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:

a1(c1(d1(x))) -> c1(x)
u1(b1(d1(d1(x)))) -> b1(x)
v1(a1(a1(x))) -> u1(v1(x))
v1(a1(c1(x))) -> u1(b1(d1(x)))
v1(c1(x)) -> b1(x)
w1(a1(a1(x))) -> u1(w1(x))
w1(a1(c1(x))) -> u1(b1(d1(x)))
w1(c1(x)) -> b1(x)

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

Q restricted rewrite system:
The TRS R consists of the following rules:

a1(c1(d1(x))) -> c1(x)
u1(b1(d1(d1(x)))) -> b1(x)
v1(a1(a1(x))) -> u1(v1(x))
v1(a1(c1(x))) -> u1(b1(d1(x)))
v1(c1(x)) -> b1(x)
w1(a1(a1(x))) -> u1(w1(x))
w1(a1(c1(x))) -> u1(b1(d1(x)))
w1(c1(x)) -> b1(x)

Q is empty.

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

V1(a1(c1(x))) -> U1(b1(d1(x)))
W1(a1(a1(x))) -> U1(w1(x))
V1(a1(a1(x))) -> V1(x)
V1(a1(a1(x))) -> U1(v1(x))
W1(a1(c1(x))) -> U1(b1(d1(x)))
W1(a1(a1(x))) -> W1(x)

The TRS R consists of the following rules:

a1(c1(d1(x))) -> c1(x)
u1(b1(d1(d1(x)))) -> b1(x)
v1(a1(a1(x))) -> u1(v1(x))
v1(a1(c1(x))) -> u1(b1(d1(x)))
v1(c1(x)) -> b1(x)
w1(a1(a1(x))) -> u1(w1(x))
w1(a1(c1(x))) -> u1(b1(d1(x)))
w1(c1(x)) -> b1(x)

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof

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

V1(a1(c1(x))) -> U1(b1(d1(x)))
W1(a1(a1(x))) -> U1(w1(x))
V1(a1(a1(x))) -> V1(x)
V1(a1(a1(x))) -> U1(v1(x))
W1(a1(c1(x))) -> U1(b1(d1(x)))
W1(a1(a1(x))) -> W1(x)

The TRS R consists of the following rules:

a1(c1(d1(x))) -> c1(x)
u1(b1(d1(d1(x)))) -> b1(x)
v1(a1(a1(x))) -> u1(v1(x))
v1(a1(c1(x))) -> u1(b1(d1(x)))
v1(c1(x)) -> b1(x)
w1(a1(a1(x))) -> u1(w1(x))
w1(a1(c1(x))) -> u1(b1(d1(x)))
w1(c1(x)) -> b1(x)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 2 SCCs with 4 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPAfsSolverProof
          ↳ QDP

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

W1(a1(a1(x))) -> W1(x)

The TRS R consists of the following rules:

a1(c1(d1(x))) -> c1(x)
u1(b1(d1(d1(x)))) -> b1(x)
v1(a1(a1(x))) -> u1(v1(x))
v1(a1(c1(x))) -> u1(b1(d1(x)))
v1(c1(x)) -> b1(x)
w1(a1(a1(x))) -> u1(w1(x))
w1(a1(c1(x))) -> u1(b1(d1(x)))
w1(c1(x)) -> b1(x)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

W1(a1(a1(x))) -> W1(x)
Used argument filtering: W1(x1)  =  x1
a1(x1)  =  a1(x1)
Used ordering: Precedence:
trivial



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPAfsSolverProof
              ↳ QDP
                ↳ PisEmptyProof
          ↳ QDP

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

a1(c1(d1(x))) -> c1(x)
u1(b1(d1(d1(x)))) -> b1(x)
v1(a1(a1(x))) -> u1(v1(x))
v1(a1(c1(x))) -> u1(b1(d1(x)))
v1(c1(x)) -> b1(x)
w1(a1(a1(x))) -> u1(w1(x))
w1(a1(c1(x))) -> u1(b1(d1(x)))
w1(c1(x)) -> b1(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
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof

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

V1(a1(a1(x))) -> V1(x)

The TRS R consists of the following rules:

a1(c1(d1(x))) -> c1(x)
u1(b1(d1(d1(x)))) -> b1(x)
v1(a1(a1(x))) -> u1(v1(x))
v1(a1(c1(x))) -> u1(b1(d1(x)))
v1(c1(x)) -> b1(x)
w1(a1(a1(x))) -> u1(w1(x))
w1(a1(c1(x))) -> u1(b1(d1(x)))
w1(c1(x)) -> b1(x)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

V1(a1(a1(x))) -> V1(x)
Used argument filtering: V1(x1)  =  x1
a1(x1)  =  a1(x1)
Used ordering: Precedence:
trivial



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
              ↳ QDP
                ↳ PisEmptyProof

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

a1(c1(d1(x))) -> c1(x)
u1(b1(d1(d1(x)))) -> b1(x)
v1(a1(a1(x))) -> u1(v1(x))
v1(a1(c1(x))) -> u1(b1(d1(x)))
v1(c1(x)) -> b1(x)
w1(a1(a1(x))) -> u1(w1(x))
w1(a1(c1(x))) -> u1(b1(d1(x)))
w1(c1(x)) -> b1(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.