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, 1) -> x
*2(1, y) -> y
*2(i1(x), x) -> 1
*2(x, i1(x)) -> 1
*2(x, *2(y, z)) -> *2(*2(x, y), z)
i1(1) -> 1
*2(*2(x, y), i1(y)) -> x
*2(*2(x, i1(y)), y) -> x
i1(i1(x)) -> x
i1(*2(x, y)) -> *2(i1(y), i1(x))
k2(x, 1) -> 1
k2(x, x) -> 1
*2(k2(x, y), k2(y, x)) -> 1
*2(*2(i1(x), k2(y, z)), x) -> k2(*2(*2(i1(x), y), x), *2(*2(i1(x), z), x))
k2(*2(x, i1(y)), *2(y, i1(x))) -> 1

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

*2(x, 1) -> x
*2(1, y) -> y
*2(i1(x), x) -> 1
*2(x, i1(x)) -> 1
*2(x, *2(y, z)) -> *2(*2(x, y), z)
i1(1) -> 1
*2(*2(x, y), i1(y)) -> x
*2(*2(x, i1(y)), y) -> x
i1(i1(x)) -> x
i1(*2(x, y)) -> *2(i1(y), i1(x))
k2(x, 1) -> 1
k2(x, x) -> 1
*2(k2(x, y), k2(y, x)) -> 1
*2(*2(i1(x), k2(y, z)), x) -> k2(*2(*2(i1(x), y), x), *2(*2(i1(x), z), x))
k2(*2(x, i1(y)), *2(y, i1(x))) -> 1

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:

I1(*2(x, y)) -> I1(x)
*12(x, *2(y, z)) -> *12(x, y)
*12(*2(i1(x), k2(y, z)), x) -> *12(*2(i1(x), z), x)
*12(*2(i1(x), k2(y, z)), x) -> *12(i1(x), z)
I1(*2(x, y)) -> *12(i1(y), i1(x))
*12(*2(i1(x), k2(y, z)), x) -> *12(i1(x), y)
*12(*2(i1(x), k2(y, z)), x) -> K2(*2(*2(i1(x), y), x), *2(*2(i1(x), z), x))
*12(*2(i1(x), k2(y, z)), x) -> *12(*2(i1(x), y), x)
I1(*2(x, y)) -> I1(y)
*12(x, *2(y, z)) -> *12(*2(x, y), z)

The TRS R consists of the following rules:

*2(x, 1) -> x
*2(1, y) -> y
*2(i1(x), x) -> 1
*2(x, i1(x)) -> 1
*2(x, *2(y, z)) -> *2(*2(x, y), z)
i1(1) -> 1
*2(*2(x, y), i1(y)) -> x
*2(*2(x, i1(y)), y) -> x
i1(i1(x)) -> x
i1(*2(x, y)) -> *2(i1(y), i1(x))
k2(x, 1) -> 1
k2(x, x) -> 1
*2(k2(x, y), k2(y, x)) -> 1
*2(*2(i1(x), k2(y, z)), x) -> k2(*2(*2(i1(x), y), x), *2(*2(i1(x), z), x))
k2(*2(x, i1(y)), *2(y, i1(x))) -> 1

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:

I1(*2(x, y)) -> I1(x)
*12(x, *2(y, z)) -> *12(x, y)
*12(*2(i1(x), k2(y, z)), x) -> *12(*2(i1(x), z), x)
*12(*2(i1(x), k2(y, z)), x) -> *12(i1(x), z)
I1(*2(x, y)) -> *12(i1(y), i1(x))
*12(*2(i1(x), k2(y, z)), x) -> *12(i1(x), y)
*12(*2(i1(x), k2(y, z)), x) -> K2(*2(*2(i1(x), y), x), *2(*2(i1(x), z), x))
*12(*2(i1(x), k2(y, z)), x) -> *12(*2(i1(x), y), x)
I1(*2(x, y)) -> I1(y)
*12(x, *2(y, z)) -> *12(*2(x, y), z)

