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
↳ Overlay + Local Confluence
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.
The TRS is overlay and locally confluent. By [15] we can switch to innermost.
↳ QTRS
↳ Overlay + Local Confluence
↳ 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))
The set Q consists of the following terms:
dfib(s(s(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:
DFIB(s(s(x)), y) → DFIB(x, y)
DFIB(s(s(x)), y) → DFIB(s(x), dfib(x, y))
The TRS R consists of the following rules:
dfib(s(s(x)), y) → dfib(s(x), dfib(x, y))
The set Q consists of the following terms:
dfib(s(s(x0)), x1)
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ QDPOrderProof
Q DP problem:
The TRS P consists of the following rules:
DFIB(s(s(x)), y) → DFIB(x, y)
DFIB(s(s(x)), y) → DFIB(s(x), dfib(x, y))
The TRS R consists of the following rules:
dfib(s(s(x)), y) → dfib(s(x), dfib(x, y))
The set Q consists of the following terms:
dfib(s(s(x0)), x1)
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.
DFIB(s(s(x)), y) → DFIB(x, y)
DFIB(s(s(x)), y) → DFIB(s(x), dfib(x, y))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
DFIB(x1, x2) = DFIB(x1, x2)
s(x1) = s(x1)
dfib(x1, x2) = dfib
Recursive path order with status [2].
Precedence:
s1 > DFIB2 > dfib
Status:
s1: multiset
dfib: []
DFIB2: multiset
The following usable rules [14] were oriented:
dfib(s(s(x)), y) → dfib(s(x), dfib(x, y))
↳ QTRS
↳ Overlay + Local Confluence
↳ 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))
The set Q consists of the following terms:
dfib(s(s(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.