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(0, y) -> 0
-2(x, 0) -> x
-2(x, s1(y)) -> if3(greater2(x, s1(y)), s1(-2(x, p1(s1(y)))), 0)
p1(0) -> 0
p1(s1(x)) -> x

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

-2(0, y) -> 0
-2(x, 0) -> x
-2(x, s1(y)) -> if3(greater2(x, s1(y)), s1(-2(x, p1(s1(y)))), 0)
p1(0) -> 0
p1(s1(x)) -> x

Q is empty.

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

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

The TRS R consists of the following rules:

-2(0, y) -> 0
-2(x, 0) -> x
-2(x, s1(y)) -> if3(greater2(x, s1(y)), s1(-2(x, p1(s1(y)))), 0)
p1(0) -> 0
p1(s1(x)) -> 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:

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

The TRS R consists of the following rules:

-2(0, y) -> 0
-2(x, 0) -> x
-2(x, s1(y)) -> if3(greater2(x, s1(y)), s1(-2(x, p1(s1(y)))), 0)
p1(0) -> 0
p1(s1(x)) -> x

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 1 SCC with 1 less node.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
QDP

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

-12(x, s1(y)) -> -12(x, p1(s1(y)))

The TRS R consists of the following rules:

-2(0, y) -> 0
-2(x, 0) -> x
-2(x, s1(y)) -> if3(greater2(x, s1(y)), s1(-2(x, p1(s1(y)))), 0)
p1(0) -> 0
p1(s1(x)) -> x

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