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 PisEmptyProof (⇔)
    ↳9 TRUE
  ↳10 QDP
    ↳11 QDPOrderProof (⇔)
    ↳12 QDP
    ↳13 PisEmptyProof (⇔)
    ↳14 TRUE

(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)  =  W(x1)
a(x1)  =  a(x1)
c(x1)  =  x1
d(x1)  =  d(x1)
u(x1)  =  u
b(x1)  =  b
v(x1)  =  v
w(x1)  =  w(x1)

Lexicographic path order with status [LPO].
Precedence:
W1 > b
a1 > u > b
a1 > w1 > b
v > d1 > b
v > u > b

Status:
v: []
w1: [1]
u: []
a1: [1]
b: []
d1: [1]
W1: [1]

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)  =  V(x1)
a(x1)  =  a(x1)
c(x1)  =  x1
d(x1)  =  d(x1)
u(x1)  =  u
b(x1)  =  b
v(x1)  =  v
w(x1)  =  w(x1)

Lexicographic path order with status [LPO].
Precedence:
V1 > b
a1 > u > b
a1 > w1 > b
v > d1 > b
v > u > b

Status:
v: []
w1: [1]
u: []
a1: [1]
b: []
d1: [1]
V1: [1]

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