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(a(x1)) → b(c(x1))
b(b(x1)) → c(d(x1))
c(c(x1)) → d(d(d(x1)))
d(d(d(x1))) → a(c(x1))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

a(a(x1)) → b(c(x1))
b(b(x1)) → c(d(x1))
c(c(x1)) → d(d(d(x1)))
d(d(d(x1))) → a(c(x1))

Q is empty.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

A(a(x1)) → B(c(x1))
C(c(x1)) → D(d(x1))
C(c(x1)) → D(d(d(x1)))
B(b(x1)) → D(x1)
A(a(x1)) → C(x1)
D(d(d(x1))) → C(x1)
C(c(x1)) → D(x1)
D(d(d(x1))) → A(c(x1))
B(b(x1)) → C(d(x1))

The TRS R consists of the following rules:

a(a(x1)) → b(c(x1))
b(b(x1)) → c(d(x1))
c(c(x1)) → d(d(d(x1)))
d(d(d(x1))) → a(c(x1))

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof

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

A(a(x1)) → B(c(x1))
C(c(x1)) → D(d(x1))
C(c(x1)) → D(d(d(x1)))
B(b(x1)) → D(x1)
A(a(x1)) → C(x1)
D(d(d(x1))) → C(x1)
C(c(x1)) → D(x1)
D(d(d(x1))) → A(c(x1))
B(b(x1)) → C(d(x1))

The TRS R consists of the following rules:

a(a(x1)) → b(c(x1))
b(b(x1)) → c(d(x1))
c(c(x1)) → d(d(d(x1)))
d(d(d(x1))) → a(c(x1))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


A(a(x1)) → B(c(x1))
C(c(x1)) → D(d(x1))
C(c(x1)) → D(d(d(x1)))
B(b(x1)) → D(x1)
A(a(x1)) → C(x1)
D(d(d(x1))) → C(x1)
C(c(x1)) → D(x1)
D(d(d(x1))) → A(c(x1))
The remaining pairs can at least be oriented weakly.

B(b(x1)) → C(d(x1))
Used ordering: Polynomial interpretation [25,35]:

POL(C(x1)) = 1/2 + (2)x_1   
POL(c(x1)) = 1/2 + (4)x_1   
POL(B(x1)) = 1/4 + (3/2)x_1   
POL(D(x1)) = 1/4 + (5/4)x_1   
POL(a(x1)) = 1/2 + (15/4)x_1   
POL(A(x1)) = 1/4 + (7/4)x_1   
POL(d(x1)) = 1/4 + (5/2)x_1   
POL(b(x1)) = 1/2 + (7/2)x_1   
The value of delta used in the strict ordering is 1/8.
The following usable rules [17] were oriented:

d(d(d(x1))) → a(c(x1))
b(b(x1)) → c(d(x1))
a(a(x1)) → b(c(x1))
c(c(x1)) → d(d(d(x1)))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof
        ↳ QDP
          ↳ DependencyGraphProof

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

B(b(x1)) → C(d(x1))

The TRS R consists of the following rules:

a(a(x1)) → b(c(x1))
b(b(x1)) → c(d(x1))
c(c(x1)) → d(d(d(x1)))
d(d(d(x1))) → a(c(x1))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 1 less node.