Termination w.r.t. Q of the given QTRS could be proven:

(0) Obligation:

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))
+(x, 0) → x
+(x, s(y)) → s(+(x, y))

Q is empty.

(1) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Recursive path order with status [RPO].
Precedence:

fib₁ > 0
fib₁ > +₂ > s₁

Status:

fib₁: [1]
s₁: multiset
0: multiset
+₂: [1,2]

With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

fib(0) → 0
fib(s(0)) → s(0)
fib(s(s(x))) → +(fib(s(x)), fib(x))
+(x, 0) → x
+(x, s(y)) → s(+(x, y))

(2) Obligation:

Q restricted rewrite system:
R is empty.
Q is empty.

(3) RisEmptyProof (EQUIVALENT transformation)

The TRS R is empty. Hence, termination is trivially proven.

(4) TRUE