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:

g(f(x), y) → f(h(x, y))
h(x, y) → g(x, f(y))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

g(f(x), y) → f(h(x, y))
h(x, y) → g(x, f(y))

Q is empty.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

H(x, y) → G(x, f(y))
G(f(x), y) → H(x, y)

The TRS R consists of the following rules:

g(f(x), y) → f(h(x, y))
h(x, y) → g(x, f(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:

H(x, y) → G(x, f(y))
G(f(x), y) → H(x, y)

The TRS R consists of the following rules:

g(f(x), y) → f(h(x, y))
h(x, y) → g(x, f(y))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


H(x, y) → G(x, f(y))
The remaining pairs can at least be oriented weakly.

G(f(x), y) → H(x, y)
Used ordering: Polynomial interpretation [25,35]:

POL(f(x1)) = 2 + (2)x_1   
POL(G(x1, x2)) = x_1   
POL(H(x1, x2)) = 2 + x_1   
The value of delta used in the strict ordering is 2.
The following usable rules [17] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof
        ↳ QDP
          ↳ DependencyGraphProof

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

G(f(x), y) → H(x, y)

The TRS R consists of the following rules:

g(f(x), y) → f(h(x, y))
h(x, y) → g(x, f(y))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 1 less node.