The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(n^1, INF).

(0) Obligation:

Runtime Complexity TRS:
The TRS R consists of the following rules:

plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
lt(0, s(y)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
fib(x) → fibiter(x, 0, 0, s(0))
fibiter(b, c, x, y) → if(lt(c, b), b, c, x, y)
if(false, b, c, x, y) → x
if(true, b, c, x, y) → fibiter(b, s(c), y, plus(x, y))

Rewrite Strategy: FULL

(1) CpxTrsToCpxRelTrsProof (BOTH BOUNDS(ID, ID) transformation)

Transformed TRS to relative TRS where S is empty.

(2) Obligation:

Runtime Complexity Relative TRS:
The TRS R consists of the following rules:

plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
lt(0, s(y)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
fib(x) → fibiter(x, 0, 0, s(0))
fibiter(b, c, x, y) → if(lt(c, b), b, c, x, y)
if(false, b, c, x, y) → x
if(true, b, c, x, y) → fibiter(b, s(c), y, plus(x, y))

S is empty.
Rewrite Strategy: FULL

(3) DecreasingLoopProof (EQUIVALENT transformation)

The following loop(s) give(s) rise to the lower bound Ω(n¹):
The rewrite sequence
plus(s(x), y) →⁺ s(plus(x, y))
gives rise to a decreasing loop by considering the right hand sides subterm at position [0].
The pumping substitution is [x / s(x)].
The result substitution is [ ].

(4) BOUNDS(n^1, INF)