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:

a(c(d(x))) → c(x)
u(b(d(d(x)))) → b(x)
v(a(a(x))) → u(v(x))
v(a(c(x))) → u(b(d(x)))
v(c(x)) → b(x)
w(a(a(x))) → u(w(x))
w(a(c(x))) → u(b(d(x)))
w(c(x)) → b(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:

V(a(a(x))) → U(v(x))
V(a(a(x))) → V(x)
V(a(c(x))) → U(b(d(x)))
W(a(a(x))) → U(w(x))
W(a(a(x))) → W(x)
W(a(c(x))) → U(b(d(x)))

The TRS R consists of the following rules:

a(c(d(x))) → c(x)
u(b(d(d(x)))) → b(x)
v(a(a(x))) → u(v(x))
v(a(c(x))) → u(b(d(x)))
v(c(x)) → b(x)
w(a(a(x))) → u(w(x))
w(a(c(x))) → u(b(d(x)))
w(c(x)) → b(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 4 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

W(a(a(x))) → W(x)

The TRS R consists of the following rules:

a(c(d(x))) → c(x)
u(b(d(d(x)))) → b(x)
v(a(a(x))) → u(v(x))
v(a(c(x))) → u(b(d(x)))
v(c(x)) → b(x)
w(a(a(x))) → u(w(x))
w(a(c(x))) → u(b(d(x)))
w(c(x)) → b(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].

The following pairs can be oriented strictly and are deleted.

W(a(a(x))) → W(x)

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

Recursive Path Order [RPO].
Precedence:

v > u > b
w > u > b

The following usable rules [FROCOS05] were oriented:

a(c(d(x))) → c(x)
u(b(d(d(x)))) → b(x)
v(a(a(x))) → u(v(x))
v(a(c(x))) → u(b(d(x)))
v(c(x)) → b(x)
w(a(a(x))) → u(w(x))
w(a(c(x))) → u(b(d(x)))
w(c(x)) → b(x)

(7) Obligation:

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

a(c(d(x))) → c(x)
u(b(d(d(x)))) → b(x)
v(a(a(x))) → u(v(x))
v(a(c(x))) → u(b(d(x)))
v(c(x)) → b(x)
w(a(a(x))) → u(w(x))
w(a(c(x))) → u(b(d(x)))
w(c(x)) → b(x)

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

(8) PisEmptyProof (EQUIVALENT transformation)

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

(9) TRUE

(10) Obligation:

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

V(a(a(x))) → V(x)

The TRS R consists of the following rules:

a(c(d(x))) → c(x)
u(b(d(d(x)))) → b(x)
v(a(a(x))) → u(v(x))
v(a(c(x))) → u(b(d(x)))
v(c(x)) → b(x)
w(a(a(x))) → u(w(x))
w(a(c(x))) → u(b(d(x)))
w(c(x)) → b(x)

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

(11) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

V(a(a(x))) → V(x)

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

Recursive Path Order [RPO].
Precedence:

v > u > b
w > u > b

The following usable rules [FROCOS05] were oriented:

a(c(d(x))) → c(x)
u(b(d(d(x)))) → b(x)
v(a(a(x))) → u(v(x))
v(a(c(x))) → u(b(d(x)))
v(c(x)) → b(x)
w(a(a(x))) → u(w(x))
w(a(c(x))) → u(b(d(x)))
w(c(x)) → b(x)

(12) Obligation:

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

a(c(d(x))) → c(x)
u(b(d(d(x)))) → b(x)
v(a(a(x))) → u(v(x))
v(a(c(x))) → u(b(d(x)))
v(c(x)) → b(x)
w(a(a(x))) → u(w(x))
w(a(c(x))) → u(b(d(x)))
w(c(x)) → b(x)

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

(13) PisEmptyProof (EQUIVALENT transformation)

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

(14) TRUE