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(0) → s(0)
f(s(0)) → s(s(0))
f(s(0)) → *(s(s(0)), f(0))
f(+(x, s(0))) → +(s(s(0)), f(x))
f(+(x, y)) → *(f(x), f(y))

Q is empty.


↳ QTRS
  ↳ RRRPoloQTRSProof

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

f(0) → s(0)
f(s(0)) → s(s(0))
f(s(0)) → *(s(s(0)), f(0))
f(+(x, s(0))) → +(s(s(0)), f(x))
f(+(x, y)) → *(f(x), f(y))

Q is empty.

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

f(0) → s(0)
f(s(0)) → s(s(0))
f(s(0)) → *(s(s(0)), f(0))
f(+(x, s(0))) → +(s(s(0)), f(x))
f(+(x, y)) → *(f(x), f(y))

Q is empty.
The following rules can be removed by the rule removal processor [15] because they are oriented strictly by a polynomial ordering:

f(0) → s(0)
f(s(0)) → s(s(0))
f(+(x, s(0))) → +(s(s(0)), f(x))
f(+(x, y)) → *(f(x), f(y))
Used ordering:
Polynomial interpretation [25]:

POL(*(x1, x2)) = x1 + x2   
POL(+(x1, x2)) = 2 + 2·x1 + 2·x2   
POL(0) = 0   
POL(f(x1)) = 1 + 2·x1   
POL(s(x1)) = 2·x1   




↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ Overlay + Local Confluence

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

f(s(0)) → *(s(s(0)), f(0))

Q is empty.

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

↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ Overlay + Local Confluence
        ↳ QTRS
          ↳ DependencyPairsProof

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

f(s(0)) → *(s(s(0)), f(0))

The set Q consists of the following terms:

f(s(0))


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(0)) → F(0)

The TRS R consists of the following rules:

f(s(0)) → *(s(s(0)), f(0))

The set Q consists of the following terms:

f(s(0))

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

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

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

F(s(0)) → F(0)

The TRS R consists of the following rules:

f(s(0)) → *(s(s(0)), f(0))

The set Q consists of the following terms:

f(s(0))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 1 less node.