Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar18/AG01SLASHHASH3.21.trs

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:

times₂(x, plus₂(y, 1)) -> plus₂(times₂(x, plus₂(y, times₂(1, 0))), x)
times₂(x, 1) -> x
plus₂(x, 0) -> x
times₂(x, 0) -> 0

Q is empty.

↳ QTRS

  ↳ Non-Overlap Check

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

times₂(x, plus₂(y, 1)) -> plus₂(times₂(x, plus₂(y, times₂(1, 0))), x)
times₂(x, 1) -> x
plus₂(x, 0) -> x
times₂(x, 0) -> 0

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:

times₂(x, plus₂(y, 1)) -> plus₂(times₂(x, plus₂(y, times₂(1, 0))), x)
times₂(x, 1) -> x
plus₂(x, 0) -> x
times₂(x, 0) -> 0

The set Q consists of the following terms:

times₂(x0, plus₂(x1, 1))
times₂(x0, 1)
plus₂(x0, 0)
times₂(x0, 0)

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

TIMES₂(x, plus₂(y, 1)) -> TIMES₂(x, plus₂(y, times₂(1, 0)))
TIMES₂(x, plus₂(y, 1)) -> TIMES₂(1, 0)
TIMES₂(x, plus₂(y, 1)) -> PLUS₂(times₂(x, plus₂(y, times₂(1, 0))), x)
TIMES₂(x, plus₂(y, 1)) -> PLUS₂(y, times₂(1, 0))

The TRS R consists of the following rules:

times₂(x, plus₂(y, 1)) -> plus₂(times₂(x, plus₂(y, times₂(1, 0))), x)
times₂(x, 1) -> x
plus₂(x, 0) -> x
times₂(x, 0) -> 0

The set Q consists of the following terms:

times₂(x0, plus₂(x1, 1))
times₂(x0, 1)
plus₂(x0, 0)
times₂(x0, 0)

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:

TIMES₂(x, plus₂(y, 1)) -> TIMES₂(x, plus₂(y, times₂(1, 0)))
TIMES₂(x, plus₂(y, 1)) -> TIMES₂(1, 0)
TIMES₂(x, plus₂(y, 1)) -> PLUS₂(times₂(x, plus₂(y, times₂(1, 0))), x)
TIMES₂(x, plus₂(y, 1)) -> PLUS₂(y, times₂(1, 0))

The TRS R consists of the following rules:

times₂(x, plus₂(y, 1)) -> plus₂(times₂(x, plus₂(y, times₂(1, 0))), x)
times₂(x, 1) -> x
plus₂(x, 0) -> x
times₂(x, 0) -> 0

The set Q consists of the following terms:

times₂(x0, plus₂(x1, 1))
times₂(x0, 1)
plus₂(x0, 0)
times₂(x0, 0)

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 3 less nodes.

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ QDPOrderProof

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

TIMES₂(x, plus₂(y, 1)) -> TIMES₂(x, plus₂(y, times₂(1, 0)))

The TRS R consists of the following rules:

times₂(x, plus₂(y, 1)) -> plus₂(times₂(x, plus₂(y, times₂(1, 0))), x)
times₂(x, 1) -> x
plus₂(x, 0) -> x
times₂(x, 0) -> 0

The set Q consists of the following terms:

times₂(x0, plus₂(x1, 1))
times₂(x0, 1)
plus₂(x0, 0)
times₂(x0, 0)

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.

TIMES₂(x, plus₂(y, 1)) -> TIMES₂(x, plus₂(y, times₂(1, 0)))

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
TIMES₂(x1, x2) = TIMES₁(x2)
plus₂(x1, x2) = plus₂(x1, x2)
1 = 1
times₂(x1, x2) = times
0 = 0

Lexicographic Path Order [19].
Precedence:

1 > TIMES₁ > 0
1 > plus₂ > 0
1 > times > 0

The following usable rules [14] were oriented:

times₂(x, 0) -> 0
plus₂(x, 0) -> x

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ QDPOrderProof

                ↳ QDP

                  ↳ PisEmptyProof

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

times₂(x, plus₂(y, 1)) -> plus₂(times₂(x, plus₂(y, times₂(1, 0))), x)
times₂(x, 1) -> x
plus₂(x, 0) -> x
times₂(x, 0) -> 0

The set Q consists of the following terms:

times₂(x0, plus₂(x1, 1))
times₂(x0, 1)
plus₂(x0, 0)
times₂(x0, 0)

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