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(0, f2(x, x)) -> x
g2(x, s1(y)) -> g2(f2(x, y), 0)
g2(s1(x), y) -> g2(f2(x, y), 0)
g2(f2(x, y), 0) -> f2(g2(x, 0), g2(y, 0))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

g2(0, f2(x, x)) -> x
g2(x, s1(y)) -> g2(f2(x, y), 0)
g2(s1(x), y) -> g2(f2(x, y), 0)
g2(f2(x, y), 0) -> f2(g2(x, 0), g2(y, 0))

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, s1(y)) -> G2(f2(x, y), 0)
G2(f2(x, y), 0) -> G2(y, 0)
G2(s1(x), y) -> G2(f2(x, y), 0)
G2(f2(x, y), 0) -> G2(x, 0)

The TRS R consists of the following rules:

g2(0, f2(x, x)) -> x
g2(x, s1(y)) -> g2(f2(x, y), 0)
g2(s1(x), y) -> g2(f2(x, y), 0)
g2(f2(x, y), 0) -> f2(g2(x, 0), g2(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:

G2(x, s1(y)) -> G2(f2(x, y), 0)
G2(f2(x, y), 0) -> G2(y, 0)
G2(s1(x), y) -> G2(f2(x, y), 0)
G2(f2(x, y), 0) -> G2(x, 0)

The TRS R consists of the following rules:

g2(0, f2(x, x)) -> x
g2(x, s1(y)) -> g2(f2(x, y), 0)
g2(s1(x), y) -> g2(f2(x, y), 0)
g2(f2(x, y), 0) -> f2(g2(x, 0), g2(y, 0))

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

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

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

G2(f2(x, y), 0) -> G2(y, 0)
G2(s1(x), y) -> G2(f2(x, y), 0)
G2(f2(x, y), 0) -> G2(x, 0)

The TRS R consists of the following rules:

g2(0, f2(x, x)) -> x
g2(x, s1(y)) -> g2(f2(x, y), 0)
g2(s1(x), y) -> g2(f2(x, y), 0)
g2(f2(x, y), 0) -> f2(g2(x, 0), g2(y, 0))

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(f2(x, y), 0) -> G2(y, 0)
G2(s1(x), y) -> G2(f2(x, y), 0)
G2(f2(x, y), 0) -> G2(x, 0)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(0) = 0   
POL(G2(x1, x2)) = 3·x1 + 3·x2   
POL(f2(x1, x2)) = 1 + 2·x1 + x2   
POL(s1(x1)) = 2 + 3·x1   

The following usable rules [14] were oriented: none



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

Q DP problem:
P is empty.
The TRS R consists of the following rules:

g2(0, f2(x, x)) -> x
g2(x, s1(y)) -> g2(f2(x, y), 0)
g2(s1(x), y) -> g2(f2(x, y), 0)
g2(f2(x, y), 0) -> f2(g2(x, 0), g2(y, 0))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.