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
  ↳ 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))

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:

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))

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

↳ 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))

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


The following pairs can be oriented strictly and are deleted.


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

H2(x, y) -> G2(x, f1(y))
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( G2(x1, x2) ) = 2x1 + 3


POL( H2(x1, x2) ) = 3x1 + 3


POL( f1(x1) ) = 3x1 + 1



The following usable rules [14] were oriented: none



↳ 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))

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 1 less node.