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:

perfectp(0) → false
perfectp(s(x)) → f(x, s(0), s(x), s(x))
f(0, y, 0, u) → true
f(0, y, s(z), u) → false
f(s(x), 0, z, u) → f(x, u, minus(z, s(x)), u)
f(s(x), s(y), z, u) → if(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u))

Rewrite Strategy: FULL

(1) DecreasingLoopProof (EQUIVALENT transformation)

The following loop(s) give(s) rise to the lower bound Ω(n¹):
The rewrite sequence
f(s(s(x368661_1)), 0, z, 0) →⁺ f(x368661_1, 0, minus(minus(z, s(s(x368661_1))), s(x368661_1)), 0)
gives rise to a decreasing loop by considering the right hand sides subterm at position [].
The pumping substitution is [x368661_1 / s(s(x368661_1))].
The result substitution is [z / minus(minus(z, s(s(x368661_1))), s(x368661_1))].

(2) BOUNDS(n^1, INF)

(3) RenamingProof (EQUIVALENT transformation)

Renamed function symbols to avoid clashes with predefined symbol.

(4) Obligation:

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

perfectp(0') → false
perfectp(s(x)) → f(x, s(0'), s(x), s(x))
f(0', y, 0', u) → true
f(0', y, s(z), u) → false
f(s(x), 0', z, u) → f(x, u, minus(z, s(x)), u)
f(s(x), s(y), z, u) → if(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u))

S is empty.
Rewrite Strategy: FULL

(5) TypeInferenceProof (BOTH BOUNDS(ID, ID) transformation)

Infered types.

(6) Obligation:

TRS:
Rules:
perfectp(0') → false
perfectp(s(x)) → f(x, s(0'), s(x), s(x))
f(0', y, 0', u) → true
f(0', y, s(z), u) → false
f(s(x), 0', z, u) → f(x, u, minus(z, s(x)), u)
f(s(x), s(y), z, u) → if(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u))

Types:
perfectp :: 0':s:minus → false:true:if
0' :: 0':s:minus
false :: false:true:if
s :: 0':s:minus → 0':s:minus
f :: 0':s:minus → 0':s:minus → 0':s:minus → 0':s:minus → false:true:if
true :: false:true:if
minus :: 0':s:minus → 0':s:minus → 0':s:minus
if :: le → false:true:if → false:true:if → false:true:if
le :: 0':s:minus → 0':s:minus → le
hole_false:true:if1_0 :: false:true:if
hole_0':s:minus2_0 :: 0':s:minus
hole_le3_0 :: le
gen_false:true:if4_0 :: Nat → false:true:if
gen_0':s:minus5_0 :: Nat → 0':s:minus

(7) OrderProof (LOWER BOUND(ID) transformation)

Heuristically decided to analyse the following defined symbols:
f

(8) Obligation:

TRS:
Rules:
perfectp(0') → false
perfectp(s(x)) → f(x, s(0'), s(x), s(x))
f(0', y, 0', u) → true
f(0', y, s(z), u) → false
f(s(x), 0', z, u) → f(x, u, minus(z, s(x)), u)
f(s(x), s(y), z, u) → if(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u))

Types:
perfectp :: 0':s:minus → false:true:if
0' :: 0':s:minus
false :: false:true:if
s :: 0':s:minus → 0':s:minus
f :: 0':s:minus → 0':s:minus → 0':s:minus → 0':s:minus → false:true:if
true :: false:true:if
minus :: 0':s:minus → 0':s:minus → 0':s:minus
if :: le → false:true:if → false:true:if → false:true:if
le :: 0':s:minus → 0':s:minus → le
hole_false:true:if1_0 :: false:true:if
hole_0':s:minus2_0 :: 0':s:minus
hole_le3_0 :: le
gen_false:true:if4_0 :: Nat → false:true:if
gen_0':s:minus5_0 :: Nat → 0':s:minus

Generator Equations:
gen_false:true:if4_0(0) ⇔ false
gen_false:true:if4_0(+(x, 1)) ⇔ if(le(0', 0'), false, gen_false:true:if4_0(x))
gen_0':s:minus5_0(0) ⇔ 0'
gen_0':s:minus5_0(+(x, 1)) ⇔ s(gen_0':s:minus5_0(x))

The following defined symbols remain to be analysed:
f