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:

f(a, f(b, x)) → f(b, f(a, x))
f(b, f(c, x)) → f(c, f(b, x))
f(c, f(a, x)) → f(a, f(c, x))

Q is empty.

(1) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(2) Obligation:

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

F(a, f(b, x)) → F(b, f(a, x))
F(a, f(b, x)) → F(a, x)
F(b, f(c, x)) → F(c, f(b, x))
F(b, f(c, x)) → F(b, x)
F(c, f(a, x)) → F(a, f(c, x))
F(c, f(a, x)) → F(c, x)

The TRS R consists of the following rules:

f(a, f(b, x)) → f(b, f(a, x))
f(b, f(c, x)) → f(c, f(b, x))
f(c, f(a, x)) → f(a, f(c, x))

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

(3) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04]. Here, we combined the reduction pair processor with the A-transformation [FROCOS05] which results in the following intermediate Q-DP Problem.
The a-transformed P is

a1(b(x)) → b1(a(x))
a1(b(x)) → a1(x)
b1(c(x)) → c1(b(x))
b1(c(x)) → b1(x)
c1(a(x)) → a1(c(x))
c1(a(x)) → c1(x)

The a-transformed usable rules are

a(b(x)) → b(a(x))
b(c(x)) → c(b(x))
c(a(x)) → a(c(x))

The following pairs can be oriented strictly and are deleted.

F(a, f(b, x)) → F(a, x)
F(b, f(c, x)) → F(b, x)
F(c, f(a, x)) → F(c, x)

The remaining pairs can at least be oriented weakly.
Used ordering: Lexicographic Path Order [LPO].
Precedence:

[a1₁, b1₁, c1₁] > [b₁, a₁, c₁]

The following usable rules [FROCOS05] were oriented:

f(a, f(b, x)) → f(b, f(a, x))
f(b, f(c, x)) → f(c, f(b, x))
f(c, f(a, x)) → f(a, f(c, x))

(4) Obligation:

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

F(a, f(b, x)) → F(b, f(a, x))
F(b, f(c, x)) → F(c, f(b, x))
F(c, f(a, x)) → F(a, f(c, x))

The TRS R consists of the following rules:

f(a, f(b, x)) → f(b, f(a, x))
f(b, f(c, x)) → f(c, f(b, x))
f(c, f(a, x)) → f(a, f(c, x))

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

(5) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04]. Here, we combined the reduction pair processor with the A-transformation [FROCOS05] which results in the following intermediate Q-DP Problem.
The a-transformed P is

a1(b(x)) → b1(a(x))
b1(c(x)) → c1(b(x))
c1(a(x)) → a1(c(x))

The a-transformed usable rules are

a(b(x)) → b(a(x))
b(c(x)) → c(b(x))
c(a(x)) → a(c(x))

The following pairs can be oriented strictly and are deleted.

F(a, f(b, x)) → F(b, f(a, x))
F(b, f(c, x)) → F(c, f(b, x))
F(c, f(a, x)) → F(a, f(c, x))

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
a1(x1)  =  a1
b(x1)  =  x1
b1(x1)  =  b1(x1)
a(x1)  =  a
c(x1)  =  c(x1)
c1(x1)  =  c1(x1)

Lexicographic Path Order [LPO].
Precedence:

[c₁, c1₁] > a1 > [b1₁, a]

The following usable rules [FROCOS05] were oriented:

f(a, f(b, x)) → f(b, f(a, x))
f(b, f(c, x)) → f(c, f(b, x))
f(c, f(a, x)) → f(a, f(c, x))

(6) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

f(a, f(b, x)) → f(b, f(a, x))
f(b, f(c, x)) → f(c, f(b, x))
f(c, f(a, x)) → f(a, f(c, x))

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

(7) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(8) TRUE