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:

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
p(s(x)) → x
f(s(x), y) → f(p(-(s(x), y)), p(-(y, s(x))))
f(x, s(y)) → f(p(-(x, s(y))), p(-(s(y), x)))

Q is empty.


↳ QTRS
  ↳ AAECC Innermost

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

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
p(s(x)) → x
f(s(x), y) → f(p(-(s(x), y)), p(-(y, s(x))))
f(x, s(y)) → f(p(-(x, s(y))), p(-(s(y), x)))

Q is empty.

We have applied [15,7] to switch to innermost. The TRS R 1 is

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
p(s(x)) → x

The TRS R 2 is

f(s(x), y) → f(p(-(s(x), y)), p(-(y, s(x))))
f(x, s(y)) → f(p(-(x, s(y))), p(-(s(y), x)))

The signature Sigma is {f}

↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof

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

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
p(s(x)) → x
f(s(x), y) → f(p(-(s(x), y)), p(-(y, s(x))))
f(x, s(y)) → f(p(-(x, s(y))), p(-(s(y), x)))

The set Q consists of the following terms:

-(x0, 0)
-(s(x0), s(x1))
p(s(x0))
f(s(x0), x1)
f(x0, s(x1))


Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

F(x, s(y)) → -1(s(y), x)
F(x, s(y)) → -1(x, s(y))
F(s(x), y) → P(-(y, s(x)))
F(s(x), y) → P(-(s(x), y))
-1(s(x), s(y)) → -1(x, y)
F(s(x), y) → F(p(-(s(x), y)), p(-(y, s(x))))
F(x, s(y)) → F(p(-(x, s(y))), p(-(s(y), x)))
F(x, s(y)) → P(-(s(y), x))
F(x, s(y)) → P(-(x, s(y)))
F(s(x), y) → -1(y, s(x))
F(s(x), y) → -1(s(x), y)

The TRS R consists of the following rules:

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
p(s(x)) → x
f(s(x), y) → f(p(-(s(x), y)), p(-(y, s(x))))
f(x, s(y)) → f(p(-(x, s(y))), p(-(s(y), x)))

The set Q consists of the following terms:

-(x0, 0)
-(s(x0), s(x1))
p(s(x0))
f(s(x0), x1)
f(x0, s(x1))

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

↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof

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

F(x, s(y)) → -1(s(y), x)
F(x, s(y)) → -1(x, s(y))
F(s(x), y) → P(-(y, s(x)))
F(s(x), y) → P(-(s(x), y))
-1(s(x), s(y)) → -1(x, y)
F(s(x), y) → F(p(-(s(x), y)), p(-(y, s(x))))
F(x, s(y)) → F(p(-(x, s(y))), p(-(s(y), x)))
F(x, s(y)) → P(-(s(y), x))
F(x, s(y)) → P(-(x, s(y)))
F(s(x), y) → -1(y, s(x))
F(s(x), y) → -1(s(x), y)

The TRS R consists of the following rules:

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
p(s(x)) → x
f(s(x), y) → f(p(-(s(x), y)), p(-(y, s(x))))
f(x, s(y)) → f(p(-(x, s(y))), p(-(s(y), x)))

The set Q consists of the following terms:

-(x0, 0)
-(s(x0), s(x1))
p(s(x0))
f(s(x0), x1)
f(x0, s(x1))

We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof

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

F(x, s(y)) → -1(x, s(y))
F(x, s(y)) → -1(s(y), x)
F(s(x), y) → P(-(s(x), y))
F(s(x), y) → P(-(y, s(x)))
-1(s(x), s(y)) → -1(x, y)
F(s(x), y) → F(p(-(s(x), y)), p(-(y, s(x))))
F(x, s(y)) → F(p(-(x, s(y))), p(-(s(y), x)))
F(x, s(y)) → P(-(x, s(y)))
F(x, s(y)) → P(-(s(y), x))
F(s(x), y) → -1(s(x), y)
F(s(x), y) → -1(y, s(x))

The TRS R consists of the following rules:

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
p(s(x)) → x
f(s(x), y) → f(p(-(s(x), y)), p(-(y, s(x))))
f(x, s(y)) → f(p(-(x, s(y))), p(-(s(y), x)))

The set Q consists of the following terms:

-(x0, 0)
-(s(x0), s(x1))
p(s(x0))
f(s(x0), x1)
f(x0, s(x1))

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
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                    ↳ QDPOrderProof
                  ↳ QDP

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

-1(s(x), s(y)) → -1(x, y)

The TRS R consists of the following rules:

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
p(s(x)) → x
f(s(x), y) → f(p(-(s(x), y)), p(-(y, s(x))))
f(x, s(y)) → f(p(-(x, s(y))), p(-(s(y), x)))

The set Q consists of the following terms:

-(x0, 0)
-(s(x0), s(x1))
p(s(x0))
f(s(x0), x1)
f(x0, s(x1))

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.


-1(s(x), s(y)) → -1(x, y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
-1(x1, x2)  =  x2
s(x1)  =  s(x1)

Recursive path order with status [2].
Quasi-Precedence:
trivial

Status:
trivial


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                    ↳ QDPOrderProof
                      ↳ QDP
                        ↳ PisEmptyProof
                  ↳ QDP

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

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
p(s(x)) → x
f(s(x), y) → f(p(-(s(x), y)), p(-(y, s(x))))
f(x, s(y)) → f(p(-(x, s(y))), p(-(s(y), x)))

The set Q consists of the following terms:

-(x0, 0)
-(s(x0), s(x1))
p(s(x0))
f(s(x0), x1)
f(x0, s(x1))

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP

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

F(s(x), y) → F(p(-(s(x), y)), p(-(y, s(x))))
F(x, s(y)) → F(p(-(x, s(y))), p(-(s(y), x)))

The TRS R consists of the following rules:

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
p(s(x)) → x
f(s(x), y) → f(p(-(s(x), y)), p(-(y, s(x))))
f(x, s(y)) → f(p(-(x, s(y))), p(-(s(y), x)))

The set Q consists of the following terms:

-(x0, 0)
-(s(x0), s(x1))
p(s(x0))
f(s(x0), x1)
f(x0, s(x1))

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