Termination w.r.t. Q proof of TRS/AG01#3.21.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

  ↳ 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

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:

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

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:

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

Q is empty.
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

  ↳ 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

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.

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

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(0) = 0
POL(1) = 2
POL(TIMES₂(x₁, x₂)) = x₁·x₂ + 3·x₂
POL(plus₂(x₁, x₂)) = x₁ + x₁·x₂ + x₂
POL(times₂(x₁, x₂)) = 1

The following usable rules [14] were oriented:

plus₂(x, 0) -> x
times₂(x, 0) -> 0

↳ 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

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.