Termination Proof using AProVE

Term Rewriting System R:
[X, XS, X1, X2]
a_{_zeros} -> cons(0, zeros)
a_{_zeros} -> zeros
a_{_tail}(cons(X, XS)) -> mark(XS)
a_{_tail}(X) -> tail(X)
mark(zeros) -> a_{_zeros}
mark(tail(X)) -> a_{_tail}(mark(X))
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(0) -> 0

Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

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)

Furthermore, R contains one SCC.

     ↳DPs

       →DP Problem 1

         ↳Narrowing Transformation

Dependency Pairs:

MARK(cons(X1, X2)) -> MARK(X1)
MARK(tail(X)) -> MARK(X)
MARK(tail(X)) -> A_{_TAIL}(mark(X))
A_{_TAIL}(cons(X, XS)) -> MARK(XS)

Rules:

a_{_zeros} -> cons(0, zeros)
a_{_zeros} -> zeros
a_{_tail}(cons(X, XS)) -> mark(XS)
a_{_tail}(X) -> tail(X)
mark(zeros) -> a_{_zeros}
mark(tail(X)) -> a_{_tail}(mark(X))
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(0) -> 0

On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

MARK(tail(X)) -> A_{_TAIL}(mark(X))

four new Dependency Pairs are created:

MARK(tail(zeros)) -> A_{_TAIL}(a_{_zeros})
MARK(tail(tail(X''))) -> A_{_TAIL}(a_{_tail}(mark(X'')))
MARK(tail(cons(X1', X2'))) -> A_{_TAIL}(cons(mark(X1'), X2'))
MARK(tail(0)) -> A_{_TAIL}(0)

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Narrowing Transformation

Dependency Pairs:

MARK(tail(cons(X1', X2'))) -> A_{_TAIL}(cons(mark(X1'), X2'))
MARK(tail(tail(X''))) -> A_{_TAIL}(a_{_tail}(mark(X'')))
A_{_TAIL}(cons(X, XS)) -> MARK(XS)
MARK(tail(zeros)) -> A_{_TAIL}(a_{_zeros})
MARK(tail(X)) -> MARK(X)
MARK(cons(X1, X2)) -> MARK(X1)

Rules:

a_{_zeros} -> cons(0, zeros)
a_{_zeros} -> zeros
a_{_tail}(cons(X, XS)) -> mark(XS)
a_{_tail}(X) -> tail(X)
mark(zeros) -> a_{_zeros}
mark(tail(X)) -> a_{_tail}(mark(X))
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(0) -> 0

On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

MARK(tail(zeros)) -> A_{_TAIL}(a_{_zeros})

two new Dependency Pairs are created:

MARK(tail(zeros)) -> A_{_TAIL}(cons(0, zeros))
MARK(tail(zeros)) -> A_{_TAIL}(zeros)

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 3

                 ↳Narrowing Transformation

Dependency Pairs:

MARK(tail(zeros)) -> A_{_TAIL}(cons(0, zeros))
MARK(tail(tail(X''))) -> A_{_TAIL}(a_{_tail}(mark(X'')))
MARK(cons(X1, X2)) -> MARK(X1)
MARK(tail(X)) -> MARK(X)
A_{_TAIL}(cons(X, XS)) -> MARK(XS)
MARK(tail(cons(X1', X2'))) -> A_{_TAIL}(cons(mark(X1'), X2'))

Rules:

a_{_zeros} -> cons(0, zeros)
a_{_zeros} -> zeros
a_{_tail}(cons(X, XS)) -> mark(XS)
a_{_tail}(X) -> tail(X)
mark(zeros) -> a_{_zeros}
mark(tail(X)) -> a_{_tail}(mark(X))
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(0) -> 0

On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

MARK(tail(tail(X''))) -> A_{_TAIL}(a_{_tail}(mark(X'')))

five new Dependency Pairs are created:

