The Runtime Complexity (full) 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:

fst(0, Z) → nil
fst(s(X), cons(Y, Z)) → cons(Y, fst(X, Z))
from(X) → cons(X, from(s(X)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
len(nil) → 0
len(cons(X, Z)) → s(len(Z))

Rewrite Strategy: FULL

(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:

fst(0, Z) → nil
fst(s(X), cons(Y, Z)) → cons(Y, fst(X, Z))
from(X) → cons(X, from(s(X)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
len(nil) → 0
len(cons(X, Z)) → s(len(Z))

S is empty.
Rewrite Strategy: FULL

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

Sliced the following arguments:
cons/0
from/0

(4) Obligation:

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

fst(0, Z) → nil
fst(s(X), cons(Z)) → cons(fst(X, Z))
from → cons(from)
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
len(nil) → 0
len(cons(Z)) → s(len(Z))

S is empty.
Rewrite Strategy: FULL

(5) InfiniteLowerBoundProof (EQUIVALENT transformation)

The loop following loop proves infinite runtime complexity:
The rewrite sequence
from →⁺ cons(from)
gives rise to a decreasing loop by considering the right hand sides subterm at position [0].
The pumping substitution is [ ].
The result substitution is [ ].

(6) BOUNDS(INF, INF)