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

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

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

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

The TRS R consists of the following rules:

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

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:

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

The TRS R consists of the following rules:

a(x1) → g(d(x1))
b(b(b(x1))) → c(d(c(x1)))
b(b(x1)) → a(g(g(x1)))
c(d(x1)) → g(g(x1))
g(g(g(x1))) → b(b(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
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof

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

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

The TRS R consists of the following rules:

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

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

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

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



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

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

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

The TRS R consists of the following rules:

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