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

0 QTRS
↳1 DependencyPairsProof (⇔)
↳2 QDP
↳3 DependencyGraphProof (⇔)
↳4 QDP
↳5 QDPOrderProof (⇔)
↳6 QDP
↳7 DependencyGraphProof (⇔)
↳8 TRUE

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

Recursive path order with status [RPO].
Quasi-Precedence:
[tail1, mark1, atail1] > azeros > cons2
[tail1, mark1, atail1] > azeros > 0
[tail1, mark1, atail1] > azeros > zeros

Status:
tail1: multiset
mark1: multiset
cons2: multiset
azeros: multiset
0: multiset
zeros: multiset
atail1: multiset


The following usable rules [FROCOS05] were oriented:

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)

(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