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

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

(0) Obligation:

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

f(0) → cons(0, n__f(n__s(n__0)))
f(s(0)) → f(p(s(0)))
p(s(0)) → 0
f(X) → n__f(X)
s(X) → n__s(X)
0 → n__0
activate(n__f(X)) → f(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__0) → 0
activate(X) → X

Q is empty.

(1) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(0) = 1   
POL(activate(x1)) = 2 + x1   
POL(cons(x1, x2)) = x1 + x2   
POL(f(x1)) = x1   
POL(n__0) = 0   
POL(n__f(x1)) = x1   
POL(n__s(x1)) = x1   
POL(p(x1)) = x1   
POL(s(x1)) = x1   
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

0 → n__0
activate(n__0) → 0
activate(X) → X


(2) Obligation:

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

f(0) → cons(0, n__f(n__s(n__0)))
f(s(0)) → f(p(s(0)))
p(s(0)) → 0
f(X) → n__f(X)
s(X) → n__s(X)
activate(n__f(X)) → f(activate(X))
activate(n__s(X)) → s(activate(X))

Q is empty.

(3) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(0) = 1   
POL(activate(x1)) = 2·x1   
POL(cons(x1, x2)) = 1 + x1 + x2   
POL(f(x1)) = 2·x1   
POL(n__0) = 0   
POL(n__f(x1)) = 2·x1   
POL(n__s(x1)) = 2·x1   
POL(p(x1)) = 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:

p(s(0)) → 0


(4) Obligation:

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

f(0) → cons(0, n__f(n__s(n__0)))
f(s(0)) → f(p(s(0)))
f(X) → n__f(X)
s(X) → n__s(X)
activate(n__f(X)) → f(activate(X))
activate(n__s(X)) → s(activate(X))

Q is empty.

(5) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(0) = 0   
POL(activate(x1)) = 2 + 2·x1   
POL(cons(x1, x2)) = 2·x1 + x2   
POL(f(x1)) = 2 + x1   
POL(n__0) = 0   
POL(n__f(x1)) = 1 + x1   
POL(n__s(x1)) = x1   
POL(p(x1)) = 2·x1   
POL(s(x1)) = 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(n__s(n__0)))
f(X) → n__f(X)


(6) Obligation:

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

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

Q is empty.

(7) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

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

activate(n__f(X)) → f(activate(X))


(8) Obligation:

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

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

Q is empty.

(9) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(0) = 0   
POL(activate(x1)) = 2·x1   
POL(f(x1)) = 2·x1   
POL(n__s(x1)) = 1 + 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:

activate(n__s(X)) → s(activate(X))


(10) Obligation:

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

f(s(0)) → f(p(s(0)))
s(X) → n__s(X)

Q is empty.

(11) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(0) = 0   
POL(f(x1)) = x1   
POL(n__s(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:

s(X) → n__s(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.

(13) 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}

(14) 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))

(15) DependencyPairsProof (EQUIVALENT transformation)

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

(16) 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.

(17) DependencyGraphProof (EQUIVALENT transformation)

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

(18) TRUE