(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
0(0(1(2(x)))) → 2(3(0(x)))
4(2(5(2(x)))) → 1(4(0(x)))
4(5(3(4(x)))) → 4(4(2(4(x))))
1(3(5(5(2(x))))) → 1(3(2(3(x))))
5(2(2(1(2(x))))) → 0(0(2(3(x))))
2(2(5(3(2(2(x)))))) → 5(1(1(0(3(x)))))
2(5(1(2(1(1(x)))))) → 5(2(1(2(4(1(x))))))
3(4(1(4(2(4(x)))))) → 1(3(3(3(x))))
3(5(2(2(4(5(x)))))) → 3(2(4(3(0(x)))))
5(2(1(0(1(5(x)))))) → 5(4(2(4(5(1(x))))))
1(3(5(4(1(2(2(x))))))) → 3(3(3(4(4(0(x))))))
4(5(4(3(0(5(1(x))))))) → 4(3(3(5(4(1(x))))))
2(1(5(2(1(3(4(4(x)))))))) → 4(0(3(4(0(1(2(x)))))))
5(4(0(2(2(4(0(4(x)))))))) → 3(1(5(1(3(0(4(x)))))))
3(4(2(1(1(2(2(5(4(x))))))))) → 3(3(3(1(3(3(4(x)))))))
5(4(4(5(0(1(4(5(4(x))))))))) → 1(5(5(0(4(1(4(5(4(x)))))))))
5(2(1(3(1(5(2(5(4(4(x)))))))))) → 5(3(4(5(0(1(4(0(3(x)))))))))
2(4(1(2(5(2(4(1(3(2(0(3(x)))))))))))) → 4(3(0(4(2(3(4(3(4(2(0(x)))))))))))
0(2(3(5(4(2(2(1(0(3(3(5(0(x))))))))))))) → 3(3(1(2(3(0(4(0(0(0(2(0(x))))))))))))
2(1(0(2(1(4(0(0(2(0(0(0(5(2(x)))))))))))))) → 2(2(4(2(1(4(3(0(5(1(3(3(0(x)))))))))))))
4(5(0(3(1(3(2(2(5(2(2(4(1(3(2(x))))))))))))))) → 4(5(3(2(1(4(5(0(0(0(4(5(4(0(0(x)))))))))))))))
5(2(2(5(2(4(4(1(2(0(1(1(0(1(1(x))))))))))))))) → 5(0(5(4(3(2(1(0(3(3(5(0(4(1(x))))))))))))))
5(4(2(0(3(3(0(0(4(0(3(2(0(5(1(x))))))))))))))) → 5(0(1(0(0(2(1(1(0(3(2(2(1(3(x))))))))))))))
5(4(1(5(1(5(4(4(2(2(0(4(3(1(5(4(4(3(1(x))))))))))))))))))) → 3(0(4(5(1(1(3(5(3(4(4(4(5(1(4(3(3(1(x))))))))))))))))))
2(0(3(0(2(2(2(0(1(4(2(1(0(4(4(3(3(1(4(4(x)))))))))))))))))))) → 5(2(4(1(1(4(5(1(0(1(2(0(3(0(1(2(3(4(3(1(x))))))))))))))))))))
Q is empty.