The TRS R consists of the following rules:

*2(x, 1) -> x
*2(1, y) -> y
*2(i1(x), x) -> 1
*2(x, i1(x)) -> 1
*2(x, *2(y, z)) -> *2(*2(x, y), z)
i1(1) -> 1
*2(*2(x, y), i1(y)) -> x
*2(*2(x, i1(y)), y) -> x
i1(i1(x)) -> x
i1(*2(x, y)) -> *2(i1(y), i1(x))
k2(x, 1) -> 1
k2(x, x) -> 1
*2(k2(x, y), k2(y, x)) -> 1
*2(*2(i1(x), k2(y, z)), x) -> k2(*2(*2(i1(x), y), x), *2(*2(i1(x), z), x))
k2(*2(x, i1(y)), *2(y, i1(x))) -> 1

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 2 less nodes.

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

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

*12(x, *2(y, z)) -> *12(x, y)
*12(*2(i1(x), k2(y, z)), x) -> *12(*2(i1(x), z), x)
*12(*2(i1(x), k2(y, z)), x) -> *12(i1(x), z)
*12(*2(i1(x), k2(y, z)), x) -> *12(i1(x), y)
*12(*2(i1(x), k2(y, z)), x) -> *12(*2(i1(x), y), x)
*12(x, *2(y, z)) -> *12(*2(x, y), z)

The TRS R consists of the following rules:

*2(x, 1) -> x
*2(1, y) -> y
*2(i1(x), x) -> 1
*2(x, i1(x)) -> 1
*2(x, *2(y, z)) -> *2(*2(x, y), z)
i1(1) -> 1
*2(*2(x, y), i1(y)) -> x
*2(*2(x, i1(y)), y) -> x
i1(i1(x)) -> x
i1(*2(x, y)) -> *2(i1(y), i1(x))
k2(x, 1) -> 1
k2(x, x) -> 1
*2(k2(x, y), k2(y, x)) -> 1
*2(*2(i1(x), k2(y, z)), x) -> k2(*2(*2(i1(x), y), x), *2(*2(i1(x), z), x))
k2(*2(x, i1(y)), *2(y, i1(x))) -> 1

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.


*12(*2(i1(x), k2(y, z)), x) -> *12(*2(i1(x), z), x)
*12(*2(i1(x), k2(y, z)), x) -> *12(i1(x), z)
*12(*2(i1(x), k2(y, z)), x) -> *12(i1(x), y)
*12(*2(i1(x), k2(y, z)), x) -> *12(*2(i1(x), y), x)
The remaining pairs can at least be oriented weakly.

*12(x, *2(y, z)) -> *12(x, y)
*12(x, *2(y, z)) -> *12(*2(x, y), z)
Used ordering: Polynomial interpretation [21]:

POL(*2(x1, x2)) = x1 + x1·x2 + x2   
POL(*12(x1, x2)) = 1 + x1 + x1·x2 + x2   
POL(1) = 0   
POL(i1(x1)) = x1   
POL(k2(x1, x2)) = 1 + x1 + x2   

The following usable rules [14] were oriented:

*2(x, i1(x)) -> 1
*2(*2(x, y), i1(y)) -> x
*2(*2(i1(x), k2(y, z)), x) -> k2(*2(*2(i1(x), y), x), *2(*2(i1(x), z), x))
i1(*2(x, y)) -> *2(i1(y), i1(x))
*2(i1(x), x) -> 1
k2(x, 1) -> 1
k2(*2(x, i1(y)), *2(y, i1(x))) -> 1
*2(1, y) -> y
k2(x, x) -> 1
i1(1) -> 1
*2(x, 1) -> x
*2(k2(x, y), k2(y, x)) -> 1
*2(*2(x, i1(y)), y) -> x
i1(i1(x)) -> x
*2(x, *2(y, z)) -> *2(*2(x, y), z)



