Termination w.r.t. Q proof of TRS/AG01#3.22.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, s₁(z))) -> plus₂(times₂(x, plus₂(y, times₂(s₁(z), 0))), times₂(x, s₁(z)))
times₂(x, 0) -> 0
times₂(x, s₁(y)) -> plus₂(times₂(x, y), x)
plus₂(x, 0) -> x
plus₂(x, s₁(y)) -> s₁(plus₂(x, y))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

times₂(x, plus₂(y, s₁(z))) -> plus₂(times₂(x, plus₂(y, times₂(s₁(z), 0))), times₂(x, s₁(z)))
times₂(x, 0) -> 0
times₂(x, s₁(y)) -> plus₂(times₂(x, y), x)
plus₂(x, 0) -> x
plus₂(x, s₁(y)) -> s₁(plus₂(x, y))

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, s₁(z))) -> TIMES₂(x, s₁(z))
TIMES₂(x, plus₂(y, s₁(z))) -> TIMES₂(x, plus₂(y, times₂(s₁(z), 0)))
TIMES₂(x, plus₂(y, s₁(z))) -> TIMES₂(s₁(z), 0)
TIMES₂(x, s₁(y)) -> TIMES₂(x, y)
TIMES₂(x, s₁(y)) -> PLUS₂(times₂(x, y), x)
TIMES₂(x, plus₂(y, s₁(z))) -> PLUS₂(y, times₂(s₁(z), 0))
PLUS₂(x, s₁(y)) -> PLUS₂(x, y)
TIMES₂(x, plus₂(y, s₁(z))) -> PLUS₂(times₂(x, plus₂(y, times₂(s₁(z), 0))), times₂(x, s₁(z)))

The TRS R consists of the following rules:

times₂(x, plus₂(y, s₁(z))) -> plus₂(times₂(x, plus₂(y, times₂(s₁(z), 0))), times₂(x, s₁(z)))
times₂(x, 0) -> 0
times₂(x, s₁(y)) -> plus₂(times₂(x, y), x)
plus₂(x, 0) -> x
plus₂(x, s₁(y)) -> s₁(plus₂(x, y))

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, s₁(z))) -> TIMES₂(x, s₁(z))
TIMES₂(x, plus₂(y, s₁(z))) -> TIMES₂(x, plus₂(y, times₂(s₁(z), 0)))
TIMES₂(x, plus₂(y, s₁(z))) -> TIMES₂(s₁(z), 0)
TIMES₂(x, s₁(y)) -> TIMES₂(x, y)
TIMES₂(x, s₁(y)) -> PLUS₂(times₂(x, y), x)
TIMES₂(x, plus₂(y, s₁(z))) -> PLUS₂(y, times₂(s₁(z), 0))
PLUS₂(x, s₁(y)) -> PLUS₂(x, y)
TIMES₂(x, plus₂(y, s₁(z))) -> PLUS₂(times₂(x, plus₂(y, times₂(s₁(z), 0))), times₂(x, s₁(z)))

The TRS R consists of the following rules:

times₂(x, plus₂(y, s₁(z))) -> plus₂(times₂(x, plus₂(y, times₂(s₁(z), 0))), times₂(x, s₁(z)))
times₂(x, 0) -> 0
times₂(x, s₁(y)) -> plus₂(times₂(x, y), x)
plus₂(x, 0) -> x
plus₂(x, s₁(y)) -> s₁(plus₂(x, y))

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 4 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

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

PLUS₂(x, s₁(y)) -> PLUS₂(x, y)

The TRS R consists of the following rules:

times₂(x, plus₂(y, s₁(z))) -> plus₂(times₂(x, plus₂(y, times₂(s₁(z), 0))), times₂(x, s₁(z)))
times₂(x, 0) -> 0
times₂(x, s₁(y)) -> plus₂(times₂(x, y), x)
plus₂(x, 0) -> x
plus₂(x, s₁(y)) -> s₁(plus₂(x, y))

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.

PLUS₂(x, s₁(y)) -> PLUS₂(x, y)

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

POL(PLUS₂(x₁, x₂)) = x₂
POL(s₁(x₁)) = 1 + x₁

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:

times₂(x, plus₂(y, s₁(z))) -> plus₂(times₂(x, plus₂(y, times₂(s₁(z), 0))), times₂(x, s₁(z)))
times₂(x, 0) -> 0
times₂(x, s₁(y)) -> plus₂(times₂(x, y), x)
plus₂(x, 0) -> x
plus₂(x, s₁(y)) -> s₁(plus₂(x, y))

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:

TIMES₂(x, plus₂(y, s₁(z))) -> TIMES₂(x, s₁(z))
TIMES₂(x, plus₂(y, s₁(z))) -> TIMES₂(x, plus₂(y, times₂(s₁(z), 0)))
TIMES₂(x, s₁(y)) -> TIMES₂(x, y)

The TRS R consists of the following rules:

times₂(x, plus₂(y, s₁(z))) -> plus₂(times₂(x, plus₂(y, times₂(s₁(z), 0))), times₂(x, s₁(z)))
times₂(x, 0) -> 0
times₂(x, s₁(y)) -> plus₂(times₂(x, y), x)
plus₂(x, 0) -> x
plus₂(x, s₁(y)) -> s₁(plus₂(x, y))

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, s₁(z))) -> TIMES₂(x, plus₂(y, times₂(s₁(z), 0)))
TIMES₂(x, s₁(y)) -> TIMES₂(x, y)

The remaining pairs can at least be oriented weakly.

TIMES₂(x, plus₂(y, s₁(z))) -> TIMES₂(x, s₁(z))

Used ordering: Polynomial interpretation [21]:

POL(0) = 0
POL(TIMES₂(x₁, x₂)) = x₂
POL(plus₂(x₁, x₂)) = x₁ + x₂
POL(s₁(x₁)) = 1 + x₁
POL(times₂(x₁, x₂)) = 0

The following usable rules [14] were oriented:

times₂(x, 0) -> 0
plus₂(x, s₁(y)) -> s₁(plus₂(x, y))
plus₂(x, 0) -> x

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ DependencyGraphProof

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

TIMES₂(x, plus₂(y, s₁(z))) -> TIMES₂(x, s₁(z))

The TRS R consists of the following rules:

times₂(x, plus₂(y, s₁(z))) -> plus₂(times₂(x, plus₂(y, times₂(s₁(z), 0))), times₂(x, s₁(z)))
times₂(x, 0) -> 0
times₂(x, s₁(y)) -> plus₂(times₂(x, y), x)
plus₂(x, 0) -> x
plus₂(x, s₁(y)) -> s₁(plus₂(x, y))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 0 SCCs with 1 less node.