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

0 QTRS
↳1 DependencyPairsProof (⇔)
↳2 QDP
↳3 DependencyGraphProof (⇔)
↳4 AND
  ↳5 QDP
    ↳6 QDPOrderProof (⇔)
    ↳7 QDP
    ↳8 QDPOrderProof (⇔)
    ↳9 QDP
    ↳10 PisEmptyProof (⇔)
    ↳11 TRUE
  ↳12 QDP
    ↳13 QDPOrderProof (⇔)
    ↳14 QDP
    ↳15 PisEmptyProof (⇔)
    ↳16 TRUE

(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)  =  x1
c(x1)  =  x1
b(x1)  =  b(x1)
a(x1)  =  a(x1)
d(x1)  =  d

Recursive Path Order [RPO].
Precedence:
a1 > b1


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)  =  x1
b(x1)  =  b
d(x1)  =  d

Recursive Path Order [RPO].
Precedence:
c1 > b
d > b


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 [RPO].
Precedence:
c1 > b
d > b


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