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:
dfib(s(s(x)), y) → dfib(s(x), dfib(x, y))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
dfib(s(s(x)), y) → dfib(s(x), dfib(x, y))
Q is empty.
Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:
DFIB(s(s(x)), y) → DFIB(s(x), dfib(x, y))
DFIB(s(s(x)), y) → DFIB(x, y)
The TRS R consists of the following rules:
dfib(s(s(x)), y) → dfib(s(x), dfib(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:
DFIB(s(s(x)), y) → DFIB(s(x), dfib(x, y))
DFIB(s(s(x)), y) → DFIB(x, y)
The TRS R consists of the following rules:
dfib(s(s(x)), y) → dfib(s(x), dfib(x, y))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].
The following pairs can be oriented strictly and are deleted.
DFIB(s(s(x)), y) → DFIB(s(x), dfib(x, y))
DFIB(s(s(x)), y) → DFIB(x, y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:
POL(dfib(x1, x2)) = 1/4
POL(s(x1)) = 1/4 + (2)x_1
POL(DFIB(x1, x2)) = (1/4)x_1 + (1/4)x_2
The value of delta used in the strict ordering is 1/16.
The following usable rules [17] were oriented:
dfib(s(s(x)), y) → dfib(s(x), dfib(x, y))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ PisEmptyProof
Q DP problem:
P is empty.
The TRS R consists of the following rules:
dfib(s(s(x)), y) → dfib(s(x), dfib(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.