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:

active(U11(tt, N)) → mark(N)
active(U21(tt, M, N)) → mark(s(plus(N, M)))
active(and(tt, X)) → mark(X)
active(isNat(0)) → mark(tt)
active(isNat(plus(V1, V2))) → mark(and(isNat(V1), isNat(V2)))
active(isNat(s(V1))) → mark(isNat(V1))
active(plus(N, 0)) → mark(U11(isNat(N), N))
active(plus(N, s(M))) → mark(U21(and(isNat(M), isNat(N)), M, N))
active(U11(X1, X2)) → U11(active(X1), X2)
active(U21(X1, X2, X3)) → U21(active(X1), X2, X3)
active(s(X)) → s(active(X))
active(plus(X1, X2)) → plus(active(X1), X2)
active(plus(X1, X2)) → plus(X1, active(X2))
active(and(X1, X2)) → and(active(X1), X2)
U11(mark(X1), X2) → mark(U11(X1, X2))
U21(mark(X1), X2, X3) → mark(U21(X1, X2, X3))
s(mark(X)) → mark(s(X))
plus(mark(X1), X2) → mark(plus(X1, X2))
plus(X1, mark(X2)) → mark(plus(X1, X2))
and(mark(X1), X2) → mark(and(X1, X2))
proper(U11(X1, X2)) → U11(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(U21(X1, X2, X3)) → U21(proper(X1), proper(X2), proper(X3))
proper(s(X)) → s(proper(X))
proper(plus(X1, X2)) → plus(proper(X1), proper(X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(isNat(X)) → isNat(proper(X))
proper(0) → ok(0)
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
U21(ok(X1), ok(X2), ok(X3)) → ok(U21(X1, X2, X3))
s(ok(X)) → ok(s(X))
plus(ok(X1), ok(X2)) → ok(plus(X1, X2))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
isNat(ok(X)) → ok(isNat(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Rewrite Strategy: INNERMOST

(1) DecreasingLoopProof (EQUIVALENT transformation)

The following loop(s) give(s) rise to the lower bound Ω(n¹):
The rewrite sequence
U11(mark(X1), X2) →⁺ mark(U11(X1, X2))
gives rise to a decreasing loop by considering the right hand sides subterm at position [0].
The pumping substitution is [X1 / mark(X1)].
The result substitution is [ ].

(2) BOUNDS(n^1, INF)