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:

lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
plus(x, 0) → x
plus(x, s(y)) → s(plus(x, y))
quot(x, s(y)) → help(x, s(y), 0)
help(x, s(y), c) → if(lt(c, x), x, s(y), c)
if(true, x, s(y), c) → s(help(x, s(y), plus(c, s(y))))
if(false, x, s(y), c) → 0

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:

lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
plus(x, 0) → x
plus(x, s(y)) → s(plus(x, y))
quot(x, s(y)) → help(x, s(y), 0)
help(x, s(y), c) → if(lt(c, x), x, s(y), c)
if(true, x, s(y), c) → s(help(x, s(y), plus(c, s(y))))
if(false, x, s(y), c) → 0

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
lt(s(x), s(y)) →⁺ lt(x, y)
gives rise to a decreasing loop by considering the right hand sides subterm at position [].
The pumping substitution is [x / s(x), y / s(y)].
The result substitution is [ ].

(4) BOUNDS(n^1, INF)