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 -> cons2(0, zeros)
a__tail1(cons2(X, XS)) -> mark1(XS)
mark1(zeros) -> a__zeros
mark1(tail1(X)) -> a__tail1(mark1(X))
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
mark1(0) -> 0
a__zeros -> zeros
a__tail1(X) -> tail1(X)

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

a__zeros -> cons2(0, zeros)
a__tail1(cons2(X, XS)) -> mark1(XS)
mark1(zeros) -> a__zeros
mark1(tail1(X)) -> a__tail1(mark1(X))
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
mark1(0) -> 0
a__zeros -> zeros
a__tail1(X) -> tail1(X)

Q is empty.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

MARK1(tail1(X)) -> A__TAIL1(mark1(X))
A__TAIL1(cons2(X, XS)) -> MARK1(XS)
MARK1(tail1(X)) -> MARK1(X)
MARK1(zeros) -> A__ZEROS
MARK1(cons2(X1, X2)) -> MARK1(X1)

The TRS R consists of the following rules:

a__zeros -> cons2(0, zeros)
a__tail1(cons2(X, XS)) -> mark1(XS)
mark1(zeros) -> a__zeros
mark1(tail1(X)) -> a__tail1(mark1(X))
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
mark1(0) -> 0
a__zeros -> zeros
a__tail1(X) -> tail1(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof

Q DP problem:
The TRS P consists of the following rules:

MARK1(tail1(X)) -> A__TAIL1(mark1(X))
A__TAIL1(cons2(X, XS)) -> MARK1(XS)
MARK1(tail1(X)) -> MARK1(X)
MARK1(zeros) -> A__ZEROS
MARK1(cons2(X1, X2)) -> MARK1(X1)

The TRS R consists of the following rules:

a__zeros -> cons2(0, zeros)
a__tail1(cons2(X, XS)) -> mark1(XS)
mark1(zeros) -> a__zeros
mark1(tail1(X)) -> a__tail1(mark1(X))
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
mark1(0) -> 0
a__zeros -> zeros
a__tail1(X) -> tail1(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 1 less node.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof

Q DP problem:
The TRS P consists of the following rules:

MARK1(tail1(X)) -> A__TAIL1(mark1(X))
A__TAIL1(cons2(X, XS)) -> MARK1(XS)
MARK1(tail1(X)) -> MARK1(X)
MARK1(cons2(X1, X2)) -> MARK1(X1)

The TRS R consists of the following rules:

a__zeros -> cons2(0, zeros)
a__tail1(cons2(X, XS)) -> mark1(XS)
mark1(zeros) -> a__zeros
mark1(tail1(X)) -> a__tail1(mark1(X))
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
mark1(0) -> 0
a__zeros -> zeros
a__tail1(X) -> tail1(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


A__TAIL1(cons2(X, XS)) -> MARK1(XS)
MARK1(cons2(X1, X2)) -> MARK1(X1)
The remaining pairs can at least be oriented weakly.

MARK1(tail1(X)) -> A__TAIL1(mark1(X))
MARK1(tail1(X)) -> MARK1(X)
Used ordering: Polynomial interpretation [21]:

POL(0) = 0   
POL(A__TAIL1(x1)) = x12   
POL(MARK1(x1)) = x12   
POL(a__tail1(x1)) = x1 + x12   
POL(a__zeros) = 3   
POL(cons2(x1, x2)) = 1 + x1 + x2   
POL(mark1(x1)) = x12   
POL(tail1(x1)) = x1 + x12   
POL(zeros) = 2   

The following usable rules [14] were oriented:

a__zeros -> zeros
a__zeros -> cons2(0, zeros)
mark1(zeros) -> a__zeros
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
a__tail1(X) -> tail1(X)
mark1(0) -> 0
mark1(tail1(X)) -> a__tail1(mark1(X))
a__tail1(cons2(X, XS)) -> mark1(XS)



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ DependencyGraphProof

Q DP problem:
The TRS P consists of the following rules:

MARK1(tail1(X)) -> A__TAIL1(mark1(X))
MARK1(tail1(X)) -> MARK1(X)

The TRS R consists of the following rules:

a__zeros -> cons2(0, zeros)
a__tail1(cons2(X, XS)) -> mark1(XS)
mark1(zeros) -> a__zeros
mark1(tail1(X)) -> a__tail1(mark1(X))
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
mark1(0) -> 0
a__zeros -> zeros
a__tail1(X) -> tail1(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 1 less node.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ QDP
                  ↳ QDPOrderProof

Q DP problem:
The TRS P consists of the following rules:

MARK1(tail1(X)) -> MARK1(X)

The TRS R consists of the following rules:

a__zeros -> cons2(0, zeros)
a__tail1(cons2(X, XS)) -> mark1(XS)
mark1(zeros) -> a__zeros
mark1(tail1(X)) -> a__tail1(mark1(X))
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
mark1(0) -> 0
a__zeros -> zeros
a__tail1(X) -> tail1(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


MARK1(tail1(X)) -> MARK1(X)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(MARK1(x1)) = x12   
POL(tail1(x1)) = 1 + x12   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ QDP
                  ↳ QDPOrderProof
                    ↳ QDP
                      ↳ PisEmptyProof

Q DP problem:
P is empty.
The TRS R consists of the following rules:

a__zeros -> cons2(0, zeros)
a__tail1(cons2(X, XS)) -> mark1(XS)
mark1(zeros) -> a__zeros
mark1(tail1(X)) -> a__tail1(mark1(X))
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
mark1(0) -> 0
a__zeros -> zeros
a__tail1(X) -> tail1(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.