(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
0(1(0(1(x)))) → 0(2(3(1(x))))
2(1(4(0(1(x))))) → 3(0(0(1(x))))
2(3(3(3(1(x))))) → 1(1(3(1(x))))
4(1(2(3(4(x))))) → 1(5(3(0(x))))
5(3(2(4(4(x))))) → 0(1(3(0(x))))
5(5(0(4(3(3(x)))))) → 0(3(2(1(3(x)))))
5(0(3(1(4(1(0(1(0(x))))))))) → 2(2(4(1(3(2(2(3(0(x)))))))))
5(5(5(4(4(0(4(0(2(1(x)))))))))) → 5(4(5(4(0(5(0(4(0(3(x))))))))))
2(2(2(0(3(3(2(0(4(5(0(x))))))))))) → 0(4(0(0(0(0(2(5(1(x)))))))))
1(0(2(0(3(5(1(4(5(2(4(4(x)))))))))))) → 1(0(1(0(3(2(3(2(1(0(3(3(x))))))))))))
4(0(4(2(3(5(0(1(4(4(5(0(x)))))))))))) → 3(5(0(2(5(1(2(5(1(4(1(x)))))))))))
4(1(0(2(3(5(4(1(4(4(1(1(x)))))))))))) → 2(5(4(4(5(5(1(2(4(4(2(x)))))))))))
2(0(0(5(5(0(2(2(2(3(4(1(4(x))))))))))))) → 3(3(3(3(1(3(1(4(1(4(5(4(3(x)))))))))))))
4(5(1(3(0(0(3(5(2(1(2(5(2(x))))))))))))) → 4(5(0(3(5(3(5(5(3(5(4(0(4(x)))))))))))))
2(1(4(5(5(0(2(4(0(5(4(4(1(4(x)))))))))))))) → 0(0(2(1(4(3(2(5(3(2(5(2(1(x)))))))))))))
3(1(3(5(4(1(1(4(5(1(1(3(5(1(x)))))))))))))) → 3(2(1(2(4(3(5(1(5(5(1(3(4(x)))))))))))))
4(4(5(3(0(2(0(0(0(3(5(0(2(1(x)))))))))))))) → 2(2(0(4(1(0(0(1(4(2(4(1(3(5(x))))))))))))))
5(0(5(4(4(4(4(2(5(1(2(3(3(1(1(x))))))))))))))) → 1(3(0(5(3(1(3(1(0(3(3(1(0(3(x))))))))))))))
5(3(5(5(4(0(2(2(4(3(2(2(1(4(1(x))))))))))))))) → 5(1(4(2(3(5(4(0(4(4(2(1(0(0(x))))))))))))))
5(5(2(1(3(3(2(5(0(1(0(5(0(5(0(x))))))))))))))) → 2(3(5(0(2(4(1(0(5(4(3(0(5(0(x))))))))))))))
2(1(2(3(2(1(2(5(5(5(0(0(3(2(1(1(x)))))))))))))))) → 4(4(1(1(3(3(1(4(2(1(3(0(3(2(3(x)))))))))))))))
3(3(3(1(4(2(0(0(3(0(1(1(3(4(1(2(x)))))))))))))))) → 3(0(4(1(5(0(5(3(5(0(4(4(3(0(4(x)))))))))))))))
2(1(1(3(5(4(0(1(4(2(5(1(4(5(4(5(1(x))))))))))))))))) → 2(3(3(3(3(2(1(4(5(5(2(2(4(4(4(3(0(x)))))))))))))))))
5(0(2(4(4(4(4(3(1(2(2(2(2(2(2(5(2(x))))))))))))))))) → 5(4(0(4(4(3(5(5(4(3(0(5(5(5(0(0(x))))))))))))))))
2(5(5(1(0(4(4(1(4(0(5(1(1(5(0(0(1(4(x)))))))))))))))))) → 3(3(1(0(0(1(1(1(2(1(2(0(2(4(4(3(5(0(x))))))))))))))))))
0(0(1(1(4(5(3(2(5(3(4(4(3(0(5(4(4(4(2(x))))))))))))))))))) → 0(0(4(5(1(4(1(5(3(0(4(0(3(3(4(0(4(0(4(x)))))))))))))))))))
1(4(1(4(1(5(2(4(4(3(3(4(5(4(4(1(2(0(2(3(x)))))))))))))))))))) → 1(2(1(5(0(1(0(4(4(4(2(5(0(1(5(1(1(4(3(3(x))))))))))))))))))))
2(1(0(0(4(2(5(5(1(3(3(4(3(3(5(5(3(3(5(2(x)))))))))))))))))))) → 4(4(3(3(5(0(4(2(2(1(1(3(1(5(2(5(2(4(5(x)))))))))))))))))))
4(2(4(2(4(5(4(5(2(1(5(3(1(0(2(5(2(1(4(2(4(x))))))))))))))))))))) → 1(5(3(1(5(0(5(2(0(5(4(3(5(2(1(2(5(4(3(2(1(x)))))))))))))))))))))
5(0(2(1(0(3(4(3(4(1(5(3(3(3(4(2(0(5(5(3(5(x))))))))))))))))))))) → 3(5(1(1(0(5(4(5(2(1(5(0(4(4(4(3(4(4(1(5(x))))))))))))))))))))
Q is empty.
