Termination w.r.t. Q of the given QTRS could be proven:

0 QTRS
  ↳1 QTRSRRRProof (⇔)
    ↳2 QTRS
    ↳3 QTRSRRRProof (⇔)
    ↳4 QTRS
    ↳5 AAECC Innermost (⇔)
    ↳6 QTRS
    ↳7 DependencyPairsProof (⇔)
    ↳8 QDP
    ↳9 DependencyGraphProof (⇔)
    ↳10 TRUE
  ↳11 QTRSRRRProof (⇔)
    ↳12 QTRS

(0) Obligation:

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.

(1) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

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   
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

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


(2) Obligation:

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.

(3) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

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   
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

p(s(X)) → X
activate(n__f(X)) → f(X)


(4) Obligation:

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

f(s(0)) → f(p(s(0)))

Q is empty.

(5) AAECC Innermost (EQUIVALENT transformation)

We have applied [NOC,AAECCNOC] 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}

(6) Obligation:

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))

(7) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(8) Obligation:

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.

(9) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.

(10) TRUE

(11) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(0) = 0   
POL(activate(x1)) = 3 + x1   
POL(cons(x1, x2)) = x1 + x2   
POL(f(x1)) = 2 + x1   
POL(n__f(x1)) = x1   
POL(p(x1)) = x1   
POL(s(x1)) = 1 + x1   
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

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


(12) Obligation:

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

f(s(0)) → f(p(s(0)))

Q is empty.