YES Termination w.r.t. Q proof of SK90_4.16.ari

(0) Obligation:

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

f(0) → s(0)
f(s(0)) → s(s(0))
f(s(0)) → *(s(s(0)), f(0))
f(+(x, s(0))) → +(s(s(0)), f(x))
f(+(x, y)) → *(f(x), f(y))

Q is empty.

(1) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(*(x1, x2)) = x1 + x2   
POL(+(x1, x2)) = 2 + 2·x1 + 2·x2   
POL(0) = 0   
POL(f(x1)) = 1 + 2·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(s(0)) → s(s(0))
f(+(x, s(0))) → +(s(s(0)), f(x))
f(+(x, y)) → *(f(x), f(y))


(2) Obligation:

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

f(0) → s(0)
f(s(0)) → *(s(s(0)), f(0))

Q is empty.

(3) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(*(x1, x2)) = x1 + x2   
POL(0) = 0   
POL(f(x1)) = 1 + 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) → s(0)


(4) Obligation:

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

f(s(0)) → *(s(s(0)), f(0))

Q is empty.

(5) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Recursive path order with status [RPO].
Quasi-Precedence:
[f1, 0] > [s1, *2]

Status:
f1: multiset
s1: multiset
0: multiset
*2: [1,2]

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

f(s(0)) → *(s(s(0)), f(0))


(6) Obligation:

Q restricted rewrite system:
R is empty.
Q is empty.

(7) RisEmptyProof (EQUIVALENT transformation)

The TRS R is empty. Hence, termination is trivially proven.

(8) YES