The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(n^1, INF).

(0) Obligation:

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

active(f(x)) → mark(f(f(x)))
chk(no(f(x))) → f(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))
mat(f(x), f(y)) → f(mat(x, y))
chk(no(c)) → active(c)
mat(f(x), c) → no(c)
f(active(x)) → active(f(x))
f(no(x)) → no(f(x))
f(mark(x)) → mark(f(x))
tp(mark(x)) → tp(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))

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:

active(f(x)) → mark(f(f(x)))
chk(no(f(x))) → f(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))
mat(f(x), f(y)) → f(mat(x, y))
chk(no(c)) → active(c)
mat(f(x), c) → no(c)
f(active(x)) → active(f(x))
f(no(x)) → no(f(x))
f(mark(x)) → mark(f(x))
tp(mark(x)) → tp(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))

S is empty.
Rewrite Strategy: FULL

(3) DecreasingLoopProof (EQUIVALENT transformation)

The following loop(s) give(s) rise to the lower bound Ω(n¹):
The rewrite sequence
f(no(x)) →⁺ no(f(x))
gives rise to a decreasing loop by considering the right hand sides subterm at position [0].
The pumping substitution is [x / no(x)].
The result substitution is [ ].

(4) BOUNDS(n^1, INF)