → 31(3(5(3(4(1(2(x)))))))
31(5(3(3(1(2(4(0(4(3(5(4(2(1(x)))))))))))))) → 31(5(3(4(1(2(x))))))
31(5(3(3(1(2(4(0(4(3(5(4(2(1(x)))))))))))))) → 31(4(1(2(x))))
31(5(3(3(1(2(4(0(4(3(5(4(2(1(x)))))))))))))) → 21(x)
The TRS R consists of the following rules:
2(0(4(2(3(3(3(5(4(2(x)))))))))) → 3(4(4(2(0(2(3(3(5(2(x))))))))))
3(5(3(3(1(2(4(0(4(3(5(4(2(1(x)))))))))))))) → 5(2(0(1(4(4(3(3(3(5(3(4(1(2(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 0 SCCs with 11 less nodes.
(6) TRUE