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:
fib(0) → 0
fib(s(0)) → s(0)
fib(s(s(x))) → +(fib(s(x)), fib(x))
Q is empty.
↳ QTRS
↳ RRRPoloQTRSProof
Q restricted rewrite system:
The TRS R consists of the following rules:
fib(0) → 0
fib(s(0)) → s(0)
fib(s(s(x))) → +(fib(s(x)), fib(x))
Q is empty.
The following Q TRS is given: Q restricted rewrite system:
The TRS R consists of the following rules:
fib(0) → 0
fib(s(0)) → s(0)
fib(s(s(x))) → +(fib(s(x)), fib(x))
Q is empty.
The following rules can be removed by the rule removal processor [15] because they are oriented strictly by a polynomial ordering:
fib(0) → 0
fib(s(0)) → s(0)
fib(s(s(x))) → +(fib(s(x)), fib(x))
Used ordering:
Polynomial interpretation [25]:
POL(+(x1, x2)) = 1 + x1 + x2
POL(0) = 0
POL(fib(x1)) = 2 + 2·x1
POL(s(x1)) = 1 + 2·x1
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RisEmptyProof
Q restricted rewrite system:
R is empty.
Q is empty.
The TRS R is empty. Hence, termination is trivially proven.