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:
a2(f, a2(f, x)) -> a2(x, g)
a2(x, g) -> a2(f, a2(g, a2(f, x)))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
a2(f, a2(f, x)) -> a2(x, g)
a2(x, g) -> a2(f, a2(g, a2(f, x)))
Q is empty.
Q DP problem:
The TRS P consists of the following rules:
A2(f, a2(f, x)) -> A2(x, g)
A2(x, g) -> A2(g, a2(f, x))
A2(x, g) -> A2(f, x)
A2(x, g) -> A2(f, a2(g, a2(f, x)))
The TRS R consists of the following rules:
a2(f, a2(f, x)) -> a2(x, g)
a2(x, g) -> a2(f, a2(g, a2(f, 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:
A2(f, a2(f, x)) -> A2(x, g)
A2(x, g) -> A2(g, a2(f, x))
A2(x, g) -> A2(f, x)
A2(x, g) -> A2(f, a2(g, a2(f, x)))
The TRS R consists of the following rules:
a2(f, a2(f, x)) -> a2(x, g)
a2(x, g) -> a2(f, a2(g, a2(f, x)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 1 SCC with 1 less node.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
A2(f, a2(f, x)) -> A2(x, g)
A2(x, g) -> A2(f, x)
A2(x, g) -> A2(f, a2(g, a2(f, x)))
The TRS R consists of the following rules:
a2(f, a2(f, x)) -> a2(x, g)
a2(x, g) -> a2(f, a2(g, a2(f, x)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.