Termination w.r.t. Q of the given QTRS could be proven:

(0) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

active(f(f(X))) → mark(c(f(g(f(X)))))
active(c(X)) → mark(d(X))
active(h(X)) → mark(c(d(X)))
active(f(X)) → f(active(X))
active(h(X)) → h(active(X))
f(mark(X)) → mark(f(X))
h(mark(X)) → mark(h(X))
proper(f(X)) → f(proper(X))
proper(c(X)) → c(proper(X))
proper(g(X)) → g(proper(X))
proper(d(X)) → d(proper(X))
proper(h(X)) → h(proper(X))
f(ok(X)) → ok(f(X))
c(ok(X)) → ok(c(X))
g(ok(X)) → ok(g(X))
d(ok(X)) → ok(d(X))
h(ok(X)) → ok(h(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Q is empty.

(1) DirectTerminationProof (EQUIVALENT transformation)

We use [DIRECT_TERMINATION] with the following order to prove termination.

Knuth-Bendix order [KBO] with precedence:

proper₁ > g₁ > c₁ > active₁ > f₁ > top₁ > h₁ > d₁ > ok₁ > mark₁

and weight map:

active_1=13
f_1=15
mark_1=3
c_1=8
g_1=1
d_1=17
h_1=15
proper_1=2
ok_1=14
top_1=1

The variable weight is 1

(2) TRUE