(1) QTRSRRRProof (EQUIVALENT transformation)
Used ordering:
Polynomial interpretation [POLO]:
POL(0(x1)) = 22 + x1
POL(1(x1)) = 13 + x1
POL(2(x1)) = 19 + x1
POL(3(x1)) = 20 + x1
POL(4(x1)) = 13 + x1
POL(5(x1)) = 13 + x1
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
0(0(1(2(x)))) → 2(3(0(x)))
4(2(5(2(x)))) → 1(4(0(x)))
4(5(3(4(x)))) → 4(4(2(4(x))))
1(3(5(5(2(x))))) → 1(3(2(3(x))))
2(2(5(3(2(2(x)))))) → 5(1(1(0(3(x)))))
3(4(1(4(2(4(x)))))) → 1(3(3(3(x))))
3(5(2(2(4(5(x)))))) → 3(2(4(3(0(x)))))
5(2(1(0(1(5(x)))))) → 5(4(2(4(5(1(x))))))
1(3(5(4(1(2(2(x))))))) → 3(3(3(4(4(0(x))))))
4(5(4(3(0(5(1(x))))))) → 4(3(3(5(4(1(x))))))
2(1(5(2(1(3(4(4(x)))))))) → 4(0(3(4(0(1(2(x)))))))
5(4(0(2(2(4(0(4(x)))))))) → 3(1(5(1(3(0(4(x)))))))
3(4(2(1(1(2(2(5(4(x))))))))) → 3(3(3(1(3(3(4(x)))))))
2(4(1(2(5(2(4(1(3(2(0(3(x)))))))))))) → 4(3(0(4(2(3(4(3(4(2(0(x)))))))))))
0(2(3(5(4(2(2(1(0(3(3(5(0(x))))))))))))) → 3(3(1(2(3(0(4(0(0(0(2(0(x))))))))))))
2(1(0(2(1(4(0(0(2(0(0(0(5(2(x)))))))))))))) → 2(2(4(2(1(4(3(0(5(1(3(3(0(x)))))))))))))
4(5(0(3(1(3(2(2(5(2(2(4(1(3(2(x))))))))))))))) → 4(5(3(2(1(4(5(0(0(0(4(5(4(0(0(x)))))))))))))))
5(2(2(5(2(4(4(1(2(0(1(1(0(1(1(x))))))))))))))) → 5(0(5(4(3(2(1(0(3(3(5(0(4(1(x))))))))))))))
5(4(2(0(3(3(0(0(4(0(3(2(0(5(1(x))))))))))))))) → 5(0(1(0(0(2(1(1(0(3(2(2(1(3(x))))))))))))))
5(4(1(5(1(5(4(4(2(2(0(4(3(1(5(4(4(3(1(x))))))))))))))))))) → 3(0(4(5(1(1(3(5(3(4(4(4(5(1(4(3(3(1(x))))))))))))))))))
2(0(3(0(2(2(2(0(1(4(2(1(0(4(4(3(3(1(4(4(x)))))))))))))))))))) → 5(2(4(1(1(4(5(1(0(1(2(0(3(0(1(2(3(4(3(1(x))))))))))))))))))))
(2) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
5(2(2(1(2(x))))) → 0(0(2(3(x))))
2(5(1(2(1(1(x)))))) → 5(2(1(2(4(1(x))))))
5(4(4(5(0(1(4(5(4(x))))))))) → 1(5(5(0(4(1(4(5(4(x)))))))))
5(2(1(3(1(5(2(5(4(4(x)))))))))) → 5(3(4(5(0(1(4(0(3(x)))))))))
Q is empty.
(3) QTRSRRRProof (EQUIVALENT transformation)
Used ordering:
Polynomial interpretation [POLO]:
POL(0(x1)) = x1
POL(1(x1)) = x1
POL(2(x1)) = x1
POL(3(x1)) = x1
POL(4(x1)) = x1
POL(5(x1)) = 1 + x1
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
5(2(2(1(2(x))))) → 0(0(2(3(x))))
5(2(1(3(1(5(2(5(4(4(x)))))))))) → 5(3(4(5(0(1(4(0(3(x)))))))))
(4) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
2(5(1(2(1(1(x)))))) → 5(2(1(2(4(1(x))))))
5(4(4(5(0(1(4(5(4(x))))))))) → 1(5(5(0(4(1(4(5(4(x)))))))))
Q is empty.
(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:
21(5(1(2(1(1(x)))))) → 51(2(1(2(4(1(x))))))
21(5(1(2(1(1(x)))))) → 21(1(2(4(1(x)))))
21(5(1(2(1(1(x)))))) → 21(4(1(x)))
51(4(4(5(0(1(4(5(4(x))))))))) → 51(5(0(4(1(4(5(4(x))))))))
51(4(4(5(0(1(4(5(4(x))))))))) → 51(0(4(1(4(5(4(x)))))))
The TRS R consists of the following rules:
2(5(1(2(1(1(x)))))) → 5(2(1(2(4(1(x))))))
5(4(4(5(0(1(4(5(4(x))))))))) → 1(5(5(0(4(1(4(5(4(x)))))))))
Q is empty.
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 5 less nodes.
(8) TRUE