(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
0(1(1(1(x)))) → 2(1(3(x)))
1(2(3(1(x)))) → 1(1(1(1(x))))
1(2(0(3(4(x))))) → 3(0(2(2(x))))
3(2(0(5(3(5(x)))))) → 3(1(2(4(2(5(x))))))
1(3(4(3(1(5(3(x))))))) → 5(2(0(0(4(0(3(x)))))))
4(2(2(4(2(4(4(x))))))) → 4(2(2(3(0(4(x))))))
1(3(3(3(4(4(1(1(x)))))))) → 1(2(4(4(3(4(0(1(x))))))))
3(0(0(2(5(0(0(1(x)))))))) → 0(2(4(1(4(2(0(1(x))))))))
4(0(1(0(0(1(2(2(1(x))))))))) → 2(4(0(0(4(5(5(5(x))))))))
2(1(3(3(0(1(3(2(5(1(x)))))))))) → 2(2(0(1(2(1(2(1(1(3(x))))))))))
3(3(3(5(3(0(2(1(4(3(x)))))))))) → 4(3(5(2(3(3(1(3(0(3(x))))))))))
3(4(3(5(1(3(1(2(2(5(x)))))))))) → 1(0(4(0(1(5(3(5(2(1(x))))))))))
5(0(1(1(0(2(2(0(1(1(x)))))))))) → 4(5(5(5(2(3(2(0(3(x)))))))))
4(0(1(1(0(0(1(5(3(5(0(1(x)))))))))))) → 2(3(4(4(5(4(5(5(0(4(2(x)))))))))))
5(5(4(2(0(5(2(4(3(2(5(5(x)))))))))))) → 3(4(0(0(1(1(3(5(1(0(1(4(x))))))))))))
3(4(0(1(1(5(3(5(0(4(4(3(4(3(x)))))))))))))) → 3(5(5(4(0(1(1(1(1(4(4(0(0(3(x))))))))))))))
4(0(0(2(4(5(4(2(1(4(3(4(4(3(x)))))))))))))) → 4(4(0(1(5(2(2(1(1(0(1(3(1(3(x))))))))))))))
1(5(2(4(4(0(2(4(5(5(1(2(3(3(1(x))))))))))))))) → 3(1(2(4(5(4(0(3(2(1(5(5(1(3(1(x)))))))))))))))
2(0(2(0(5(5(5(1(4(4(4(3(3(3(0(x))))))))))))))) → 5(3(3(3(2(5(0(5(4(4(5(3(3(0(x))))))))))))))
5(0(4(2(1(1(2(4(1(0(3(1(0(3(5(x))))))))))))))) → 4(5(3(2(3(1(1(0(1(3(5(3(0(1(x))))))))))))))
5(0(4(5(4(0(4(0(4(5(2(0(1(2(1(x))))))))))))))) → 1(2(2(2(0(4(0(5(1(3(4(4(1(1(4(x)))))))))))))))
5(3(1(2(3(2(2(2(4(1(0(4(3(3(4(0(x)))))))))))))))) → 5(5(4(0(2(5(0(0(1(3(0(3(4(3(4(0(x))))))))))))))))
1(3(4(1(3(3(0(2(3(3(3(2(5(4(4(4(4(x))))))))))))))))) → 1(1(3(2(5(0(4(2(4(3(4(0(3(3(3(0(4(x)))))))))))))))))
3(4(3(3(4(3(2(4(1(1(2(1(4(5(3(0(0(x))))))))))))))))) → 1(1(3(0(5(3(2(0(4(4(1(3(5(0(0(2(0(x)))))))))))))))))
5(3(3(4(5(3(5(1(2(5(5(0(5(5(1(5(3(x))))))))))))))))) → 5(5(3(5(2(1(3(3(5(1(3(5(0(5(4(3(x))))))))))))))))
2(0(1(2(0(3(5(1(4(5(2(3(5(2(4(1(0(3(x)))))))))))))))))) → 1(4(5(4(0(5(5(1(3(0(1(4(4(0(3(3(4(3(x))))))))))))))))))
5(1(1(0(2(2(1(1(5(5(2(2(4(0(4(3(2(2(x)))))))))))))))))) → 3(1(1(1(2(2(1(5(4(3(0(2(3(5(4(2(2(x)))))))))))))))))
2(2(3(0(5(2(3(4(4(0(3(3(0(2(0(2(4(1(1(x))))))))))))))))))) → 2(1(5(2(1(1(5(4(5(4(3(3(5(4(5(5(3(4(x))))))))))))))))))
1(3(5(3(3(3(4(0(1(2(3(3(0(4(3(0(3(4(4(5(x)))))))))))))))))))) → 4(2(4(5(4(5(1(4(4(5(5(0(2(1(1(1(1(4(2(5(1(x)))))))))))))))))))))
5(0(4(2(0(0(0(3(3(1(5(2(2(5(1(3(4(3(0(5(1(x))))))))))))))))))))) → 0(3(1(5(4(3(1(1(3(0(3(2(0(0(2(2(2(1(4(4(1(x)))))))))))))))))))))
Q is empty.
