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(0, f(x, x)) → x
g(x, s(y)) → g(f(x, y), 0)
g(s(x), y) → g(f(x, y), 0)
g(f(x, y), 0) → f(g(x, 0), g(y, 0))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

g(0, f(x, x)) → x
g(x, s(y)) → g(f(x, y), 0)
g(s(x), y) → g(f(x, y), 0)
g(f(x, y), 0) → f(g(x, 0), g(y, 0))

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:

G(s(x), y) → G(f(x, y), 0)
G(x, s(y)) → G(f(x, y), 0)
G(f(x, y), 0) → G(y, 0)
G(f(x, y), 0) → G(x, 0)

The TRS R consists of the following rules:

g(0, f(x, x)) → x
g(x, s(y)) → g(f(x, y), 0)
g(s(x), y) → g(f(x, y), 0)
g(f(x, y), 0) → f(g(x, 0), g(y, 0))

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof

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

G(s(x), y) → G(f(x, y), 0)
G(x, s(y)) → G(f(x, y), 0)
G(f(x, y), 0) → G(y, 0)
G(f(x, y), 0) → G(x, 0)

The TRS R consists of the following rules:

g(0, f(x, x)) → x
g(x, s(y)) → g(f(x, y), 0)
g(s(x), y) → g(f(x, y), 0)
g(f(x, y), 0) → f(g(x, 0), g(y, 0))

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

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

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

G(s(x), y) → G(f(x, y), 0)
G(f(x, y), 0) → G(y, 0)
G(f(x, y), 0) → G(x, 0)

The TRS R consists of the following rules:

g(0, f(x, x)) → x
g(x, s(y)) → g(f(x, y), 0)
g(s(x), y) → g(f(x, y), 0)
g(f(x, y), 0) → f(g(x, 0), g(y, 0))

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.


G(f(x, y), 0) → G(y, 0)
G(f(x, y), 0) → G(x, 0)
The remaining pairs can at least be oriented weakly.

G(s(x), y) → G(f(x, y), 0)
Used ordering: Polynomial interpretation [25,35]:

POL(f(x1, x2)) = 1/4 + x_1 + x_2   
POL(s(x1)) = 1/4 + x_1   
POL(G(x1, x2)) = (4)x_1 + (4)x_2   
POL(0) = 0   
The value of delta used in the strict ordering is 1.
The following usable rules [17] were oriented: none



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

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

G(s(x), y) → G(f(x, y), 0)

The TRS R consists of the following rules:

g(0, f(x, x)) → x
g(x, s(y)) → g(f(x, y), 0)
g(s(x), y) → g(f(x, y), 0)
g(f(x, y), 0) → f(g(x, 0), g(y, 0))

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.