MARK(tail(tail(X'''))) -> A_{_TAIL}(tail(mark(X''')))
MARK(tail(tail(zeros))) -> A_{_TAIL}(a_{_tail}(a_{_zeros}))
MARK(tail(tail(tail(X')))) -> A_{_TAIL}(a_{_tail}(a_{_tail}(mark(X'))))
MARK(tail(tail(cons(X1', X2')))) -> A_{_TAIL}(a_{_tail}(cons(mark(X1'), X2')))
MARK(tail(tail(0))) -> A_{_TAIL}(a_{_tail}(0))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 4

                 ↳Argument Filtering and Ordering

Dependency Pairs:

MARK(tail(tail(0))) -> A_{_TAIL}(a_{_tail}(0))
MARK(tail(tail(cons(X1', X2')))) -> A_{_TAIL}(a_{_tail}(cons(mark(X1'), X2')))
MARK(tail(tail(tail(X')))) -> A_{_TAIL}(a_{_tail}(a_{_tail}(mark(X'))))
MARK(tail(tail(zeros))) -> A_{_TAIL}(a_{_tail}(a_{_zeros}))
MARK(tail(cons(X1', X2'))) -> A_{_TAIL}(cons(mark(X1'), X2'))
MARK(cons(X1, X2)) -> MARK(X1)
MARK(tail(X)) -> MARK(X)
A_{_TAIL}(cons(X, XS)) -> MARK(XS)
MARK(tail(zeros)) -> A_{_TAIL}(cons(0, zeros))

Rules:

a_{_zeros} -> cons(0, zeros)
a_{_zeros} -> zeros
a_{_tail}(cons(X, XS)) -> mark(XS)
a_{_tail}(X) -> tail(X)
mark(zeros) -> a_{_zeros}
mark(tail(X)) -> a_{_tail}(mark(X))
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(0) -> 0

The following dependency pairs can be strictly oriented:

MARK(tail(tail(0))) -> A_{_TAIL}(a_{_tail}(0))
MARK(tail(tail(cons(X1', X2')))) -> A_{_TAIL}(a_{_tail}(cons(mark(X1'), X2')))
MARK(tail(tail(tail(X')))) -> A_{_TAIL}(a_{_tail}(a_{_tail}(mark(X'))))
MARK(tail(tail(zeros))) -> A_{_TAIL}(a_{_tail}(a_{_zeros}))
MARK(tail(cons(X1', X2'))) -> A_{_TAIL}(cons(mark(X1'), X2'))
MARK(cons(X1, X2)) -> MARK(X1)
MARK(tail(X)) -> MARK(X)
A_{_TAIL}(cons(X, XS)) -> MARK(XS)
MARK(tail(zeros)) -> A_{_TAIL}(cons(0, zeros))

The following rules can be oriented:

a_{_tail}(cons(X, XS)) -> mark(XS)
a_{_tail}(X) -> tail(X)
mark(zeros) -> a_{_zeros}
mark(tail(X)) -> a_{_tail}(mark(X))
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(0) -> 0
a_{_zeros} -> cons(0, zeros)
a_{_zeros} -> zeros

Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:

{a_{_zeros}, tail, mark, a_{_tail}} > cons > MARK > A_{_TAIL}
{a_{_zeros}, tail, mark, a_{_tail}} > 0
{a_{_zeros}, tail, mark, a_{_tail}} > zeros

resulting in one new DP problem.
Used Argument Filtering System:

MARK(x₁) -> MARK(x₁)
A_{_TAIL}(x₁) -> A_{_TAIL}(x₁)
tail(x₁) -> tail(x₁)
a_{_tail}(x₁) -> a_{_tail}(x₁)
cons(x₁, x₂) -> cons(x₁, x₂)
mark(x₁) -> mark(x₁)
a_{_zeros} -> a_{_zeros}

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 5

                 ↳Dependency Graph

Dependency Pair:

Rules:

a_{_zeros} -> cons(0, zeros)
a_{_zeros} -> zeros
a_{_tail}(cons(X, XS)) -> mark(XS)
a_{_tail}(X) -> tail(X)
mark(zeros) -> a_{_zeros}
mark(tail(X)) -> a_{_tail}(mark(X))
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(0) -> 0

Using the Dependency Graph resulted in no new DP problems.

Termination of R successfully shown.
Duration:
0:00 minutes