Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar14/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.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
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

      ↳ QDPOrderProof

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.
We use the reduction pair processor [13].

The following pairs can be strictly oriented and are deleted.

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)

The remaining pairs can at least by weakly be 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 ordering: Combined order from the following AFS and order.
PLUS₂(x1, x2) = PLUS₁(x2)
s₁(x1) = x1
plus₂(x1, x2) = plus₁(x2)

Lexicographic Path Order [19].
Precedence:

plus₁ > PLUS₁

The following usable rules [14] were 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))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ QDPOrderProof

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.
We use the reduction pair processor [13].

The following pairs can be strictly oriented and are deleted.

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 remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
PLUS₂(x1, x2) = x1
s₁(x1) = s₁(x1)
plus₂(x1, x2) = plus₂(x1, x2)

Lexicographic Path Order [19].
Precedence:

[s₁, plus_2]

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:

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.