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 + x2
POL(dfib2(x1, x2)) = 3 + 3·x1
POL(s1(x1)) = 3 + x1 + 3·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.