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(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 Order [17,21] with Interpretation:

POL( s1(x1) ) = x1 + 3


POL( -12(x1, x2) ) = 3x2 + 3



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
            ↳ QDPOrderProof

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.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


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 remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( -2(x1, x2) ) = x1


POL( s1(x1) ) = x1 + 3


POL( 0 ) = 0


POL( F2(x1, x2) ) = max{0, 3x1 + 2x2 - 3}


POL( p1(x1) ) = max{0, x1 - 3}



The following usable rules [14] were oriented:

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



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

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.