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:

minus(minus(x)) → x
minus(+(x, y)) → *(minus(minus(minus(x))), minus(minus(minus(y))))
minus(*(x, y)) → +(minus(minus(minus(x))), minus(minus(minus(y))))
f(minus(x)) → minus(minus(minus(f(x))))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

minus(minus(x)) → x
minus(+(x, y)) → *(minus(minus(minus(x))), minus(minus(minus(y))))
minus(*(x, y)) → +(minus(minus(minus(x))), minus(minus(minus(y))))
f(minus(x)) → minus(minus(minus(f(x))))

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:

MINUS(*(x, y)) → MINUS(minus(minus(x)))
F(minus(x)) → F(x)
MINUS(+(x, y)) → MINUS(minus(y))
MINUS(*(x, y)) → MINUS(minus(minus(y)))
MINUS(*(x, y)) → MINUS(y)
MINUS(+(x, y)) → MINUS(minus(minus(y)))
MINUS(+(x, y)) → MINUS(minus(x))
F(minus(x)) → MINUS(f(x))
F(minus(x)) → MINUS(minus(f(x)))
MINUS(+(x, y)) → MINUS(y)
MINUS(*(x, y)) → MINUS(minus(y))
MINUS(+(x, y)) → MINUS(minus(minus(x)))
F(minus(x)) → MINUS(minus(minus(f(x))))
MINUS(+(x, y)) → MINUS(x)
MINUS(*(x, y)) → MINUS(minus(x))
MINUS(*(x, y)) → MINUS(x)

The TRS R consists of the following rules:

minus(minus(x)) → x
minus(+(x, y)) → *(minus(minus(minus(x))), minus(minus(minus(y))))
minus(*(x, y)) → +(minus(minus(minus(x))), minus(minus(minus(y))))
f(minus(x)) → minus(minus(minus(f(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:

MINUS(*(x, y)) → MINUS(minus(minus(x)))
F(minus(x)) → F(x)
MINUS(+(x, y)) → MINUS(minus(y))
MINUS(*(x, y)) → MINUS(minus(minus(y)))
MINUS(*(x, y)) → MINUS(y)
MINUS(+(x, y)) → MINUS(minus(minus(y)))
MINUS(+(x, y)) → MINUS(minus(x))
F(minus(x)) → MINUS(f(x))
F(minus(x)) → MINUS(minus(f(x)))
MINUS(+(x, y)) → MINUS(y)
MINUS(*(x, y)) → MINUS(minus(y))
MINUS(+(x, y)) → MINUS(minus(minus(x)))
F(minus(x)) → MINUS(minus(minus(f(x))))
MINUS(+(x, y)) → MINUS(x)
MINUS(*(x, y)) → MINUS(minus(x))
MINUS(*(x, y)) → MINUS(x)

The TRS R consists of the following rules:

minus(minus(x)) → x
minus(+(x, y)) → *(minus(minus(minus(x))), minus(minus(minus(y))))
minus(*(x, y)) → +(minus(minus(minus(x))), minus(minus(minus(y))))
f(minus(x)) → minus(minus(minus(f(x))))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] 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:

MINUS(*(x, y)) → MINUS(minus(minus(x)))
MINUS(+(x, y)) → MINUS(minus(x))
MINUS(+(x, y)) → MINUS(minus(y))
MINUS(*(x, y)) → MINUS(y)
MINUS(*(x, y)) → MINUS(minus(minus(y)))
MINUS(+(x, y)) → MINUS(minus(minus(y)))
MINUS(+(x, y)) → MINUS(y)
MINUS(*(x, y)) → MINUS(minus(y))
MINUS(+(x, y)) → MINUS(minus(minus(x)))
MINUS(*(x, y)) → MINUS(minus(x))
MINUS(+(x, y)) → MINUS(x)
MINUS(*(x, y)) → MINUS(x)

The TRS R consists of the following rules:

minus(minus(x)) → x
minus(+(x, y)) → *(minus(minus(minus(x))), minus(minus(minus(y))))
minus(*(x, y)) → +(minus(minus(minus(x))), minus(minus(minus(y))))
f(minus(x)) → minus(minus(minus(f(x))))

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.


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

POL(minus(x1)) = x_1   
POL(*(x1, x2)) = 1 + x_1 + x_2   
POL(MINUS(x1)) = (4)x_1   
POL(+(x1, x2)) = 1 + x_1 + x_2   
The value of delta used in the strict ordering is 4.
The following usable rules [17] were oriented:

minus(minus(x)) → x
minus(*(x, y)) → +(minus(minus(minus(x))), minus(minus(minus(y))))
minus(+(x, y)) → *(minus(minus(minus(x))), minus(minus(minus(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:

minus(minus(x)) → x
minus(+(x, y)) → *(minus(minus(minus(x))), minus(minus(minus(y))))
minus(*(x, y)) → +(minus(minus(minus(x))), minus(minus(minus(y))))
f(minus(x)) → minus(minus(minus(f(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:

F(minus(x)) → F(x)

The TRS R consists of the following rules:

minus(minus(x)) → x
minus(+(x, y)) → *(minus(minus(minus(x))), minus(minus(minus(y))))
minus(*(x, y)) → +(minus(minus(minus(x))), minus(minus(minus(y))))
f(minus(x)) → minus(minus(minus(f(x))))

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.


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

POL(minus(x1)) = 1 + (4)x_1   
POL(F(x1)) = (4)x_1   
The value of delta used in the strict ordering is 4.
The following usable rules [17] 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:

minus(minus(x)) → x
minus(+(x, y)) → *(minus(minus(minus(x))), minus(minus(minus(y))))
minus(*(x, y)) → +(minus(minus(minus(x))), minus(minus(minus(y))))
f(minus(x)) → minus(minus(minus(f(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.