The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(INF, INF).

(0) Obligation:

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

nesteql(Nil) → Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Nil)))))))))))))))))
nesteql(Cons(x, xs)) → nesteql(eql(Cons(x, xs)))
eql(Nil) → Nil
eql(Cons(x, xs)) → eql(Cons(x, xs))
number17(n) → Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Nil)))))))))))))))))
goal(x) → nesteql(x)

Rewrite Strategy: INNERMOST

(1) CpxTrsToCpxRelTrsProof (BOTH BOUNDS(ID, ID) transformation)

Transformed TRS to relative TRS where S is empty.

(2) Obligation:

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

nesteql(Nil) → Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Nil)))))))))))))))))
nesteql(Cons(x, xs)) → nesteql(eql(Cons(x, xs)))
eql(Nil) → Nil
eql(Cons(x, xs)) → eql(Cons(x, xs))
number17(n) → Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Cons(Nil, Nil)))))))))))))))))
goal(x) → nesteql(x)

S is empty.
Rewrite Strategy: INNERMOST

(3) SlicingProof (LOWER BOUND(ID) transformation)

Sliced the following arguments:
Cons/0
Cons/1
number17/0

(4) Obligation:

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

nesteql(Nil) → Cons
nesteql(Cons) → nesteql(eql(Cons))
eql(Nil) → Nil
eql(Cons) → eql(Cons)
number17 → Cons
goal(x) → nesteql(x)

S is empty.
Rewrite Strategy: INNERMOST

(5) InfiniteLowerBoundProof (EQUIVALENT transformation)

The loop following loop proves infinite runtime complexity:
The rewrite sequence
eql(Cons) →⁺ eql(Cons)
gives rise to a decreasing loop by considering the right hand sides subterm at position [].
The pumping substitution is [ ].
The result substitution is [ ].

(6) BOUNDS(INF, INF)