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:

f(s(x)) → s(s(f(p(s(x)))))
f(0) → 0
p(s(x)) → x

Q is empty.


↳ QTRS
  ↳ Overlay + Local Confluence

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

f(s(x)) → s(s(f(p(s(x)))))
f(0) → 0
p(s(x)) → x

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:

f(s(x)) → s(s(f(p(s(x)))))
f(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

f(s(x0))
f(0)
p(s(x0))


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(s(x)) → P(s(x))
F(s(x)) → F(p(s(x)))

The TRS R consists of the following rules:

f(s(x)) → s(s(f(p(s(x)))))
f(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

f(s(x0))
f(0)
p(s(x0))

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

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

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

F(s(x)) → P(s(x))
F(s(x)) → F(p(s(x)))

The TRS R consists of the following rules:

f(s(x)) → s(s(f(p(s(x)))))
f(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

f(s(x0))
f(0)
p(s(x0))

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

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof

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

F(s(x)) → P(s(x))
F(s(x)) → F(p(s(x)))

The TRS R consists of the following rules:

f(s(x)) → s(s(f(p(s(x)))))
f(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

f(s(x0))
f(0)
p(s(x0))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 1 less node.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ QDP

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

F(s(x)) → F(p(s(x)))

The TRS R consists of the following rules:

f(s(x)) → s(s(f(p(s(x)))))
f(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

f(s(x0))
f(0)
p(s(x0))

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