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:
f(X, X) → f(a, n__b)
b → a
b → n__b
activate(n__b) → b
activate(X) → X
Q is empty.
↳ QTRS
↳ RRRPoloQTRSProof
Q restricted rewrite system:
The TRS R consists of the following rules:
f(X, X) → f(a, n__b)
b → a
b → n__b
activate(n__b) → b
activate(X) → X
Q is empty.
The following Q TRS is given: Q restricted rewrite system:
The TRS R consists of the following rules:
f(X, X) → f(a, n__b)
b → a
b → n__b
activate(n__b) → b
activate(X) → X
Q is empty.
The following rules can be removed by the rule removal processor [15] because they are oriented strictly by a polynomial ordering:
b → a
b → n__b
activate(n__b) → b
activate(X) → X
Used ordering:
Polynomial interpretation [25]:
POL(a) = 0
POL(activate(x1)) = 2 + x1
POL(b) = 1
POL(f(x1, x2)) = x1 + x2
POL(n__b) = 0
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ AAECC Innermost
Q restricted rewrite system:
The TRS R consists of the following rules:
f(X, X) → f(a, n__b)
Q is empty.
We have applied [19,8] to switch to innermost. The TRS R 1 is none
The TRS R 2 is
f(X, X) → f(a, n__b)
The signature Sigma is {f}
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
f(X, X) → f(a, n__b)
The set Q consists of the following terms:
f(x0, x0)
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(X, X) → F(a, n__b)
The TRS R consists of the following rules:
f(X, X) → f(a, n__b)
The set Q consists of the following terms:
f(x0, x0)
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
F(X, X) → F(a, n__b)
The TRS R consists of the following rules:
f(X, X) → f(a, n__b)
The set Q consists of the following terms:
f(x0, x0)
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 1 less node.