(1) QTRS Reverse (EQUIVALENT transformation)
We applied the QTRS Reverse Processor [REVERSE].
(2) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
1(0(1(0(x)))) → 1(3(2(0(x))))
1(0(4(1(2(x))))) → 1(0(0(3(x))))
1(3(3(3(2(x))))) → 1(3(1(1(x))))
4(3(2(1(4(x))))) → 0(3(5(1(x))))
4(4(2(3(5(x))))) → 0(3(1(0(x))))
3(3(4(0(5(5(x)))))) → 3(1(2(3(0(x)))))
0(1(0(1(4(1(3(0(5(x))))))))) → 0(3(2(2(3(1(4(2(2(x)))))))))
1(2(0(4(0(4(4(5(5(5(x)))))))))) → 3(0(4(0(5(0(4(5(4(5(x))))))))))
0(5(4(0(2(3(3(0(2(2(2(x))))))))))) → 1(5(2(0(0(0(0(4(0(x)))))))))
4(4(2(5(4(1(5(3(0(2(0(1(x)))))))))))) → 3(3(0(1(2(3(2(3(0(1(0(1(x))))))))))))
0(5(4(4(1(0(5(3(2(4(0(4(x)))))))))))) → 1(4(1(5(2(1(5(2(0(5(3(x)))))))))))
1(1(4(4(1(4(5(3(2(0(1(4(x)))))))))))) → 2(4(4(2(1(5(5(4(4(5(2(x)))))))))))
4(1(4(3(2(2(2(0(5(5(0(0(2(x))))))))))))) → 3(4(5(4(1(4(1(3(1(3(3(3(3(x)))))))))))))
2(5(2(1(2(5(3(0(0(3(1(5(4(x))))))))))))) → 4(0(4(5(3(5(5(3(5(3(0(5(4(x)))))))))))))
4(1(4(4(5(0(4(2(0(5(5(4(1(2(x)))))))))))))) → 1(2(5(2(3(5(2(3(4(1(2(0(0(x)))))))))))))
1(5(3(1(1(5(4(1(1(4(5(3(1(3(x)))))))))))))) → 4(3(1(5(5(1(5(3(4(2(1(2(3(x)))))))))))))
1(2(0(5(3(0(0(0(2(0(3(5(4(4(x)))))))))))))) → 5(3(1(4(2(4(1(0(0(1(4(0(2(2(x))))))))))))))
1(1(3(3(2(1(5(2(4(4(4(4(5(0(5(x))))))))))))))) → 3(0(1(3(3(0(1(3(1(3(5(0(3(1(x))))))))))))))
1(4(1(2(2(3(4(2(2(0(4(5(5(3(5(x))))))))))))))) → 0(0(1(2(4(4(0(4(5(3(2(4(1(5(x))))))))))))))
0(5(0(5(0(1(0(5(2(3(3(1(2(5(5(x))))))))))))))) → 0(5(0(3(4(5(0(1(4(2(0(5(3(2(x))))))))))))))
1(1(2(3(0(0(5(5(5(2(1(2(3(2(1(2(x)))))))))))))))) → 3(2(3(0(3(1(2(4(1(3(3(1(1(4(4(x)))))))))))))))
2(1(4(3(1(1(0(3(0(0(2(4(1(3(3(3(x)))))))))))))))) → 4(0(3(4(4(0(5(3(5(0(5(1(4(0(3(x)))))))))))))))
1(5(4(5(4(1(5(2(4(1(0(4(5(3(1(1(2(x))))))))))))))))) → 0(3(4(4(4(2(2(5(5(4(1(2(3(3(3(3(2(x)))))))))))))))))
2(5(2(2(2(2(2(2(1(3(4(4(4(4(2(0(5(x))))))))))))))))) → 0(0(5(5(5(0(3(4(5(5(3(4(4(0(4(5(x))))))))))))))))
