Termination w.r.t. Q of the following Term Rewriting System could be proven:

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
  ↳ 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))))

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:

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))))

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:

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))))

Q is empty.
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
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ 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))
bits1(0) -> 0
bits1(s1(x)) -> s1(bits1(half1(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 Order [17,21] with Interpretation:

POL( HALF1(x1) ) = max{0, x1 - 1}


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



The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ 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))
bits1(0) -> 0
bits1(s1(x)) -> s1(bits1(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
            ↳ QDPOrderProof

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))))

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.


BITS1(s1(x)) -> BITS1(half1(s1(x)))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( 0 ) = max{0, -1}


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


POL( half1(x1) ) = max{0, x1 - 2}


POL( BITS1(x1) ) = x1 + 2



The following usable rules [14] were oriented:

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



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof

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))))

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.