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:

p(m, n, s(r)) → p(m, r, n)
p(m, s(n), 0) → p(0, n, m)
p(m, 0, 0) → m

Q is empty.


QTRS
  ↳ Overlay + Local Confluence

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

p(m, n, s(r)) → p(m, r, n)
p(m, s(n), 0) → p(0, n, m)
p(m, 0, 0) → m

Q is empty.

The TRS is overlay and locally confluent. By [15] we can switch to innermost.

↳ QTRS
  ↳ Overlay + Local Confluence
QTRS
      ↳ DependencyPairsProof

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

p(m, n, s(r)) → p(m, r, n)
p(m, s(n), 0) → p(0, n, m)
p(m, 0, 0) → m

The set Q consists of the following terms:

p(x0, x1, s(x2))
p(x0, s(x1), 0)
p(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:

P(m, n, s(r)) → P(m, r, n)
P(m, s(n), 0) → P(0, n, m)

The TRS R consists of the following rules:

p(m, n, s(r)) → p(m, r, n)
p(m, s(n), 0) → p(0, n, m)
p(m, 0, 0) → m

The set Q consists of the following terms:

p(x0, x1, s(x2))
p(x0, s(x1), 0)
p(x0, 0, 0)

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

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
QDP
          ↳ QDPOrderProof

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

P(m, n, s(r)) → P(m, r, n)
P(m, s(n), 0) → P(0, n, m)

The TRS R consists of the following rules:

p(m, n, s(r)) → p(m, r, n)
p(m, s(n), 0) → p(0, n, m)
p(m, 0, 0) → m

The set Q consists of the following terms:

p(x0, x1, s(x2))
p(x0, s(x1), 0)
p(x0, 0, 0)

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.


P(m, n, s(r)) → P(m, r, n)
P(m, s(n), 0) → P(0, n, m)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
P(x1, x2, x3)  =  P(x1, x2, x3)
s(x1)  =  s(x1)
0  =  0

Recursive path order with status [2].
Precedence:
s1 > P3
0 > P3

Status:
s1: multiset
0: multiset
P3: multiset

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ QDPOrderProof
QDP
              ↳ PisEmptyProof

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

p(m, n, s(r)) → p(m, r, n)
p(m, s(n), 0) → p(0, n, m)
p(m, 0, 0) → m

The set Q consists of the following terms:

p(x0, x1, s(x2))
p(x0, s(x1), 0)
p(x0, 0, 0)

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