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:
f(X) → g(n__h(f(X)))
h(X) → n__h(X)
activate(n__h(X)) → h(X)
activate(X) → X
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
f(X) → g(n__h(f(X)))
h(X) → n__h(X)
activate(n__h(X)) → h(X)
activate(X) → X
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:
ACTIVATE(n__h(X)) → H(X)
F(X) → F(X)
The TRS R consists of the following rules:
f(X) → g(n__h(f(X)))
h(X) → n__h(X)
activate(n__h(X)) → h(X)
activate(X) → 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:
ACTIVATE(n__h(X)) → H(X)
F(X) → F(X)
The TRS R consists of the following rules:
f(X) → g(n__h(f(X)))
h(X) → n__h(X)
activate(n__h(X)) → h(X)
activate(X) → X
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
F(X) → F(X)
The TRS R consists of the following rules:
f(X) → g(n__h(f(X)))
h(X) → n__h(X)
activate(n__h(X)) → h(X)
activate(X) → X
Q is empty.
We have to consider all minimal (P,Q,R)-chains.