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:

p3(m, n, s1(r)) -> p3(m, r, n)
p3(m, s1(n), 0) -> p3(0, n, m)
p3(m, 0, 0) -> m

Q is empty.


QTRS
  ↳ Non-Overlap Check

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

p3(m, n, s1(r)) -> p3(m, r, n)
p3(m, s1(n), 0) -> p3(0, n, m)
p3(m, 0, 0) -> m

Q is empty.

The TRS is non-overlapping. Hence, we can switch to innermost.

↳ QTRS
  ↳ Non-Overlap Check
QTRS
      ↳ DependencyPairsProof

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

p3(m, n, s1(r)) -> p3(m, r, n)
p3(m, s1(n), 0) -> p3(0, n, m)
p3(m, 0, 0) -> m

The set Q consists of the following terms:

p3(x0, x1, s1(x2))
p3(x0, s1(x1), 0)
p3(x0, 0, 0)


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

P3(m, n, s1(r)) -> P3(m, r, n)
P3(m, s1(n), 0) -> P3(0, n, m)

The TRS R consists of the following rules:

p3(m, n, s1(r)) -> p3(m, r, n)
p3(m, s1(n), 0) -> p3(0, n, m)
p3(m, 0, 0) -> m

The set Q consists of the following terms:

p3(x0, x1, s1(x2))
p3(x0, s1(x1), 0)
p3(x0, 0, 0)

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

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
QDP

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

P3(m, n, s1(r)) -> P3(m, r, n)
P3(m, s1(n), 0) -> P3(0, n, m)

The TRS R consists of the following rules:

p3(m, n, s1(r)) -> p3(m, r, n)
p3(m, s1(n), 0) -> p3(0, n, m)
p3(m, 0, 0) -> m

The set Q consists of the following terms:

p3(x0, x1, s1(x2))
p3(x0, s1(x1), 0)
p3(x0, 0, 0)

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