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, x) → x
plus(s(x), y) → s(plus(x, y))
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
p(s(x)) → x
p(0) → 0
minus(x, 0) → x
minus(0, x) → 0
minus(x, s(y)) → p(minus(x, y))
isZero(0) → true
isZero(s(x)) → false
facIter(x, y) → if(isZero(x), minus(x, s(0)), y, times(y, x))
if(true, x, y, z) → y
if(false, x, y, z) → facIter(x, z)
factorial(x) → facIter(x, s(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:

plus(0, x) → x
plus(s(x), y) → s(plus(x, y))
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
p(s(x)) → x
p(0) → 0
minus(x, 0) → x
minus(0, x) → 0
minus(x, s(y)) → p(minus(x, y))
isZero(0) → true
isZero(s(x)) → false
facIter(x, y) → if(isZero(x), minus(x, s(0)), y, times(y, x))
if(true, x, y, z) → y
if(false, x, y, z) → facIter(x, z)
factorial(x) → facIter(x, s(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
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)