Termination w.r.t. Q of the given QTRS could be proven:

(0) Obligation:

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.

(1) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(2) Obligation:

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

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

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.

(4) Obligation:

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.

(5) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

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.
Used ordering: Lexicographic Path Order [LPO].
Precedence:

[MARK₁, A_TAIL1] > [tail₁, mark₁, a_tail1] > a_zeros > zeros > cons₂
[MARK₁, A_TAIL1] > [tail₁, mark₁, a_tail1] > a_zeros > 0 > cons₂

The following usable rules [FROCOS05] were oriented:

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

(6) Obligation:

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.

(7) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.

(8) TRUE