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
  ↳ 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)))

Q is empty.

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)))

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:

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)))

Q is empty.
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
  ↳ 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)))

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


The following pairs can be oriented strictly and are deleted.


LASTBIT1(s1(s1(x))) -> LASTBIT1(x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(LASTBIT1(x1)) = 2·x1   
POL(s1(x1)) = 1 + 2·x1   

The following usable rules [14] were oriented: none



↳ 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)))

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
            ↳ 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)))

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


The following pairs can be oriented strictly and are deleted.


HALF1(s1(s1(x))) -> HALF1(x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(HALF1(x1)) = 2·x1   
POL(s1(x1)) = 1 + 2·x1   

The following usable rules [14] were oriented: none



↳ 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)))

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
          ↳ 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)))

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