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(isNatKind(activate(V1)), activate(V1), activate(V2))
U12(tt, V1, V2) → U13(isNatKind(activate(V2)), activate(V1), activate(V2))
U13(tt, V1, V2) → U14(isNatKind(activate(V2)), activate(V1), activate(V2))
U14(tt, V1, V2) → U15(isNat(activate(V1)), activate(V2))
U15(tt, V2) → U16(isNat(activate(V2)))
U16(tt) → tt
U21(tt, V1) → U22(isNatKind(activate(V1)), activate(V1))
U22(tt, V1) → U23(isNat(activate(V1)))
U23(tt) → tt
U31(tt, V2) → U32(isNatKind(activate(V2)))
U32(tt) → tt
U41(tt) → tt
U51(tt, N) → U52(isNatKind(activate(N)), activate(N))
U52(tt, N) → activate(N)
U61(tt, M, N) → U62(isNatKind(activate(M)), activate(M), activate(N))
U62(tt, M, N) → U63(isNat(activate(N)), activate(M), activate(N))
U63(tt, M, N) → U64(isNatKind(activate(N)), activate(M), activate(N))
U64(tt, M, N) → s(plus(activate(N), activate(M)))
isNat(n__0) → tt
isNat(n__plus(V1, V2)) → U11(isNatKind(activate(V1)), activate(V1), activate(V2))
isNat(n__s(V1)) → U21(isNatKind(activate(V1)), activate(V1))
isNatKind(n__0) → tt
isNatKind(n__plus(V1, V2)) → U31(isNatKind(activate(V1)), activate(V2))
isNatKind(n__s(V1)) → U41(isNatKind(activate(V1)))
plus(N, 0) → U51(isNat(N), N)
plus(N, s(M)) → U61(isNat(M), M, N)
0 → n__0
plus(X1, X2) → n__plus(X1, X2)
s(X) → n__s(X)
activate(n__0) → 0
activate(n__plus(X1, X2)) → plus(activate(X1), activate(X2))
activate(n__s(X)) → s(activate(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
activate(n__plus(X1, n__0)) →⁺ U51(isNat(activate(X1)), activate(X1))
gives rise to a decreasing loop by considering the right hand sides subterm at position [0,0].
The pumping substitution is [X1 / n__plus(X1, n__0)].
The result substitution is [ ].

The rewrite sequence
activate(n__plus(X1, n__0)) →⁺ U51(isNat(activate(X1)), activate(X1))
gives rise to a decreasing loop by considering the right hand sides subterm at position [1].
The pumping substitution is [X1 / n__plus(X1, n__0)].
The result substitution is [ ].

(2) BOUNDS(2^n, INF)