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 strictly oriented and are deleted.


A__TAIL1(cons2(X, XS)) -> MARK1(XS)
MARK1(tail1(X)) -> MARK1(X)
MARK1(cons2(X1, X2)) -> MARK1(X1)
The remaining pairs can at least by weakly be oriented.

MARK1(tail1(X)) -> A__TAIL1(mark1(X))
Used ordering: Combined order from the following AFS and order.
MARK1(x1)  =  x1
tail1(x1)  =  tail1(x1)
A__TAIL1(x1)  =  x1
mark1(x1)  =  mark1(x1)
cons2(x1, x2)  =  cons2(x1, x2)
a__tail1(x1)  =  a__tail1(x1)
zeros  =  zeros
a__zeros  =  a__zeros
0  =  0

Lexicographic Path Order [19].
Precedence:
[tail1, mark1, atail1, azeros] > cons2
[tail1, mark1, atail1, azeros] > zeros
[tail1, mark1, atail1, azeros] > 0


The following usable rules [14] were oriented:

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



↳ 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))

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 0 SCCs with 1 less node.