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(x, 0) → x
minus(0, s(y)) → 0
minus(s(x), s(y)) → minus(x, y)
plus(x, 0) → x
plus(x, s(y)) → s(plus(x, y))
mod(s(x), 0) → 0
mod(x, s(y)) → help(x, s(y), 0)
help(x, s(y), c) → if(le(c, x), x, s(y), c)
if(true, x, s(y), c) → help(x, s(y), plus(c, s(y)))
if(false, x, s(y), c) → minus(x, minus(c, 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(x, 0) → x
minus(0, s(y)) → 0
minus(s(x), s(y)) → minus(x, y)
plus(x, 0) → x
plus(x, s(y)) → s(plus(x, y))
mod(s(x), 0) → 0
mod(x, s(y)) → help(x, s(y), 0)
help(x, s(y), c) → if(le(c, x), x, s(y), c)
if(true, x, s(y), c) → help(x, s(y), plus(c, s(y)))
if(false, x, s(y), c) → minus(x, minus(c, 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)