Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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

*2(0, x) -> 0
*2(1, x) -> x
*2(2, 2) -> .2(1, 0)
*2(3, x) -> .2(x, *2(min, x))
*2(min, min) -> 1
*2(2, min) -> .2(min, 2)
*2(.2(x, y), z) -> .2(*2(x, z), *2(y, z))
*2(+2(y, z), x) -> +2(*2(x, y), *2(x, z))
+2(0, x) -> x
+2(x, x) -> *2(2, x)
+2(1, 2) -> 3
+2(1, min) -> 0
+2(2, min) -> 1
+2(3, x) -> .2(1, +2(min, x))
+2(.2(x, y), z) -> .2(x, +2(y, z))
+2(*2(2, x), x) -> *2(3, x)
+2(*2(min, x), x) -> 0
+2(*2(2, v), *2(min, v)) -> v
.2(min, 3) -> min
.2(x, min) -> .2(+2(min, x), 3)
.2(0, x) -> x
.2(x, .2(y, z)) -> .2(+2(x, y), z)

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

*2(0, x) -> 0
*2(1, x) -> x
*2(2, 2) -> .2(1, 0)
*2(3, x) -> .2(x, *2(min, x))
*2(min, min) -> 1
*2(2, min) -> .2(min, 2)
*2(.2(x, y), z) -> .2(*2(x, z), *2(y, z))
*2(+2(y, z), x) -> +2(*2(x, y), *2(x, z))
+2(0, x) -> x
+2(x, x) -> *2(2, x)
+2(1, 2) -> 3
+2(1, min) -> 0
+2(2, min) -> 1
+2(3, x) -> .2(1, +2(min, x))
+2(.2(x, y), z) -> .2(x, +2(y, z))
+2(*2(2, x), x) -> *2(3, x)
+2(*2(min, x), x) -> 0
+2(*2(2, v), *2(min, v)) -> v
.2(min, 3) -> min
.2(x, min) -> .2(+2(min, x), 3)
.2(0, x) -> x
.2(x, .2(y, z)) -> .2(+2(x, y), 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, 2) -> .12(1, 0)
+12(3, x) -> .12(1, +2(min, x))
*12(+2(y, z), x) -> +12(*2(x, y), *2(x, z))
*12(.2(x, y), z) -> *12(y, z)
+12(3, x) -> +12(min, x)
+12(.2(x, y), z) -> .12(x, +2(y, z))
.12(x, .2(y, z)) -> .12(+2(x, y), z)
.12(x, min) -> .12(+2(min, x), 3)
*12(2, min) -> .12(min, 2)
.12(x, .2(y, z)) -> +12(x, y)
+12(*2(2, x), x) -> *12(3, x)
*12(.2(x, y), z) -> .12(*2(x, z), *2(y, z))
*12(3, x) -> *12(min, x)
*12(+2(y, z), x) -> *12(x, y)
+12(x, x) -> *12(2, x)
*12(3, x) -> .12(x, *2(min, x))
.12(x, min) -> +12(min, x)
+12(.2(x, y), z) -> +12(y, z)
*12(.2(x, y), z) -> *12(x, z)
*12(+2(y, z), x) -> *12(x, z)

The TRS R consists of the following rules:

*2(0, x) -> 0
*2(1, x) -> x
*2(2, 2) -> .2(1, 0)
*2(3, x) -> .2(x, *2(min, x))
*2(min, min) -> 1
*2(2, min) -> .2(min, 2)
*2(.2(x, y), z) -> .2(*2(x, z), *2(y, z))
*2(+2(y, z), x) -> +2(*2(x, y), *2(x, z))
+2(0, x) -> x
+2(x, x) -> *2(2, x)
+2(1, 2) -> 3
+2(1, min) -> 0
+2(2, min) -> 1
+2(3, x) -> .2(1, +2(min, x))
+2(.2(x, y), z) -> .2(x, +2(y, z))
+2(*2(2, x), x) -> *2(3, x)
+2(*2(min, x), x) -> 0
+2(*2(2, v), *2(min, v)) -> v
.2(min, 3) -> min
.2(x, min) -> .2(+2(min, x), 3)
.2(0, x) -> x
.2(x, .2(y, z)) -> .2(+2(x, y), z)

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:

*12(2, 2) -> .12(1, 0)
+12(3, x) -> .12(1, +2(min, x))
*12(+2(y, z), x) -> +12(*2(x, y), *2(x, z))
*12(.2(x, y), z) -> *12(y, z)
+12(3, x) -> +12(min, x)
+12(.2(x, y), z) -> .12(x, +2(y, z))
.12(x, .2(y, z)) -> .12(+2(x, y), z)
.12(x, min) -> .12(+2(min, x), 3)
*12(2, min) -> .12(min, 2)
.12(x, .2(y, z)) -> +12(x, y)
+12(*2(2, x), x) -> *12(3, x)
*12(.2(x, y), z) -> .12(*2(x, z), *2(y, z))
*12(3, x) -> *12(min, x)
*12(+2(y, z), x) -> *12(x, y)
+12(x, x) -> *12(2, x)
*12(3, x) -> .12(x, *2(min, x))
.12(x, min) -> +12(min, x)
+12(.2(x, y), z) -> +12(y, z)
*12(.2(x, y), z) -> *12(x, z)
*12(+2(y, z), x) -> *12(x, z)

The TRS R consists of the following rules:

*2(0, x) -> 0
*2(1, x) -> x
*2(2, 2) -> .2(1, 0)
*2(3, x) -> .2(x, *2(min, x))
*2(min, min) -> 1
*2(2, min) -> .2(min, 2)
*2(.2(x, y), z) -> .2(*2(x, z), *2(y, z))
*2(+2(y, z), x) -> +2(*2(x, y), *2(x, z))
+2(0, x) -> x
+2(x, x) -> *2(2, x)
+2(1, 2) -> 3
+2(1, min) -> 0
+2(2, min) -> 1
+2(3, x) -> .2(1, +2(min, x))
+2(.2(x, y), z) -> .2(x, +2(y, z))
+2(*2(2, x), x) -> *2(3, x)
+2(*2(min, x), x) -> 0
+2(*2(2, v), *2(min, v)) -> v
.2(min, 3) -> min
.2(x, min) -> .2(+2(min, x), 3)
.2(0, x) -> x
.2(x, .2(y, z)) -> .2(+2(x, y), 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 with 9 less nodes.

