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.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
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
      ↳ QDPOrderProof

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.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


*12(x, oplus2(y, z)) -> *12(x, y)
*12(x, oplus2(y, z)) -> *12(x, z)
The remaining pairs can at least by weakly be oriented.

*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)
Used ordering: Combined order from the following AFS and order.
*12(x1, x2)  =  x2
+2(x1, x2)  =  +1(x2)
oplus2(x1, x2)  =  oplus2(x1, x2)
*2(x1, x2)  =  x2
otimes2(x1, x2)  =  otimes1(x2)

Lexicographic Path Order [19].
Precedence:
trivial


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof
        ↳ 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 [13,14,18] contains 2 SCCs.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ QDPOrderProof
              ↳ 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.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


*12(x, *2(y, z)) -> *12(otimes2(x, y), z)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
*12(x1, x2)  =  *11(x2)
*2(x1, x2)  =  *1(x2)
otimes2(x1, x2)  =  otimes2(x1, x2)

Lexicographic Path Order [19].
Precedence:
[*^11, otimes2]


The following usable rules [14] were oriented: none



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

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.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


*12(+2(x, y), z) -> *12(x, z)
*12(+2(x, y), z) -> *12(y, z)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
*12(x1, x2)  =  *11(x1)
+2(x1, x2)  =  +2(x1, x2)

Lexicographic Path Order [19].
Precedence:
+2 > *^11


The following usable rules [14] were oriented: none



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