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:

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

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

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

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

The TRS R consists of the following rules:

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

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

The TRS R consists of the following rules:

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


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

G2(x, y) -> H2(x, y)
Used ordering: Polynomial interpretation [21]:

POL(G2(x1, x2)) = 1 + x1   
POL(H2(x1, x2)) = 1 + x1   
POL(f1(x1)) = 1 + x1   

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:

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

The TRS R consists of the following rules:

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