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)


Using Dependency Pairs [1,13] we result in the following initial DP problem:
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
          ↳ QDPOrderProof

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.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


DFIB2(s1(s1(x)), y) -> DFIB2(x, y)
DFIB2(s1(s1(x)), y) -> DFIB2(s1(x), dfib2(x, y))
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
DFIB2(x1, x2)  =  x1
s1(x1)  =  s1(x1)
dfib2(x1, x2)  =  dfib1(x2)

Lexicographic Path Order [19].
Precedence:
dfib1 > s1


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ QDPOrderProof
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.