Termination w.r.t. Q proof of /tmp/tmp4d40F6/torpa4.srs

Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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

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

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

a(b(c(a(x1)))) → b(a(c(b(a(b(x1))))))
a(d(x1)) → c(x1)
a(f(f(x1))) → g(x1)
b(g(x1)) → g(b(x1))
c(x1) → f(f(x1))
c(a(c(x1))) → b(c(a(b(c(x1)))))
c(d(x1)) → a(a(x1))
g(x1) → c(a(x1))
g(x1) → d(d(d(d(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:

C(a(c(x1))) → A(b(c(x1)))
C(d(x1)) → A(a(x1))
A(b(c(a(x1)))) → B(a(c(b(a(b(x1))))))
C(a(c(x1))) → C(a(b(c(x1))))
A(d(x1)) → C(x1)
G(x1) → C(a(x1))
A(b(c(a(x1)))) → A(b(x1))
C(d(x1)) → A(x1)
B(g(x1)) → G(b(x1))
A(b(c(a(x1)))) → A(c(b(a(b(x1)))))
A(f(f(x1))) → G(x1)
G(x1) → A(x1)
A(b(c(a(x1)))) → B(x1)
C(a(c(x1))) → B(c(x1))
A(b(c(a(x1)))) → B(a(b(x1)))
A(b(c(a(x1)))) → C(b(a(b(x1))))
C(a(c(x1))) → B(c(a(b(c(x1)))))
B(g(x1)) → B(x1)

The TRS R consists of the following rules:

a(b(c(a(x1)))) → b(a(c(b(a(b(x1))))))
a(d(x1)) → c(x1)
a(f(f(x1))) → g(x1)
b(g(x1)) → g(b(x1))
c(x1) → f(f(x1))
c(a(c(x1))) → b(c(a(b(c(x1)))))
c(d(x1)) → a(a(x1))
g(x1) → c(a(x1))
g(x1) → d(d(d(d(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:

C(a(c(x1))) → A(b(c(x1)))
C(d(x1)) → A(a(x1))
A(b(c(a(x1)))) → B(a(c(b(a(b(x1))))))
C(a(c(x1))) → C(a(b(c(x1))))
A(d(x1)) → C(x1)
G(x1) → C(a(x1))
A(b(c(a(x1)))) → A(b(x1))
C(d(x1)) → A(x1)
B(g(x1)) → G(b(x1))
A(b(c(a(x1)))) → A(c(b(a(b(x1)))))
A(f(f(x1))) → G(x1)
G(x1) → A(x1)
A(b(c(a(x1)))) → B(x1)
C(a(c(x1))) → B(c(x1))
A(b(c(a(x1)))) → B(a(b(x1)))
A(b(c(a(x1)))) → C(b(a(b(x1))))
C(a(c(x1))) → B(c(a(b(c(x1)))))
B(g(x1)) → B(x1)

The TRS R consists of the following rules:

a(b(c(a(x1)))) → b(a(c(b(a(b(x1))))))
a(d(x1)) → c(x1)
a(f(f(x1))) → g(x1)
b(g(x1)) → g(b(x1))
c(x1) → f(f(x1))
c(a(c(x1))) → b(c(a(b(c(x1)))))
c(d(x1)) → a(a(x1))
g(x1) → c(a(x1))
g(x1) → d(d(d(d(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.

C(a(c(x1))) → A(b(c(x1)))
A(b(c(a(x1)))) → B(a(c(b(a(b(x1))))))
A(d(x1)) → C(x1)
A(b(c(a(x1)))) → A(b(x1))
C(d(x1)) → A(x1)
A(f(f(x1))) → G(x1)
G(x1) → A(x1)
A(b(c(a(x1)))) → B(x1)
C(a(c(x1))) → B(c(x1))
A(b(c(a(x1)))) → B(a(b(x1)))
A(b(c(a(x1)))) → C(b(a(b(x1))))
B(g(x1)) → B(x1)

The remaining pairs can at least be oriented weakly.

C(d(x1)) → A(a(x1))
C(a(c(x1))) → C(a(b(c(x1))))
G(x1) → C(a(x1))
B(g(x1)) → G(b(x1))
A(b(c(a(x1)))) → A(c(b(a(b(x1)))))
C(a(c(x1))) → B(c(a(b(c(x1)))))

Used ordering: Polynomial interpretation [25,35]:

POL(C(x₁)) = 5/2 + x_1
POL(c(x₁)) = 5/2 + x_1
POL(f(x₁)) = 5/4 + x_1
POL(B(x₁)) = x_1
POL(a(x₁)) = 3/2 + x_1
POL(g(x₁)) = 4 + x_1
POL(A(x₁)) = 2 + x_1
POL(G(x₁)) = 4 + x_1
POL(b(x₁)) = x_1
POL(d(x₁)) = 1 + x_1

The value of delta used in the strict ordering is 1/2.
The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ DependencyGraphProof

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

B(g(x1)) → G(b(x1))
A(b(c(a(x1)))) → A(c(b(a(b(x1)))))
C(a(c(x1))) → C(a(b(c(x1))))
C(d(x1)) → A(a(x1))
C(a(c(x1))) → B(c(a(b(c(x1)))))
G(x1) → C(a(x1))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 2 SCCs with 1 less node.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

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

A(b(c(a(x1)))) → A(c(b(a(b(x1)))))

The TRS R consists of the following rules:

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

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

B(g(x1)) → G(b(x1))
C(a(c(x1))) → C(a(b(c(x1))))
G(x1) → C(a(x1))
C(a(c(x1))) → B(c(a(b(c(x1)))))

The TRS R consists of the following rules:

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

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