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:
h(X) → g(X, X)
g(a, X) → f(b, activate(X))
f(X, X) → h(a)
a → b
activate(X) → X
The set Q consists of the following terms:
h(x0)
activate(x0)
f(x0, x0)
a
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
h(X) → g(X, X)
g(a, X) → f(b, activate(X))
f(X, X) → h(a)
a → b
activate(X) → X
The set Q consists of the following terms:
h(x0)
activate(x0)
f(x0, x0)
a
Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:
G(a, X) → F(b, activate(X))
F(X, X) → H(a)
F(X, X) → A
G(a, X) → ACTIVATE(X)
H(X) → G(X, X)
The TRS R consists of the following rules:
h(X) → g(X, X)
g(a, X) → f(b, activate(X))
f(X, X) → h(a)
a → b
activate(X) → X
The set Q consists of the following terms:
h(x0)
activate(x0)
f(x0, x0)
a
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:
G(a, X) → F(b, activate(X))
F(X, X) → H(a)
F(X, X) → A
G(a, X) → ACTIVATE(X)
H(X) → G(X, X)
The TRS R consists of the following rules:
h(X) → g(X, X)
g(a, X) → f(b, activate(X))
f(X, X) → h(a)
a → b
activate(X) → X
The set Q consists of the following terms:
h(x0)
activate(x0)
f(x0, x0)
a
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 5 less nodes.