4(1(0(0(5(1(1(5(0(4(1(4(4(0(1(5(5(2(x)))))))))))))))))) → 0(5(3(4(4(2(0(2(1(2(1(1(1(0(0(1(3(3(x))))))))))))))))))
2(4(4(4(5(0(3(4(4(3(5(2(3(5(4(1(1(0(0(x))))))))))))))))))) → 4(0(4(0(4(3(3(0(4(0(3(5(1(4(1(5(4(0(0(x)))))))))))))))))))
3(2(0(2(1(4(4(5(4(3(3(4(4(2(5(1(4(1(4(1(x)))))))))))))))))))) → 3(3(4(1(1(5(1(0(5(2(4(4(4(0(1(0(5(1(2(1(x))))))))))))))))))))
2(5(3(3(5(5(3(3(4(3(3(1(5(5(2(4(0(0(1(2(x)))))))))))))))))))) → 5(4(2(5(2(5(1(3(1(1(2(2(4(0(5(3(3(4(4(x)))))))))))))))))))
4(2(4(1(2(5(2(0(1(3(5(1(2(5(4(5(4(2(4(2(4(x))))))))))))))))))))) → 1(2(3(4(5(2(1(2(5(3(4(5(0(2(5(0(5(1(3(5(1(x)))))))))))))))))))))
5(3(5(5(0(2(4(3(3(3(5(1(4(3(4(3(0(1(2(0(5(x))))))))))))))))))))) → 5(1(4(4(3(4(4(4(0(5(1(2(5(4(5(0(1(1(5(3(x))))))))))))))))))))
Q is empty.
(3) QTRSRRRProof (EQUIVALENT transformation)
Used ordering:
Polynomial interpretation [POLO]:
POL(0(x1)) = 8 + x1
POL(1(x1)) = 7 + x1
POL(2(x1)) = 8 + x1
POL(3(x1)) = 6 + x1
POL(4(x1)) = 8 + 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:
1(0(1(0(x)))) → 1(3(2(0(x))))
1(0(4(1(2(x))))) → 1(0(0(3(x))))
1(3(3(3(2(x))))) → 1(3(1(1(x))))
4(3(2(1(4(x))))) → 0(3(5(1(x))))
4(4(2(3(5(x))))) → 0(3(1(0(x))))
3(3(4(0(5(5(x)))))) → 3(1(2(3(0(x)))))
1(2(0(4(0(4(4(5(5(5(x)))))))))) → 3(0(4(0(5(0(4(5(4(5(x))))))))))
0(5(4(0(2(3(3(0(2(2(2(x))))))))))) → 1(5(2(0(0(0(0(4(0(x)))))))))
4(4(2(5(4(1(5(3(0(2(0(1(x)))))))))))) → 3(3(0(1(2(3(2(3(0(1(0(1(x))))))))))))
0(5(4(4(1(0(5(3(2(4(0(4(x)))))))))))) → 1(4(1(5(2(1(5(2(0(5(3(x)))))))))))
1(1(4(4(1(4(5(3(2(0(1(4(x)))))))))))) → 2(4(4(2(1(5(5(4(4(5(2(x)))))))))))
4(1(4(3(2(2(2(0(5(5(0(0(2(x))))))))))))) → 3(4(5(4(1(4(1(3(1(3(3(3(3(x)))))))))))))
4(1(4(4(5(0(4(2(0(5(5(4(1(2(x)))))))))))))) → 1(2(5(2(3(5(2(3(4(1(2(0(0(x)))))))))))))
1(5(3(1(1(5(4(1(1(4(5(3(1(3(x)))))))))))))) → 4(3(1(5(5(1(5(3(4(2(1(2(3(x)))))))))))))
1(1(3(3(2(1(5(2(4(4(4(4(5(0(5(x))))))))))))))) → 3(0(1(3(3(0(1(3(1(3(5(0(3(1(x))))))))))))))
