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:

*(x, *(y, z)) → *(otimes(x, y), z)
*(1, y) → y
*(+(x, y), z) → oplus(*(x, z), *(y, z))
*(x, oplus(y, z)) → oplus(*(x, y), *(x, z))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

*(x, *(y, z)) → *(otimes(x, y), z)
*(1, y) → y
*(+(x, y), z) → oplus(*(x, z), *(y, z))
*(x, oplus(y, z)) → oplus(*(x, y), *(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:

*1(+(x, y), z) → *1(x, z)
*1(x, oplus(y, z)) → *1(x, y)
*1(x, oplus(y, z)) → *1(x, z)
*1(x, *(y, z)) → *1(otimes(x, y), z)
*1(+(x, y), z) → *1(y, z)

The TRS R consists of the following rules:

*(x, *(y, z)) → *(otimes(x, y), z)
*(1, y) → y
*(+(x, y), z) → oplus(*(x, z), *(y, z))
*(x, oplus(y, z)) → oplus(*(x, y), *(x, z))

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:

*1(+(x, y), z) → *1(x, z)
*1(x, oplus(y, z)) → *1(x, y)
*1(x, oplus(y, z)) → *1(x, z)
*1(x, *(y, z)) → *1(otimes(x, y), z)
*1(+(x, y), z) → *1(y, z)

The TRS R consists of the following rules:

*(x, *(y, z)) → *(otimes(x, y), z)
*(1, y) → y
*(+(x, y), z) → oplus(*(x, z), *(y, z))
*(x, oplus(y, z)) → oplus(*(x, y), *(x, z))

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
          ↳ QDPOrderProof

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

*1(+(x, y), z) → *1(x, z)
*1(x, oplus(y, z)) → *1(x, y)
*1(x, oplus(y, z)) → *1(x, z)
*1(x, *(y, z)) → *1(otimes(x, y), z)
*1(+(x, y), z) → *1(y, z)

The TRS R consists of the following rules:

*(x, *(y, z)) → *(otimes(x, y), z)
*(1, y) → y
*(+(x, y), z) → oplus(*(x, z), *(y, z))
*(x, oplus(y, z)) → oplus(*(x, y), *(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 oriented strictly and are deleted.


*1(+(x, y), z) → *1(x, z)
*1(x, oplus(y, z)) → *1(x, y)
*1(x, oplus(y, z)) → *1(x, z)
*1(x, *(y, z)) → *1(otimes(x, y), z)
*1(+(x, y), z) → *1(y, z)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
*1(x1, x2)  =  *1(x1, x2)
+(x1, x2)  =  +(x1, x2)
oplus(x1, x2)  =  oplus(x1, x2)
*(x1, x2)  =  *(x1, x2)
otimes(x1, x2)  =  x2

Lexicographic path order with status [19].
Quasi-Precedence:
[*^12, +2, oplus2, *2]

Status:
*^12: [1,2]
oplus2: [1,2]
+2: [2,1]
*2: [1,2]


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ PisEmptyProof

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

*(x, *(y, z)) → *(otimes(x, y), z)
*(1, y) → y
*(+(x, y), z) → oplus(*(x, z), *(y, z))
*(x, oplus(y, z)) → oplus(*(x, y), *(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.