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, 0) -> x
+2(minus1(x), x) -> 0
minus1(0) -> 0
minus1(minus1(x)) -> x
minus1(+2(x, y)) -> +2(minus1(y), minus1(x))
*2(x, 1) -> x
*2(x, 0) -> 0
*2(x, +2(y, z)) -> +2(*2(x, y), *2(x, z))
*2(x, minus1(y)) -> minus1(*2(x, y))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

+2(x, 0) -> x
+2(minus1(x), x) -> 0
minus1(0) -> 0
minus1(minus1(x)) -> x
minus1(+2(x, y)) -> +2(minus1(y), minus1(x))
*2(x, 1) -> x
*2(x, 0) -> 0
*2(x, +2(y, z)) -> +2(*2(x, y), *2(x, z))
*2(x, minus1(y)) -> minus1(*2(x, y))

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(x, minus1(y)) -> MINUS1(*2(x, y))
*12(x, +2(y, z)) -> +12(*2(x, y), *2(x, z))
MINUS1(+2(x, y)) -> MINUS1(y)
*12(x, minus1(y)) -> *12(x, y)
*12(x, +2(y, z)) -> *12(x, z)
MINUS1(+2(x, y)) -> MINUS1(x)
*12(x, +2(y, z)) -> *12(x, y)
MINUS1(+2(x, y)) -> +12(minus1(y), minus1(x))

The TRS R consists of the following rules:

+2(x, 0) -> x
+2(minus1(x), x) -> 0
minus1(0) -> 0
minus1(minus1(x)) -> x
minus1(+2(x, y)) -> +2(minus1(y), minus1(x))
*2(x, 1) -> x
*2(x, 0) -> 0
*2(x, +2(y, z)) -> +2(*2(x, y), *2(x, z))
*2(x, minus1(y)) -> minus1(*2(x, y))

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(x, minus1(y)) -> MINUS1(*2(x, y))
*12(x, +2(y, z)) -> +12(*2(x, y), *2(x, z))
MINUS1(+2(x, y)) -> MINUS1(y)
*12(x, minus1(y)) -> *12(x, y)
*12(x, +2(y, z)) -> *12(x, z)
MINUS1(+2(x, y)) -> MINUS1(x)
*12(x, +2(y, z)) -> *12(x, y)
MINUS1(+2(x, y)) -> +12(minus1(y), minus1(x))

The TRS R consists of the following rules:

+2(x, 0) -> x
+2(minus1(x), x) -> 0
minus1(0) -> 0
minus1(minus1(x)) -> x
minus1(+2(x, y)) -> +2(minus1(y), minus1(x))
*2(x, 1) -> x
*2(x, 0) -> 0
*2(x, +2(y, z)) -> +2(*2(x, y), *2(x, z))
*2(x, minus1(y)) -> minus1(*2(x, y))

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 3 less nodes.

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

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

MINUS1(+2(x, y)) -> MINUS1(y)
MINUS1(+2(x, y)) -> MINUS1(x)

The TRS R consists of the following rules:

+2(x, 0) -> x
+2(minus1(x), x) -> 0
minus1(0) -> 0
minus1(minus1(x)) -> x
minus1(+2(x, y)) -> +2(minus1(y), minus1(x))
*2(x, 1) -> x
*2(x, 0) -> 0
*2(x, +2(y, z)) -> +2(*2(x, y), *2(x, z))
*2(x, minus1(y)) -> minus1(*2(x, y))

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.


MINUS1(+2(x, y)) -> MINUS1(y)
MINUS1(+2(x, y)) -> MINUS1(x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(+2(x1, x2)) = 1 + 2·x1 + 2·x2   
POL(MINUS1(x1)) = 2·x1   

The following usable rules [14] were oriented: none



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

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

+2(x, 0) -> x
+2(minus1(x), x) -> 0
minus1(0) -> 0
minus1(minus1(x)) -> x
minus1(+2(x, y)) -> +2(minus1(y), minus1(x))
*2(x, 1) -> x
*2(x, 0) -> 0
*2(x, +2(y, z)) -> +2(*2(x, y), *2(x, z))
*2(x, minus1(y)) -> minus1(*2(x, y))

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:

*12(x, minus1(y)) -> *12(x, y)
*12(x, +2(y, z)) -> *12(x, z)
*12(x, +2(y, z)) -> *12(x, y)

The TRS R consists of the following rules:

+2(x, 0) -> x
+2(minus1(x), x) -> 0
minus1(0) -> 0
minus1(minus1(x)) -> x
minus1(+2(x, y)) -> +2(minus1(y), minus1(x))
*2(x, 1) -> x
*2(x, 0) -> 0
*2(x, +2(y, z)) -> +2(*2(x, y), *2(x, z))
*2(x, minus1(y)) -> minus1(*2(x, y))

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

*12(x, minus1(y)) -> *12(x, y)
Used ordering: Polynomial interpretation [21]:

POL(*12(x1, x2)) = 2·x2   
POL(+2(x1, x2)) = 2 + 2·x1 + 2·x2   
POL(minus1(x1)) = 2·x1   

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(x, minus1(y)) -> *12(x, y)

The TRS R consists of the following rules:

+2(x, 0) -> x
+2(minus1(x), x) -> 0
minus1(0) -> 0
minus1(minus1(x)) -> x
minus1(+2(x, y)) -> +2(minus1(y), minus1(x))
*2(x, 1) -> x
*2(x, 0) -> 0
*2(x, +2(y, z)) -> +2(*2(x, y), *2(x, z))
*2(x, minus1(y)) -> minus1(*2(x, y))

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, minus1(y)) -> *12(x, y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(*12(x1, x2)) = 2·x2   
POL(minus1(x1)) = 1 + 2·x1   

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(x, 0) -> x
+2(minus1(x), x) -> 0
minus1(0) -> 0
minus1(minus1(x)) -> x
minus1(+2(x, y)) -> +2(minus1(y), minus1(x))
*2(x, 1) -> x
*2(x, 0) -> 0
*2(x, +2(y, z)) -> +2(*2(x, y), *2(x, z))
*2(x, minus1(y)) -> minus1(*2(x, y))

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.