YES Termination w.r.t. Q proof of AG01_3.26.ari

(0) Obligation:

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

f(x) → s(x)
f(s(s(x))) → s(f(f(x)))

Q is empty.

(1) DependencyPairsProof (EQUIVALENT transformation)

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

(2) Obligation:

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

F(s(s(x))) → F(f(x))
F(s(s(x))) → F(x)

The TRS R consists of the following rules:

f(x) → s(x)
f(s(s(x))) → s(f(f(x)))

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

(3) MNOCProof (EQUIVALENT transformation)

We use the modular non-overlap check [LPAR04] to enlarge Q to all left-hand sides of R.

(4) Obligation:

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

F(s(s(x))) → F(f(x))
F(s(s(x))) → F(x)

The TRS R consists of the following rules:

f(x) → s(x)
f(s(s(x))) → s(f(f(x)))

The set Q consists of the following terms:

f(x0)

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

(5) MRRProof (EQUIVALENT transformation)

By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

F(s(s(x))) → F(f(x))
F(s(s(x))) → F(x)


Used ordering: Polynomial interpretation [POLO]:

POL(F(x1)) = 3·x1   
POL(f(x1)) = 3 + x1   
POL(s(x1)) = 3 + x1   

(6) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

f(x) → s(x)
f(s(s(x))) → s(f(f(x)))

The set Q consists of the following terms:

f(x0)

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

(7) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(8) YES