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(plus(X, Y), Z) → plus(X, plus(Y, Z))
times(X, s(Y)) → plus(X, times(Y, X))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

Q restricted rewrite system:
The TRS R consists of the following rules:

plus(plus(X, Y), Z) → plus(X, plus(Y, Z))
times(X, s(Y)) → plus(X, times(Y, X))

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:

TIMES(X, s(Y)) → TIMES(Y, X)
TIMES(X, s(Y)) → PLUS(X, times(Y, X))
PLUS(plus(X, Y), Z) → PLUS(Y, Z)
PLUS(plus(X, Y), Z) → PLUS(X, plus(Y, Z))

The TRS R consists of the following rules:

plus(plus(X, Y), Z) → plus(X, plus(Y, Z))
times(X, s(Y)) → plus(X, times(Y, X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof

Q DP problem:
The TRS P consists of the following rules:

TIMES(X, s(Y)) → TIMES(Y, X)
TIMES(X, s(Y)) → PLUS(X, times(Y, X))
PLUS(plus(X, Y), Z) → PLUS(Y, Z)
PLUS(plus(X, Y), Z) → PLUS(X, plus(Y, Z))

The TRS R consists of the following rules:

plus(plus(X, Y), Z) → plus(X, plus(Y, Z))
times(X, s(Y)) → plus(X, times(Y, X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 2 SCCs with 1 less node.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

PLUS(plus(X, Y), Z) → PLUS(Y, Z)
PLUS(plus(X, Y), Z) → PLUS(X, plus(Y, Z))

The TRS R consists of the following rules:

plus(plus(X, Y), Z) → plus(X, plus(Y, Z))
times(X, s(Y)) → plus(X, times(Y, X))

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(plus(X, Y), Z) → PLUS(Y, Z)
PLUS(plus(X, Y), Z) → PLUS(X, plus(Y, Z))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(PLUS(x1, x2)) = x_1   
POL(plus(x1, x2)) = 3 + (2)x_1 + x_2   
The value of delta used in the strict ordering is 3.
The following usable rules [17] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof
          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

plus(plus(X, Y), Z) → plus(X, plus(Y, Z))
times(X, s(Y)) → plus(X, times(Y, X))

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.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof

Q DP problem:
The TRS P consists of the following rules:

TIMES(X, s(Y)) → TIMES(Y, X)

The TRS R consists of the following rules:

plus(plus(X, Y), Z) → plus(X, plus(Y, Z))
times(X, s(Y)) → plus(X, times(Y, X))

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.


TIMES(X, s(Y)) → TIMES(Y, X)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(TIMES(x1, x2)) = (2)x_1 + x_2   
POL(s(x1)) = 1 + (4)x_1   
The value of delta used in the strict ordering is 1.
The following usable rules [17] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof

Q DP problem:
P is empty.
The TRS R consists of the following rules:

plus(plus(X, Y), Z) → plus(X, plus(Y, Z))
times(X, s(Y)) → plus(X, times(Y, X))

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.