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(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
p1(s1(x)) -> x
f2(s1(x), y) -> f2(p1(-2(s1(x), y)), p1(-2(y, s1(x))))
f2(x, s1(y)) -> f2(p1(-2(x, s1(y))), p1(-2(s1(y), x)))

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
p1(s1(x)) -> x
f2(s1(x), y) -> f2(p1(-2(s1(x), y)), p1(-2(y, s1(x))))
f2(x, s1(y)) -> f2(p1(-2(x, s1(y))), p1(-2(s1(y), 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:

F2(x, s1(y)) -> -12(s1(y), x)
F2(x, s1(y)) -> -12(x, s1(y))
F2(s1(x), y) -> P1(-2(y, s1(x)))
F2(s1(x), y) -> P1(-2(s1(x), y))
-12(s1(x), s1(y)) -> -12(x, y)
F2(s1(x), y) -> F2(p1(-2(s1(x), y)), p1(-2(y, s1(x))))
F2(x, s1(y)) -> F2(p1(-2(x, s1(y))), p1(-2(s1(y), x)))
F2(x, s1(y)) -> P1(-2(s1(y), x))
F2(x, s1(y)) -> P1(-2(x, s1(y)))
F2(s1(x), y) -> -12(y, s1(x))
F2(s1(x), y) -> -12(s1(x), y)

The TRS R consists of the following rules:

-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
p1(s1(x)) -> x
f2(s1(x), y) -> f2(p1(-2(s1(x), y)), p1(-2(y, s1(x))))
f2(x, s1(y)) -> f2(p1(-2(x, s1(y))), p1(-2(s1(y), 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:

F2(x, s1(y)) -> -12(s1(y), x)
F2(x, s1(y)) -> -12(x, s1(y))
F2(s1(x), y) -> P1(-2(y, s1(x)))
F2(s1(x), y) -> P1(-2(s1(x), y))
-12(s1(x), s1(y)) -> -12(x, y)
F2(s1(x), y) -> F2(p1(-2(s1(x), y)), p1(-2(y, s1(x))))
F2(x, s1(y)) -> F2(p1(-2(x, s1(y))), p1(-2(s1(y), x)))
F2(x, s1(y)) -> P1(-2(s1(y), x))
F2(x, s1(y)) -> P1(-2(x, s1(y)))
F2(s1(x), y) -> -12(y, s1(x))
F2(s1(x), y) -> -12(s1(x), y)

The TRS R consists of the following rules:

-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
p1(s1(x)) -> x
f2(s1(x), y) -> f2(p1(-2(s1(x), y)), p1(-2(y, s1(x))))
f2(x, s1(y)) -> f2(p1(-2(x, s1(y))), p1(-2(s1(y), 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 8 less nodes.

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

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

-12(s1(x), s1(y)) -> -12(x, y)

The TRS R consists of the following rules:

-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
p1(s1(x)) -> x
f2(s1(x), y) -> f2(p1(-2(s1(x), y)), p1(-2(y, s1(x))))
f2(x, s1(y)) -> f2(p1(-2(x, s1(y))), p1(-2(s1(y), 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.


-12(s1(x), s1(y)) -> -12(x, y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(-12(x1, x2)) = x1   
POL(s1(x1)) = 1 + 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(s1(x), s1(y)) -> -2(x, y)
p1(s1(x)) -> x
f2(s1(x), y) -> f2(p1(-2(s1(x), y)), p1(-2(y, s1(x))))
f2(x, s1(y)) -> f2(p1(-2(x, s1(y))), p1(-2(s1(y), 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

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

F2(s1(x), y) -> F2(p1(-2(s1(x), y)), p1(-2(y, s1(x))))
F2(x, s1(y)) -> F2(p1(-2(x, s1(y))), p1(-2(s1(y), x)))

The TRS R consists of the following rules:

-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
p1(s1(x)) -> x
f2(s1(x), y) -> f2(p1(-2(s1(x), y)), p1(-2(y, s1(x))))
f2(x, s1(y)) -> f2(p1(-2(x, s1(y))), p1(-2(s1(y), x)))

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