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:

g2(f1(x), y) -> f1(h2(x, y))
h2(x, y) -> g2(x, f1(y))

Q is empty.


↳ QTRS
  ↳ Non-Overlap Check

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

g2(f1(x), y) -> f1(h2(x, y))
h2(x, y) -> g2(x, f1(y))

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:

g2(f1(x), y) -> f1(h2(x, y))
h2(x, y) -> g2(x, f1(y))

The set Q consists of the following terms:

g2(f1(x0), x1)
h2(x0, x1)


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

H2(x, y) -> G2(x, f1(y))
G2(f1(x), y) -> H2(x, y)

The TRS R consists of the following rules:

g2(f1(x), y) -> f1(h2(x, y))
h2(x, y) -> g2(x, f1(y))

The set Q consists of the following terms:

g2(f1(x0), x1)
h2(x0, x1)

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

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

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

H2(x, y) -> G2(x, f1(y))
G2(f1(x), y) -> H2(x, y)

The TRS R consists of the following rules:

g2(f1(x), y) -> f1(h2(x, y))
h2(x, y) -> g2(x, f1(y))

The set Q consists of the following terms:

g2(f1(x0), x1)
h2(x0, x1)

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


G2(f1(x), y) -> H2(x, y)
The remaining pairs can at least by weakly be oriented.

H2(x, y) -> G2(x, f1(y))
Used ordering: Combined order from the following AFS and order.
H2(x1, x2)  =  H1(x1)
G2(x1, x2)  =  G1(x1)
f1(x1)  =  f1(x1)

Lexicographic Path Order [19].
Precedence:
[H1, G1, f1]


The following usable rules [14] were oriented: none



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

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

H2(x, y) -> G2(x, f1(y))

The TRS R consists of the following rules:

g2(f1(x), y) -> f1(h2(x, y))
h2(x, y) -> g2(x, f1(y))

The set Q consists of the following terms:

g2(f1(x0), x1)
h2(x0, x1)

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 0 SCCs with 1 less node.