Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar4/RubioSLASHbn129.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:

plus₂(s₁(X), plus₂(Y, Z)) -> plus₂(X, plus₂(s₁(s₁(Y)), Z))
plus₂(s₁(X1), plus₂(X2, plus₂(X3, X4))) -> plus₂(X1, plus₂(X3, plus₂(X2, X4)))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

plus₂(s₁(X), plus₂(Y, Z)) -> plus₂(X, plus₂(s₁(s₁(Y)), Z))
plus₂(s₁(X1), plus₂(X2, plus₂(X3, X4))) -> plus₂(X1, plus₂(X3, plus₂(X2, X4)))

Q is empty.

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

PLUS₂(s₁(X1), plus₂(X2, plus₂(X3, X4))) -> PLUS₂(X3, plus₂(X2, X4))
PLUS₂(s₁(X1), plus₂(X2, plus₂(X3, X4))) -> PLUS₂(X1, plus₂(X3, plus₂(X2, X4)))
PLUS₂(s₁(X), plus₂(Y, Z)) -> PLUS₂(X, plus₂(s₁(s₁(Y)), Z))
PLUS₂(s₁(X), plus₂(Y, Z)) -> PLUS₂(s₁(s₁(Y)), Z)
PLUS₂(s₁(X1), plus₂(X2, plus₂(X3, X4))) -> PLUS₂(X2, X4)

The TRS R consists of the following rules:

plus₂(s₁(X), plus₂(Y, Z)) -> plus₂(X, plus₂(s₁(s₁(Y)), Z))
plus₂(s₁(X1), plus₂(X2, plus₂(X3, X4))) -> plus₂(X1, plus₂(X3, plus₂(X2, X4)))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPAfsSolverProof

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

PLUS₂(s₁(X1), plus₂(X2, plus₂(X3, X4))) -> PLUS₂(X3, plus₂(X2, X4))
PLUS₂(s₁(X1), plus₂(X2, plus₂(X3, X4))) -> PLUS₂(X1, plus₂(X3, plus₂(X2, X4)))
PLUS₂(s₁(X), plus₂(Y, Z)) -> PLUS₂(X, plus₂(s₁(s₁(Y)), Z))
PLUS₂(s₁(X), plus₂(Y, Z)) -> PLUS₂(s₁(s₁(Y)), Z)
PLUS₂(s₁(X1), plus₂(X2, plus₂(X3, X4))) -> PLUS₂(X2, X4)

The TRS R consists of the following rules:

plus₂(s₁(X), plus₂(Y, Z)) -> plus₂(X, plus₂(s₁(s₁(Y)), Z))
plus₂(s₁(X1), plus₂(X2, plus₂(X3, X4))) -> plus₂(X1, plus₂(X3, plus₂(X2, X4)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

PLUS₂(s₁(X1), plus₂(X2, plus₂(X3, X4))) -> PLUS₂(X3, plus₂(X2, X4))
PLUS₂(s₁(X), plus₂(Y, Z)) -> PLUS₂(s₁(s₁(Y)), Z)
PLUS₂(s₁(X1), plus₂(X2, plus₂(X3, X4))) -> PLUS₂(X2, X4)

Used argument filtering: PLUS₂(x1, x2) = x2
plus₂(x1, x2) = plus₁(x2)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPAfsSolverProof

        ↳ QDP

          ↳ QDPAfsSolverProof

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

PLUS₂(s₁(X1), plus₂(X2, plus₂(X3, X4))) -> PLUS₂(X1, plus₂(X3, plus₂(X2, X4)))
PLUS₂(s₁(X), plus₂(Y, Z)) -> PLUS₂(X, plus₂(s₁(s₁(Y)), Z))

The TRS R consists of the following rules:

plus₂(s₁(X), plus₂(Y, Z)) -> plus₂(X, plus₂(s₁(s₁(Y)), Z))
plus₂(s₁(X1), plus₂(X2, plus₂(X3, X4))) -> plus₂(X1, plus₂(X3, plus₂(X2, X4)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

PLUS₂(s₁(X1), plus₂(X2, plus₂(X3, X4))) -> PLUS₂(X1, plus₂(X3, plus₂(X2, X4)))
PLUS₂(s₁(X), plus₂(Y, Z)) -> PLUS₂(X, plus₂(s₁(s₁(Y)), Z))

Used argument filtering: PLUS₂(x1, x2) = x1
s₁(x1) = s₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPAfsSolverProof

        ↳ QDP

          ↳ QDPAfsSolverProof

            ↳ QDP

              ↳ PisEmptyProof

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

plus₂(s₁(X), plus₂(Y, Z)) -> plus₂(X, plus₂(s₁(s₁(Y)), Z))
plus₂(s₁(X1), plus₂(X2, plus₂(X3, X4))) -> plus₂(X1, plus₂(X3, plus₂(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.