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

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

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

The TRS R consists of the following rules:

*2(x, 1) -> x
*2(1, y) -> y
*2(i1(x), x) -> 1
*2(x, i1(x)) -> 1
*2(x, *2(y, z)) -> *2(*2(x, y), z)
i1(1) -> 1
*2(*2(x, y), i1(y)) -> x
*2(*2(x, i1(y)), y) -> x
i1(i1(x)) -> x
i1(*2(x, y)) -> *2(i1(y), i1(x))
k2(x, 1) -> 1
k2(x, x) -> 1
*2(k2(x, y), k2(y, x)) -> 1
*2(*2(i1(x), k2(y, z)), x) -> k2(*2(*2(i1(x), y), x), *2(*2(i1(x), z), x))
k2(*2(x, i1(y)), *2(y, i1(x))) -> 1

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.


*12(x, *2(y, z)) -> *12(x, y)
*12(x, *2(y, z)) -> *12(*2(x, y), z)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(*2(x1, x2)) = 1 + x1 + x2   
POL(*12(x1, x2)) = x2   
POL(1) = 0   
POL(i1(x1)) = 0   
POL(k2(x1, x2)) = 0   

The following usable rules [14] were oriented: none



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

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

*2(x, 1) -> x
*2(1, y) -> y
*2(i1(x), x) -> 1
*2(x, i1(x)) -> 1
*2(x, *2(y, z)) -> *2(*2(x, y), z)
i1(1) -> 1
*2(*2(x, y), i1(y)) -> x
*2(*2(x, i1(y)), y) -> x
i1(i1(x)) -> x
i1(*2(x, y)) -> *2(i1(y), i1(x))
k2(x, 1) -> 1
k2(x, x) -> 1
*2(k2(x, y), k2(y, x)) -> 1
*2(*2(i1(x), k2(y, z)), x) -> k2(*2(*2(i1(x), y), x), *2(*2(i1(x), z), x))
k2(*2(x, i1(y)), *2(y, i1(x))) -> 1

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:

I1(*2(x, y)) -> I1(x)
I1(*2(x, y)) -> I1(y)

The TRS R consists of the following rules:

*2(x, 1) -> x
*2(1, y) -> y
*2(i1(x), x) -> 1
*2(x, i1(x)) -> 1
*2(x, *2(y, z)) -> *2(*2(x, y), z)
i1(1) -> 1
*2(*2(x, y), i1(y)) -> x
*2(*2(x, i1(y)), y) -> x
i1(i1(x)) -> x
i1(*2(x, y)) -> *2(i1(y), i1(x))
k2(x, 1) -> 1
k2(x, x) -> 1
*2(k2(x, y), k2(y, x)) -> 1
*2(*2(i1(x), k2(y, z)), x) -> k2(*2(*2(i1(x), y), x), *2(*2(i1(x), z), x))
k2(*2(x, i1(y)), *2(y, i1(x))) -> 1

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.


I1(*2(x, y)) -> I1(x)
I1(*2(x, y)) -> I1(y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(*2(x1, x2)) = 1 + x1 + x2   
POL(I1(x1)) = 1 + x12   

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:

*2(x, 1) -> x
*2(1, y) -> y
*2(i1(x), x) -> 1
*2(x, i1(x)) -> 1
*2(x, *2(y, z)) -> *2(*2(x, y), z)
i1(1) -> 1
*2(*2(x, y), i1(y)) -> x
*2(*2(x, i1(y)), y) -> x
i1(i1(x)) -> x
i1(*2(x, y)) -> *2(i1(y), i1(x))
k2(x, 1) -> 1
k2(x, x) -> 1
*2(k2(x, y), k2(y, x)) -> 1
*2(*2(i1(x), k2(y, z)), x) -> k2(*2(*2(i1(x), y), x), *2(*2(i1(x), z), x))
k2(*2(x, i1(y)), *2(y, i1(x))) -> 1

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.