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(s(s(s(s(s(s(s(x)))))))), y, y) → f(id(s(s(s(s(s(s(s(s(x))))))))), y, y)
id(s(x)) → s(id(x))
id(0) → 0

Q is empty.

(1) AAECC Innermost (EQUIVALENT transformation)

We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is

id(s(x)) → s(id(x))
id(0) → 0

The TRS R 2 is

f(s(s(s(s(s(s(s(s(x)))))))), y, y) → f(id(s(s(s(s(s(s(s(s(x))))))))), y, y)

The signature Sigma is {f}

(2) Obligation:

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

f(s(s(s(s(s(s(s(s(x)))))))), y, y) → f(id(s(s(s(s(s(s(s(s(x))))))))), y, y)
id(s(x)) → s(id(x))
id(0) → 0

The set Q consists of the following terms:

f(s(s(s(s(s(s(s(s(x0)))))))), x1, x1)
id(s(x0))
id(0)

(3) DependencyPairsProof (EQUIVALENT transformation)

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

(4) Obligation:

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

F(s(s(s(s(s(s(s(s(x)))))))), y, y) → F(id(s(s(s(s(s(s(s(s(x))))))))), y, y)
F(s(s(s(s(s(s(s(s(x)))))))), y, y) → ID(s(s(s(s(s(s(s(s(x)))))))))
ID(s(x)) → ID(x)

The TRS R consists of the following rules:

f(s(s(s(s(s(s(s(s(x)))))))), y, y) → f(id(s(s(s(s(s(s(s(s(x))))))))), y, y)
id(s(x)) → s(id(x))
id(0) → 0

The set Q consists of the following terms:

f(s(s(s(s(s(s(s(s(x0)))))))), x1, x1)
id(s(x0))
id(0)

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

(5) DependencyGraphProof (EQUIVALENT transformation)

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

(6) Complex Obligation (AND)

(7) Obligation:

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

ID(s(x)) → ID(x)

The TRS R consists of the following rules:

f(s(s(s(s(s(s(s(s(x)))))))), y, y) → f(id(s(s(s(s(s(s(s(s(x))))))))), y, y)
id(s(x)) → s(id(x))
id(0) → 0

The set Q consists of the following terms:

f(s(s(s(s(s(s(s(s(x0)))))))), x1, x1)
id(s(x0))
id(0)

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

(8) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

ID(s(x)) → ID(x)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ID(x1)  =  ID(x1)
s(x1)  =  s(x1)
f(x1, x2, x3)  =  f(x2)
id(x1)  =  id(x1)
0  =  0

Recursive Path Order [RPO].
Precedence:

ID₁ > s₁
f₁ > id₁ > 0 > s₁

The following usable rules [FROCOS05] were oriented:

f(s(s(s(s(s(s(s(s(x)))))))), y, y) → f(id(s(s(s(s(s(s(s(s(x))))))))), y, y)
id(s(x)) → s(id(x))
id(0) → 0

(9) Obligation:

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

f(s(s(s(s(s(s(s(s(x)))))))), y, y) → f(id(s(s(s(s(s(s(s(s(x))))))))), y, y)
id(s(x)) → s(id(x))
id(0) → 0

The set Q consists of the following terms:

f(s(s(s(s(s(s(s(s(x0)))))))), x1, x1)
id(s(x0))
id(0)

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

(10) PisEmptyProof (EQUIVALENT transformation)

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

(11) TRUE

(12) Obligation:

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

F(s(s(s(s(s(s(s(s(x)))))))), y, y) → F(id(s(s(s(s(s(s(s(s(x))))))))), y, y)

The TRS R consists of the following rules:

f(s(s(s(s(s(s(s(s(x)))))))), y, y) → f(id(s(s(s(s(s(s(s(s(x))))))))), y, y)
id(s(x)) → s(id(x))
id(0) → 0

The set Q consists of the following terms:

f(s(s(s(s(s(s(s(s(x0)))))))), x1, x1)
id(s(x0))
id(0)

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