Termination w.r.t. Q of the following Term Rewriting System could be proven:

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.


↳ QTRS
  ↳ DependencyPairsProof

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.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof

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

V(a(c(x))) → U(b(d(x)))
W(a(a(x))) → U(w(x))
V(a(a(x))) → V(x)
V(a(a(x))) → U(v(x))
W(a(c(x))) → U(b(d(x)))
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.
We deleted some edges using various graph approximations

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof

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

V(a(c(x))) → U(b(d(x)))
V(a(a(x))) → V(x)
W(a(a(x))) → U(w(x))
V(a(a(x))) → U(v(x))
W(a(c(x))) → U(b(d(x)))
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.
The approximation of the Dependency Graph [13,14,18] contains 2 SCCs with 4 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ QDPOrderProof
              ↳ QDP

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.
We use the reduction pair processor [13].


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.
none
Used ordering: Combined order from the following AFS and order.
W(x1)  =  x1
a(x1)  =  a(x1)

Recursive Path Order [2].
Precedence:
trivial


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ PisEmptyProof
              ↳ QDP

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ QDPOrderProof

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.
We use the reduction pair processor [13].


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.
none
Used ordering: Combined order from the following AFS and order.
V(x1)  =  x1
a(x1)  =  a(x1)

Recursive Path Order [2].
Precedence:
trivial


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ PisEmptyProof

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