The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(2^n, INF).

(0) Obligation:

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

U11(tt, V1, V2) → U12(isNat(activate(V1)), activate(V2))
U12(tt, V2) → U13(isNat(activate(V2)))
U13(tt) → tt
U21(tt, V1) → U22(isNat(activate(V1)))
U22(tt) → tt
U31(tt, N) → activate(N)
U41(tt, M, N) → s(plus(activate(N), activate(M)))
and(tt, X) → activate(X)
isNat(n__0) → tt
isNat(n__plus(V1, V2)) → U11(and(isNatKind(activate(V1)), n__isNatKind(activate(V2))), activate(V1), activate(V2))
isNat(n__s(V1)) → U21(isNatKind(activate(V1)), activate(V1))
isNatKind(n__0) → tt
isNatKind(n__plus(V1, V2)) → and(isNatKind(activate(V1)), n__isNatKind(activate(V2)))
isNatKind(n__s(V1)) → isNatKind(activate(V1))
plus(N, 0) → U31(and(isNat(N), n__isNatKind(N)), N)
plus(N, s(M)) → U41(and(and(isNat(M), n__isNatKind(M)), n__and(n__isNat(N), n__isNatKind(N))), M, N)
0 → n__0
plus(X1, X2) → n__plus(X1, X2)
isNatKind(X) → n__isNatKind(X)
s(X) → n__s(X)
and(X1, X2) → n__and(X1, X2)
isNat(X) → n__isNat(X)
activate(n__0) → 0
activate(n__plus(X1, X2)) → plus(activate(X1), activate(X2))
activate(n__isNatKind(X)) → isNatKind(X)
activate(n__s(X)) → s(activate(X))
activate(n__and(X1, X2)) → and(activate(X1), X2)
activate(n__isNat(X)) → isNat(X)
activate(X) → X

Rewrite Strategy: FULL

(1) DecreasingLoopProof (EQUIVALENT transformation)

The following loop(s) give(s) rise to the lower bound Ω(2ⁿ):
The rewrite sequence
isNat(n__plus(n__isNat(X30719_4), V2)) →⁺ U11(and(isNatKind(isNat(X30719_4)), n__isNatKind(activate(V2))), isNat(X30719_4), activate(V2))
gives rise to a decreasing loop by considering the right hand sides subterm at position [0,0,0].
The pumping substitution is [X30719_4 / n__plus(n__isNat(X30719_4), V2)].
The result substitution is [ ].

The rewrite sequence
isNat(n__plus(n__isNat(X30719_4), V2)) →⁺ U11(and(isNatKind(isNat(X30719_4)), n__isNatKind(activate(V2))), isNat(X30719_4), activate(V2))
gives rise to a decreasing loop by considering the right hand sides subterm at position [1].
The pumping substitution is [X30719_4 / n__plus(n__isNat(X30719_4), V2)].
The result substitution is [ ].

(2) BOUNDS(2^n, INF)