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:

-(0, y) → 0
-(x, 0) → x
-(x, s(y)) → if(greater(x, s(y)), s(-(x, p(s(y)))), 0)
p(0) → 0
p(s(x)) → x

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:

-(0, y) → 0
-(x, 0) → x
-(x, s(y)) → if(greater(x, s(y)), s(-(x, p(s(y)))), 0)
p(0) → 0
p(s(x)) → x

S is empty.
Rewrite Strategy: FULL

(3) SlicingProof (LOWER BOUND(ID) transformation)

Sliced the following arguments:
if/0
if/2
greater/0
greater/1

(4) Obligation:

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

-(0, y) → 0
-(x, 0) → x
-(x, s(y)) → if(s(-(x, p(s(y)))))
p(0) → 0
p(s(x)) → x

S is empty.
Rewrite Strategy: FULL

(5) DecreasingLoopProof (EQUIVALENT transformation)

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

(6) BOUNDS(n^1, INF)