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
      ↳ RuleRemovalProof

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.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

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)


Used ordering: POLO with Polynomial interpretation [25]:

POL(PLUS(x1, x2)) = 2·x1 + x2   
POL(plus(x1, x2)) = 1 + 2·x1 + x2   
POL(s(x1)) = x1   



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ RuleRemovalProof
        ↳ QDP
          ↳ QDPSizeChangeProof

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.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs: