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
  ↳ DependencyPairsProof

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.

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))

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

↳ 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))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly 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 be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(DFIB2(x1, x2)) = 3·x1 + 2·x1·x2 + 2·x2   
POL(dfib2(x1, x2)) = 2 + 3·x1 + 3·x1·x2   
POL(s1(x1)) = 3 + 3·x1 + 2·x12   

The following usable rules [14] were oriented:

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



↳ 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))

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