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:

f(x, y) → g1(x, x, y)
f(x, y) → g1(y, x, x)
f(x, y) → g2(x, y, y)
f(x, y) → g2(y, y, x)
g1(x, x, y) → h(x, y)
g1(y, x, x) → h(x, y)
g2(x, y, y) → h(x, y)
g2(y, y, x) → h(x, y)
h(x, x) → x

Q is empty.


↳ QTRS
  ↳ RRRPoloQTRSProof

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

f(x, y) → g1(x, x, y)
f(x, y) → g1(y, x, x)
f(x, y) → g2(x, y, y)
f(x, y) → g2(y, y, x)
g1(x, x, y) → h(x, y)
g1(y, x, x) → h(x, y)
g2(x, y, y) → h(x, y)
g2(y, y, x) → h(x, y)
h(x, x) → x

Q is empty.

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

f(x, y) → g1(x, x, y)
f(x, y) → g1(y, x, x)
f(x, y) → g2(x, y, y)
f(x, y) → g2(y, y, x)
g1(x, x, y) → h(x, y)
g1(y, x, x) → h(x, y)
g2(x, y, y) → h(x, y)
g2(y, y, x) → h(x, y)
h(x, 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:

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

POL(f(x1, x2)) = 1 + 2·x1 + 2·x2   
POL(g1(x1, x2, x3)) = 1 + x1 + x2 + x3   
POL(g2(x1, x2, x3)) = 1 + x1 + x2 + x3   
POL(h(x1, x2)) = 1 + x1 + x2   




↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof

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

f(x, y) → g1(x, x, y)
f(x, y) → g1(y, x, x)
f(x, y) → g2(x, y, y)
f(x, y) → g2(y, y, x)
g1(x, x, y) → h(x, y)
g1(y, x, x) → h(x, y)
g2(x, y, y) → h(x, y)
g2(y, y, x) → h(x, y)

Q is empty.

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

f(x, y) → g1(x, x, y)
f(x, y) → g1(y, x, x)
f(x, y) → g2(x, y, y)
f(x, y) → g2(y, y, x)
g1(x, x, y) → h(x, y)
g1(y, x, x) → h(x, y)
g2(x, y, y) → h(x, y)
g2(y, y, x) → h(x, y)

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

f(x, y) → g1(x, x, y)
f(x, y) → g1(y, x, x)
f(x, y) → g2(x, y, y)
f(x, y) → g2(y, y, x)
Used ordering:
Polynomial interpretation [25]:

POL(f(x1, x2)) = 2 + 2·x1 + 2·x2   
POL(g1(x1, x2, x3)) = 1 + x1 + x2 + x3   
POL(g2(x1, x2, x3)) = 1 + x1 + x2 + x3   
POL(h(x1, x2)) = 1 + x1 + x2   




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

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

g1(x, x, y) → h(x, y)
g1(y, x, x) → h(x, y)
g2(x, y, y) → h(x, y)
g2(y, y, x) → h(x, y)

Q is empty.

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

g1(x, x, y) → h(x, y)
g1(y, x, x) → h(x, y)
g2(x, y, y) → h(x, y)
g2(y, y, x) → h(x, y)

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

g1(x, x, y) → h(x, y)
g1(y, x, x) → h(x, y)
g2(x, y, y) → h(x, y)
g2(y, y, x) → h(x, y)
Used ordering:
Polynomial interpretation [25]:

POL(g1(x1, x2, x3)) = 2 + 2·x1 + 2·x2 + 2·x3   
POL(g2(x1, x2, x3)) = 2 + 2·x1 + 2·x2 + 2·x3   
POL(h(x1, x2)) = 1 + x1 + x2   




↳ 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.