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
  ↳ DependencyPairsProof

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.

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

MARK(tail(X)) → A__TAIL(mark(X))
A__TAIL(cons(X, XS)) → MARK(XS)
MARK(tail(X)) → MARK(X)
MARK(zeros) → A__ZEROS
MARK(cons(X1, X2)) → MARK(X1)

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof

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

MARK(tail(X)) → A__TAIL(mark(X))
A__TAIL(cons(X, XS)) → MARK(XS)
MARK(tail(X)) → MARK(X)
MARK(zeros) → A__ZEROS
MARK(cons(X1, X2)) → MARK(X1)

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.
We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof

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

MARK(tail(X)) → A__TAIL(mark(X))
A__TAIL(cons(X, XS)) → MARK(XS)
MARK(tail(X)) → MARK(X)
MARK(zeros) → A__ZEROS
MARK(cons(X1, X2)) → MARK(X1)

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.
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
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ QDPOrderProof

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

MARK(tail(X)) → A__TAIL(mark(X))
A__TAIL(cons(X, XS)) → MARK(XS)
MARK(tail(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)

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.
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__TAIL(cons(X, XS)) → MARK(XS)
MARK(tail(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
The remaining pairs can at least be oriented weakly.

MARK(tail(X)) → A__TAIL(mark(X))
Used ordering: Combined order from the following AFS and order.
MARK(x1)  =  MARK(x1)
tail(x1)  =  tail(x1)
A__TAIL(x1)  =  A__TAIL(x1)
mark(x1)  =  mark(x1)
cons(x1, x2)  =  cons(x1, x2)
a__zeros  =  a__zeros
zeros  =  zeros
0  =  0
a__tail(x1)  =  a__tail(x1)

Recursive Path Order [2].
Precedence:
[MARK1, tail1, ATAIL1, mark1, azeros, atail1] > zeros > cons2
[MARK1, tail1, ATAIL1, mark1, azeros, atail1] > 0 > cons2


The following usable rules [14] were oriented:

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



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

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

MARK(tail(X)) → A__TAIL(mark(X))

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.
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.