(1) QTRSRRRProof (EQUIVALENT transformation)
Used ordering:
Polynomial interpretation [POLO]:
POL(0(x1)) = 23 + x1
POL(1(x1)) = 21 + x1
POL(2(x1)) = 26 + x1
POL(3(x1)) = 27 + x1
POL(4(x1)) = 25 + x1
POL(5(x1)) = 25 + x1
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
0(1(1(1(x)))) → 2(1(3(x)))
1(2(3(1(x)))) → 1(1(1(1(x))))
1(2(0(3(4(x))))) → 3(0(2(2(x))))
3(2(0(5(3(5(x)))))) → 3(1(2(4(2(5(x))))))
1(3(4(3(1(5(3(x))))))) → 5(2(0(0(4(0(3(x)))))))
4(2(2(4(2(4(4(x))))))) → 4(2(2(3(0(4(x))))))
1(3(3(3(4(4(1(1(x)))))))) → 1(2(4(4(3(4(0(1(x))))))))
3(0(0(2(5(0(0(1(x)))))))) → 0(2(4(1(4(2(0(1(x))))))))
4(0(1(0(0(1(2(2(1(x))))))))) → 2(4(0(0(4(5(5(5(x))))))))
2(1(3(3(0(1(3(2(5(1(x)))))))))) → 2(2(0(1(2(1(2(1(1(3(x))))))))))
3(4(3(5(1(3(1(2(2(5(x)))))))))) → 1(0(4(0(1(5(3(5(2(1(x))))))))))
5(0(1(1(0(2(2(0(1(1(x)))))))))) → 4(5(5(5(2(3(2(0(3(x)))))))))
4(0(1(1(0(0(1(5(3(5(0(1(x)))))))))))) → 2(3(4(4(5(4(5(5(0(4(2(x)))))))))))
5(5(4(2(0(5(2(4(3(2(5(5(x)))))))))))) → 3(4(0(0(1(1(3(5(1(0(1(4(x))))))))))))
3(4(0(1(1(5(3(5(0(4(4(3(4(3(x)))))))))))))) → 3(5(5(4(0(1(1(1(1(4(4(0(0(3(x))))))))))))))
4(0(0(2(4(5(4(2(1(4(3(4(4(3(x)))))))))))))) → 4(4(0(1(5(2(2(1(1(0(1(3(1(3(x))))))))))))))
1(5(2(4(4(0(2(4(5(5(1(2(3(3(1(x))))))))))))))) → 3(1(2(4(5(4(0(3(2(1(5(5(1(3(1(x)))))))))))))))
2(0(2(0(5(5(5(1(4(4(4(3(3(3(0(x))))))))))))))) → 5(3(3(3(2(5(0(5(4(4(5(3(3(0(x))))))))))))))
5(0(4(2(1(1(2(4(1(0(3(1(0(3(5(x))))))))))))))) → 4(5(3(2(3(1(1(0(1(3(5(3(0(1(x))))))))))))))
5(0(4(5(4(0(4(0(4(5(2(0(1(2(1(x))))))))))))))) → 1(2(2(2(0(4(0(5(1(3(4(4(1(1(4(x)))))))))))))))
5(3(1(2(3(2(2(2(4(1(0(4(3(3(4(0(x)))))))))))))))) → 5(5(4(0(2(5(0(0(1(3(0(3(4(3(4(0(x))))))))))))))))
1(3(4(1(3(3(0(2(3(3(3(2(5(4(4(4(4(x))))))))))))))))) → 1(1(3(2(5(0(4(2(4(3(4(0(3(3(3(0(4(x)))))))))))))))))
3(4(3(3(4(3(2(4(1(1(2(1(4(5(3(0(0(x))))))))))))))))) → 1(1(3(0(5(3(2(0(4(4(1(3(5(0(0(2(0(x)))))))))))))))))
5(3(3(4(5(3(5(1(2(5(5(0(5(5(1(5(3(x))))))))))))))))) → 5(5(3(5(2(1(3(3(5(1(3(5(0(5(4(3(x))))))))))))))))
2(0(1(2(0(3(5(1(4(5(2(3(5(2(4(1(0(3(x)))))))))))))))))) → 1(4(5(4(0(5(5(1(3(0(1(4(4(0(3(3(4(3(x))))))))))))))))))
5(1(1(0(2(2(1(1(5(5(2(2(4(0(4(3(2(2(x)))))))))))))))))) → 3(1(1(1(2(2(1(5(4(3(0(2(3(5(4(2(2(x)))))))))))))))))
2(2(3(0(5(2(3(4(4(0(3(3(0(2(0(2(4(1(1(x))))))))))))))))))) → 2(1(5(2(1(1(5(4(5(4(3(3(5(4(5(5(3(4(x))))))))))))))))))
1(3(5(3(3(3(4(0(1(2(3(3(0(4(3(0(3(4(4(5(x)))))))))))))))))))) → 4(2(4(5(4(5(1(4(4(5(5(0(2(1(1(1(1(4(2(5(1(x)))))))))))))))))))))
5(0(4(2(0(0(0(3(3(1(5(2(2(5(1(3(4(3(0(5(1(x))))))))))))))))))))) → 0(3(1(5(4(3(1(1(3(0(3(2(0(0(2(2(2(1(4(4(1(x)))))))))))))))))))))
(2) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
3(3(3(5(3(0(2(1(4(3(x)))))))))) → 4(3(5(2(3(3(1(3(0(3(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:
31(3(3(5(3(0(2(1(4(3(x)))))))))) → 31(5(2(3(3(1(3(0(3(x)))))))))
31(3(3(5(3(0(2(1(4(3(x)))))))))) → 31(3(1(3(0(3(x))))))
31(3(3(5(3(0(2(1(4(3(x)))))))))) → 31(1(3(0(3(x)))))
31(3(3(5(3(0(2(1(4(3(x)))))))))) → 31(0(3(x)))
The TRS R consists of the following rules:
3(3(3(5(3(0(2(1(4(3(x)))))))))) → 4(3(5(2(3(3(1(3(0(3(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 4 less nodes.
(6) TRUE