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:

d1(x) -> e1(u1(x))
d1(u1(x)) -> c1(x)
c1(u1(x)) -> b1(x)
v1(e1(x)) -> x
b1(u1(x)) -> a1(e1(x))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

d1(x) -> e1(u1(x))
d1(u1(x)) -> c1(x)
c1(u1(x)) -> b1(x)
v1(e1(x)) -> x
b1(u1(x)) -> a1(e1(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:

C1(u1(x)) -> B1(x)
D1(u1(x)) -> C1(x)

The TRS R consists of the following rules:

d1(x) -> e1(u1(x))
d1(u1(x)) -> c1(x)
c1(u1(x)) -> b1(x)
v1(e1(x)) -> x
b1(u1(x)) -> a1(e1(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:

C1(u1(x)) -> B1(x)
D1(u1(x)) -> C1(x)

The TRS R consists of the following rules:

d1(x) -> e1(u1(x))
d1(u1(x)) -> c1(x)
c1(u1(x)) -> b1(x)
v1(e1(x)) -> x
b1(u1(x)) -> a1(e1(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 0 SCCs with 2 less nodes.