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(a) → f(a)
a → b

The set Q consists of the following terms:

a



↳ QTRS
  ↳ RRRPoloQTRSProof

Q restricted rewrite system:
The TRS R consists of the following rules:

f(a) → f(a)
a → b

The set Q consists of the following terms:

a


The following Q TRS is given: Q restricted rewrite system:
The TRS R consists of the following rules:

f(a) → f(a)
a → b

The set Q consists of the following terms:

a

The following rules can be removed by the rule removal processor [15] because they are oriented strictly by a polynomial ordering:

a → b
Used ordering:
Polynomial interpretation [25]:

POL(a) = 2   
POL(b) = 1   
POL(f(x1)) = x1   




↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ DependencyPairsProof

Q restricted rewrite system:
The TRS R consists of the following rules:

f(a) → f(a)

The set Q consists of the following terms:

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:

F(a) → F(a)

The TRS R consists of the following rules:

f(a) → f(a)

The set Q consists of the following terms:

a

We have to consider all minimal (P,Q,R)-chains.

↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof

Q DP problem:
The TRS P consists of the following rules:

F(a) → F(a)

The TRS R consists of the following rules:

f(a) → f(a)

The set Q consists of the following terms:

a

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.