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:

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

Q is empty.


↳ QTRS
  ↳ RRRPoloQTRSProof

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

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

Q is empty.

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

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

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

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




↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof

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

minus(minus(x)) → x

Q is empty.

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

minus(minus(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:

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

POL(minus(x1)) = 2 + 2·x1   




↳ 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.