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:

g1(a) -> g1(b)
b -> f2(a, a)
f2(a, a) -> g1(d)

Q is empty.


QTRS
  ↳ Non-Overlap Check

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

g1(a) -> g1(b)
b -> f2(a, a)
f2(a, a) -> g1(d)

Q is empty.

The TRS is non-overlapping. Hence, we can switch to innermost.

↳ QTRS
  ↳ Non-Overlap Check
QTRS
      ↳ DependencyPairsProof

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

g1(a) -> g1(b)
b -> f2(a, a)
f2(a, a) -> g1(d)

The set Q consists of the following terms:

g1(a)
b
f2(a, a)


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

F2(a, a) -> G1(d)
B -> F2(a, a)
G1(a) -> B
G1(a) -> G1(b)

The TRS R consists of the following rules:

g1(a) -> g1(b)
b -> f2(a, a)
f2(a, a) -> g1(d)

The set Q consists of the following terms:

g1(a)
b
f2(a, a)

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

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
QDP
          ↳ DependencyGraphProof

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

F2(a, a) -> G1(d)
B -> F2(a, a)
G1(a) -> B
G1(a) -> G1(b)

The TRS R consists of the following rules:

g1(a) -> g1(b)
b -> f2(a, a)
f2(a, a) -> g1(d)

The set Q consists of the following terms:

g1(a)
b
f2(a, a)

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.