The Runtime Complexity (innermost) 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:

a__p(0) → 0
a__p(s(X)) → mark(X)
a__leq(0, Y) → true
a__leq(s(X), 0) → false
a__leq(s(X), s(Y)) → a__leq(mark(X), mark(Y))
a__if(true, X, Y) → mark(X)
a__if(false, X, Y) → mark(Y)
a__diff(X, Y) → a__if(a__leq(mark(X), mark(Y)), 0, s(diff(p(X), Y)))
mark(p(X)) → a__p(mark(X))
mark(leq(X1, X2)) → a__leq(mark(X1), mark(X2))
mark(if(X1, X2, X3)) → a__if(mark(X1), X2, X3)
mark(diff(X1, X2)) → a__diff(mark(X1), mark(X2))
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(true) → true
mark(false) → false
a__p(X) → p(X)
a__leq(X1, X2) → leq(X1, X2)
a__if(X1, X2, X3) → if(X1, X2, X3)
a__diff(X1, X2) → diff(X1, X2)

Rewrite Strategy: INNERMOST

(1) DecreasingLoopProof (EQUIVALENT transformation)

The following loop(s) give(s) rise to the lower bound Ω(n¹):
The rewrite sequence
a__if(true, if(true, X24580_4, X34581_4), Y) →⁺ a__if(true, X24580_4, X34581_4)
gives rise to a decreasing loop by considering the right hand sides subterm at position [].
The pumping substitution is [X24580_4 / if(true, X24580_4, X34581_4)].
The result substitution is [Y / X34581_4].

(2) BOUNDS(n^1, INF)