Termination w.r.t. Q proof of TRS/Rubiobn129.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:

plus(s(X), plus(Y, Z)) → plus(X, plus(s(s(Y)), Z))
plus(s(X1), plus(X2, plus(X3, X4))) → plus(X1, plus(X3, plus(X2, X4)))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

plus(s(X), plus(Y, Z)) → plus(X, plus(s(s(Y)), Z))
plus(s(X1), plus(X2, plus(X3, X4))) → plus(X1, plus(X3, plus(X2, X4)))

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:

PLUS(s(X1), plus(X2, plus(X3, X4))) → PLUS(X3, plus(X2, X4))
PLUS(s(X1), plus(X2, plus(X3, X4))) → PLUS(X1, plus(X3, plus(X2, X4)))
PLUS(s(X), plus(Y, Z)) → PLUS(X, plus(s(s(Y)), Z))
PLUS(s(X), plus(Y, Z)) → PLUS(s(s(Y)), Z)
PLUS(s(X1), plus(X2, plus(X3, X4))) → PLUS(X2, X4)

The TRS R consists of the following rules:

plus(s(X), plus(Y, Z)) → plus(X, plus(s(s(Y)), Z))
plus(s(X1), plus(X2, plus(X3, X4))) → plus(X1, plus(X3, plus(X2, X4)))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

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

PLUS(s(X1), plus(X2, plus(X3, X4))) → PLUS(X3, plus(X2, X4))
PLUS(s(X1), plus(X2, plus(X3, X4))) → PLUS(X1, plus(X3, plus(X2, X4)))
PLUS(s(X), plus(Y, Z)) → PLUS(X, plus(s(s(Y)), Z))
PLUS(s(X), plus(Y, Z)) → PLUS(s(s(Y)), Z)
PLUS(s(X1), plus(X2, plus(X3, X4))) → PLUS(X2, X4)

The TRS R consists of the following rules:

plus(s(X), plus(Y, Z)) → plus(X, plus(s(s(Y)), Z))
plus(s(X1), plus(X2, plus(X3, X4))) → plus(X1, plus(X3, plus(X2, X4)))

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.

PLUS(s(X1), plus(X2, plus(X3, X4))) → PLUS(X3, plus(X2, X4))
PLUS(s(X), plus(Y, Z)) → PLUS(s(s(Y)), Z)
PLUS(s(X1), plus(X2, plus(X3, X4))) → PLUS(X2, X4)

The remaining pairs can at least be oriented weakly.

PLUS(s(X1), plus(X2, plus(X3, X4))) → PLUS(X1, plus(X3, plus(X2, X4)))
PLUS(s(X), plus(Y, Z)) → PLUS(X, plus(s(s(Y)), Z))

Used ordering: Combined order from the following AFS and order.
PLUS(x1, x2) = PLUS(x2)
s(x1) = x1
plus(x1, x2) = plus(x2)

Lexicographic path order with status [19].
Quasi-Precedence:

[PLUS₁, plus_1]

Status:

plus₁: [1]
PLUS₁: [1]

The following usable rules [14] were oriented:

plus(s(X1), plus(X2, plus(X3, X4))) → plus(X1, plus(X3, plus(X2, X4)))
plus(s(X), plus(Y, Z)) → plus(X, plus(s(s(Y)), Z))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ QDPOrderProof

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

PLUS(s(X1), plus(X2, plus(X3, X4))) → PLUS(X1, plus(X3, plus(X2, X4)))
PLUS(s(X), plus(Y, Z)) → PLUS(X, plus(s(s(Y)), Z))

The TRS R consists of the following rules:

plus(s(X), plus(Y, Z)) → plus(X, plus(s(s(Y)), Z))
plus(s(X1), plus(X2, plus(X3, X4))) → plus(X1, plus(X3, plus(X2, X4)))

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.

PLUS(s(X1), plus(X2, plus(X3, X4))) → PLUS(X1, plus(X3, plus(X2, X4)))
PLUS(s(X), plus(Y, Z)) → PLUS(X, plus(s(s(Y)), Z))

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
PLUS(x1, x2) = PLUS(x1)
s(x1) = s(x1)
plus(x1, x2) = plus

Lexicographic path order with status [19].
Quasi-Precedence:

plus > PLUS₁ > s₁

Status:

plus: []
s₁: [1]
PLUS₁: [1]

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ PisEmptyProof

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

plus(s(X), plus(Y, Z)) → plus(X, plus(s(s(Y)), Z))
plus(s(X1), plus(X2, plus(X3, X4))) → plus(X1, plus(X3, plus(X2, X4)))

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.