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:

s(a) → a
s(s(x)) → x
s(f(x, y)) → f(s(y), s(x))
s(g(x, y)) → g(s(x), s(y))
f(x, a) → x
f(a, y) → y
f(g(x, y), g(u, v)) → g(f(x, u), f(y, v))
g(a, a) → a

Q is empty.


↳ QTRS
  ↳ RRRPoloQTRSProof

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

s(a) → a
s(s(x)) → x
s(f(x, y)) → f(s(y), s(x))
s(g(x, y)) → g(s(x), s(y))
f(x, a) → x
f(a, y) → y
f(g(x, y), g(u, v)) → g(f(x, u), f(y, v))
g(a, a) → a

Q is empty.

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

s(a) → a
s(s(x)) → x
s(f(x, y)) → f(s(y), s(x))
s(g(x, y)) → g(s(x), s(y))
f(x, a) → x
f(a, y) → y
f(g(x, y), g(u, v)) → g(f(x, u), f(y, v))
g(a, a) → a

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

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

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




↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof

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

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

Q is empty.

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

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

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

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

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




↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ RRRPoloQTRSProof

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

s(s(x)) → x

Q is empty.

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

s(s(x)) → x

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

s(s(x)) → x
Used ordering:
Polynomial interpretation [25]:

POL(s(x1)) = 2 + 2·x1   




↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ 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.