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 oriented strictly 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 be oriented weakly.

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: Polynomial interpretation [21]:

POL(PLUS2(x1, x2)) = 2·x2   
POL(plus2(x1, x2)) = 2 + x2   
POL(s1(x1)) = 0   

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 oriented strictly 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 be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(PLUS2(x1, x2)) = 2·x1   
POL(plus2(x1, x2)) = 0   
POL(s1(x1)) = 2 + 2·x1   

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.