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:

f(1) → f(g(1))
f(f(x)) → f(x)
g(0) → g(f(0))
g(g(x)) → g(x)

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

f(1) → f(g(1))
f(f(x)) → f(x)
g(0) → g(f(0))
g(g(x)) → g(x)

Q is empty.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

F(1) → F(g(1))
G(0) → G(f(0))
F(1) → G(1)
G(0) → F(0)

The TRS R consists of the following rules:

f(1) → f(g(1))
f(f(x)) → f(x)
g(0) → g(f(0))
g(g(x)) → g(x)

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof

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

F(1) → F(g(1))
G(0) → G(f(0))
F(1) → G(1)
G(0) → F(0)

The TRS R consists of the following rules:

f(1) → f(g(1))
f(f(x)) → f(x)
g(0) → g(f(0))
g(g(x)) → g(x)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof

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

F(1) → F(g(1))
F(1) → G(1)
G(0) → G(f(0))
G(0) → F(0)

The TRS R consists of the following rules:

f(1) → f(g(1))
f(f(x)) → f(x)
g(0) → g(f(0))
g(g(x)) → g(x)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 0 SCCs with 4 less nodes.