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))
s1(log1(0)) -> s1(0)
log1(s1(x)) -> s1(log1(half1(s1(x))))

Q is empty.


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))
s1(log1(0)) -> s1(0)
log1(s1(x)) -> s1(log1(half1(s1(x))))

Q is empty.

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

LOG1(s1(x)) -> S1(log1(half1(s1(x))))
LOG1(s1(x)) -> LOG1(half1(s1(x)))
HALF1(s1(s1(x))) -> S1(half1(x))
LOG1(s1(x)) -> HALF1(s1(x))
S1(log1(0)) -> S1(0)
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))
s1(log1(0)) -> s1(0)
log1(s1(x)) -> s1(log1(half1(s1(x))))

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

↳ QTRS
  ↳ DependencyPairsProof
QDP
      ↳ DependencyGraphProof

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

LOG1(s1(x)) -> S1(log1(half1(s1(x))))
LOG1(s1(x)) -> LOG1(half1(s1(x)))
HALF1(s1(s1(x))) -> S1(half1(x))
LOG1(s1(x)) -> HALF1(s1(x))
S1(log1(0)) -> S1(0)
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))
s1(log1(0)) -> s1(0)
log1(s1(x)) -> s1(log1(half1(s1(x))))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 2 SCCs with 4 less nodes.

↳ 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))
s1(log1(0)) -> s1(0)
log1(s1(x)) -> s1(log1(half1(s1(x))))

Q is empty.
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: Quasi Precedence: trivial


↳ 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))
s1(log1(0)) -> s1(0)
log1(s1(x)) -> s1(log1(half1(s1(x))))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
QDP

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

LOG1(s1(x)) -> LOG1(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))
s1(log1(0)) -> s1(0)
log1(s1(x)) -> s1(log1(half1(s1(x))))

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