Termination w.r.t. Q of the given QTRS could not be shown:

(0) Obligation:

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

f(s(x), x) → f(s(x), round(x))
round(0) → 0
round(0) → s(0)
round(s(0)) → s(0)
round(s(s(x))) → s(s(round(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(x), x) → F(s(x), round(x))
F(s(x), x) → ROUND(x)
ROUND(s(s(x))) → ROUND(x)

The TRS R consists of the following rules:

f(s(x), x) → f(s(x), round(x))
round(0) → 0
round(0) → s(0)
round(s(0)) → s(0)
round(s(s(x))) → s(s(round(x)))

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

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 1 less node.

(4) Complex Obligation (AND)

(5) Obligation:

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

ROUND(s(s(x))) → ROUND(x)

The TRS R consists of the following rules:

f(s(x), x) → f(s(x), round(x))
round(0) → 0
round(0) → s(0)
round(s(0)) → s(0)
round(s(s(x))) → s(s(round(x)))

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

(6) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

ROUND(s(s(x))) → ROUND(x)

The remaining pairs can at least be oriented weakly.
Used ordering: Recursive path order with status [RPO].
Quasi-Precedence:

s₁ > ROUND₁

Status:

ROUND₁: multiset
s₁: multiset

The following usable rules [FROCOS05] were oriented: none

(7) Obligation:

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

f(s(x), x) → f(s(x), round(x))
round(0) → 0
round(0) → s(0)
round(s(0)) → s(0)
round(s(s(x))) → s(s(round(x)))

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

(8) PisEmptyProof (EQUIVALENT transformation)

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

(9) TRUE

(10) Obligation:

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

F(s(x), x) → F(s(x), round(x))

The TRS R consists of the following rules:

f(s(x), x) → f(s(x), round(x))
round(0) → 0
round(0) → s(0)
round(s(0)) → s(0)
round(s(s(x))) → s(s(round(x)))

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