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))
lastbit1(0) -> 0
lastbit1(s1(0)) -> s1(0)
lastbit1(s1(s1(x))) -> lastbit1(x)
conv1(0) -> cons2(nil, 0)
conv1(s1(x)) -> cons2(conv1(half1(s1(x))), lastbit1(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))
lastbit1(0) -> 0
lastbit1(s1(0)) -> s1(0)
lastbit1(s1(s1(x))) -> lastbit1(x)
conv1(0) -> cons2(nil, 0)
conv1(s1(x)) -> cons2(conv1(half1(s1(x))), lastbit1(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))
lastbit1(0) -> 0
lastbit1(s1(0)) -> s1(0)
lastbit1(s1(s1(x))) -> lastbit1(x)
conv1(0) -> cons2(nil, 0)
conv1(s1(x)) -> cons2(conv1(half1(s1(x))), lastbit1(s1(x)))

The set Q consists of the following terms:

half1(0)
half1(s1(0))
half1(s1(s1(x0)))
lastbit1(0)
lastbit1(s1(0))
lastbit1(s1(s1(x0)))
conv1(0)
conv1(s1(x0))


Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

CONV1(s1(x)) -> CONV1(half1(s1(x)))
LASTBIT1(s1(s1(x))) -> LASTBIT1(x)
CONV1(s1(x)) -> LASTBIT1(s1(x))
CONV1(s1(x)) -> 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))
lastbit1(0) -> 0
lastbit1(s1(0)) -> s1(0)
lastbit1(s1(s1(x))) -> lastbit1(x)
conv1(0) -> cons2(nil, 0)
conv1(s1(x)) -> cons2(conv1(half1(s1(x))), lastbit1(s1(x)))

The set Q consists of the following terms:

half1(0)
half1(s1(0))
half1(s1(s1(x0)))
lastbit1(0)
lastbit1(s1(0))
lastbit1(s1(s1(x0)))
conv1(0)
conv1(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:

CONV1(s1(x)) -> CONV1(half1(s1(x)))
LASTBIT1(s1(s1(x))) -> LASTBIT1(x)
CONV1(s1(x)) -> LASTBIT1(s1(x))
CONV1(s1(x)) -> 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))
lastbit1(0) -> 0
lastbit1(s1(0)) -> s1(0)
lastbit1(s1(s1(x))) -> lastbit1(x)
conv1(0) -> cons2(nil, 0)
conv1(s1(x)) -> cons2(conv1(half1(s1(x))), lastbit1(s1(x)))

The set Q consists of the following terms:

half1(0)
half1(s1(0))
half1(s1(s1(x0)))
lastbit1(0)
lastbit1(s1(0))
lastbit1(s1(s1(x0)))
conv1(0)
conv1(s1(x0))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 3 SCCs with 2 less nodes.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ QDPOrderProof
              ↳ QDP
              ↳ QDP

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

LASTBIT1(s1(s1(x))) -> LASTBIT1(x)

The TRS R consists of the following rules:

half1(0) -> 0
half1(s1(0)) -> 0
half1(s1(s1(x))) -> s1(half1(x))
lastbit1(0) -> 0
lastbit1(s1(0)) -> s1(0)
lastbit1(s1(s1(x))) -> lastbit1(x)
conv1(0) -> cons2(nil, 0)
conv1(s1(x)) -> cons2(conv1(half1(s1(x))), lastbit1(s1(x)))

The set Q consists of the following terms:

half1(0)
half1(s1(0))
half1(s1(s1(x0)))
lastbit1(0)
lastbit1(s1(0))
lastbit1(s1(s1(x0)))
conv1(0)
conv1(s1(x0))

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


LASTBIT1(s1(s1(x))) -> LASTBIT1(x)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
LASTBIT1(x1)  =  LASTBIT1(x1)
s1(x1)  =  s1(x1)

Lexicographic Path Order [19].
Precedence:
s1 > LASTBIT1

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ PisEmptyProof
              ↳ QDP
              ↳ 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))
lastbit1(0) -> 0
lastbit1(s1(0)) -> s1(0)
lastbit1(s1(s1(x))) -> lastbit1(x)
conv1(0) -> cons2(nil, 0)
conv1(s1(x)) -> cons2(conv1(half1(s1(x))), lastbit1(s1(x)))

The set Q consists of the following terms:

half1(0)
half1(s1(0))
half1(s1(s1(x0)))
lastbit1(0)
lastbit1(s1(0))
lastbit1(s1(s1(x0)))
conv1(0)
conv1(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
                ↳ QDPOrderProof
              ↳ 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))
lastbit1(0) -> 0
lastbit1(s1(0)) -> s1(0)
lastbit1(s1(s1(x))) -> lastbit1(x)
conv1(0) -> cons2(nil, 0)
conv1(s1(x)) -> cons2(conv1(half1(s1(x))), lastbit1(s1(x)))

The set Q consists of the following terms:

half1(0)
half1(s1(0))
half1(s1(s1(x0)))
lastbit1(0)
lastbit1(s1(0))
lastbit1(s1(s1(x0)))
conv1(0)
conv1(s1(x0))

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


HALF1(s1(s1(x))) -> HALF1(x)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
HALF1(x1)  =  HALF1(x1)
s1(x1)  =  s1(x1)

Lexicographic Path Order [19].
Precedence:
s1 > HALF1

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ 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))
lastbit1(0) -> 0
lastbit1(s1(0)) -> s1(0)
lastbit1(s1(s1(x))) -> lastbit1(x)
conv1(0) -> cons2(nil, 0)
conv1(s1(x)) -> cons2(conv1(half1(s1(x))), lastbit1(s1(x)))

The set Q consists of the following terms:

half1(0)
half1(s1(0))
half1(s1(s1(x0)))
lastbit1(0)
lastbit1(s1(0))
lastbit1(s1(s1(x0)))
conv1(0)
conv1(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
              ↳ QDP

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

CONV1(s1(x)) -> CONV1(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))
lastbit1(0) -> 0
lastbit1(s1(0)) -> s1(0)
lastbit1(s1(s1(x))) -> lastbit1(x)
conv1(0) -> cons2(nil, 0)
conv1(s1(x)) -> cons2(conv1(half1(s1(x))), lastbit1(s1(x)))

The set Q consists of the following terms:

half1(0)
half1(s1(0))
half1(s1(s1(x0)))
lastbit1(0)
lastbit1(s1(0))
lastbit1(s1(s1(x0)))
conv1(0)
conv1(s1(x0))

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