Termination w.r.t. Q of the following Term Rewriting System could not be shown:
Q restricted rewrite system:
The TRS R consists of the following rules:
1(1(x1)) → 4(3(x1))
1(2(x1)) → 2(1(x1))
2(2(x1)) → 1(1(1(x1)))
3(3(x1)) → 5(6(x1))
3(4(x1)) → 1(1(x1))
4(4(x1)) → 3(x1)
5(5(x1)) → 6(2(x1))
5(6(x1)) → 1(2(x1))
6(6(x1)) → 2(1(x1))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
1(1(x1)) → 4(3(x1))
1(2(x1)) → 2(1(x1))
2(2(x1)) → 1(1(1(x1)))
3(3(x1)) → 5(6(x1))
3(4(x1)) → 1(1(x1))
4(4(x1)) → 3(x1)
5(5(x1)) → 6(2(x1))
5(6(x1)) → 1(2(x1))
6(6(x1)) → 2(1(x1))
Q is empty.
Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:
51(6(x1)) → 21(x1)
51(5(x1)) → 61(2(x1))
11(2(x1)) → 21(1(x1))
31(4(x1)) → 11(x1)
51(6(x1)) → 11(2(x1))
21(2(x1)) → 11(1(1(x1)))
61(6(x1)) → 11(x1)
31(3(x1)) → 51(6(x1))
11(2(x1)) → 11(x1)
11(1(x1)) → 31(x1)
21(2(x1)) → 11(1(x1))
41(4(x1)) → 31(x1)
51(5(x1)) → 21(x1)
21(2(x1)) → 11(x1)
61(6(x1)) → 21(1(x1))
31(3(x1)) → 61(x1)
31(4(x1)) → 11(1(x1))
11(1(x1)) → 41(3(x1))
The TRS R consists of the following rules:
1(1(x1)) → 4(3(x1))
1(2(x1)) → 2(1(x1))
2(2(x1)) → 1(1(1(x1)))
3(3(x1)) → 5(6(x1))
3(4(x1)) → 1(1(x1))
4(4(x1)) → 3(x1)
5(5(x1)) → 6(2(x1))
5(6(x1)) → 1(2(x1))
6(6(x1)) → 2(1(x1))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
51(6(x1)) → 21(x1)
51(5(x1)) → 61(2(x1))
11(2(x1)) → 21(1(x1))
31(4(x1)) → 11(x1)
51(6(x1)) → 11(2(x1))
21(2(x1)) → 11(1(1(x1)))
61(6(x1)) → 11(x1)
31(3(x1)) → 51(6(x1))
11(2(x1)) → 11(x1)
11(1(x1)) → 31(x1)
21(2(x1)) → 11(1(x1))
41(4(x1)) → 31(x1)
51(5(x1)) → 21(x1)
21(2(x1)) → 11(x1)
61(6(x1)) → 21(1(x1))
31(3(x1)) → 61(x1)
31(4(x1)) → 11(1(x1))
11(1(x1)) → 41(3(x1))
The TRS R consists of the following rules:
1(1(x1)) → 4(3(x1))
1(2(x1)) → 2(1(x1))
2(2(x1)) → 1(1(1(x1)))
3(3(x1)) → 5(6(x1))
3(4(x1)) → 1(1(x1))
4(4(x1)) → 3(x1)
5(5(x1)) → 6(2(x1))
5(6(x1)) → 1(2(x1))
6(6(x1)) → 2(1(x1))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.