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 [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof

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

PLUS(plus(X, Y), Z) → PLUS(X, plus(Y, Z))
PLUS(plus(X, Y), Z) → PLUS(Y, Z)
TIMES(X, s(Y)) → TIMES(Y, X)
TIMES(X, s(Y)) → PLUS(X, 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 deleted some edges using various graph approximations

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof

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

PLUS(plus(X, Y), Z) → PLUS(X, plus(Y, Z))
PLUS(plus(X, Y), Z) → PLUS(Y, Z)
TIMES(X, s(Y)) → TIMES(Y, X)
TIMES(X, s(Y)) → PLUS(X, 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.
The approximation of the Dependency Graph [13,14,18] contains 2 SCCs with 1 less node.

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

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

PLUS(plus(X, Y), Z) → PLUS(X, plus(Y, Z))
PLUS(plus(X, Y), Z) → 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 [13].


The following pairs can be oriented strictly and are deleted.


PLUS(plus(X, Y), Z) → PLUS(X, plus(Y, Z))
PLUS(plus(X, Y), Z) → PLUS(Y, Z)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
PLUS(x1, x2)  =  x1
plus(x1, x2)  =  plus(x1, x2)

Recursive Path Order [2].
Precedence:
trivial

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ 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
      ↳ EdgeDeletionProof
        ↳ 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 [13].


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: Combined order from the following AFS and order.
TIMES(x1, x2)  =  TIMES(x1, x2)
s(x1)  =  s(x1)

Recursive Path Order [2].
Precedence:
trivial

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ 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.