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(h(x)) → f(i(x))
g(i(x)) → g(h(x))
h(a) → b
i(a) → b

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

f(h(x)) → f(i(x))
g(i(x)) → g(h(x))
h(a) → b
i(a) → b

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:

G(i(x)) → H(x)
F(h(x)) → F(i(x))
F(h(x)) → I(x)
G(i(x)) → G(h(x))

The TRS R consists of the following rules:

f(h(x)) → f(i(x))
g(i(x)) → g(h(x))
h(a) → b
i(a) → b

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:

G(i(x)) → H(x)
F(h(x)) → F(i(x))
F(h(x)) → I(x)
G(i(x)) → G(h(x))

The TRS R consists of the following rules:

f(h(x)) → f(i(x))
g(i(x)) → g(h(x))
h(a) → b
i(a) → b

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:

G(i(x)) → H(x)
F(h(x)) → F(i(x))
F(h(x)) → I(x)
G(i(x)) → G(h(x))

The TRS R consists of the following rules:

f(h(x)) → f(i(x))
g(i(x)) → g(h(x))
h(a) → b
i(a) → b

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.