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

The TRS R consists of the following rules:

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

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

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 5 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(6) 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(c(x)) → a1(x)
a1(b(x)) → a1(x)

The a-transformed usable rules are

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

The following pairs can be oriented strictly and are deleted.

F(a, f(b, x)) → F(a, x)

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

Recursive path order with status [RPO].
Quasi-Precedence:

[a1₁, b₁, a₁]

Status:

a1₁: multiset
b₁: multiset
a₁: multiset
d: multiset

The following usable rules [FROCOS05] were oriented:

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

(7) Obligation:

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

F(a, f(c, x)) → F(a, x)

The TRS R consists of the following rules:

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

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

(8) 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(c(x)) → a1(x)

The a-transformed usable rules are

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

The following pairs can be oriented strictly and are deleted.

F(a, f(c, x)) → F(a, x)

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

Recursive path order with status [RPO].
Quasi-Precedence:

d > [c₁, a₁]

Status:

c₁: multiset
a₁: multiset
d: multiset

The following usable rules [FROCOS05] were oriented:

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

(9) Obligation:

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

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

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

(10) PisEmptyProof (EQUIVALENT transformation)

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

(11) TRUE

(12) Obligation:

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

F(d, f(c, x)) → F(d, f(a, x))

The TRS R consists of the following rules:

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

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

(13) 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

d1(c(x)) → d1(a(x))

The a-transformed usable rules are

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

The following pairs can be oriented strictly and are deleted.

F(d, f(c, x)) → F(d, f(a, x))

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
d1(x1)  =  x1
c(x1)  =  c(x1)
a(x1)  =  x1
b(x1)  =  b
d(x1)  =  d

Recursive path order with status [RPO].
Quasi-Precedence:

[c₁, d] > b

Status:

c₁: multiset
b: []
d: multiset

The following usable rules [FROCOS05] were oriented:

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

(14) Obligation:

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

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

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

(15) PisEmptyProof (EQUIVALENT transformation)

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

(16) TRUE