(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
0(1(2(3(x)))) → 4(4(2(3(x))))
5(3(3(5(4(x))))) → 5(1(0(2(x))))
2(3(1(5(0(5(x)))))) → 2(1(3(5(0(5(x))))))
5(3(3(5(5(4(x)))))) → 4(2(4(3(2(x)))))
5(1(4(5(1(1(5(x))))))) → 1(4(0(2(3(2(5(x)))))))
3(3(4(3(1(3(0(5(x)))))))) → 3(5(2(4(5(0(5(2(x))))))))
3(1(2(2(2(1(3(1(3(x))))))))) → 1(4(3(1(5(0(2(2(x))))))))
3(4(2(0(5(2(3(5(3(x))))))))) → 3(5(4(4(2(2(0(5(1(x)))))))))
5(5(1(3(3(5(4(0(0(x))))))))) → 3(1(0(1(4(2(4(3(x))))))))
3(0(2(5(1(5(0(1(5(0(x)))))))))) → 1(2(2(0(0(4(3(4(4(x)))))))))
3(5(5(4(4(4(2(0(0(3(x)))))))))) → 1(1(2(3(2(3(4(1(x))))))))
3(0(4(3(3(5(0(4(4(0(4(2(x)))))))))))) → 3(4(5(5(3(2(0(5(1(4(2(x)))))))))))
5(2(0(4(5(0(2(1(1(1(2(0(x)))))))))))) → 3(0(0(2(2(4(5(1(3(1(0(x)))))))))))
5(5(4(3(3(4(5(4(5(0(0(4(5(x))))))))))))) → 5(0(1(0(3(1(4(1(2(3(1(x)))))))))))
5(2(1(3(0(2(2(4(5(2(2(0(0(1(x)))))))))))))) → 3(4(5(1(4(3(3(5(0(3(0(1(x))))))))))))
3(1(5(2(5(5(3(3(4(4(5(2(3(2(4(x))))))))))))))) → 3(0(5(4(4(4(2(0(0(1(4(3(2(4(x))))))))))))))
4(5(5(4(3(4(4(2(4(2(4(3(3(3(3(x))))))))))))))) → 4(5(0(0(4(4(5(4(4(3(4(0(0(0(x))))))))))))))
0(1(2(4(3(1(1(4(1(5(0(2(5(3(2(4(3(x))))))))))))))))) → 4(2(2(1(3(1(3(0(4(5(1(2(2(5(5(4(1(x)))))))))))))))))
2(4(3(0(4(2(0(0(2(5(1(0(2(0(0(4(4(x))))))))))))))))) → 5(4(1(2(1(2(1(0(2(0(4(3(1(0(0(2(x))))))))))))))))
3(3(3(1(0(2(1(1(5(2(4(0(0(4(5(2(2(0(2(x))))))))))))))))))) → 3(2(2(3(1(5(5(5(3(0(3(1(4(3(2(3(1(x)))))))))))))))))
5(3(2(2(5(2(1(3(0(2(4(3(2(5(3(3(0(5(4(x))))))))))))))))))) → 1(3(0(3(3(4(5(5(0(5(5(4(0(2(1(1(0(0(2(x)))))))))))))))))))
5(4(5(5(5(2(0(1(2(1(0(1(2(1(5(3(1(3(1(x))))))))))))))))))) → 0(0(3(5(3(0(2(0(1(4(0(5(4(3(0(2(4(1(x))))))))))))))))))
4(0(4(0(5(1(0(3(2(5(3(1(3(0(2(5(3(5(0(0(x)))))))))))))))))))) → 1(5(3(5(2(0(5(4(4(5(0(1(4(4(3(1(3(2(5(1(x))))))))))))))))))))
5(4(2(1(3(2(5(4(2(2(0(0(5(5(1(0(5(1(3(0(x)))))))))))))))))))) → 4(4(2(4(0(1(3(2(5(1(3(4(4(0(0(1(1(1(2(0(x))))))))))))))))))))
3(0(4(5(4(1(4(3(5(5(3(5(4(0(1(4(3(5(0(3(2(x))))))))))))))))))))) → 1(2(4(1(1(2(5(4(2(4(0(4(2(5(1(4(2(1(3(1(2(x)))))))))))))))))))))
Q is empty.
