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(0) → cons(0, n__f(s(0)))
f(s(0)) → f(p(s(0)))
p(s(X)) → X
f(X) → n__f(X)
activate(n__f(X)) → f(X)
activate(X) → X

Q is empty.


↳ QTRS
  ↳ RRRPoloQTRSProof

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

f(0) → cons(0, n__f(s(0)))
f(s(0)) → f(p(s(0)))
p(s(X)) → X
f(X) → n__f(X)
activate(n__f(X)) → f(X)
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(0) → cons(0, n__f(s(0)))
f(s(0)) → f(p(s(0)))
p(s(X)) → X
f(X) → n__f(X)
activate(n__f(X)) → f(X)
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:

f(0) → cons(0, n__f(s(0)))
f(X) → n__f(X)
activate(X) → X
Used ordering:
Polynomial interpretation [25]:

POL(0) = 0   
POL(activate(x1)) = 2 + x1   
POL(cons(x1, x2)) = x1 + 2·x2   
POL(f(x1)) = 2 + 2·x1   
POL(n__f(x1)) = 2·x1   
POL(p(x1)) = 2·x1   
POL(s(x1)) = 2·x1   




↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof

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

f(s(0)) → f(p(s(0)))
p(s(X)) → X
activate(n__f(X)) → f(X)

Q is empty.

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

f(s(0)) → f(p(s(0)))
p(s(X)) → X
activate(n__f(X)) → f(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:

p(s(X)) → X
activate(n__f(X)) → f(X)
Used ordering:
Polynomial interpretation [25]:

POL(0) = 0   
POL(activate(x1)) = 2 + 2·x1   
POL(f(x1)) = 1 + x1   
POL(n__f(x1)) = 2 + 2·x1   
POL(p(x1)) = x1   
POL(s(x1)) = 1 + 2·x1   




↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ AAECC Innermost

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

f(s(0)) → f(p(s(0)))

Q is empty.

We have applied [19,8] to switch to innermost. The TRS R 1 is none

The TRS R 2 is

f(s(0)) → f(p(s(0)))

The signature Sigma is {f}

↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ AAECC Innermost
            ↳ QTRS
              ↳ DependencyPairsProof

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

f(s(0)) → f(p(s(0)))

The set Q consists of the following terms:

f(s(0))


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(s(0)) → F(p(s(0)))

The TRS R consists of the following rules:

f(s(0)) → f(p(s(0)))

The set Q consists of the following terms:

f(s(0))

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

↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ AAECC Innermost
            ↳ QTRS
              ↳ DependencyPairsProof
                ↳ QDP
                  ↳ DependencyGraphProof

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

F(s(0)) → F(p(s(0)))

The TRS R consists of the following rules:

f(s(0)) → f(p(s(0)))

The set Q consists of the following terms:

f(s(0))

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.