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)  =  plus1(x1)

Lexicographic Path Order [19].
Precedence:
[s1, 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
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.