Termination w.r.t. Q of the following Term Rewriting System could be proven:
Q restricted rewrite system:
The TRS R consists of the following rules:
f2(a, f2(b, x)) -> f2(a, f2(a, f2(a, x)))
f2(b, f2(a, x)) -> f2(b, f2(b, f2(b, x)))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
f2(a, f2(b, x)) -> f2(a, f2(a, f2(a, x)))
f2(b, f2(a, x)) -> f2(b, f2(b, f2(b, x)))
Q is empty.
Q DP problem:
The TRS P consists of the following rules:
F2(b, f2(a, x)) -> F2(b, f2(b, f2(b, x)))
F2(a, f2(b, x)) -> F2(a, x)
F2(a, f2(b, x)) -> F2(a, f2(a, x))
F2(b, f2(a, x)) -> F2(b, x)
F2(b, f2(a, x)) -> F2(b, f2(b, x))
F2(a, f2(b, x)) -> F2(a, f2(a, f2(a, x)))
The TRS R consists of the following rules:
f2(a, f2(b, x)) -> f2(a, f2(a, f2(a, x)))
f2(b, f2(a, x)) -> f2(b, f2(b, f2(b, x)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
F2(b, f2(a, x)) -> F2(b, f2(b, f2(b, x)))
F2(a, f2(b, x)) -> F2(a, x)
F2(a, f2(b, x)) -> F2(a, f2(a, x))
F2(b, f2(a, x)) -> F2(b, x)
F2(b, f2(a, x)) -> F2(b, f2(b, x))
F2(a, f2(b, x)) -> F2(a, f2(a, f2(a, x)))
The TRS R consists of the following rules:
f2(a, f2(b, x)) -> f2(a, f2(a, f2(a, x)))
f2(b, f2(a, x)) -> f2(b, f2(b, f2(b, x)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 2 SCCs.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
F2(b, f2(a, x)) -> F2(b, f2(b, f2(b, x)))
F2(b, f2(a, x)) -> F2(b, x)
F2(b, f2(a, x)) -> F2(b, f2(b, x))
The TRS R consists of the following rules:
f2(a, f2(b, x)) -> f2(a, f2(a, f2(a, x)))
f2(b, f2(a, x)) -> f2(b, f2(b, f2(b, x)))
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(b, f2(a, x)) -> F2(b, f2(b, f2(b, x)))
F2(b, f2(a, x)) -> F2(b, x)
F2(b, f2(a, x)) -> F2(b, f2(b, x))
Used argument filtering: F2(x1, x2) = x2
f2(x1, x2) = f2(x1, x2)
a = a
b = b
Used ordering: Quasi Precedence:
a > b
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
f2(a, f2(b, x)) -> f2(a, f2(a, f2(a, x)))
f2(b, f2(a, x)) -> f2(b, f2(b, f2(b, x)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
Q DP problem:
The TRS P consists of the following rules:
F2(a, f2(b, x)) -> F2(a, x)
F2(a, f2(b, x)) -> F2(a, f2(a, x))
F2(a, f2(b, x)) -> F2(a, f2(a, f2(a, x)))
The TRS R consists of the following rules:
f2(a, f2(b, x)) -> f2(a, f2(a, f2(a, x)))
f2(b, f2(a, x)) -> f2(b, f2(b, f2(b, x)))
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(a, f2(b, x)) -> F2(a, x)
F2(a, f2(b, x)) -> F2(a, f2(a, x))
F2(a, f2(b, x)) -> F2(a, f2(a, f2(a, x)))
Used argument filtering: F2(x1, x2) = x2
f2(x1, x2) = f2(x1, x2)
b = b
a = a
Used ordering: Quasi Precedence:
b > a
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
Q DP problem:
P is empty.
The TRS R consists of the following rules:
f2(a, f2(b, x)) -> f2(a, f2(a, f2(a, x)))
f2(b, f2(a, x)) -> f2(b, f2(b, f2(b, x)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.