Termination w.r.t. Q proof of /tmp/tmpyzpY5K/z115.srs

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

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

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

The TRS R consists of the following rules:

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

The TRS R consists of the following rules:

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

The remaining pairs can at least be oriented weakly.

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

Used ordering: Polynomial interpretation [25,35]:

POL(C(x₁)) = x_1
POL(c(x₁)) = 1 + x_1
POL(D(x₁)) = x_1
POL(B(x₁)) = x_1
POL(a(x₁)) = 2 + x_1
POL(A(x₁)) = x_1
POL(b(x₁)) = 2 + x_1
POL(d(x₁)) = 3 + x_1

The value of delta used in the strict ordering is 1.
The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ DependencyGraphProof

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

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

The TRS R consists of the following rules:

a(a(x1)) → d(c(x1))
a(b(x1)) → c(c(c(x1)))
b(b(x1)) → a(c(c(x1)))
c(c(x1)) → b(x1)
c(d(x1)) → a(a(x1))
d(d(x1)) → b(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 3 less nodes.