1(4(1(2(2(3(4(2(2(0(4(5(5(3(5(x))))))))))))))) → 0(0(1(2(4(4(0(4(5(3(2(4(1(5(x))))))))))))))
0(5(0(5(0(1(0(5(2(3(3(1(2(5(5(x))))))))))))))) → 0(5(0(3(4(5(0(1(4(2(0(5(3(2(x))))))))))))))
1(1(2(3(0(0(5(5(5(2(1(2(3(2(1(2(x)))))))))))))))) → 3(2(3(0(3(1(2(4(1(3(3(1(1(4(4(x)))))))))))))))
2(1(4(3(1(1(0(3(0(0(2(4(1(3(3(3(x)))))))))))))))) → 4(0(3(4(4(0(5(3(5(0(5(1(4(0(3(x)))))))))))))))
1(5(4(5(4(1(5(2(4(1(0(4(5(3(1(1(2(x))))))))))))))))) → 0(3(4(4(4(2(2(5(5(4(1(2(3(3(3(3(2(x)))))))))))))))))
2(5(2(2(2(2(2(2(1(3(4(4(4(4(2(0(5(x))))))))))))))))) → 0(0(5(5(5(0(3(4(5(5(3(4(4(0(4(5(x))))))))))))))))
4(1(0(0(5(1(1(5(0(4(1(4(4(0(1(5(5(2(x)))))))))))))))))) → 0(5(3(4(4(2(0(2(1(2(1(1(1(0(0(1(3(3(x))))))))))))))))))
2(5(3(3(5(5(3(3(4(3(3(1(5(5(2(4(0(0(1(2(x)))))))))))))))))))) → 5(4(2(5(2(5(1(3(1(1(2(2(4(0(5(3(3(4(4(x)))))))))))))))))))
4(2(4(1(2(5(2(0(1(3(5(1(2(5(4(5(4(2(4(2(4(x))))))))))))))))))))) → 1(2(3(4(5(2(1(2(5(3(4(5(0(2(5(0(5(1(3(5(1(x)))))))))))))))))))))
5(3(5(5(0(2(4(3(3(3(5(1(4(3(4(3(0(1(2(0(5(x))))))))))))))))))))) → 5(1(4(4(3(4(4(4(0(5(1(2(5(4(5(0(1(1(5(3(x))))))))))))))))))))
(4) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
0(1(0(1(4(1(3(0(5(x))))))))) → 0(3(2(2(3(1(4(2(2(x)))))))))
2(5(2(1(2(5(3(0(0(3(1(5(4(x))))))))))))) → 4(0(4(5(3(5(5(3(5(3(0(5(4(x)))))))))))))
1(2(0(5(3(0(0(0(2(0(3(5(4(4(x)))))))))))))) → 5(3(1(4(2(4(1(0(0(1(4(0(2(2(x))))))))))))))
2(4(4(4(5(0(3(4(4(3(5(2(3(5(4(1(1(0(0(x))))))))))))))))))) → 4(0(4(0(4(3(3(0(4(0(3(5(1(4(1(5(4(0(0(x)))))))))))))))))))
3(2(0(2(1(4(4(5(4(3(3(4(4(2(5(1(4(1(4(1(x)))))))))))))))))))) → 3(3(4(1(1(5(1(0(5(2(4(4(4(0(1(0(5(1(2(1(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:
01(1(0(1(4(1(3(0(5(x))))))))) → 01(3(2(2(3(1(4(2(2(x)))))))))
01(1(0(1(4(1(3(0(5(x))))))))) → 31(2(2(3(1(4(2(2(x))))))))
01(1(0(1(4(1(3(0(5(x))))))))) → 21(2(3(1(4(2(2(x)))))))
01(1(0(1(4(1(3(0(5(x))))))))) → 21(3(1(4(2(2(x))))))
01(1(0(1(4(1(3(0(5(x))))))))) → 31(1(4(2(2(x)))))
