(0) Obligation:
Runtime Complexity TRS:
The TRS R consists of the following rules:
a__incr(nil) → nil
a__incr(cons(X, L)) → cons(s(mark(X)), incr(L))
a__adx(nil) → nil
a__adx(cons(X, L)) → a__incr(cons(mark(X), adx(L)))
a__nats → a__adx(a__zeros)
a__zeros → cons(0, zeros)
a__head(cons(X, L)) → mark(X)
a__tail(cons(X, L)) → mark(L)
mark(incr(X)) → a__incr(mark(X))
mark(adx(X)) → a__adx(mark(X))
mark(nats) → a__nats
mark(zeros) → a__zeros
mark(head(X)) → a__head(mark(X))
mark(tail(X)) → a__tail(mark(X))
mark(nil) → nil
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(0) → 0
a__incr(X) → incr(X)
a__adx(X) → adx(X)
a__nats → nats
a__zeros → zeros
a__head(X) → head(X)
a__tail(X) → tail(X)
Rewrite Strategy: FULL
(1) DecreasingLoopProof (EQUIVALENT transformation)
The following loop(s) give(s) rise to the lower bound Ω(n1):
The rewrite sequence
a__tail(cons(X, tail(cons(X114187_3, X214188_3)))) →+ a__tail(cons(mark(X114187_3), X214188_3))
gives rise to a decreasing loop by considering the right hand sides subterm at position [].
The pumping substitution is [X214188_3 / tail(cons(X114187_3, X214188_3))].
The result substitution is [X / mark(X114187_3)].
(2) BOUNDS(n^1, INF)