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

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

+12(3, x) -> .12(1, +2(min, x))
*12(3, x) -> .12(x, *2(min, x))
+12(.2(x, y), z) -> +12(y, z)
+12(.2(x, y), z) -> .12(x, +2(y, z))
.12(x, .2(y, z)) -> .12(+2(x, y), z)
.12(x, .2(y, z)) -> +12(x, y)
+12(*2(2, x), x) -> *12(3, x)

The TRS R consists of the following rules:

*2(0, x) -> 0
*2(1, x) -> x
*2(2, 2) -> .2(1, 0)
*2(3, x) -> .2(x, *2(min, x))
*2(min, min) -> 1
*2(2, min) -> .2(min, 2)
*2(.2(x, y), z) -> .2(*2(x, z), *2(y, z))
*2(+2(y, z), x) -> +2(*2(x, y), *2(x, z))
+2(0, x) -> x
+2(x, x) -> *2(2, x)
+2(1, 2) -> 3
+2(1, min) -> 0
+2(2, min) -> 1
+2(3, x) -> .2(1, +2(min, x))
+2(.2(x, y), z) -> .2(x, +2(y, z))
+2(*2(2, x), x) -> *2(3, x)
+2(*2(min, x), x) -> 0
+2(*2(2, v), *2(min, v)) -> v
.2(min, 3) -> min
.2(x, min) -> .2(+2(min, x), 3)
.2(0, x) -> x
.2(x, .2(y, z)) -> .2(+2(x, y), z)

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

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

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

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

The TRS R consists of the following rules:

*2(0, x) -> 0
*2(1, x) -> x
*2(2, 2) -> .2(1, 0)
*2(3, x) -> .2(x, *2(min, x))
*2(min, min) -> 1
*2(2, min) -> .2(min, 2)
*2(.2(x, y), z) -> .2(*2(x, z), *2(y, z))
*2(+2(y, z), x) -> +2(*2(x, y), *2(x, z))
+2(0, x) -> x
+2(x, x) -> *2(2, x)
+2(1, 2) -> 3
+2(1, min) -> 0
+2(2, min) -> 1
+2(3, x) -> .2(1, +2(min, x))
+2(.2(x, y), z) -> .2(x, +2(y, z))
+2(*2(2, x), x) -> *2(3, x)
+2(*2(min, x), x) -> 0
+2(*2(2, v), *2(min, v)) -> v
.2(min, 3) -> min
.2(x, min) -> .2(+2(min, x), 3)
.2(0, x) -> x
.2(x, .2(y, z)) -> .2(+2(x, y), 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.


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

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

POL(*12(x1, x2)) = x1 + 2·x2   
POL(+2(x1, x2)) = 1 + 2·x1 + 2·x2   
POL(.2(x1, x2)) = x1 + x2   

The following usable rules [14] were oriented: none



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

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

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

The TRS R consists of the following rules:

*2(0, x) -> 0
*2(1, x) -> x
*2(2, 2) -> .2(1, 0)
*2(3, x) -> .2(x, *2(min, x))
*2(min, min) -> 1
*2(2, min) -> .2(min, 2)
*2(.2(x, y), z) -> .2(*2(x, z), *2(y, z))
*2(+2(y, z), x) -> +2(*2(x, y), *2(x, z))
+2(0, x) -> x
+2(x, x) -> *2(2, x)
+2(1, 2) -> 3
+2(1, min) -> 0
+2(2, min) -> 1
+2(3, x) -> .2(1, +2(min, x))
+2(.2(x, y), z) -> .2(x, +2(y, z))
+2(*2(2, x), x) -> *2(3, x)
+2(*2(min, x), x) -> 0
+2(*2(2, v), *2(min, v)) -> v
.2(min, 3) -> min
.2(x, min) -> .2(+2(min, x), 3)
.2(0, x) -> x
.2(x, .2(y, z)) -> .2(+2(x, y), 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.


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

POL(*12(x1, x2)) = 2·x1   
POL(.2(x1, x2)) = 1 + x1 + 2·x2   

The following usable rules [14] were oriented: none



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

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

*2(0, x) -> 0
*2(1, x) -> x
*2(2, 2) -> .2(1, 0)
*2(3, x) -> .2(x, *2(min, x))
*2(min, min) -> 1
*2(2, min) -> .2(min, 2)
*2(.2(x, y), z) -> .2(*2(x, z), *2(y, z))
*2(+2(y, z), x) -> +2(*2(x, y), *2(x, z))
+2(0, x) -> x
+2(x, x) -> *2(2, x)
+2(1, 2) -> 3
+2(1, min) -> 0
+2(2, min) -> 1
+2(3, x) -> .2(1, +2(min, x))
+2(.2(x, y), z) -> .2(x, +2(y, z))
+2(*2(2, x), x) -> *2(3, x)
+2(*2(min, x), x) -> 0
+2(*2(2, v), *2(min, v)) -> v
.2(min, 3) -> min
.2(x, min) -> .2(+2(min, x), 3)
.2(0, x) -> x
.2(x, .2(y, z)) -> .2(+2(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.