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 [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

PLUS(s(X), plus(Y, Z)) → PLUS(s(s(Y)), Z)
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(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(X), plus(Y, Z)) → PLUS(s(s(Y)), Z)
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(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 [15].


The following pairs can be oriented strictly and are deleted.


PLUS(s(X), plus(Y, Z)) → PLUS(s(s(Y)), Z)
PLUS(s(X1), plus(X2, plus(X3, X4))) → PLUS(X3, plus(X2, X4))
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: Polynomial interpretation [25,35]:

POL(PLUS(x1, x2)) = (1/4)x_2   
POL(plus(x1, x2)) = 1/4 + (4)x_2   
POL(s(x1)) = 0   
The value of delta used in the strict ordering is 1/16.
The following usable rules [17] 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 [15].


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: Polynomial interpretation [25,35]:

POL(PLUS(x1, x2)) = (1/2)x_1   
POL(plus(x1, x2)) = 0   
POL(s(x1)) = 1/4 + (4)x_1   
The value of delta used in the strict ordering is 1/8.
The following usable rules [17] 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.