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:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
bits(0) → 0
bits(s(x)) → s(bits(half(s(x))))

Q is empty.


↳ QTRS
  ↳ Overlay + Local Confluence

Q restricted rewrite system:
The TRS R consists of the following rules:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
bits(0) → 0
bits(s(x)) → s(bits(half(s(x))))

Q is empty.

The TRS is overlay and locally confluent. By [15] we can switch to innermost.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof

Q restricted rewrite system:
The TRS R consists of the following rules:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
bits(0) → 0
bits(s(x)) → s(bits(half(s(x))))

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
bits(0)
bits(s(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:

BITS(s(x)) → HALF(s(x))
BITS(s(x)) → BITS(half(s(x)))
HALF(s(s(x))) → HALF(x)

The TRS R consists of the following rules:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
bits(0) → 0
bits(s(x)) → s(bits(half(s(x))))

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
bits(0)
bits(s(x0))

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

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof

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

BITS(s(x)) → HALF(s(x))
BITS(s(x)) → BITS(half(s(x)))
HALF(s(s(x))) → HALF(x)

The TRS R consists of the following rules:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
bits(0) → 0
bits(s(x)) → s(bits(half(s(x))))

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
bits(0)
bits(s(x0))

We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof

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

BITS(s(x)) → HALF(s(x))
BITS(s(x)) → BITS(half(s(x)))
HALF(s(s(x))) → HALF(x)

The TRS R consists of the following rules:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
bits(0) → 0
bits(s(x)) → s(bits(half(s(x))))

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
bits(0)
bits(s(x0))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 2 SCCs with 1 less node.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                    ↳ QDPOrderProof
                  ↳ QDP

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

HALF(s(s(x))) → HALF(x)

The TRS R consists of the following rules:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
bits(0) → 0
bits(s(x)) → s(bits(half(s(x))))

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
bits(0)
bits(s(x0))

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.


HALF(s(s(x))) → HALF(x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
HALF(x1)  =  x1
s(x1)  =  s(x1)

Recursive path order with status [2].
Quasi-Precedence:
trivial

Status:
s1: [1]


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                    ↳ QDPOrderProof
                      ↳ QDP
                        ↳ PisEmptyProof
                  ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
bits(0) → 0
bits(s(x)) → s(bits(half(s(x))))

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
bits(0)
bits(s(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
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP

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

BITS(s(x)) → BITS(half(s(x)))

The TRS R consists of the following rules:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
bits(0) → 0
bits(s(x)) → s(bits(half(s(x))))

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
bits(0)
bits(s(x0))

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