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:

rev(nil) → nil
rev(rev(x)) → x
rev(++(x, y)) → ++(rev(y), rev(x))
++(nil, y) → y
++(x, nil) → x
++(.(x, y), z) → .(x, ++(y, z))
++(x, ++(y, z)) → ++(++(x, y), z)
make(x) → .(x, nil)

Q is empty.


↳ QTRS
  ↳ RRRPoloQTRSProof

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

rev(nil) → nil
rev(rev(x)) → x
rev(++(x, y)) → ++(rev(y), rev(x))
++(nil, y) → y
++(x, nil) → x
++(.(x, y), z) → .(x, ++(y, z))
++(x, ++(y, z)) → ++(++(x, y), z)
make(x) → .(x, nil)

Q is empty.

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

rev(nil) → nil
rev(rev(x)) → x
rev(++(x, y)) → ++(rev(y), rev(x))
++(nil, y) → y
++(x, nil) → x
++(.(x, y), z) → .(x, ++(y, z))
++(x, ++(y, z)) → ++(++(x, y), z)
make(x) → .(x, nil)

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

rev(nil) → nil
++(nil, y) → y
++(x, nil) → x
make(x) → .(x, nil)
Used ordering:
Polynomial interpretation [25]:

POL(++(x1, x2)) = x1 + x2   
POL(.(x1, x2)) = 2·x1 + x2   
POL(make(x1)) = 2 + 2·x1   
POL(nil) = 1   
POL(rev(x1)) = 2·x1   




↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof

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

rev(rev(x)) → x
rev(++(x, y)) → ++(rev(y), rev(x))
++(.(x, y), z) → .(x, ++(y, z))
++(x, ++(y, z)) → ++(++(x, y), z)

Q is empty.

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

rev(rev(x)) → x
rev(++(x, y)) → ++(rev(y), rev(x))
++(.(x, y), z) → .(x, ++(y, z))
++(x, ++(y, z)) → ++(++(x, y), z)

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

rev(++(x, y)) → ++(rev(y), rev(x))
Used ordering:
Polynomial interpretation [25]:

POL(++(x1, x2)) = 2 + x1 + x2   
POL(.(x1, x2)) = 2·x1 + x2   
POL(rev(x1)) = 2·x1   




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

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

rev(rev(x)) → x
++(.(x, y), z) → .(x, ++(y, z))
++(x, ++(y, z)) → ++(++(x, y), z)

Q is empty.

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

rev(rev(x)) → x
++(.(x, y), z) → .(x, ++(y, z))
++(x, ++(y, z)) → ++(++(x, y), z)

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

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

POL(++(x1, x2)) = x1 + 2·x2   
POL(.(x1, x2)) = 2·x1 + x2   
POL(rev(x1)) = 2 + 2·x1   




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

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

++(.(x, y), z) → .(x, ++(y, z))
++(x, ++(y, z)) → ++(++(x, y), z)

Q is empty.

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

++(.(x, y), z) → .(x, ++(y, z))
++(x, ++(y, z)) → ++(++(x, y), z)

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

++(x, ++(y, z)) → ++(++(x, y), z)
Used ordering:
Polynomial interpretation [25]:

POL(++(x1, x2)) = 2 + x1 + 2·x2   
POL(.(x1, x2)) = 2·x1 + x2   




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

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

++(.(x, y), z) → .(x, ++(y, z))

Q is empty.

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

++(.(x, y), z) → .(x, ++(y, z))

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

++(.(x, y), z) → .(x, ++(y, z))
Used ordering:
Polynomial interpretation [25]:

POL(++(x1, x2)) = 2 + 2·x1 + 2·x2   
POL(.(x1, x2)) = 1 + 2·x1 + x2   




↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ 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.