Termination w.r.t. Q proof of /tmp/tmpq3DJvU/torpa3.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:

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

Q is empty.

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

  ↳ QTRS Reverse

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

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

Q is empty.

We have reversed the following QTRS:
The set of rules R is

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

The set Q is empty.
We have obtained the following QTRS:

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

The set Q is empty.

↳ QTRS

  ↳ QTRS Reverse

    ↳ QTRS

  ↳ DependencyPairsProof

  ↳ QTRS Reverse

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

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

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

The TRS R consists of the following rules:

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

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

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

b(b(x1)) → c(d(x1))
c(c(x1)) → d(d(d(x1)))
c(x1) → g(x1)
d(d(x1)) → c(f(x1))
d(d(d(x1))) → g(c(x1))
f(x1) → a(g(x1))
g(x1) → d(a(b(x1)))
g(g(x1)) → b(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 1 SCC with 1 less node.

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

  ↳ QTRS Reverse

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

C(c(x1)) → D(d(d(x1)))
G(x1) → B(x1)
G(g(x1)) → B(c(x1))
C(c(x1)) → D(x1)
G(g(x1)) → C(x1)
B(b(x1)) → C(d(x1))
C(c(x1)) → D(d(x1))
F(x1) → G(x1)
B(b(x1)) → D(x1)
D(d(x1)) → C(f(x1))
D(d(d(x1))) → G(c(x1))
C(x1) → G(x1)
D(d(x1)) → F(x1)
D(d(d(x1))) → C(x1)

The TRS R consists of the following rules:

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

G(x1) → B(x1)
G(g(x1)) → B(c(x1))
C(c(x1)) → D(x1)
G(g(x1)) → C(x1)
B(b(x1)) → C(d(x1))
C(c(x1)) → D(d(x1))
F(x1) → G(x1)
B(b(x1)) → D(x1)
D(d(x1)) → F(x1)
D(d(d(x1))) → C(x1)

The remaining pairs can at least be oriented weakly.

C(c(x1)) → D(d(d(x1)))
D(d(x1)) → C(f(x1))
D(d(d(x1))) → G(c(x1))
C(x1) → G(x1)

Used ordering: Polynomial interpretation [25]:

POL(B(x₁)) = 6 + 4·x₁
POL(C(x₁)) = 8 + 4·x₁
POL(D(x₁)) = 4·x₁
POL(F(x₁)) = 13 + 4·x₁
POL(G(x₁)) = 8 + 4·x₁
POL(a(x₁)) = 2
POL(b(x₁)) = 6 + x₁
POL(c(x₁)) = 6 + x₁
POL(d(x₁)) = 4 + x₁
POL(f(x₁)) = 2
POL(g(x₁)) = 6 + x₁

The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

  ↳ QTRS Reverse

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

C(c(x1)) → D(d(d(x1)))
C(x1) → G(x1)
D(d(d(x1))) → G(c(x1))
D(d(x1)) → C(f(x1))

The TRS R consists of the following rules:

b(b(x1)) → c(d(x1))
c(c(x1)) → d(d(d(x1)))
c(x1) → g(x1)
d(d(x1)) → c(f(x1))
d(d(d(x1))) → g(c(x1))
f(x1) → a(g(x1))
g(x1) → d(a(b(x1)))
g(g(x1)) → b(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 1 SCC with 2 less nodes.

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ Narrowing

  ↳ QTRS Reverse

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

C(c(x1)) → D(d(d(x1)))
D(d(x1)) → C(f(x1))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule D(d(x1)) → C(f(x1)) at position [0] we obtained the following new rules:

D(d(x0)) → C(a(g(x0)))

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ Narrowing

                    ↳ QDP

                      ↳ DependencyGraphProof

  ↳ QTRS Reverse

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

C(c(x1)) → D(d(d(x1)))
D(d(x0)) → C(a(g(x0)))

The TRS R consists of the following rules:

b(b(x1)) → c(d(x1))
c(c(x1)) → d(d(d(x1)))
c(x1) → g(x1)
d(d(x1)) → c(f(x1))
d(d(d(x1))) → g(c(x1))
f(x1) → a(g(x1))
g(x1) → d(a(b(x1)))
g(g(x1)) → b(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 2 less nodes.
We have reversed the following QTRS:
The set of rules R is

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

The set Q is empty.
We have obtained the following QTRS:

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

The set Q is empty.

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

  ↳ QTRS Reverse

    ↳ QTRS

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

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

Q is empty.