(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) =
x1
s(
x1) =
s(
x1)
f(
x1,
x2,
x3) =
f
id(
x1) =
x1
0 =
0
Recursive Path Order [RPO].
Precedence:
trivial
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.