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:

dfib2(s1(s1(x)), y) -> dfib2(s1(x), dfib2(x, y))

Q is empty.


QTRS
  ↳ Non-Overlap Check

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

dfib2(s1(s1(x)), y) -> dfib2(s1(x), dfib2(x, y))

Q is empty.

The TRS is non-overlapping. Hence, we can switch to innermost.

↳ QTRS
  ↳ Non-Overlap Check
QTRS
      ↳ DependencyPairsProof

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

dfib2(s1(s1(x)), y) -> dfib2(s1(x), dfib2(x, y))

The set Q consists of the following terms:

dfib2(s1(s1(x0)), x1)


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

DFIB2(s1(s1(x)), y) -> DFIB2(x, y)
DFIB2(s1(s1(x)), y) -> DFIB2(s1(x), dfib2(x, y))

The TRS R consists of the following rules:

dfib2(s1(s1(x)), y) -> dfib2(s1(x), dfib2(x, y))

The set Q consists of the following terms:

dfib2(s1(s1(x0)), x1)

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

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
QDP
          ↳ QDPAfsSolverProof

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

DFIB2(s1(s1(x)), y) -> DFIB2(x, y)
DFIB2(s1(s1(x)), y) -> DFIB2(s1(x), dfib2(x, y))

The TRS R consists of the following rules:

dfib2(s1(s1(x)), y) -> dfib2(s1(x), dfib2(x, y))

The set Q consists of the following terms:

dfib2(s1(s1(x0)), x1)

We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

DFIB2(s1(s1(x)), y) -> DFIB2(x, y)
DFIB2(s1(s1(x)), y) -> DFIB2(s1(x), dfib2(x, y))
Used argument filtering: DFIB2(x1, x2)  =  x1
s1(x1)  =  s1(x1)
dfib2(x1, x2)  =  dfib
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ QDPAfsSolverProof
QDP
              ↳ PisEmptyProof

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

dfib2(s1(s1(x)), y) -> dfib2(s1(x), dfib2(x, y))

The set Q consists of the following terms:

dfib2(s1(s1(x0)), x1)

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