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:

*2(x, *2(y, z)) -> *2(otimes2(x, y), z)
*2(1, y) -> y
*2(+2(x, y), z) -> oplus2(*2(x, z), *2(y, z))
*2(x, oplus2(y, z)) -> oplus2(*2(x, y), *2(x, z))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

*2(x, *2(y, z)) -> *2(otimes2(x, y), z)
*2(1, y) -> y
*2(+2(x, y), z) -> oplus2(*2(x, z), *2(y, z))
*2(x, oplus2(y, z)) -> oplus2(*2(x, y), *2(x, z))

Q is empty.

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

*12(+2(x, y), z) -> *12(x, z)
*12(x, oplus2(y, z)) -> *12(x, y)
*12(x, oplus2(y, z)) -> *12(x, z)
*12(x, *2(y, z)) -> *12(otimes2(x, y), z)
*12(+2(x, y), z) -> *12(y, z)

The TRS R consists of the following rules:

*2(x, *2(y, z)) -> *2(otimes2(x, y), z)
*2(1, y) -> y
*2(+2(x, y), z) -> oplus2(*2(x, z), *2(y, z))
*2(x, oplus2(y, z)) -> oplus2(*2(x, y), *2(x, z))

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPAfsSolverProof

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

*12(+2(x, y), z) -> *12(x, z)
*12(x, oplus2(y, z)) -> *12(x, y)
*12(x, oplus2(y, z)) -> *12(x, z)
*12(x, *2(y, z)) -> *12(otimes2(x, y), z)
*12(+2(x, y), z) -> *12(y, z)

The TRS R consists of the following rules:

*2(x, *2(y, z)) -> *2(otimes2(x, y), z)
*2(1, y) -> y
*2(+2(x, y), z) -> oplus2(*2(x, z), *2(y, z))
*2(x, oplus2(y, z)) -> oplus2(*2(x, y), *2(x, z))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

*12(x, oplus2(y, z)) -> *12(x, y)
*12(x, oplus2(y, z)) -> *12(x, z)
Used argument filtering: *12(x1, x2)  =  x2
oplus2(x1, x2)  =  oplus2(x1, x2)
*2(x1, x2)  =  x2
Used ordering: Precedence:
trivial



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPAfsSolverProof
        ↳ QDP
          ↳ DependencyGraphProof

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

*12(+2(x, y), z) -> *12(x, z)
*12(x, *2(y, z)) -> *12(otimes2(x, y), z)
*12(+2(x, y), z) -> *12(y, z)

The TRS R consists of the following rules:

*2(x, *2(y, z)) -> *2(otimes2(x, y), z)
*2(1, y) -> y
*2(+2(x, y), z) -> oplus2(*2(x, z), *2(y, z))
*2(x, oplus2(y, z)) -> oplus2(*2(x, y), *2(x, z))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 2 SCCs.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPAfsSolverProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ QDPAfsSolverProof
              ↳ QDP

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

*12(x, *2(y, z)) -> *12(otimes2(x, y), z)

The TRS R consists of the following rules:

*2(x, *2(y, z)) -> *2(otimes2(x, y), z)
*2(1, y) -> y
*2(+2(x, y), z) -> oplus2(*2(x, z), *2(y, z))
*2(x, oplus2(y, z)) -> oplus2(*2(x, y), *2(x, z))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

*12(x, *2(y, z)) -> *12(otimes2(x, y), z)
Used argument filtering: *12(x1, x2)  =  x2
*2(x1, x2)  =  *1(x2)
Used ordering: Precedence:
trivial



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPAfsSolverProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ QDPAfsSolverProof
                  ↳ QDP
                    ↳ PisEmptyProof
              ↳ QDP

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

*2(x, *2(y, z)) -> *2(otimes2(x, y), z)
*2(1, y) -> y
*2(+2(x, y), z) -> oplus2(*2(x, z), *2(y, z))
*2(x, oplus2(y, z)) -> oplus2(*2(x, y), *2(x, z))

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
      ↳ QDPAfsSolverProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ QDPAfsSolverProof

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

*12(+2(x, y), z) -> *12(x, z)
*12(+2(x, y), z) -> *12(y, z)

The TRS R consists of the following rules:

*2(x, *2(y, z)) -> *2(otimes2(x, y), z)
*2(1, y) -> y
*2(+2(x, y), z) -> oplus2(*2(x, z), *2(y, z))
*2(x, oplus2(y, z)) -> oplus2(*2(x, y), *2(x, z))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

*12(+2(x, y), z) -> *12(x, z)
*12(+2(x, y), z) -> *12(y, z)
Used argument filtering: *12(x1, x2)  =  x1
+2(x1, x2)  =  +2(x1, x2)
Used ordering: Precedence:
trivial



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPAfsSolverProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ QDPAfsSolverProof
                  ↳ QDP
                    ↳ PisEmptyProof

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

*2(x, *2(y, z)) -> *2(otimes2(x, y), z)
*2(1, y) -> y
*2(+2(x, y), z) -> oplus2(*2(x, z), *2(y, z))
*2(x, oplus2(y, z)) -> oplus2(*2(x, y), *2(x, z))

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.