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:

minus1(minus1(x)) -> x
minus1(+2(x, y)) -> *2(minus1(minus1(minus1(x))), minus1(minus1(minus1(y))))
minus1(*2(x, y)) -> +2(minus1(minus1(minus1(x))), minus1(minus1(minus1(y))))
f1(minus1(x)) -> minus1(minus1(minus1(f1(x))))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

minus1(minus1(x)) -> x
minus1(+2(x, y)) -> *2(minus1(minus1(minus1(x))), minus1(minus1(minus1(y))))
minus1(*2(x, y)) -> +2(minus1(minus1(minus1(x))), minus1(minus1(minus1(y))))
f1(minus1(x)) -> minus1(minus1(minus1(f1(x))))

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:

F1(minus1(x)) -> MINUS1(f1(x))
MINUS1(+2(x, y)) -> MINUS1(minus1(x))
MINUS1(+2(x, y)) -> MINUS1(minus1(minus1(y)))
MINUS1(+2(x, y)) -> MINUS1(y)
MINUS1(+2(x, y)) -> MINUS1(x)
MINUS1(*2(x, y)) -> MINUS1(minus1(minus1(y)))
F1(minus1(x)) -> F1(x)
MINUS1(*2(x, y)) -> MINUS1(x)
MINUS1(*2(x, y)) -> MINUS1(minus1(y))
MINUS1(*2(x, y)) -> MINUS1(y)
F1(minus1(x)) -> MINUS1(minus1(f1(x)))
F1(minus1(x)) -> MINUS1(minus1(minus1(f1(x))))
MINUS1(*2(x, y)) -> MINUS1(minus1(minus1(x)))
MINUS1(+2(x, y)) -> MINUS1(minus1(minus1(x)))
MINUS1(*2(x, y)) -> MINUS1(minus1(x))
MINUS1(+2(x, y)) -> MINUS1(minus1(y))

The TRS R consists of the following rules:

minus1(minus1(x)) -> x
minus1(+2(x, y)) -> *2(minus1(minus1(minus1(x))), minus1(minus1(minus1(y))))
minus1(*2(x, y)) -> +2(minus1(minus1(minus1(x))), minus1(minus1(minus1(y))))
f1(minus1(x)) -> minus1(minus1(minus1(f1(x))))

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:

F1(minus1(x)) -> MINUS1(f1(x))
MINUS1(+2(x, y)) -> MINUS1(minus1(x))
MINUS1(+2(x, y)) -> MINUS1(minus1(minus1(y)))
MINUS1(+2(x, y)) -> MINUS1(y)
MINUS1(+2(x, y)) -> MINUS1(x)
MINUS1(*2(x, y)) -> MINUS1(minus1(minus1(y)))
F1(minus1(x)) -> F1(x)
MINUS1(*2(x, y)) -> MINUS1(x)
MINUS1(*2(x, y)) -> MINUS1(minus1(y))
MINUS1(*2(x, y)) -> MINUS1(y)
F1(minus1(x)) -> MINUS1(minus1(f1(x)))
F1(minus1(x)) -> MINUS1(minus1(minus1(f1(x))))
MINUS1(*2(x, y)) -> MINUS1(minus1(minus1(x)))
MINUS1(+2(x, y)) -> MINUS1(minus1(minus1(x)))
MINUS1(*2(x, y)) -> MINUS1(minus1(x))
MINUS1(+2(x, y)) -> MINUS1(minus1(y))

The TRS R consists of the following rules:

minus1(minus1(x)) -> x
minus1(+2(x, y)) -> *2(minus1(minus1(minus1(x))), minus1(minus1(minus1(y))))
minus1(*2(x, y)) -> +2(minus1(minus1(minus1(x))), minus1(minus1(minus1(y))))
f1(minus1(x)) -> minus1(minus1(minus1(f1(x))))

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

The TRS R consists of the following rules:

minus1(minus1(x)) -> x
minus1(+2(x, y)) -> *2(minus1(minus1(minus1(x))), minus1(minus1(minus1(y))))
minus1(*2(x, y)) -> +2(minus1(minus1(minus1(x))), minus1(minus1(minus1(y))))
f1(minus1(x)) -> minus1(minus1(minus1(f1(x))))

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(minus1(x))
MINUS1(+2(x, y)) -> MINUS1(y)
MINUS1(+2(x, y)) -> MINUS1(minus1(minus1(y)))
MINUS1(+2(x, y)) -> MINUS1(x)
MINUS1(*2(x, y)) -> MINUS1(minus1(minus1(y)))
MINUS1(*2(x, y)) -> MINUS1(x)
MINUS1(*2(x, y)) -> MINUS1(minus1(minus1(x)))
MINUS1(+2(x, y)) -> MINUS1(minus1(minus1(x)))
MINUS1(*2(x, y)) -> MINUS1(minus1(x))
MINUS1(*2(x, y)) -> MINUS1(y)
MINUS1(*2(x, y)) -> MINUS1(minus1(y))
MINUS1(+2(x, y)) -> MINUS1(minus1(y))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(*2(x1, x2)) = 1 + x1 + 2·x2   
POL(+2(x1, x2)) = 1 + x1 + 2·x2   
POL(MINUS1(x1)) = 2·x1   
POL(minus1(x1)) = x1   

The following usable rules [14] were oriented:

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



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

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

minus1(minus1(x)) -> x
minus1(+2(x, y)) -> *2(minus1(minus1(minus1(x))), minus1(minus1(minus1(y))))
minus1(*2(x, y)) -> +2(minus1(minus1(minus1(x))), minus1(minus1(minus1(y))))
f1(minus1(x)) -> minus1(minus1(minus1(f1(x))))

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:

F1(minus1(x)) -> F1(x)

The TRS R consists of the following rules:

minus1(minus1(x)) -> x
minus1(+2(x, y)) -> *2(minus1(minus1(minus1(x))), minus1(minus1(minus1(y))))
minus1(*2(x, y)) -> +2(minus1(minus1(minus1(x))), minus1(minus1(minus1(y))))
f1(minus1(x)) -> minus1(minus1(minus1(f1(x))))

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.


F1(minus1(x)) -> F1(x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(F1(x1)) = 2·x1   
POL(minus1(x1)) = 1 + 2·x1   

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:

minus1(minus1(x)) -> x
minus1(+2(x, y)) -> *2(minus1(minus1(minus1(x))), minus1(minus1(minus1(y))))
minus1(*2(x, y)) -> +2(minus1(minus1(minus1(x))), minus1(minus1(minus1(y))))
f1(minus1(x)) -> minus1(minus1(minus1(f1(x))))

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.