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:

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
  ↳ 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

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:

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

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof

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

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.


P3(m, n, s1(r)) -> P3(m, r, n)
P3(m, s1(n), 0) -> P3(0, n, m)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( P3(x1, ..., x3) ) = max{0, x1 + x2 + x3 - 2}


POL( s1(x1) ) = x1 + 3


POL( 0 ) = 0



The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof
        ↳ QDP
          ↳ PisEmptyProof

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