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(plus2(X, Y), Z) -> plus2(X, plus2(Y, Z))
times2(X, s1(Y)) -> plus2(X, times2(Y, X))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

plus2(plus2(X, Y), Z) -> plus2(X, plus2(Y, Z))
times2(X, s1(Y)) -> plus2(X, times2(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:

PLUS2(plus2(X, Y), Z) -> PLUS2(X, plus2(Y, Z))
PLUS2(plus2(X, Y), Z) -> PLUS2(Y, Z)
TIMES2(X, s1(Y)) -> TIMES2(Y, X)
TIMES2(X, s1(Y)) -> PLUS2(X, times2(Y, X))

The TRS R consists of the following rules:

plus2(plus2(X, Y), Z) -> plus2(X, plus2(Y, Z))
times2(X, s1(Y)) -> plus2(X, times2(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:

PLUS2(plus2(X, Y), Z) -> PLUS2(X, plus2(Y, Z))
PLUS2(plus2(X, Y), Z) -> PLUS2(Y, Z)
TIMES2(X, s1(Y)) -> TIMES2(Y, X)
TIMES2(X, s1(Y)) -> PLUS2(X, times2(Y, X))

The TRS R consists of the following rules:

plus2(plus2(X, Y), Z) -> plus2(X, plus2(Y, Z))
times2(X, s1(Y)) -> plus2(X, times2(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
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP

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

PLUS2(plus2(X, Y), Z) -> PLUS2(X, plus2(Y, Z))
PLUS2(plus2(X, Y), Z) -> PLUS2(Y, Z)

The TRS R consists of the following rules:

plus2(plus2(X, Y), Z) -> plus2(X, plus2(Y, Z))
times2(X, s1(Y)) -> plus2(X, times2(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.


PLUS2(plus2(X, Y), Z) -> PLUS2(X, plus2(Y, Z))
PLUS2(plus2(X, Y), Z) -> PLUS2(Y, Z)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( plus2(x1, x2) ) = x1 + x2 + 1


POL( PLUS2(x1, x2) ) = x1 + 1



The following usable rules [14] 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:

plus2(plus2(X, Y), Z) -> plus2(X, plus2(Y, Z))
times2(X, s1(Y)) -> plus2(X, times2(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:

TIMES2(X, s1(Y)) -> TIMES2(Y, X)

The TRS R consists of the following rules:

plus2(plus2(X, Y), Z) -> plus2(X, plus2(Y, Z))
times2(X, s1(Y)) -> plus2(X, times2(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.


TIMES2(X, s1(Y)) -> TIMES2(Y, X)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( s1(x1) ) = x1 + 1


POL( TIMES2(x1, x2) ) = x1 + x2 + 1



The following usable rules [14] 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:

plus2(plus2(X, Y), Z) -> plus2(X, plus2(Y, Z))
times2(X, s1(Y)) -> plus2(X, times2(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.