Termination w.r.t. Q of the following Term Rewriting System could not be shown:
Q restricted rewrite system:
The TRS R consists of the following rules:
half1(0) -> 0
half1(s1(0)) -> 0
half1(s1(s1(x))) -> s1(half1(x))
bits1(0) -> 0
bits1(s1(x)) -> s1(bits1(half1(s1(x))))
Q is empty.
↳ QTRS
↳ Non-Overlap Check
Q restricted rewrite system:
The TRS R consists of the following rules:
half1(0) -> 0
half1(s1(0)) -> 0
half1(s1(s1(x))) -> s1(half1(x))
bits1(0) -> 0
bits1(s1(x)) -> s1(bits1(half1(s1(x))))
Q is empty.
The TRS is non-overlapping. Hence, we can switch to innermost.
↳ QTRS
↳ Non-Overlap Check
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
half1(0) -> 0
half1(s1(0)) -> 0
half1(s1(s1(x))) -> s1(half1(x))
bits1(0) -> 0
bits1(s1(x)) -> s1(bits1(half1(s1(x))))
The set Q consists of the following terms:
half1(0)
half1(s1(0))
half1(s1(s1(x0)))
bits1(0)
bits1(s1(x0))
Q DP problem:
The TRS P consists of the following rules:
BITS1(s1(x)) -> HALF1(s1(x))
BITS1(s1(x)) -> BITS1(half1(s1(x)))
HALF1(s1(s1(x))) -> HALF1(x)
The TRS R consists of the following rules:
half1(0) -> 0
half1(s1(0)) -> 0
half1(s1(s1(x))) -> s1(half1(x))
bits1(0) -> 0
bits1(s1(x)) -> s1(bits1(half1(s1(x))))
The set Q consists of the following terms:
half1(0)
half1(s1(0))
half1(s1(s1(x0)))
bits1(0)
bits1(s1(x0))
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ Non-Overlap Check
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
BITS1(s1(x)) -> HALF1(s1(x))
BITS1(s1(x)) -> BITS1(half1(s1(x)))
HALF1(s1(s1(x))) -> HALF1(x)
The TRS R consists of the following rules:
half1(0) -> 0
half1(s1(0)) -> 0
half1(s1(s1(x))) -> s1(half1(x))
bits1(0) -> 0
bits1(s1(x)) -> s1(bits1(half1(s1(x))))
The set Q consists of the following terms:
half1(0)
half1(s1(0))
half1(s1(s1(x0)))
bits1(0)
bits1(s1(x0))
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 2 SCCs with 1 less node.
↳ QTRS
↳ Non-Overlap Check
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
HALF1(s1(s1(x))) -> HALF1(x)
The TRS R consists of the following rules:
half1(0) -> 0
half1(s1(0)) -> 0
half1(s1(s1(x))) -> s1(half1(x))
bits1(0) -> 0
bits1(s1(x)) -> s1(bits1(half1(s1(x))))
The set Q consists of the following terms:
half1(0)
half1(s1(0))
half1(s1(s1(x0)))
bits1(0)
bits1(s1(x0))
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.
HALF1(s1(s1(x))) -> HALF1(x)
Used argument filtering: HALF1(x1) = x1
s1(x1) = s1(x1)
Used ordering: Precedence:
trivial
↳ QTRS
↳ Non-Overlap Check
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
half1(0) -> 0
half1(s1(0)) -> 0
half1(s1(s1(x))) -> s1(half1(x))
bits1(0) -> 0
bits1(s1(x)) -> s1(bits1(half1(s1(x))))
The set Q consists of the following terms:
half1(0)
half1(s1(0))
half1(s1(s1(x0)))
bits1(0)
bits1(s1(x0))
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.
↳ QTRS
↳ Non-Overlap Check
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
BITS1(s1(x)) -> BITS1(half1(s1(x)))
The TRS R consists of the following rules:
half1(0) -> 0
half1(s1(0)) -> 0
half1(s1(s1(x))) -> s1(half1(x))
bits1(0) -> 0
bits1(s1(x)) -> s1(bits1(half1(s1(x))))
The set Q consists of the following terms:
half1(0)
half1(s1(0))
half1(s1(s1(x0)))
bits1(0)
bits1(s1(x0))
We have to consider all minimal (P,Q,R)-chains.