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:

a__a → a__c
a__b → a__c
a__c → e
a__k → l
a__d → m
a__a → a__d
a__b → a__d
a__c → l
a__k → m
a__A → a__h(a__f(a__a), a__f(a__b))
a__h(X, X) → a__g(mark(X), mark(X), a__f(a__k))
a__g(d, X, X) → a__A
a__f(X) → a__z(mark(X), X)
a__z(e, X) → mark(X)
mark(A) → a__A
mark(a) → a__a
mark(b) → a__b
mark(c) → a__c
mark(d) → a__d
mark(k) → a__k
mark(z(X1, X2)) → a__z(mark(X1), X2)
mark(f(X)) → a__f(mark(X))
mark(h(X1, X2)) → a__h(mark(X1), mark(X2))
mark(g(X1, X2, X3)) → a__g(mark(X1), mark(X2), mark(X3))
mark(e) → e
mark(l) → l
mark(m) → m
a__A → A
a__a → a
a__b → b
a__c → c
a__d → d
a__k → k
a__z(X1, X2) → z(X1, X2)
a__f(X) → f(X)
a__h(X1, X2) → h(X1, X2)
a__g(X1, X2, X3) → g(X1, X2, X3)

Rewrite Strategy: FULL

(1) DecreasingLoopProof (EQUIVALENT transformation)

The following loop(s) give(s) rise to the lower bound Ω(2ⁿ):
The rewrite sequence
mark(z(f(X35587_0), X2)) →⁺ a__z(a__z(mark(mark(X35587_0)), mark(X35587_0)), X2)
gives rise to a decreasing loop by considering the right hand sides subterm at position [0,0,0].
The pumping substitution is [X35587_0 / z(f(X35587_0), X2)].
The result substitution is [ ].

The rewrite sequence
mark(z(f(X35587_0), X2)) →⁺ a__z(a__z(mark(mark(X35587_0)), mark(X35587_0)), X2)
gives rise to a decreasing loop by considering the right hand sides subterm at position [0,1].
The pumping substitution is [X35587_0 / z(f(X35587_0), X2)].
The result substitution is [ ].

(2) BOUNDS(2^n, INF)