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:

le(0, Y) → true
le(s(X), 0) → false
le(s(X), s(Y)) → le(X, Y)
minus(0, Y) → 0
minus(s(X), Y) → ifMinus(le(s(X), Y), s(X), Y)
ifMinus(true, s(X), Y) → 0
ifMinus(false, s(X), Y) → s(minus(X, Y))
quot(0, s(Y)) → 0
quot(s(X), s(Y)) → s(quot(minus(X, Y), s(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:

le(0, Y) → true
le(s(X), 0) → false
le(s(X), s(Y)) → le(X, Y)
minus(0, Y) → 0
minus(s(X), Y) → ifMinus(le(s(X), Y), s(X), Y)
ifMinus(true, s(X), Y) → 0
ifMinus(false, s(X), Y) → s(minus(X, Y))
quot(0, s(Y)) → 0
quot(s(X), s(Y)) → s(quot(minus(X, Y), s(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
le(s(X), s(Y)) →⁺ le(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)