01(1(0(1(4(1(3(0(5(x))))))))) → 11(4(2(2(x))))
01(1(0(1(4(1(3(0(5(x))))))))) → 21(2(x))
01(1(0(1(4(1(3(0(5(x))))))))) → 21(x)
21(5(2(1(2(5(3(0(0(3(1(5(4(x))))))))))))) → 01(4(5(3(5(5(3(5(3(0(5(4(x))))))))))))
21(5(2(1(2(5(3(0(0(3(1(5(4(x))))))))))))) → 31(5(5(3(5(3(0(5(4(x)))))))))
21(5(2(1(2(5(3(0(0(3(1(5(4(x))))))))))))) → 31(5(3(0(5(4(x))))))
21(5(2(1(2(5(3(0(0(3(1(5(4(x))))))))))))) → 31(0(5(4(x))))
21(5(2(1(2(5(3(0(0(3(1(5(4(x))))))))))))) → 01(5(4(x)))
11(2(0(5(3(0(0(0(2(0(3(5(4(4(x)))))))))))))) → 31(1(4(2(4(1(0(0(1(4(0(2(2(x)))))))))))))
11(2(0(5(3(0(0(0(2(0(3(5(4(4(x)))))))))))))) → 11(4(2(4(1(0(0(1(4(0(2(2(x))))))))))))
11(2(0(5(3(0(0(0(2(0(3(5(4(4(x)))))))))))))) → 21(4(1(0(0(1(4(0(2(2(x))))))))))
11(2(0(5(3(0(0(0(2(0(3(5(4(4(x)))))))))))))) → 11(0(0(1(4(0(2(2(x))))))))
11(2(0(5(3(0(0(0(2(0(3(5(4(4(x)))))))))))))) → 01(0(1(4(0(2(2(x)))))))
11(2(0(5(3(0(0(0(2(0(3(5(4(4(x)))))))))))))) → 01(1(4(0(2(2(x))))))
11(2(0(5(3(0(0(0(2(0(3(5(4(4(x)))))))))))))) → 11(4(0(2(2(x)))))
11(2(0(5(3(0(0(0(2(0(3(5(4(4(x)))))))))))))) → 01(2(2(x)))
11(2(0(5(3(0(0(0(2(0(3(5(4(4(x)))))))))))))) → 21(2(x))
11(2(0(5(3(0(0(0(2(0(3(5(4(4(x)))))))))))))) → 21(x)
21(4(4(4(5(0(3(4(4(3(5(2(3(5(4(1(1(0(0(x))))))))))))))))))) → 01(4(0(4(3(3(0(4(0(3(5(1(4(1(5(4(0(0(x))))))))))))))))))
21(4(4(4(5(0(3(4(4(3(5(2(3(5(4(1(1(0(0(x))))))))))))))))))) → 01(4(3(3(0(4(0(3(5(1(4(1(5(4(0(0(x))))))))))))))))
21(4(4(4(5(0(3(4(4(3(5(2(3(5(4(1(1(0(0(x))))))))))))))))))) → 31(3(0(4(0(3(5(1(4(1(5(4(0(0(x))))))))))))))
21(4(4(4(5(0(3(4(4(3(5(2(3(5(4(1(1(0(0(x))))))))))))))))))) → 31(0(4(0(3(5(1(4(1(5(4(0(0(x)))))))))))))
21(4(4(4(5(0(3(4(4(3(5(2(3(5(4(1(1(0(0(x))))))))))))))))))) → 01(4(0(3(5(1(4(1(5(4(0(0(x))))))))))))
21(4(4(4(5(0(3(4(4(3(5(2(3(5(4(1(1(0(0(x))))))))))))))))))) → 01(3(5(1(4(1(5(4(0(0(x))))))))))
21(4(4(4(5(0(3(4(4(3(5(2(3(5(4(1(1(0(0(x))))))))))))))))))) → 31(5(1(4(1(5(4(0(0(x)))))))))