(1) QTRSRRRProof (EQUIVALENT transformation)
Used ordering:
Polynomial interpretation [POLO]:
POL(0(x1)) = 7 + x1
POL(1(x1)) = 7 + x1
POL(2(x1)) = 6 + x1
POL(3(x1)) = 10 + x1
POL(4(x1)) = 6 + x1
POL(5(x1)) = 8 + x1
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
0(1(2(3(x)))) → 4(4(2(3(x))))
5(3(3(5(4(x))))) → 5(1(0(2(x))))
5(3(3(5(5(4(x)))))) → 4(2(4(3(2(x)))))
5(1(4(5(1(1(5(x))))))) → 1(4(0(2(3(2(5(x)))))))
3(3(4(3(1(3(0(5(x)))))))) → 3(5(2(4(5(0(5(2(x))))))))
3(1(2(2(2(1(3(1(3(x))))))))) → 1(4(3(1(5(0(2(2(x))))))))
3(4(2(0(5(2(3(5(3(x))))))))) → 3(5(4(4(2(2(0(5(1(x)))))))))
5(5(1(3(3(5(4(0(0(x))))))))) → 3(1(0(1(4(2(4(3(x))))))))
3(0(2(5(1(5(0(1(5(0(x)))))))))) → 1(2(2(0(0(4(3(4(4(x)))))))))
3(5(5(4(4(4(2(0(0(3(x)))))))))) → 1(1(2(3(2(3(4(1(x))))))))
3(0(4(3(3(5(0(4(4(0(4(2(x)))))))))))) → 3(4(5(5(3(2(0(5(1(4(2(x)))))))))))
5(2(0(4(5(0(2(1(1(1(2(0(x)))))))))))) → 3(0(0(2(2(4(5(1(3(1(0(x)))))))))))
5(5(4(3(3(4(5(4(5(0(0(4(5(x))))))))))))) → 5(0(1(0(3(1(4(1(2(3(1(x)))))))))))
5(2(1(3(0(2(2(4(5(2(2(0(0(1(x)))))))))))))) → 3(4(5(1(4(3(3(5(0(3(0(1(x))))))))))))
3(1(5(2(5(5(3(3(4(4(5(2(3(2(4(x))))))))))))))) → 3(0(5(4(4(4(2(0(0(1(4(3(2(4(x))))))))))))))
4(5(5(4(3(4(4(2(4(2(4(3(3(3(3(x))))))))))))))) → 4(5(0(0(4(4(5(4(4(3(4(0(0(0(x))))))))))))))
0(1(2(4(3(1(1(4(1(5(0(2(5(3(2(4(3(x))))))))))))))))) → 4(2(2(1(3(1(3(0(4(5(1(2(2(5(5(4(1(x)))))))))))))))))
2(4(3(0(4(2(0(0(2(5(1(0(2(0(0(4(4(x))))))))))))))))) → 5(4(1(2(1(2(1(0(2(0(4(3(1(0(0(2(x))))))))))))))))
3(3(3(1(0(2(1(1(5(2(4(0(0(4(5(2(2(0(2(x))))))))))))))))))) → 3(2(2(3(1(5(5(5(3(0(3(1(4(3(2(3(1(x)))))))))))))))))
5(3(2(2(5(2(1(3(0(2(4(3(2(5(3(3(0(5(4(x))))))))))))))))))) → 1(3(0(3(3(4(5(5(0(5(5(4(0(2(1(1(0(0(2(x)))))))))))))))))))
5(4(5(5(5(2(0(1(2(1(0(1(2(1(5(3(1(3(1(x))))))))))))))))))) → 0(0(3(5(3(0(2(0(1(4(0(5(4(3(0(2(4(1(x))))))))))))))))))
4(0(4(0(5(1(0(3(2(5(3(1(3(0(2(5(3(5(0(0(x)))))))))))))))))))) → 1(5(3(5(2(0(5(4(4(5(0(1(4(4(3(1(3(2(5(1(x))))))))))))))))))))
5(4(2(1(3(2(5(4(2(2(0(0(5(5(1(0(5(1(3(0(x)))))))))))))))))))) → 4(4(2(4(0(1(3(2(5(1(3(4(4(0(0(1(1(1(2(0(x))))))))))))))))))))
3(0(4(5(4(1(4(3(5(5(3(5(4(0(1(4(3(5(0(3(2(x))))))))))))))))))))) → 1(2(4(1(1(2(5(4(2(4(0(4(2(5(1(4(2(1(3(1(2(x)))))))))))))))))))))
(2) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
2(3(1(5(0(5(x)))))) → 2(1(3(5(0(5(x))))))
Q is empty.
(3) AAECC Innermost (EQUIVALENT transformation)
We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is none
The TRS R 2 is
2(3(1(5(0(5(x)))))) → 2(1(3(5(0(5(x))))))
The signature Sigma is {
2}
(4) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
2(3(1(5(0(5(x)))))) → 2(1(3(5(0(5(x))))))
The set Q consists of the following terms:
2(3(1(5(0(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:
21(3(1(5(0(5(x)))))) → 21(1(3(5(0(5(x))))))
The TRS R consists of the following rules:
2(3(1(5(0(5(x)))))) → 2(1(3(5(0(5(x))))))
The set Q consists of the following terms:
2(3(1(5(0(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