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:
f2(f2(a, a), x) -> f2(a, f2(b, f2(a, x)))
f2(x, f2(y, z)) -> f2(f2(x, y), z)
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
f2(f2(a, a), x) -> f2(a, f2(b, f2(a, x)))
f2(x, f2(y, z)) -> f2(f2(x, y), z)
Q is empty.
Q DP problem:
The TRS P consists of the following rules:
F2(x, f2(y, z)) -> F2(x, y)
F2(x, f2(y, z)) -> F2(f2(x, y), z)
F2(f2(a, a), x) -> F2(b, f2(a, x))
F2(f2(a, a), x) -> F2(a, f2(b, f2(a, x)))
F2(f2(a, a), x) -> F2(a, x)
The TRS R consists of the following rules:
f2(f2(a, a), x) -> f2(a, f2(b, f2(a, x)))
f2(x, f2(y, z)) -> f2(f2(x, y), z)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ QDPAfsSolverProof
Q DP problem:
The TRS P consists of the following rules:
F2(x, f2(y, z)) -> F2(x, y)
F2(x, f2(y, z)) -> F2(f2(x, y), z)
F2(f2(a, a), x) -> F2(b, f2(a, x))
F2(f2(a, a), x) -> F2(a, f2(b, f2(a, x)))
F2(f2(a, a), x) -> F2(a, x)
The TRS R consists of the following rules:
f2(f2(a, a), x) -> f2(a, f2(b, f2(a, x)))
f2(x, f2(y, z)) -> f2(f2(x, y), z)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.
F2(f2(a, a), x) -> F2(b, f2(a, x))
Used argument filtering: F2(x1, x2) = x1
f2(x1, x2) = x1
a = a
b = b
Used ordering: Quasi Precedence:
a > b
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
F2(x, f2(y, z)) -> F2(x, y)
F2(x, f2(y, z)) -> F2(f2(x, y), z)
F2(f2(a, a), x) -> F2(a, f2(b, f2(a, x)))
F2(f2(a, a), x) -> F2(a, x)
The TRS R consists of the following rules:
f2(f2(a, a), x) -> f2(a, f2(b, f2(a, x)))
f2(x, f2(y, z)) -> f2(f2(x, y), z)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.