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:

a__zeros → cons(0, zeros)
a__tail(cons(X, XS)) → mark(XS)
mark(zeros) → a__zeros
mark(tail(X)) → a__tail(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
a__zeros → zeros
a__tail(X) → tail(X)

Q is empty.


↳ QTRS
  ↳ RRRPoloQTRSProof

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

a__zeros → cons(0, zeros)
a__tail(cons(X, XS)) → mark(XS)
mark(zeros) → a__zeros
mark(tail(X)) → a__tail(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
a__zeros → zeros
a__tail(X) → tail(X)

Q is empty.

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

a__zeros → cons(0, zeros)
a__tail(cons(X, XS)) → mark(XS)
mark(zeros) → a__zeros
mark(tail(X)) → a__tail(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
a__zeros → zeros
a__tail(X) → tail(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:

a__tail(cons(X, XS)) → mark(XS)
a__zeros → zeros
a__tail(X) → tail(X)
Used ordering:
Polynomial interpretation [25]:

POL(0) = 0   
POL(a__tail(x1)) = 2 + x1   
POL(a__zeros) = 2   
POL(cons(x1, x2)) = 2·x1 + 2·x2   
POL(mark(x1)) = 2·x1   
POL(tail(x1)) = 1 + x1   
POL(zeros) = 1   




↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof

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

a__zeros → cons(0, zeros)
mark(zeros) → a__zeros
mark(tail(X)) → a__tail(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0

Q is empty.

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

a__zeros → cons(0, zeros)
mark(zeros) → a__zeros
mark(tail(X)) → a__tail(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 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:

mark(tail(X)) → a__tail(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
Used ordering:
Polynomial interpretation [25]:

POL(0) = 1   
POL(a__tail(x1)) = 1 + x1   
POL(a__zeros) = 2   
POL(cons(x1, x2)) = 1 + x1 + 2·x2   
POL(mark(x1)) = 2 + 2·x1   
POL(tail(x1)) = 2 + 2·x1   
POL(zeros) = 0   




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

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

a__zeros → cons(0, zeros)
mark(zeros) → a__zeros

Q is empty.

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

a__zeros → cons(0, zeros)
mark(zeros) → a__zeros

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

mark(zeros) → a__zeros
Used ordering:
Polynomial interpretation [25]:

POL(0) = 0   
POL(a__zeros) = 1   
POL(cons(x1, x2)) = 2·x1 + x2   
POL(mark(x1)) = 2 + 2·x1   
POL(zeros) = 1   




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

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

a__zeros → cons(0, zeros)

Q is empty.

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

a__zeros → cons(0, zeros)

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

a__zeros → cons(0, zeros)
Used ordering:
Polynomial interpretation [25]:

POL(0) = 0   
POL(a__zeros) = 2   
POL(cons(x1, x2)) = 1 + 2·x1 + 2·x2   
POL(zeros) = 0   




↳ 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.