21(4(4(4(5(0(3(4(4(3(5(2(3(5(4(1(1(0(0(x))))))))))))))))))) → 11(4(1(5(4(0(0(x)))))))
21(4(4(4(5(0(3(4(4(3(5(2(3(5(4(1(1(0(0(x))))))))))))))))))) → 11(5(4(0(0(x)))))
31(2(0(2(1(4(4(5(4(3(3(4(4(2(5(1(4(1(4(1(x)))))))))))))))))))) → 31(3(4(1(1(5(1(0(5(2(4(4(4(0(1(0(5(1(2(1(x))))))))))))))))))))
31(2(0(2(1(4(4(5(4(3(3(4(4(2(5(1(4(1(4(1(x)))))))))))))))))))) → 31(4(1(1(5(1(0(5(2(4(4(4(0(1(0(5(1(2(1(x)))))))))))))))))))
31(2(0(2(1(4(4(5(4(3(3(4(4(2(5(1(4(1(4(1(x)))))))))))))))))))) → 11(1(5(1(0(5(2(4(4(4(0(1(0(5(1(2(1(x)))))))))))))))))
31(2(0(2(1(4(4(5(4(3(3(4(4(2(5(1(4(1(4(1(x)))))))))))))))))))) → 11(5(1(0(5(2(4(4(4(0(1(0(5(1(2(1(x))))))))))))))))
31(2(0(2(1(4(4(5(4(3(3(4(4(2(5(1(4(1(4(1(x)))))))))))))))))))) → 11(0(5(2(4(4(4(0(1(0(5(1(2(1(x))))))))))))))
31(2(0(2(1(4(4(5(4(3(3(4(4(2(5(1(4(1(4(1(x)))))))))))))))))))) → 01(5(2(4(4(4(0(1(0(5(1(2(1(x)))))))))))))
31(2(0(2(1(4(4(5(4(3(3(4(4(2(5(1(4(1(4(1(x)))))))))))))))))))) → 21(4(4(4(0(1(0(5(1(2(1(x)))))))))))
31(2(0(2(1(4(4(5(4(3(3(4(4(2(5(1(4(1(4(1(x)))))))))))))))))))) → 01(1(0(5(1(2(1(x)))))))
31(2(0(2(1(4(4(5(4(3(3(4(4(2(5(1(4(1(4(1(x)))))))))))))))))))) → 11(0(5(1(2(1(x))))))
31(2(0(2(1(4(4(5(4(3(3(4(4(2(5(1(4(1(4(1(x)))))))))))))))))))) → 01(5(1(2(1(x)))))
31(2(0(2(1(4(4(5(4(3(3(4(4(2(5(1(4(1(4(1(x)))))))))))))))))))) → 11(2(1(x)))
31(2(0(2(1(4(4(5(4(3(3(4(4(2(5(1(4(1(4(1(x)))))))))))))))))))) → 21(1(x))
The TRS R consists of the following rules:
0(1(0(1(4(1(3(0(5(x))))))))) → 0(3(2(2(3(1(4(2(2(x)))))))))
2(5(2(1(2(5(3(0(0(3(1(5(4(x))))))))))))) → 4(0(4(5(3(5(5(3(5(3(0(5(4(x)))))))))))))
1(2(0(5(3(0(0(0(2(0(3(5(4(4(x)))))))))))))) → 5(3(1(4(2(4(1(0(0(1(4(0(2(2(x))))))))))))))
2(4(4(4(5(0(3(4(4(3(5(2(3(5(4(1(1(0(0(x))))))))))))))))))) → 4(0(4(0(4(3(3(0(4(0(3(5(1(4(1(5(4(0(0(x)))))))))))))))))))
3(2(0(2(1(4(4(5(4(3(3(4(4(2(5(1(4(1(4(1(x)))))))))))))))))))) → 3(3(4(1(1(5(1(0(5(2(4(4(4(0(1(0(5(1(2(1(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 44 less nodes.
(8) TRUE