(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
f(s(x), x) → f(s(x), round(s(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(s(x)))
F(s(x), x) → ROUND(s(x))
ROUND(s(s(x))) → ROUND(x)
The TRS R consists of the following rules:
f(s(x), x) → f(s(x), round(s(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(s(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: Combined order from the following AFS and order.
ROUND(
x1) =
x1
s(
x1) =
s(
x1)
f(
x1,
x2) =
f(
x1)
round(
x1) =
round(
x1)
0 =
0
Lexicographic path order with status [LPO].
Quasi-Precedence:
round1 > [s1, f1, 0]
Status:
round1: [1]
f1: [1]
s1: [1]
0: []
The following usable rules [FROCOS05] were oriented:
f(s(x), x) → f(s(x), round(s(x)))
round(0) → 0
round(0) → s(0)
round(s(0)) → s(0)
round(s(s(x))) → s(s(round(x)))
(7) Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:
f(s(x), x) → f(s(x), round(s(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(s(x)))
The TRS R consists of the following rules:
f(s(x), x) → f(s(x), round(s(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.