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:

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

Q is empty.


QTRS
  ↳ Overlay + Local Confluence
  ↳ QTRS Reverse

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

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

Q is empty.

The TRS is overlay and locally confluent. By [19] we can switch to innermost.

↳ QTRS
  ↳ Overlay + Local Confluence
QTRS
      ↳ DependencyPairsProof
  ↳ QTRS Reverse

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

f(s(x)) → s(f(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 [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

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

The TRS R consists of the following rules:

f(s(x)) → s(f(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
          ↳ DependencyGraphProof
  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

f(s(x)) → s(f(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 [15,17,22] contains 1 SCC with 1 less node.

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

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

F(s(x)) → F(f(p(s(x))))
F(s(x)) → F(p(s(x)))

The TRS R consists of the following rules:

f(s(x)) → s(f(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 have reversed the following QTRS:
The set of rules R is

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

The set Q is empty.
We have obtained the following QTRS:

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

The set Q is empty.

↳ QTRS
  ↳ Overlay + Local Confluence
  ↳ QTRS Reverse
QTRS
      ↳ RFCMatchBoundsTRSProof

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

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

Q is empty.

Termination of the TRS R could be shown with a Match Bound [6,7] of 2. This implies Q-termination of R.
The following rules were used to construct the certificate:

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

The certificate found is represented by the following graph.

The certificate consists of the following enumerated nodes:

108, 109, 110, 111, 113, 112, 114, 115, 117, 116

Node 108 is start node and node 109 is final node.

Those nodes are connect through the following edges: