Termination w.r.t. Q proof of /tmp/tmpbL8J6j/cime4.trs

Termination w.r.t. Q of the following Term Rewriting System could be proven:

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

g(X) → u(h(X), h(X), X)
u(d, c(Y), X) → k(Y)
h(d) → c(a)
h(d) → c(b)
f(k(a), k(b), X) → f(X, X, X)

The set Q consists of the following terms:

f(k(a), k(b), x0)
g(x0)
h(d)
u(d, c(x0), x1)

↳ QTRS

  ↳ DependencyPairsProof

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

g(X) → u(h(X), h(X), X)
u(d, c(Y), X) → k(Y)
h(d) → c(a)
h(d) → c(b)
f(k(a), k(b), X) → f(X, X, X)

The set Q consists of the following terms:

f(k(a), k(b), x0)
g(x0)
h(d)
u(d, c(x0), x1)

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

G(X) → U(h(X), h(X), X)
F(k(a), k(b), X) → F(X, X, X)
G(X) → H(X)

The TRS R consists of the following rules:

g(X) → u(h(X), h(X), X)
u(d, c(Y), X) → k(Y)
h(d) → c(a)
h(d) → c(b)
f(k(a), k(b), X) → f(X, X, X)

The set Q consists of the following terms:

f(k(a), k(b), x0)
g(x0)
h(d)
u(d, c(x0), x1)

We have to consider all minimal (P,Q,R)-chains.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

Q DP problem:
The TRS P consists of the following rules:

G(X) → U(h(X), h(X), X)
F(k(a), k(b), X) → F(X, X, X)
G(X) → H(X)

The TRS R consists of the following rules:

g(X) → u(h(X), h(X), X)
u(d, c(Y), X) → k(Y)
h(d) → c(a)
h(d) → c(b)
f(k(a), k(b), X) → f(X, X, X)

The set Q consists of the following terms:

f(k(a), k(b), x0)
g(x0)
h(d)
u(d, c(x0), x1)

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 3 less nodes.