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:

*(i(x), x) → 1
*(1, y) → y
*(x, 0) → 0
*(*(x, y), z) → *(x, *(y, z))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

*(i(x), x) → 1
*(1, y) → y
*(x, 0) → 0
*(*(x, y), z) → *(x, *(y, z))

Q is empty.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

*1(*(x, y), z) → *1(y, z)
*1(*(x, y), z) → *1(x, *(y, z))

The TRS R consists of the following rules:

*(i(x), x) → 1
*(1, y) → y
*(x, 0) → 0
*(*(x, y), z) → *(x, *(y, 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:

*1(*(x, y), z) → *1(y, z)
*1(*(x, y), z) → *1(x, *(y, z))

The TRS R consists of the following rules:

*(i(x), x) → 1
*(1, y) → y
*(x, 0) → 0
*(*(x, y), z) → *(x, *(y, z))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


*1(*(x, y), z) → *1(y, z)
The remaining pairs can at least be oriented weakly.

*1(*(x, y), z) → *1(x, *(y, z))
Used ordering: Polynomial interpretation [25,35]:

POL(i(x1)) = 15/4   
POL(*1(x1, x2)) = (3/4)x_1 + (3/4)x_2   
POL(*(x1, x2)) = 3/4 + x_1 + x_2   
POL(1) = 1/4   
POL(0) = 1/4   
The value of delta used in the strict ordering is 9/16.
The following usable rules [17] were oriented:

*(i(x), x) → 1
*(1, y) → y
*(x, 0) → 0
*(*(x, y), z) → *(x, *(y, z))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof
        ↳ QDP
          ↳ QDPOrderProof

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

*1(*(x, y), z) → *1(x, *(y, z))

The TRS R consists of the following rules:

*(i(x), x) → 1
*(1, y) → y
*(x, 0) → 0
*(*(x, y), z) → *(x, *(y, z))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


*1(*(x, y), z) → *1(x, *(y, z))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(i(x1)) = 15/4   
POL(*1(x1, x2)) = (1/4)x_1   
POL(*(x1, x2)) = 4 + (4)x_1   
POL(1) = 7/4   
POL(0) = 2   
The value of delta used in the strict ordering is 1.
The following usable rules [17] were oriented: none



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

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

*(i(x), x) → 1
*(1, y) → y
*(x, 0) → 0
*(*(x, y), z) → *(x, *(y, 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.