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
  ↳ RRRPoloQTRSProof

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.

The following Q TRS is given: 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.
The following rules can be removed by the rule removal processor [15] because they are oriented strictly by a polynomial ordering:

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

POL(0) = 0   
POL(f(x1, x2)) = x1 + x2   
POL(g(x1, x2)) = x1 + x2   
POL(s(x1)) = 1 + 2·x1   




↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof

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

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

Q is empty.

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

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

Q is empty.
The following rules can be removed by the rule removal processor [15] because they are oriented strictly by a polynomial ordering:

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

POL(0) = 0   
POL(f(x1, x2)) = 2 + x1 + x2   
POL(g(x1, x2)) = 1 + 2·x1 + 2·x2   




↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ RisEmptyProof

Q restricted rewrite system:
R is empty.
Q is empty.

The TRS R is empty. Hence, termination is trivially proven.