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(n__a, X, X) → f(activate(X), b, n__b)
b → a
a → n__a
b → n__b
activate(n__a) → a
activate(n__b) → b
activate(X) → X
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
f(n__a, X, X) → f(activate(X), b, n__b)
b → a
a → n__a
b → n__b
activate(n__a) → a
activate(n__b) → b
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:
F(n__a, X, X) → B
F(n__a, X, X) → ACTIVATE(X)
ACTIVATE(n__a) → A
ACTIVATE(n__b) → B
F(n__a, X, X) → F(activate(X), b, n__b)
B → A
The TRS R consists of the following rules:
f(n__a, X, X) → f(activate(X), b, n__b)
b → a
a → n__a
b → n__b
activate(n__a) → a
activate(n__b) → b
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:
F(n__a, X, X) → B
F(n__a, X, X) → ACTIVATE(X)
ACTIVATE(n__a) → A
ACTIVATE(n__b) → B
F(n__a, X, X) → F(activate(X), b, n__b)
B → A
The TRS R consists of the following rules:
f(n__a, X, X) → f(activate(X), b, n__b)
b → a
a → n__a
b → n__b
activate(n__a) → a
activate(n__b) → b
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 5 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
F(n__a, X, X) → F(activate(X), b, n__b)
The TRS R consists of the following rules:
f(n__a, X, X) → f(activate(X), b, n__b)
b → a
a → n__a
b → n__b
activate(n__a) → a
activate(n__b) → b
activate(X) → X
Q is empty.
We have to consider all minimal (P,Q,R)-chains.